Class 10

Solving these TS 10th Class Maths Bits with Answers Chapter 6 Progressions Bits for 10th Class will help students to build their problem-solving skills.

Progressions Bits for 10th Class

Question 1.
How many terms of A.P – 6, \(\frac{-11}{2}\), -5, …………… are needed to obtain a sum – 25 ?
A) 10 or 15
B) 15 or 18
C) 5 or 20
D) 8 or 12
Answer:
C) 5 or 20

Question 2.
The sum of natural numbers from 1 to 100 is
A) 4050
B) 10100
C) 55
D) 5050
Answer:
D) 5050

Question 3.
The sum of first 50 even numbers is
A) 1250
B) 2550
C) 1275
D) 2275
Answer:
B) 2550

Question 4.
The sum of first 20 odd numbers is
A) 400
B) 210
C) 420
D) 05
Answer:
A) 400

Question 5.
If x – 1, x + 3, 3x – 1 are in A.P, then x is equal
A) 5
B) 8
C) 420
D) 405
Answer:
D) 405

Question 6.
The numbers -15, -11, -7, -3, …………… is
A) A.P. with d = 4
B) A.P. with d = – 4
C) A.P. with d = 8
D) G.P.
Answer:
A) A.P. with d = 4

Question 7.
The next term of the A.P. \(\sqrt{48}\), \(\sqrt{75}\), \(\sqrt{147}\), ………………. is
A) \(\sqrt{27}\)
B) \(\sqrt{197}\)
C) \(\sqrt{192}\)
D) \(\sqrt{243}\)
Answer:
C) \(\sqrt{192}\)

Question 8.
Which term of the A.P. 24, 21, 18, ……………… is the first negative ?
A) a₁₀
B) a₉
C) a₆
D) a₁₁
Answer:
A) a₁₀

Question 9.
Which term of the A.P. 125, 120, 115, …. is the first negative ?
A) 25^th
B) 26^th
C) 24^th
D) 27^th
Answer:
D) 27^th

Question 10.
Which term of the A.P. 100, 90, 80, …………….. is zero ?
A) 10^th
B) 9^th
C) 11^th
D) 12^th
Answer:
C) 11^th

Question 11.
(a + 3d), (a + d), (a – d), …… the next term of the A.P. is
A) a + 2d
B) a – 2d
C) a – 4d
D) a – 3d
Answer:
D) a – 3d

Question 12.
The sum of 15 terms of the A.P. 4, 7, 10, …………….. is
A) 385
B) 475
C) 375
D) 325
Answer:
C) 375

Question 13.
If a₇ – a₃ = 32, then the common difference of the A.P. is
A) 8
B) 6
C) 4
D) 6
Answer:
A) 8

Question 14.
a₂₈ – a₂₃ = 15, then the common difference of the A.P. is
A) 3
B) 5
C) 7
D) 15
Answer:
A) 3

Question 15.
If a, b, c are in A.P. then b = ………………
A) a + c
B) \(\frac{a+c}{2}\)
C) a – c
D) \(\frac{a-c}{2}\)
Answer:
B) \(\frac{a+c}{2}\)

Question 16.
The 17^th term of 1.1, 2.2, 3.3, 4.4, ………….. is
A) 18.7
B) 19.8
C) 17.6
D) 17.17
Answer:
A) 18.7

Question 17.
The 25^th term of – 300, – 290, – 280, …. is
A) – 60
B) – 80
C) 60
D) 80
Answer:
A) – 60

Question 18.
How many numbers are divisible by 4 lying between 101 and 250 ?
A) 40
B) 62
C) 38
D) 37
Answer:
D) 37

Question 19.
The common ratio of the G.P. 3, 6, 12, 24, ……………. is
A) 3
B) 2
C) \(\frac{1}{2}\)
D) \(\frac{1}{3}\)
Answer:
B) 2

Question 20.
The common ratio of the G.P. 192, 36, 9, ……………. is
A) 4
B) 2
C) 6
D) \(\frac{1}{4}\)
Answer:
D) \(\frac{1}{4}\)

Question 21.
The 103^rd term of 1, -1, 1, -1, ……….. is
A) 1
B) -1
C) 0
D) -2
Answer:
B) -1

Question 22.
Which term of the G.P. 2,6,18, 54, is 2 × 3¹⁰ ?
A) 10^th
B) 11^th
C) 12^th
D) 9^th
Answer:
B) 11^th

Question 23.
If a₇ + a₄ of a G.P. is 343, then the common ratio is
A) 11
B) 9
C) 3
D) 7
Answer:
D) 7

Question 24.
If a, b, c are in G.P, then b = …….
A) ac
B) \(\frac{a+c}{2}\)
C) a²c²
D) \(\sqrt{\mathrm{ac}}\)
Answer:
D) \(\sqrt{\mathrm{ac}}\)

Question 25.
If 4, x, 9 are in G.P, then x = …………..
A) 7
B) 6
C) 8
D) 5
Answer:
B) 6

Question 26.
If 3, x, 11 are in A.P, then x = …………..
A) \(\sqrt{21}\)
B) 14
C) 4
D) 7
Answer:
D) 7

Question 27.
If x, xy, xy², xy³, …….. forms a G.P, then its 15^th term is
A) xy¹⁵
B) xy¹⁴
C) x¹⁴y
D) x¹⁵y
Answer:
B) xy¹⁴

Question 28.
If a = 3 and a₇ = 33, then a₁₁ is
A) 55
B) 53
C) 73
D) 63
Answer:
B) 53

Question 29.
-20, -18, -16, …… which term of this A.P. is a first positive term ?
A) 10
B) 11
C) 12
D) 9
Answer:
C) 12

Question 30.
1, -1, 1, -1, 1, -1, …………. up to 131 terms, then S₁₃₁ = ………….
A) 1
B) -1
C) 131
D) 130
Answer:
A) 1

Question 31.
The 10^th term of the AP 3, 11, 19, ……….. is
A) 73
B) 16
C) 75
D) 85
Answer:
C) 75

Question 32.
…………. term of AP 21, 18, 15, ………. is – 81.
A) 35
B) 16
C) 30
D) none
Answer:
A) 35

Question 33.
The 8^th term from the end of the AP 7,10, 13, ……….. 1814 is
A) 324
B) 181
C) 163
D) 161
Answer:
C) 163

Question 34.
The n^th term of a, a + d, a + 2d, ………… is
A) a + (n – 1)d
B) a – (n + 1)d
C) a² – (n – 1)d
D) d + (n – 1)a
Answer:
A) a + (n – 1)d

Question 35.
In the AP, first term is 4 and common difference is -1 then AP is …………
A) 9, 3,-6, ….
B) 10, 12, 14,
C) 5,8,16,….
D) 4, 3, 2, ……
Answer:
D) 4, 3, 2, ……

Question 36.
The AP with first term is 8 and common difference 2 1/2 is …………
A) 8, 10 1/2, 13
B) 8, 10, 11 1/2
C) 16, 15 1/2, 10 1/2
D) none
Answer:
A) 8, 10 1/2, 13

Question 37.
In the AP -9, -14, -19, -24 ……. a₃₀ – a₂₀ = ……….
A) 80
B) -60
C) 50
D) -50
Answer:
D) -50

Question 38.
The next term of the AP 51, 59, 67, 75 is …………….
A) 12
B) 16
C) 83
D) 38
Answer:
C) 83

Question 39.
Number of terms of the AP – 5 + (- 8) + (-11) + ………. +(-230) is …………
A) 66
B) 76
C) 86
D) none
Answer:
B) 76

Question 40.
15^th term of the AP x – 7, x – 2, x + 3, ………… is
A) x + 63
B) x – 6
C) x – 63
D) x + 16
Answer:
A) x + 63

Question 41.
The common ratio of the GP 4, 20, 100, 500, …. is …….
A) 8
B) 2
C) 5
D) 9
Answer:
C) 5

Question 42.
The 16^th term of 4, -4, 4, -4, ………… is ………….
A) 16
B) 8
C) 4
D) -4
Answer:
D) -4

Question 43.
A.P 1, -1, -3, -5, ……. d = …………
A) -2
B) 1
C) 2
D) 10
Answer:
A) -2

Question 44.
In the A.P -11, -9, -7, ….. d = ……….
A) 4
B) 3
C) -2
D) 2
Answer:
D) 2

Question 45.
In the A.P 100, 103, 106, … d = ……..
A) 4
B) 8
C) 6
D) none
Answer:
D) none

Question 46.
A G.P with r = – 2 is …………
A) 5, – 10, 20, -40, …..
B) 2, 4, 8, 16
C) 3, -6, 10, 16
D) all
Answer:
A) 5, – 10, 20, -40, …..

Question 47.
A G.P with r = 2 is ………..
A) 7, 14, 28, …..
B) 8, 16, 10 ……
C) 12, 24, 19, …
D) none
Answer:
A) 7, 14, 28, …..

Question 48.
5, 10, 15, ….. 10^th term is ……
A) 20
B) 90
C) 60
D) 50
Answer:
D) 50

Question 49.
8, 16, 32, …. 6^th term is …….
A) 256
B) 156
C) 108
D) none
Answer:
A) 256

Question 50.
-1, 1, -1, … 11^th term is ………
A) 1
B) -1
C) 10
D) 9
Answer:
B) -1

Question 51.
– 8, -6, -4, …… a₇ = ……….
A) 1
B) 12
C) 10
D) 6
Answer:
D) 6

Question 52.
1, 2, 3, ………. sum to 10 terms is ….
A) 55
B) 65
C) 60
D) 90
Answer:
A) 55

Question 53.
If a, b, c are in GP then b² = ………..
A) \(\frac{\mathrm{c}}{\mathrm{a}}\)
B) \(\frac{\mathrm{a}}{\mathrm{c}}\)
C) \(\sqrt{\mathrm{ac}}\)
D) ac
Answer:
D) ac

Question 54.
In a GP a₆ = ………..
A) ar⁵
B) a⁵r
C) a⁵r⁵
D) all
Answer:
A) ar⁵

Question 55.
Which term of G.P 2, 8, 32, … is 512 ?
A) 16
B) 5
C) 9
D) 10
Answer:
B) 5

Question 56.
\(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), …… a₇ = …….
A) \(\frac{1}{1827}\)
B) \(\frac{1}{2187}\)
C) \(\frac{1}{8127}\)
D) none
Answer:
B) \(\frac{1}{2187}\)

Question 57.
3, \(\frac{3}{2}\), \(\frac{3}{4}\), ….. r = …………
A) 1
B) 2
C) -1
D) none
Answer:
D) none

Question 58.
If 2, x, 6 are in GP then x = ……..
A) 2\(\sqrt{3}\)
B) 8\(\sqrt{3}\)
C) 2\(\sqrt{2}\)
D) \(\sqrt{3}\)
Answer:
A) 2\(\sqrt{3}\)

Question 59.
4, 16, ☐, 256 then ☐ = …………
A) 161
B) 64
C) 62
D) 68
Answer:
B) 64

Question 60.
1 + 2+ 3 + ……. + n = ……..
A) \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)
B) \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)
C) \(\frac{\mathrm{n}^2+1}{2}\)
D) none
Answer:
A) \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

Question 61.
1² + 2² + 3² + ……. + n² = ……….
A) \(\frac{\mathrm{n}^2(\mathrm{n}-1)^2}{2}\)
B) \(\frac{\mathrm{n}(2 \mathrm{n}+1)}{4}\)
C) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)
D) \(\frac{(\mathrm{n}+1)^2}{2}\)
Answer:
C) \(\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}\)

Question 62.
In an AP 7a₇ = 11a₁₁ then a₁₈ = ……..
A) -1
B) 0
C) 1
D) 7
Answer:
B) 0

Question 63.
AM of 24 and 16 is …….
A) 22
B) 19
C) 16
D) 20
Answer:
D) 20

Question 64.
If a, b, c are in AP then 2b = ………..
A) a + c
B) a – c
C) \(\frac{\mathrm{a+c}}{2}\)
D) \(\frac{\mathrm{a}}{2}\)
Answer:
C) \(\frac{\mathrm{a+c}}{2}\)

Question 65.
The sum of first 40 positive integers divisible by 6 is ……………
A) 9420
B) 4920
C) 9920
D) 1290
Answer:
B) 4920

Question 66.
a_n = 9 – 5n, a₄ = …………
A) 30
B) 10
C) 11
D) -11
Answer:
D) -11

Question 67.
In a GP, a₈ = 192, r = 2 then a₁₂ = ………….
A) 3072
B) 7032
C) 1032
D) 1100
Answer:
A) 3072

Question 68.
2, – 6,18, – 54, …. r = ………..
A) -3
B) 3
C) 1
D) – 2
Answer:
A) -3

Question 69.
\(\frac{5}{2}\), \(\frac{5}{4}\), \(\frac{5}{8}\), ……. a_n = ………….
A) \(\frac{5}{2^{\mathrm{n}-1}}\)
B) \(\frac{5}{2^{\mathrm{n}}}\)
C) \(\frac{5}{2^{\mathrm{n}-2}}\)
D) none
Answer:
B) \(\frac{5}{2^{\mathrm{n}}}\)

Question 70.
\(\frac{1}{\sqrt{2}}\), -2, \(\frac{8}{\sqrt{2}}\), ……….. a₅ = …………
A) 16 \(\sqrt{2}\)
B) 32 \(\sqrt{2}\)
C) 6 \(\sqrt{2}\)
D) 31 \(\sqrt{2}\)
Answer:
B) 32 \(\sqrt{2}\)

Question 71.
3, -3², 3³, …. a₆ = ………
A) -729
B) 729
C) 829
D) 114
Answer:
A) -729

Question 72.
In a G.P, a = 81, r = –\(\frac{1}{3}\), a₃ = …………..
A) -9
B) 9
C) -3
D) none
Answer:
B) 9

Question 73.
In a G.P 25, -5, 1, –\(\frac{1}{5}\), ……. r = ………….
A) \(\frac{-3}{5}\)
B) 2
C) -1
D) \(\frac{-1}{5}\)
Answer:
D) \(\frac{-1}{5}\)

Question 74.
S_n in AP = …………..
A) \(\frac{\mathrm{n}}{2}\) [a + l]
B) \(\frac{\mathrm{n}}{3}\) [a + l]
C) 2n[a + l]
D) none
Answer:
A) \(\frac{\mathrm{n}}{2}\) [a + l]

Question 75.
In AP a₁₂ = 37, d = 3, then a = ………..
A) 8
B) – 4
C) -3
D) 4
Answer:
D) 4

Question 76.
In the above problem S₁₂ = ………….
A) 246
B) 146
C) 123
D) 112
Answer:
A) 246

Question 77.
-1, \(\frac{1}{4}\), \(\frac{3}{2}\), ….. Sum to 10 terms = ……..
A) 26.5
B) 16.25
C) 36.25
D) 46.25
Answer:
D) 46.25

Question 78.
-5 + (-8) + (-11) + ….. + (-230) = …..
A) -8930
B) 8930
C) 8390
D) none
Answer:
A) -8930

Question 79.
The sum of first 1000 positive integers is ………..
A) 500500
B) 50051
C) 8005
D) none
Answer:
A) 500500

Question 80.
16 + 11 + 6 + ……. 23 terms = ……….
A) 119
B) -987
C) 891
D) -897
Answer:
D) -897

Question 81.
Identify the 3-digit number that divisible by
A) 126
B) 128
C) 122
D) none
Answer:
B) 128

Question 82.
In an A.P a₁ = – 4, a₆ = 6 then a₂ = ………….
A) 3
B) 6
C) 1
D) -2
Answer:
D) -2

Question 83.
In the above problem a₅ = …………….
A) 4
B) -3
C) 6
D) -4
Answer:
A) 4

Question 84.
In the formula a_n = 3.6, a = -18.9, d = 2.5, n = ……………
A) 13
B) 12
C) 10
D) 20
Answer:
C) 10

Question 85.
5, 1, -3, -7, …….. a₁₀ = ………..
A) -23
B) 22
C) 31
D) -31
Answer:
D) -31

Question 86.
\(\frac{1}{4}\), \(\frac{-1}{4}\), \(\frac{-3}{4}\), \(\frac{-5}{4}\), …… d = …………
A) \(\frac{-1}{2}\)
B) \(\frac{1}{2}\)
C) 2
D) -1
Answer:
A) \(\frac{-1}{2}\)

Question 87.
If 4, x, 16 are in G.P then x = …………..
A) 12
B) 16
C) 8
D) None
Answer:
C) 8

Question 88.
1³ + 2³ + 3³ + ……….. + n³ = ………….
A) \(\frac{\mathrm{n}}{2}\)
B) \(\frac{(\mathrm{n}+1)^2}{2}\)
C) \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)
D) \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\)
Answer:
D) \(\frac{\mathrm{n}^2(\mathrm{n}+1)^2}{4}\)

Question 89.
G.M of a and b is ……
A) \(\sqrt{a b}\)
B) \(\frac{a b}{2}\)
C) \(\frac{a+b}{2}\)
D) none
Answer:
A) \(\sqrt{a b}\)

Question 90.
G.M of a and \(\frac{1}{a}\) is ……
A) -3
B) 1
C) 7
D) 8
Answer:
B) 1

Question 91.
Reciprocals of term of GP is ………..
A) AP
B) GP
C) HP
D) none
Answer:
B) GP

Question 92.
If a, b, c are in GP then \(\frac{\mathrm{b}}{\mathrm{a}}\) = ……………
A) \(\frac{\mathrm{c}}{\mathrm{b}}\)
B) \(\frac{\mathrm{b}}{\mathrm{c}}\)
C) \(\frac{1}{\mathrm{b}}\)
D) none
Answer:
A) \(\frac{\mathrm{c}}{\mathrm{b}}\)

Question 93.
a_n in GP = …………
A) ar
B) ar^n-1
C) a^n-1
D) \(\frac{\mathrm{a}}{2}\)r^n-1
Answer:
B) ar^n-1

Question 94.
1 + 1+ 1 + ………. + n terms = …………
A) \(\frac{\mathrm{n}}{2}\)
B) n
C) n – 1
D) none
Answer:
B) n

Question 95.
If a, b, c are in GP then b is called …….
A) Geometric mean
B) Arithmetic mean
C) Number
D) None
Answer:
A) Geometric mean

Question 96.
Σ n = 10, Σ n³ = ………….
A) 100
B) 1001
C) 200
D) 80
Answer:
A) 100

Question 97.
In a series a_n = \(\frac{\mathrm{n}(\mathrm{n}+1)}{3}\), a₂ = …….
A) 41
B) 3
C) 4
D) 2
Answer:
D) 2

Question 98.
AM of 10 and 20 is ……..
A) 12
B) 15
C) 25
D) 10
Answer:
B) 15

Question 99.
a_n = (n – 1) (n – 2) then a₂ = ………….
A) 1
B) 0
C) 2
D) 3
Answer:
B) 0

Question 100.
If a, b, c are in AP then b – a = …………..
A) c + b
B) a + b
C) c – b
D) none
Answer:
C) c – b

Question 101.
If \(\frac{-2}{7}\), x, \(\frac{-7}{2}\) are 3 consecutive terms of a GP then x = ………….
A) ± 1
B) ± 2
C) ± 3
D) ± 4
Answer:
A) ± 1

Question 102.
GM of 5 and 125 is ……………
A) 13
B) 16
C) 10
D) 25
Answer:
D) 25

Question 103.
S_n in AP = …………..
A) \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]
B) \(\frac{\mathrm{n}}{2}\) [a + (n + 1)d]
C) n[2a + (n – 1)d]
D) none
Answer:
A) \(\frac{\mathrm{n}}{2}\) [2a + (n – 1)d]

Question 104.
a_n = \(\frac{n}{n+2}\); a₃ = ……………
A) \(\frac{1}{2}\)
B) \(\frac{5}{3}\)
C) \(\frac{3}{5}\)
D) none
Answer:
C) \(\frac{3}{5}\)

Question 105.
The n^th term of a GP is 2(0.5)^n-1, r = ………..
A) – 2
B) \(\frac{1}{2}\)
C) 2
D) -1
Answer:
B) \(\frac{1}{2}\)

Question 106.
The common ratio of the GP 2, \(\sqrt{8}\), 4,……. is ……….
A) \(\sqrt{3}\)
B) 4
C) 3
D) \(\sqrt{2}\)
Answer:
D) \(\sqrt{2}\)

Question 107.
a, b, c are in AP then 3^a, 3^b, 3^c are in …………
A) GP
B) HP
C) AP
D) none
Answer:
A) GP

Question 108.
2, \(\frac{5}{2}\), 3, ……. S₂₅ = …………..
A) 110
B) 180
C) 100
D) none
Answer:
D) none

Question 109.
AM of M, P, C is ……..
A) \(\frac{\text { MPC }}{3}\)
B) M-P-C
C) \(\frac{\mathrm{M}+\mathrm{P}+\mathrm{C}}{3}\)
D) \(\frac{\mathrm{M}+\mathrm{P}-\mathrm{C}}{3}\)
Answer:
C) \(\frac{\mathrm{M}+\mathrm{P}+\mathrm{C}}{3}\)

Question 110.
a_n = 2ⁿ, a₅ = ……………..
A) 32
B) 23
C) 18
D) 13
Answer:
A) 32

Question 111.
AM of 5 and 95 is …………
A) 105
B) 505
C) 501
D) 50
Answer:
D) 50

Question 112.
GM of x³ and \(\frac{1}{\mathrm{x}^3}\) = ……………
A) -7
B) 1
C) 3
D) none
Answer:
B) 1

Question 113.
n – 1, n – 2, n – 3, ……….. a₁₀ = ………….
A) n – 10
B) n – 9
C) n + 9
D) n – 3
Answer:
A) n – 10

Question 114.
Product of n GMs between a and b is ………..
A) (ab)^n/2
B) (ab)ⁿ
C) \(\frac{\mathrm{a}}{\mathrm{b}}\)
D) \(\frac{\mathrm{a}^n}{\mathrm{b}}\)
Answer:
A) (ab)^n/2

Question 115.
If \(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\) is the AM of a and b then n = ………..
A) \(\frac{-1}{2}\)
B) 1
C) 0
D) 4
Answer:
C) 0

Question 116.
7, 10, 13, ……, a₅ = …………….
A) 19
B) 100
C) 131
D) 12
Answer:
A) 19

Question 117.
22, 32, 42, ………., a₇ = …………
A) 81
B) 92
C) 69
D) 82
Answer:
D) 82

Question 118.
1, 4, 7, 10, ………., d = ………..
A) 13
B) 3
C) 4
D) none
Answer:
B) 3

Question 119.
In AP a_p = q, a_q = p then a_p+q = ………….
A) q – p
B) p – q
C) 0
D) – 1
Answer:
C) 0

Question 120.
Which term of AP 7 + 4 + 1 + ……… is – 56 ?
A) 22
B) 20
C) 18
D) 19
Answer:
A) 22

Question 121.
1 + 6 + 4 + 9 + 7 + 12 + ………….. to 40 terms.
A) 20
B) 60
C) 90
D) none
Answer:
D) none

Question 122.
AM of x² + y² and x² – y² is ……….
A) \(\frac{x^2}{2}\)
B) x²
C) x
D) 2x
Answer:
B) x²

Question 123.
\(\frac{\mathrm{b}+\mathrm{c}-\mathrm{a}}{\mathrm{a}}, \frac{\mathrm{c}+\mathrm{a}-\mathrm{b}}{\mathrm{b}}, \frac{\mathrm{a}+\mathrm{b}-\mathrm{c}}{\mathrm{c}}\) are in AP then \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in ……………
A) HP
B) GP
C) AP
D) None
Answer:
C) AP

Question 124.
If there are n AM’s between a and b then d = ……………

Answer:
(A)

Question 125.
10, 100, 1000, ….., r = ……………
A) 12
B) 9
C) 8
D) 10
Answer:
D) 10

Question 126.
1 + \(\frac{1}{2}\) + \(\frac{1}{2^2}\) + …………., r = ………………
A) 3
B) \(\frac{1}{2}\)
C) 2
D) -1
Answer:
B) \(\frac{1}{2}\)

Question 127.
3, 6, 12, r = …………..
A) 1
B) 10
C) 3
D) none
Answer:
D) none

Question 128.
a, a², a³, …………., r = ………….
A) a
B) a²
C) a³
D) none
Answer:
A) a

Question 129.
a, b, c are in A.P then \(\frac{1}{\mathrm{a}}, \frac{1}{\mathrm{~b}}, \frac{1}{\mathrm{c}}\) are in ……………..
A) GP
B) AP
C) HP
D) none
Answer:
C) HP

Question 130.
GM of x, y, z is …………..
A) xyz³
B) \(\sqrt[3]{x y z}\)
C) \(\frac{x y z}{3}\)
D) \(\frac{x+y+z}{3}\)
Answer:
B) \(\sqrt[3]{x y z}\)

Question 131.
The n^th term of a GP = a_n = ar^{n – 1} here ‘r’ meters ……..
A) common difference
B) common ratio
C) first term
D) radius
Answer:
B) common ratio

Question 132.
The ‘n’ th term of an A.P. is a_n = 3 + 2n then the common difference is …………….. (A.P. Mar. ’15)
A) 2
B) 3
C) 4
D) 5
Answer:
A) 2

Question 133.
The common difference of the AP x – y, x, x + y is …………… (A.P. Mar. ’15)
A) x
B) y
C) -x
D) -y
Answer:
B) y

Question 134.
The common difference in the AP 2a – b, 4a – 3b, 6a – 5b is ………….. (A.P. June ’15)
A) 2a – 2b
B) a – b
C) 2a-b
D) 4a – 3b
Answer:
A) 2a – 2b

Question 135.
In a GP a₁ = 20 and a₄ = 540 then r = …………. (A.P. June ’15)
A) 27
B) 3
C) 520
D) 18
Answer:
B) 3

Question 136.
Formula for sum of ‘n’ terms in an A.P. = ………….. (A.P. June ’15)
A) Σ_n = \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)
B) Σ_n² = \(\frac{\mathrm{n}^2(\mathrm{n}+1)(\mathrm{n}+2)}{6}\)
C) S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d] or S_n = \(\frac{\mathrm{n}}{2}\)[a + l]
D) a_n = a + (n – 1) d
Answer:
C) S_n = \(\frac{\mathrm{n}}{2}\)[2a + (n – 1)d] or S_n = \(\frac{\mathrm{n}}{2}\)[a + l]

Question 137.
In an A.P. the 3^rd term = 5 and 7^th term = 9 then the common difference = ………….. (T.S. Mar. ’15)
A) 1
B) 2
C) 3
D) 4
Answer:
A) 1

Question 138.
The term ‘r’ refers to …. in the formula of ‘n’ ^th term of a GP a_n = ar^{n – 1}. (T.S. Mar. ’16)
A) n^th term
B) number of terms
C) common ratio
D) first term
Answer:
C) common ratio

Question 139.
Which of the following is a GP with a common ration “r”. (T.S. Mar. ’16)
A) \(\sqrt{2}\), \(\sqrt{6}\), \(\sqrt{18}\)
B) \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{12}\)
C) \(\sqrt{5}\), \(\sqrt{15}\), \(\sqrt{45}\)
D) \(\sqrt{7}\), \(\sqrt{21}\), \(\sqrt{63}\)
Answer:
B) \(\sqrt{3}\), \(\sqrt{6}\), \(\sqrt{12}\)

Question 140.
Which term of the G.P.3, 9, 22, …………. is 243?
A) 7
B) 8
C) 6
D) 5
Answer:
D) 5

Question 141.
The series (n – 1), (n – 2), (n – 3), ……….. is a type of
A) AP
B) GP
C) may be both
D) none
Answer:
A) AP

Question 142.
If log a, log b, log c are in A.P. then a, b, c are
A) A.P.
B) G.P.
C) Both A.P. and G.P.
D) neither A.P. nor G.P.
Answer:
B) G.P.

Solving these TS 10th Class Maths Bits with Answers Chapter 7 Coordinate Geometry Bits for 10th Class will help students to build their problem-solving skills.

Coordinate Geometry Bits for 10th Class

Question 1.
The scientist who introduced co-ordinate Geometry is
A) J.J. Sylvester
B) Crames
C) Rene Descartes
D) Newton
Answer:
C) Rene Descartes

Question 2.
The distance between the points (x, 1) and (1, y)is
A) \(\sqrt{(x+1)^2+(y+1)^2}\)
B) \(\sqrt{(1-x)^2+(y-1)^2}\)
C) \(\sqrt{(x-1)^2+(y+1)^2}\)
D) \(\sqrt{(x+1)^2+(y-1)^2}\)
Answer:
B) \(\sqrt{(1-x)^2+(y-1)^2}\)

Question 3.
A point on Y – axis is
A) (3, 0)
B) (1, 2)
C) (0, 0)
D) (0, 3)
Answer:
D) (0, 3)

Question 4.
The points (0, 3) (0, 0) (4, 0) form a
A) Isosceles triangle
B) Equilateral triangle
C) Right angle triangle
D) Scalene triangle
Answer:
C) Right angle triangle

Question 5.
The slope of the line joining (4, 6) and (2, -5) is
A) \(\frac{11}{2}\)
B) \(\frac{2}{11}\)
C) \(\frac{-2}{11}\)
D) \(\frac{-11}{2}\)
Answer:
A) \(\frac{11}{2}\)

Question 6.
The distance between (7, 0) and (4, k) is 5 units then k =
A) – 4
B) + 4
C) + 4
D) none
Answer:
B) + 4

Question 7.
A straight line make an angle 8 with the X-axis then the slope is
A) – sin θ
B) cos θ
C) tan θ
D) sec θ
Answer:
C) tan θ

Question 8.
Mid-point of the line joining points (-4, 2) and (2, 8) is
A) (-1, 5)
B) (-3, -3)
C) (-2, 10)
D) (-6, -6)
Answer:
A) (-1, 5)

Question 9.
(1, 2), (2, 3), (3, 4) are vertices of a triangle then its centroid is
A) (6, 9)
B) (0, -1)
C) (2, 3)
D) (-2, -3)
Answer:
C) (2, 3)

Question 10.
(1, 2), (2, 3), (3, 1) are the midpoints of the sides of a triangle then the area of the triangle is (in.sq.units)
A) 7
B) \(\frac{5}{2}\)
C) 14
D) 6
Answer:
D) 6

Question 11.
The co-ordinates of the mid point of the line joining points (3,-1) and (5, 3) is
A) (8, 4)
B) (\(\frac{8}{3}\), \(\frac{4}{3}\))
C) (4, 2)
D) (4, 1)
Answer:
D) (4, 1)

Question 12.
The equation of the line passing through origin having slope
A) 2x – 3y = 0
B) 2y = x
C) 3x – 2y = 0
D) 3y – 2x
Answer:
D) 3y – 2x

Question 13.
If a line make an angle 150° with +ve X – axis then the slope of the line
A) \(\sqrt{3}\)
B) –\(\frac{1}{\sqrt{3}}\)
C) –\(\sqrt{3}\)
D) \(\frac{1}{\sqrt{3}}\)
Answer:
B) –\(\frac{1}{\sqrt{3}}\)

Question 14.
Slope of the line 3x – 4y + 12 = 0
A) \(\frac{3}{4}\)
B) \(\frac{-4}{3}\)
C) 4
D) \(\frac{1}{\sqrt{3}}\)
Answer:
A) \(\frac{3}{4}\)

Question 15.
Slope of the line joining points (2, -3) (1, 4)
A) – 7
B) 7
C) -1
D) – \(\frac{3}{7}\)
Answer:
A) – 7

Question 16.
The angle made by the line y = x with +ve direction of Y-axis
A) 30°
B) 60°
C) 90°
D) 45°
Answer:
D) 45°

Question 17.
The slope of the line \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1
A) –\(\frac{\mathrm{a}}{\mathrm{b}}\)
B) \(\frac{\mathrm{-b}}{\mathrm{a}}\)
C) 1
D) \(\frac{\mathrm{a}}{\mathrm{b}}\)
Answer:
B) –\(\frac{\mathrm{-b}}{\mathrm{a}}\)

Question 18.
The line y = mx + c cut the Y – axis at the point
A) (0, 0)
B) (0, c)
C) (c, 0)
D) (0, m)
Answer:
B) (0, c)

Question 19.
The distance of the point (a, b) from the origin
A) \(\sqrt{a+b}\)
B) \(\sqrt{a^2-b^2}\)
C) \(\sqrt{a^2+b^2}\)
D) \(\sqrt{a}+\sqrt{b}\)
Answer:
C) \(\sqrt{a^2+b^2}\)

Question 20.
The centriod of the triangle made by the vertices (-2, 3), B(4, 1), C(1, 2) is
A) (1, 2)
B) (\(\frac{3}{2}\), 0)
C) (0, 0)
D) (\(\frac{7}{3}\), 2)
Answer:
A) (1, 2)

Question 21.
The slope of the line is \(\frac{2}{5}\) then the slope of the parallel of that line.
A) \(\frac{5}{2}\)
B) \(\frac{2}{5}\)
C) \(\frac{-5}{2}\)
D) 1
Answer:
B) \(\frac{2}{5}\)

Question 22.
If the line y = mx + c passing through the points (0, 3) (4, 0) then intercept is
A) 4
B) – 3
C) -4
D) 3
Answer:
A) 4

Question 23.
The line ax + by + c = 0 intersects X – axis at
A) (\(\frac{\mathrm{c}}{\mathrm{a}}\), 0)
B) (0, \(\frac{-\mathrm{c}}{\mathrm{b}}\))
C) (0, 0)
D) (\(\frac{-\mathrm{c}}{\mathrm{a}}\), 0)
Answer:
D) (\(\frac{-\mathrm{c}}{\mathrm{a}}\), 0)

Question 24.
The line ax + by + c = 0 intersects Y-axis at
A) (0, \(\frac{-\mathrm{c}}{\mathrm{b}}\))
B) (\(\frac{\mathrm{c}}{\mathrm{b}}\), 0)
C) (\(\frac{-\mathrm{c}}{\mathrm{b}}\), 0)
D) (0, -c)
Answer:
A) (0, \(\frac{-\mathrm{c}}{\mathrm{b}}\))

Question 25.
Distance between the points (0, 0) (a cosθ, a sinθ)
A) 2
B) a²
C) a
D) \(\frac{\mathrm{a}^2}{2}\)
Answer:
C) a

Question 26.
If ‘C is the mid point of A (0, 0) B(4, 8). then the co-ordinates of the mid-points B and C is
A) (2, 4)
B) (3, 6)
C) (1, 2)
D) (4, 6)
Answer:
B) (3, 6)

Question 27.
The slope of the line y = \(\frac{1}{2}\) x is
A) 0
B) \(\frac{1}{2}\)
C) –\(\frac{1}{2}\)
D) 1
Answer:
B) \(\frac{1}{2}\)

Question 28.
The equation of the X-axis is
A) y = 0
B) x = 0
C) x = y
D) xy = 0
Answer:
A) y = 0

Question 29.
The point of intersection of the lines x = 2 and y = 3 is
A) (0, 0)
B) (3, 0)
C) (0, 2)
D) (\(\frac{4}{3}\), 1)
Answer:
D) (\(\frac{4}{3}\), 1)

Question 30.
If the line 3x + 4y = k passing through the point (4, 2) then k
A) 10
B) – 20
C) 20
D) – 10
Answer:
C) 20

Question 31.
If the distance between two points (5, 2) (3, a) is \(\sqrt{8}\) units then a
A) – 2
B) 2
C) 8
D) 4
Answer:
D) 4

Question 32.
The ratio in which the Y-axis dividing line joining the points (5, 7) (-1, 3) is
A) 5 : 1
B) 3 : 7
C) 2 : 1
D) 4 : 3
Answer:
A) 5 : 1

Question 33.
The point on X – axis
A) (2, 3)
B) (2, 0)
C) (0, 4)
D) (0, -3)
Answer:
B) (2, 0)

Question 34.
The point on Y – axis
A) (0, 2)
B) (4, 0)
C) (-3, 4)
D) (2, -1)
Answer:
A) (0, 2)

Question 35.
The slope of the line x = 0
A) 0
B) 1
C) ∞
D) – 1
Answer:
C) ∞

Question 36.
The point (4, -7) in ……………….. quadrant
A) Q₁
B) Q₂
C) Q₃
D) Q₄
Answer:
D) Q₄

Question 37.
The line ax – by + c = 0 of slope
A) \(\frac{-\mathrm{a}}{\mathrm{b}}\)
B) \(\frac{\mathrm{a}}{\mathrm{b}}\)
C) \(\frac{-\mathrm{b}}{\mathrm{a}}\)
D) \(\frac{\mathrm{b}}{\mathrm{a}}\)
Answer:
B) \(\frac{\mathrm{a}}{\mathrm{b}}\)

Question 38.
The line y = 5 is
A) Parallel to x-axis
B) Parallel to y-axis
C) Perpendicular to x – axis
D) Perpendicular to y-axis
Answer:
A) Parallel to x-axis

Question 39.
If the line 2x – 3y = k passes through the origin then the value of k
A) – 1
B) 0
C) 1
D) –\(\frac{1}{2}\)
Answer:
D) –\(\frac{1}{2}\)

Question 40.
The centriod of the triangle made by the vertices A(-2, 3) (4, 1) (1, 2) is
A) (1, 2)
B) [\(\frac{3}{2}\), 2]
C) (0, 0)
D) [\(\frac{7}{3}\), 2]
Answer:
D) [\(\frac{7}{3}\), 2]

Question 41.
Area of the ∆ formed by the points A(0, 0), B (1, 0) and C(0, 1) is ……………….. sq. units. (A.P. Mar. ’15)
A) 0
B) \(\frac{1}{2}\)
C) 1
D) \(\frac{1}{4}\)
Answer:
B) \(\frac{1}{2}\)

Question 42.
Where do the points lie on co-ordinate axis (-4, 0), (2, 0), (6, 0), (8, 0) (A.P. Mar. ’15)
A) Q₁
B) x-axis
C) y-axis
D)Q₄
Answer:
B) x-axis

Question 43.
The distance between (0, 0) (x₁, y₁) points is ……………. units. (A.P. Mar. ’15)
A) \(\sqrt{x_1^2+y_1^2}\)
B) \(\sqrt{x_1+y_1}\)
C) \(\sqrt{x^2+y^2}\)
D) \(\sqrt{x+y}\)
Answer:
A) \(\sqrt{x_1^2+y_1^2}\)

Question 44.
The distance between the points (x₁, y₁) and (x₂, y₂) which are on the line parallel to Y-axis is ……………. (A.P. Mar.’15)
A) |Y₁ – Y₂| or |Y₂ – Y₁|
B) |y₂² – y₁²| or |y₁² – y₂²|
C) \(\sqrt{\left(x_2-x_1^2\right)+\left(y_2-y_1^2\right)}\)
D) |x₂ – x₁| or |x₁ – x₂|
Answer:
A) |Y₁ – Y₂| or |Y₂ – Y₁|

Question 45.
The mid point of line segment joined by (4, 5) and (-6, 3) is ……………… (A.P. June ’15)
A) (1, 4)
B) (-1, 4)
C) (1, -4)
D) (-1, -4)
Answer:
B) (-1, 4)

Question 46.
The distance to X-axis from the point (3, -4) is …………….. (A.P. June’15)
A) 3
B) 4
C) 5
D) 1
Answer:
B) 4

Question 47.
If the mid point of (-4, a) and (2, 8) is (-1, 5) then a = ………………. (A.P. June ’15)
A) -4
B) 2
C) 5
D) 8
Answer:
B) 2

Question 48.
The graph of y = 5 is ……………… (A.P. June ’15)
A) Parallel of X-axis
B) Perpendicular to X-axis
C) Parallel to Y-axis
D) Perpendicular to Y-axis
Answer:
A) Parallel of X-axis

Question 49.
The distance between (0, 7) and (-7, 0) is ………………… (A.P. Mar. ’16)
A) 2\(\sqrt{7}\)
B) 7\(\sqrt{2}\)
C) \(\sqrt{14}\)
D) +1
Answer:
B) 7\(\sqrt{2}\)

Question 50.
Slope of Y-axis is ……………… (A.P. Mar.’16)
A) not defined
B) 0
C) well defined
D) finite
Answer:
A) not defined

Question 51.
The distance from X-axis to (-4, 3) is ………………. units. (A.P. Mar.’16)
A) 2
B) 3
C) -4
D) -1
Answer:
B) 3

Question 52.
The distance from origin to (2, 3) is …………….. units. (A.P. Mar.’16)
A) \(\sqrt{6}\)
B) \(\sqrt{5}\)
C) \(\sqrt{1}\)
D) \(\sqrt{13}\)
Answer:
D) \(\sqrt{13}\)

Question 53.
The distance from Y-axis to (4, 0) is ……………… units. (A.P. Mar.’16)
A) 4 units
B) \(\sqrt{16}\) units
C) 16 units
D) 2\(\sqrt{2}\) units
Answer:
A) 4 units

Question 54.
The mid point of (2, 3) and (-2, 3) is ……………… (A.P. Mar.’16)
A) (0, 3)
B) (-2, 0)
C) (3, 0)
D) (-3, 2)
Answer:
A) (0, 3)

Question 55.
The centroid of the triangle formed by (0, 3); (3, 0) and (0, 0) is …………….. (A.P. Mar.’16)
A) (1, 1)
B) (0, 3)
C) (3, 3)
D) (3, 0)
Answer:
A) (1, 1)

Question 56.
Slope of the line that passes through the point P(x₁, y₁) and Q (x₂, y₂) and making an angle ‘θ’ with X-axis is ……………….. (A.P. Mar.’16)
A) \(\frac{\mathrm{y}_2+\mathrm{y}_1}{\mathrm{x}_2+\mathrm{x}_1}\)
B) θ
C) \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)
D) sin θ
Answer:
C) \(\frac{\mathrm{y}_2-\mathrm{y}_1}{\mathrm{x}_2-\mathrm{x}_1}\)

Question 57.
Slope of the line passing through the points (-1, 1) and (1, 1) is …………….. (A.P. Mar. ’16)
A) -1
B) 0
C) 1
D) not defined
Answer:
C) 1

Question 58.
A point on the Y-axis is of the form
A) (0, y)
B) (x, 0)
C) (x, y)
D) (y, y)
Answer:
A) (0, y)

Question 59.
A point on the X-axis is of the form
A) (0, y)
B) (x, 0)
C) (x, y)
D) (x, x)
Answer:
B) (x, 0)

Question 60.
The distance of the point (-8, 3) from the origin is ……………….
A) 5
B) 55
C) 73
D) 24
Answer:
C) 73

Question 61.
The distance of the point (-4, 3) from X-axis is ……………..
A) -4
B) -3
C) 4
D) 3
Answer:
D) 3

Question 62.
The distance of the point (-8, -7) from Y-axis is ………….
A) 8
B) -7
C) -8
D) 7
Answer:
A) 8

Question 63.
The points (-3,0), (0,5) and (3,0) are the vertices of a ……………… triangle.
A) scalene
B) isosceles
C) equilateral
D) right angled
Answer:
B) isosceles

Question 64.
The distance between the points (-2, 3) and (2, -3) is ……………..
A) 0
B) 52
C) \(\sqrt{52}\)
D) 16
Answer:
C) \(\sqrt{52}\)

Question 65.
If the distance between the points (4, y) and (1, 0) is 5, then y = …………
A) 0
B) 4
C) ±4
D) ±2
Answer:
C) ±4

Question 66.
The distance between the points (0, 7) and (-7, 0) is ……………….
A) \(\sqrt{14}\)
B) 49
C) 2\(\sqrt{7}\)
D) 7\(\sqrt{2}\)
Answer:
D) 7\(\sqrt{2}\)

Question 67.
A circle is drawn with origin as centre and passing through (2, 3), then its radius is ………………
A) 2
B) 3
C) 13
D) \(\sqrt{13}\)
Answer:
D) \(\sqrt{13}\)

Question 68.
The area of the triangle formed by (a, b + c), (b, c + a) and (c, a + b) is
A) 2(a + b + c)
B) abc
C) 0
D) a + b + c
Answer:
C) 0

Question 69.
If points (x, 0), (0, y) and (1, 1) are collinear, then \(\frac{1}{x}\) + \(\frac{1}{y}\) = …………….
A) 1
B) -1
C) 0
D) 2
Answer:
A) 1

Question 70.
The point which divides the line segment joining the points (3, 4) and (7, -6) internally in the ratio 1 : 2 lies in the ………….. quadrant.
A) Q₁
B) Q₂
C) Q₃
D) Q₄
Answer:
D) Q₄

Question 71.
The points (a, 2a), (3a, 3a) and (3, 1) are collinear, then k = …………….
A) \(\frac{-1}{4}\)
B) \(\frac{1}{3}\)
C) \(\frac{-2}{3}\)
D) \(\frac{-1}{3}\)
Answer:
D) \(\frac{-1}{3}\)

Question 72.
A circle drawn with origin as centre passes through (\(\frac{13}{2}\), 0). The point which doesn’t lie in the interior of the circle is ………………..
A) (-6, 3)
B) (5, \(\frac{1}{2}\))
C) (2, \(\frac{7}{3}\))
D) (\(\frac{-3}{4}\), 1)
Answer:
A) (-6, 3)

Question 73.
The distance of the point (-9, 40) from the origin is ………….
A) 9
B) 40
C) 53
D) 41
Answer:
D) 41

Question 74.
If (-2, 8) and (6, -4) are the end points of the diameter of a circle, then the centre of the circle is ………………..
A) (3, 6)
B) (4, 2)
C) (2, 2)
D) (-3, 2)
Answer:
C) (2, 2)

Question 75.
The angle between X-axis and Y-axis is ………………
A) 0°
B) 180°
C) 360°
D) 90°
Answer:
D) 90°

Question 76.
The midpoint of the line joining of (2, 3) and (-2, -3) is ………………
A) (0, 0)
B) (2, 3)
C) (1, 1\(\frac{1}{2}\))
D) (-1, -1\(\frac{1}{2}\))
Answer:
A) (0, 0)

Question 77.
The slope of line join of (5, -1), (0, 8) is ……………….
A) \(\frac{7}{5}\)
B) \(\frac{9}{5}\)
C) –\(\frac{9}{5}\)
D) –\(\frac{5}{9}\)
Answer:
C) –\(\frac{9}{5}\)

Question 78.
Slope of X-axis is ……………………
A) 0
B) 1
C) -1
D) Not defined
Answer:
A) 0

Question 79.
The centroid of the triangle whose vertices are (2, -3), (4, 6), (-2, 8) is
A) \(\left(\frac{8}{3}, \frac{17}{3}\right)\)
B) (4, 11)
C) (-3, -8)
D) \(\left(\frac{4}{3}, \frac{11}{3}\right)\)
Answer:
D) \(\left(\frac{4}{3}, \frac{11}{3}\right)\)

Question 80.
Two vertices of a triangle are (3, 5) and (-4, -5). If the centroid of the triangle is (4,3), find the third vertex.
A) (13, 9)
B) (-9, -13)
C) (9, 13)
D) (13, -9)
Answer:
A) (13, 9)

Question 81.
The ratio in which the point (4, 8) divide the line segment joining the points (8, 6) and (0, 10) is ………………..
A) 2 : 1
B) 1 : 1
C) 1 : 2
D) 3 : 1
Answer:
B) 1 : 1

Question 82.
If (-2, -1), (a, 0), (4, b) and (1,2) are the vertices of a parallelogram then a = ……………….
A) 3
B) -1
C) 4
D) 1
Answer:
A) 3

Question 83.
In the above problem b = ……………
A) 3
B) 4
C) -5
D) none
Answer:
D) none

Question 84.
(-2, 8) ∈ ………………….
A) Q₁
B) Q₄
C) Q₂
D) Q₃
Answer:
D) Q₃

Question 85.
If A, B, C are collinear then area of ∆ABC = ………………
A) 2
B) 1
C) 0
D) none
Answer:
C) 0

Question 86.
Area of triangle formed by (-4, 0), (0, 0) and (0, 5) is ……………. sq. units.
A) 12
B) 10
C) 13
D) 9
Answer:
B) 10

Question 87.
The value of p if the distance between (2, 3) and (p, 3) is 5 is ……………….
A) 7
B) 9
C) 12
D) 10
Answer:
A) 7

Question 88.
The value of k if the distance between (2, 8) and (2, k) is 3 is ……………….
A) 4.5
B) 10
C) 9
D) 5
Answer:
D) 5

Question 89.
A(0, -1), B(2,1) and C(0, 3) are the vertices of ∆ABC then median through B has a length …………… units.
A) 9.5
B) 10
C) 2
D) 9
Answer:
C) 2

Question 90.
The closed figure formed by the points (-2, 0), (2, 0), (2, 2), (0, 4) and (-2, -2) is a …………………
A) pentagon
B) triangle
C) circle
D) none
Answer:
A) pentagon

Question 91.
The co-ordinates of the midpoints joining P(x₁, y₁) and Q(x₂, y₂) is ……………
A) \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
B) \(\left(\frac{x_1-x_2}{2}, \frac{y_1+y_2}{2}\right)\)
C) \(\left(\frac{x_1+y_1}{2}, 1\right)\)
D) none
Answer:
A) \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

Question 92.
The centroid divides each median in the ……………… ratio.
A) 3 : 1
B) 1 : 3
C) 1 : 2
D) 2 : 1
Answer:
D) 2 : 1

Question 93.
If the distance between the points (3, k) and (4, 1) is \(\sqrt{10}\) then the value of k = ……………..
A) 8 or 10
B) 4 or -2
C) -1 or 2
D) none
Answer:
B) 4 or -2

Question 94.
If the points (1, 2) (-1, x) and (2, 3) are collinear then the value of x is ……………….
A) 9
B) 7
C) 0
D) -1
Answer:
C) 0

Question 95.
If the centroid of the triangle formed with (a, b), (b, c) and (c, a) is 0(0, 0) then a³ + b³ + c³ = ……………..
A) a + b + c
B) \(\frac{a+b+c}{3}\)
C) \(\frac{a b c}{3}\)
D) 3 abc
Answer:
D) 3 abc

Question 96.
The distance between the points (a cos θ, 0) and (0, a sin θ) is ……………….. units.
A) \(\frac{a}{3}\)
B) a
C) a²
D) \(\frac{a}{2}\)
Answer:
C) a²

Question 97.
Distance of (x, y) from X-axis is …………………
A) y
B) -x
C) -y
D) none
Answer:
A) y

Question 98.
Distance of (x, y) from Y-axis is …………………
A) -x
B) Y
C) x
D) none
Answer:
C) x

Question 99.
(x, 0) is a point on …………………..
A) X-axis
B) Y-axis
C) origin
D) none
Answer:
A) X-axis

Question 100.
(0, y) is a point on ………………..
A) (0, 0)
B) Y-axis
C) X-axis
D) none
Answer:
D) none

Question 101.
Distance of (x, y) from origin is ……………….
A) \(\sqrt{x}+\sqrt{y}\)
B) \(\sqrt{x+y}\)
C) \(\sqrt{xy}\)
D) \(\sqrt{x^2+y^2}\)
Answer:
D) \(\sqrt{x^2+y^2}\)

Question 102.
If a < 0 then (-a, -a) ∈ …………………..
A) Q₂
B) Q₁
C) Q₄
D) Q₃
Answer:
B) Q₁

Question 103.
Slope of the line y = mx is …………………
A) y
B) x
C) m
D) none
Answer:
C) m

Question 104.
Slope of the line joining the points (2a, 3b) and (a, -b) is ……………….
A) \(\frac{-a}{b}\)
B) \(\frac{b}{a}\)
C) \(\frac{b}{4 a}\)
D) \(\frac{4 b}{a}\)
Answer:
D) \(\frac{4 b}{a}\)

Question 105.
Slope of the line joining the points A(-1.4, -3.7) and B(-2.4, 1.3) is ……………….
A) -5
B) 5
C) 6
D) 7
Answer:
A) -5

Question 106.
(3, -5) ∈ …………………
A) Q₄
B) Q₃
C) Q₁
D) Q₂
Answer:
A) Q₄

Question 107.
The angle between the lines x = 2 and y = 3 is …………………….
A) 60°
B) 70°
C) 90°
D) 80°
Answer:
C) 90°

Question 108.
Slope of vertical line is ……………………
A) 0
B) -1
C) 3
D) not defined
Answer:
D) not defined

Question 109.
Area of triangle formed with (-5, -1), (3, -5) and (5, 2) is ………………. sq. units.
A) 28
B) 20
C) 32
D) 16
Answer:
C) 32

Question 110.
If the points (k, k), (2, 3) and (4, -1) are collinear then k = ……………………..
A) \(\frac{-1}{7}\)
B) \(\frac{1}{2}\)
C) \(\frac{3}{7}\)
D) \(\frac{7}{3}\)
Answer:
D) \(\frac{7}{3}\)

Question 111.
A(2, 0), B(1, 2), C(1, 6) then ∆ABC = …………………..
A) 10
B) 12
C) 0
D) 9
Answer:
C) 0

Question 112.
Identify collinear points.
A) (1, -6) (3, -4), (4, -3)
B) (1, -1) (2, 3), (2, 0)
C) (5, 2), (3, -5), (-5, -1)
D) all
Answer:
A) (1, -6) (3, -4), (4, -3)

Question 113.
The area of square formed with the vertices (0, -1), (2, 1), (0, 3) and (-2, 1) taken in order as vertices is ………………. sq. units.
A) 12
B) 6
C) 8
D) none
Answer:
D) none

Question 114.
The co-ordinates of centroid of the triangle formed with the vertices (-1, 3), (6, -3) and (-3, 6) is …………………….
A) (1, \(\frac{1}{2}\))
B) (\(\frac{2}{3}\), 2)
C) (8, \(\frac{-1}{2}\))
D) (0, 3)
Answer:
B) (\(\frac{2}{3}\), 2)

Question 115.
A(1, -1), B(0, 6) and C(-3, 0) then G = ………………..
A) (\(\frac{8}{9}\), \(\frac{1}{7}\))
B) (\(\frac{6}{7}\), \(\frac{1}{3}\))
C) (\(\frac{1}{2}\), \(\frac{1}{3}\))
D) (\(\frac{-2}{3}\), \(\frac{5}{3}\))
Answer:
D) (\(\frac{-2}{3}\), \(\frac{5}{3}\))

Question 116.
The point of concurrence of medians of a triangle is called ……………….
A) centroid
B) orthocentre
C) centre
D) none
Answer:
A) centroid

Question 117.
Mid point of the line joining the points (1,1) and (0, 0) is ……………….
A) (0, 9)
B) (3, 7)
C) (\(\frac{1}{2}\), \(\frac{1}{2}\))
D) (1, \(\frac{1}{2}\))
Answer:
C) (\(\frac{1}{2}\), \(\frac{1}{2}\))

Question 118.
The radius of the circle whose centre is (3, 2) and passes through (-5, 6) is ………………. units.
A) 2\(\sqrt{5}\)
B) 4\(\sqrt{7}\)
C) 4\(\sqrt{3}\)
D) 4\(\sqrt{5}\)
Answer:
D) 4\(\sqrt{5}\)

Question 119.
Area of parallelogram = ……………….. sq.units.
A) \(\frac{1}{2}\)bh
B) bh
C) b²h²
D) none
Answer:
B) bh

Question 120.
A(4, 5), B{7, 6) then AB = …………………. units.
A) \(\sqrt{10}\)
B) 10
C) 8
D) \(\sqrt{19}\)
Answer:
A) \(\sqrt{10}\)

Question 121.
In quadrilateral ABCD, AB = BC = CD = AD and AC ≠ BD then it is a ……………
A) trapezium
B) square
C) parallelogram
D) none
Answer:
C) parallelogram

Question 122.
A(a, b) and B(-a, -b) then BA = …………………. units.
A) 2\(\sqrt{a}\)
B) 2\(\sqrt{a^2+b^2}\)
C) 2\(\sqrt{b}\)
D) 2\(\sqrt{a^2+b}\)
Answer:
B) 2\(\sqrt{a^2+b^2}\)

Question 123.
If θ is the angle made by a line with X-axis then slope m = …………………..
A) tan θ
B) sec θ
C) cosec θ
D) none θ
Answer:
A) tan θ

Question 124.
A(4, 0), B(8, 0) then AB = ………………. units.
A) 6
B) 10
C) 4
D) 12
Answer:
C) 4

Question 125.
Other name for X-Coordinate of a point is …………………..
A) abscissa
B) point
C) ordinate
D) none
Answer:
A) abscissa

Question 126.
(8, 10) ∈ …………………….
A) Q₂
B) Q₁
C) Q₃
D) none
Answer:
B) Q₁

Question 127.
Slope of horizontal line is ………………….
A) 3
B) -1
C) 0
D) none
Answer:
C) 0

Question 128.
ax + by + c = 0, represents a ………………..
A) straight line
B) circle
C) curve
D) none
Answer:
A) straight line

Question 129.
In Heron’s formula S = …………….
A) \(\frac{a-b-c}{2}\)
b) \(\frac{a+b-c}{2}\)
C) \(\frac{ab}{2}\) + c
D) \(\frac{a+b+c}{2}\)
Answer:
D) \(\frac{a+b+c}{2}\)

Question 130.
Coordinates of origin are ……………………
A) (a, b)
B) (3, 7)
C) (0, 0)
D) none
Answer:
C) (0, 0)

Question 131.
A(4, 3), B(8, 6) then AB = ………………. units.
A) 9
B) 5
C) 16
D) 12
Answer:
B) 5

Question 132.
Q₁ ∩ Q₂ = …………………..
A) Φ
B) {0}
C){8, 4}
D) none
Answer:
A) Φ

Question 133.
Slope of the line \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 is ………………..
A) \(\frac{-b}{a}\)
B) \(\frac{b}{a}\)
C) \(\frac{a}{b}\)
D) none
Answer:
A) \(\frac{-b}{a}\)

Question 134.
The midpoint of the line joining the points (1, 2) and (1, p) is (1, -1) then p = ……………………
A) -31
B) -3
C) -4
D) none
Answer:
C) -4

Question 135.
The centroid of the triangle formed with the line x + y = 6 with the co-ordinate axes is ………………..
A) (4, 0)
B) (1, 3)
C) (8, 1)
D) (2, 2)
Answer:
D) (2, 2)

Question 136.
Slope of the line joining the points (2, 5) and (k, 3) is 2 then k = ………………..
A) 4
B) 1
C) -1
D) none
Answer:
B) 1

Question 137.
A point on X-axis is …………………….
A) (9, 0)
B) (0, 3)
C) (9, 3)
D) (3, -1)
Answer:
A) (9, 0)

Question 138.
The slope of a line passing through (-2, 3) and (4, a) is \(\frac{-5}{3}\) then a = ……………….
A) 1
B) 7
C) -7
D) 2
Answer:
C) -7

Question 139.
If (1, x) is at \(\sqrt{10}\) units from origin then the value of x = ………………..
A) ±31
B) ±3
C) ±2
D) ±1
Answer:
B) ±3

Question 140.
A = (\(\frac{1}{2}\), \(\frac{3}{2}\)), B = (\(\frac{3}{2}\), \(\frac{-1}{2}\)) then BA = ……………
A) \(\sqrt{5}\)
B) \(\sqrt{6}\)
C) \(\sqrt{19}\)
D) none
Answer:
A) \(\sqrt{5}\)

Question 141.
X and Y axes will intersect at …………………
A) (1, 1)
B) (2, 2)
C) (0, 0)
D) (8, 5)
Answer:
C) (0, 0)

Question 142.
In ∆ ABC, AB = AC = BC then it is ……………….. triangle.
A) scalene
B) equilateral
C) isosceles
D) none
Answer:
B) equilateral

Question 143.
Y-axis can be represented by ………………….
A) x = 0
B) y = 0
C) y = \(\frac{-1}{2}\)
D) all
Answer:
A) x = 0

Question 144.
y intercept of the line x – 2y + 1 = 0 is ……………..
A) \(\frac{-1}{2}\)
B) 1
C) -1
D) \(\frac{1}{2}\)
Answer:
B) 1

Question 145.
In the figure AD : GD = ………………..

A) 3 : 1
B) 1 : 2
C) 2 : 1
D) none
Answer:
A) 3 : 1

Question 146.
Equation of X-axis is ………………..
A) x = 0
B) x = 7
C) x = 1
D) y = 0
Answer:
D) y = 0

Question 147.
If (p, 2p), (2p, 3p) and (3,1) are collinear then P = …………………
A) \(\frac{1}{3}\)
B) -1
C) \(\frac{-1}{3}\)
D) none
Answer:
A) \(\frac{1}{3}\)

Question 148.
In ∆ABC, all the sides are different then it is called ………………….. triangle.
A) isosceles
B) scalene
C) equilateral
D) none
Answer:
B) scalene

Question 149.
In ∆PQR, PQ = QR then it is called …………………… triangle.
A) isosceles
B) right triangle
C) equilateral
D) none
Answer:
A) isosceles

Question 150.
A(1, -1), B(2 1/2, 0), C(4, 1) then area of ∆ABC = ……………… sq. units.
A) 2
B) 9
C) 0
D) none
Answer:
C) 0

Question 151.
The point of concurrence of attitudes of a triangle is called its ……………….
A) orthocentre
B) centroid
C) isosceles
D) none
Answer:
A) orthocentre

Question 152.
Angle made by the line y = x with the positive direction of X-axis is ……………………
A) 45°
B) 60°
C) 90°
D) 70°
Answer:
A) 45°

Question 153.
Number of medians of a triangle is ……………..
A) 5
B) 4
C) 7
D) 3
Answer:
D) 3

Question 154.
Slope of line y = 7 is ……………….
A) 1
B) 7
C) 0
D) none
Answer:
C) 0

Question 155.
If A(p, q), B(m, n) and C(p – m, q – n) are collinear then pn = ………………….
A) q²m
B) qm
C) \(\frac{q}{m}\)
D) none
Answer:
B) qm

Question 156.
In the problem y = …………………

A) 3
B) 7
C) -3
D) 8
Answer:
A) 3

Question 157.
Area of trapezium = ……………….. sq. units.
A) ph
B) h(a + b)
C) \(\frac{1}{2}\) h(a + b)
D) \(\frac{1}{2}\)(a + b)
Answer:
C) \(\frac{1}{2}\) h(a + b)

Question 158.
P(cos θ, -cos θ), Q(sin θ, sin θ) then PQ = ……………….
A) cos θ
B) sin² θ
C) 0
D) none
Answer:
D) none

Question 159.
A(t, 2t), B(-2, 6), C(3, 1) and ∆ABC = 5 sq. units then t = ………………..
A) 9
B) 4
C) -9
D) 2
Answer:
D) 2

Question 160.
The diagonals of a parallelogram whose vertices are (2, 3), (4, 5), (4, 9) and (2, 7) will intersect at ………………
A) (0, 0)
B) (5, 6)
C) (0, 9)
D) (3, 6)
Answer:
B) (5, 6)

Question 161.
Slope of the line 3x – 2 = 0 is …………………
A) 2
B) 3
C) 0
D) not defined
Answer:
D) not defined

Question 162.
Each angle is equilateral triangle is ………………….
A) 100°
B) 70°
C) 60°
D) 90°
Answer:
C) 60°

Question 163.
A(cot θ, 1), B(0, 0) then BA = ………………….
A) 5
B) 4
C) 1
D) none
Answer:
D) none

Question 164.
Slope of the line joining the points A(0, 0), B(\(\frac{1}{2}\), \(\frac{1}{2}\)) is ……………………
A) 4
B) 1
C) 3
D) 7
Answer:
B) 1

Question 165.
(3, 0), (8, 0), (\(\frac{1}{2}\), 0) ……………….. points lie on ………………..
A) X-axis
B) Y-axis
C) (0, 0)
D) none
Answer:
A) X-axis

Question 166.
(x, y) ∈ Q₄ then ……………….
A) x = 0, y = 0
B) x < 0, y > 0
C) x > 0, y < 0
D) none
Answer:
C) x > 0, y < 0

Question 167.
y intercept of the line y = mx + c is ……………….
A) y
B) m
C) 1
D) none
Answer:
D) none

Question 168.
The midpoint of a line segment divides it in the ratio
A) 1 : 1
B) 2 : 1
C) 1 : 2
D) 1 : 4
Answer:
A) 1 : 1

Question 169.
Diagonals in a parallelogram …………………
A) equal
B) trisect
C) bisect
D) none
Answer:
C) bisect

Question 170.
The line joining the mid point of one side of a triangle from opposite vertex in called ………………….
A) ortho centre
B) median
C) centroid
D) none
Answer:
B) median

Question 171.
x intercept of the line x – y + 1 = 0 is ………………
A) 1
B) 2
C) 7
D) -1
Answer:
A) 1

Question 172.
In rhombus all sides are ……………….
A) equal
B) not equal
C) 3 cm
D) 8 cm
Answer:
A) equal

Question 173.
If the point (4, -p) lie on X-axis then p² + 2p – 1 = ………………….
A) 0
B) 1
C) -1
D) 4
Answer:
C) -1

Question 174.
If the point (a, 5) lie on Y-axis, the value of a …………………
A) a > 0
B) a < 0
C) a = 0
D) none
Answer:
C) a = 0

Question 175.
If the distance between the points (x₁, y₁) and (x₂, y₂) is |x₁ – x₂| then they are parallel to ……………….
A) X-axis
B) XY-axis
C) XY-axis
D) Y-axis
Answer:
A) X-axis

Solving these TS 10th Class Maths Bits with Answers Chapter 14 Statistics Bits for 10th Class will help students to build their problem-solving skills.

Statistics Bits for 10th Class

Question 1.
Write upper limit of 20 – 25
A) 22.5
B) 20
C) 25
D) 45
Answer:
C) 25

Question 2.
Write lower limit of 4 – 11
A) 7
B) 4
C) 7 – 5
D) 11
Answer:
B) 4

Question 3.
What is the mid value 35 – 45
A) 40
B) 35
C) 45
D) 10
Answer:
A) 40

Question 4.
The most frequently used measure of central tendency
A) Median
B) Mode
C) Mean
D) None of these
Answer:
C) Mean

Question 5.
The value of among the observation which occurs most frequently is called
A) Median
B) Mode
C) Mean
D) None of these
Answer:
B) Mode

Question 6.
The width of the class interval 20 – 30 is
A) 10
B) 20
C) 30
D) 25
Answer:
A) 10

Question 7.
Which is the better measure of central tendency when individual observations are not important
A) Mean
B) Mode
C) Median
D) None of these
Answer:
C) Median

Question 8.
Which measure of central tendency is given by the x – coordinate of the point of intersection of the more than ogive and less than ogive
A) Mean
B) Median
C) Mode
D) Weighteel mean
Answer:
B) Median

Question 9.
The A.M of first n natural numbers is
A) \(\frac{\mathrm{n}}{2}\) + 1
B) \(\frac{\mathrm{n}-1}{2}\)
C) \(\frac{\mathrm{n}+1}{2}\)
D) \(\frac{\mathrm{n}}{2}\)
Answer:
C) \(\frac{\mathrm{n}+1}{2}\)

Question 10.
Which of the following is not a measure of’ central tendency
A) standard deviation
B) mode
C) median
D) mean
Answer:
A) standard deviation

Question 11.
If the mean of 4, x, 6, 9, 9, y and 13 is 8 then
A) x + y = 15
B) x – y = 16
C) xy = 16
D) x + y = 8
Answer:
A) x + y = 15

Question 12.
The mean of first ‘n’ odd natural number is
A) \(\frac{\mathrm{n}}{2}\)
B) n
C) n²
D) \(\frac{\mathrm{n}+1}{2}\)
Answer:
B) n

Question 13.
If the mean of the first n natural numbers is 11 then n =
A) 11
B) 12
C) 9
D) 21
Answer:
D) 21

Question 14.
The median of the first 8 prime numbers is
A) 7
B) 11
C) 9
D) 5
Answer:
C) 9

Question 15.
If the median of the data 15, 16, 17, 2x + 4, 3x – 2, 21, 22 written in ascending order is 18.5 then x
A) 6
B) 5
C) 8
D) 7
Answer:
D) 7

Question 16.
The empirical relationship between mean, median and mode is
A) Mode = 3 median – 2 mean
B) Mode = 2 median – 3 mean
C) Mode = 3 median + 2 mean
D) Mode = 2 median + 3 mean
Answer:
A) Mode = 3 median – 2 mean

Question 17.
If 35 is removed from the data; 30, 34, 35, 36, 37, 38, 39, 40 the median is increased by
A) 0.5
B) 1
C) 1.5
D) 3
Answer:
A) 0.5

Question 18.
The median and mean of a distribution of 29 and 32 respectively then mode
A) 25.5
B) 23
C) 18
D) 30.5
Answer:
B) 23

Question 19.
The mode of the distribution 2, 5, 7, 4, 5, 6, 5, 4, 3 is
A) 4
B) 6
C) 5
D) 3
Answer:
C) 5

Question 20.
The median of the observations 15, 12, 14, 20, 16, 10 is
A) 14
B) 15
C) 13.5
D) 14.5
Answer:
D) 14.5

Question 21.
The mid value of the class 10 – 19 is
A) 9
B) 10
C) 14.5
D) 29
Answer:
C) 14.5

Question 22.
Mid value of the class is used to calculate
A) A.M.
B) Median
C) Mode
D) None of these
Answer:
C) Mode

Question 23.
The mean of x – 1, x and x + 1 is
A) 3x
B) x
C) \(\frac{3 x-1}{3}\)
D) \(\frac{3 x+1}{3}\)
Answer:
B) x

Question 24.
If mean = x; median = y then mode
A) \(\frac{x+y}{2}\)
B) 3y – 2x
C) 2y – 3x
D) \(\frac{x}{2}\) + \(\frac{y}{3}\)
Answer:
B) 3y – 2x

Question 25.
The median of the observation 3, 18, 6, 16, 12 and 10 is
A) 11
B) 10
C) 12
D) 13
Answer:
A) 11

Question 26.
If mean of 8, 6, 4, x, 3, 6 and 0 is 4 then the value of x = ……………. (A.P. Mar. ’15)
A) 7
B) 6
C) 1
D) 4
Answer:
C) 1

Question 27.
The extreme values of some data influences high on ……………. (A.P. Mar. ’15)
A) AM
B) Median
C) Mode
D) Range
Answer:
A) AM

Question 28.
In a data ‘n’ scores are given and if ‘n’ is odd, then median is ……………. (A.P. Mar. ’15)
A) \(\left(\frac{n+1}{2}\right)^{\text {th }}\) event
B) n^th event
C) \(\left(\frac{n-1}{2}\right)^{\text {th }}\) event
D) (n – 1)^th event
Answer:
A) \(\left(\frac{n+1}{2}\right)^{\text {th }}\) event

Question 29.
The class mark of 10 – 25 is …………… (A.P. June. ‘ 15)
A) 10
B) 25
C) 17.5
D) 17
Answer:
C) 17.5

Question 30.
Mode of the data 5, 3, 4, -2, 3, 2, 2, 1, P is 3 then the value of ‘p’ = ……………… (A.P. June. ’15)
A) 2
B) 5
C) -2
D) 3
Answer:
D) 3

Question 31.
Class interval of the class 11 – 20 is …………………. (A.P. Mar. ’16 ) (A.P. June.’15)
A) 9
B) 10
C) 11
D) 20
Answer:
B) 10

Question 32.
The AM of the data 8, 6, 4, x, 3, 6, 0 is 4 then the value of x = …………… (A.P. Mar. ’16)
A) 7
B) 6
C) 1
D) 4
Answer:
C) 1

Question 33.
Mean of 1, 2, x, 3 is ‘0’ then the value of x = ……………. (A.P. Mar. ’16)
A) 6
B) -6
C) 3
D) -3
Answer:
B) -6

Question 34.
If the sum of 15 observations is 420 then their mean = …………… (A.P. Mar. ’16 )
A) 28
B) 26
C) 24
D) 30
Answer:
A) 28

Question 35.
In “more than ogive curve” we consider in drawing ………………. (A.P. Mar. ’15 )
A) more than cumulative frequency, lower limits
B) more than cumulative frequency, upper limits
C) lower than cumulative frequency, lower limits
D) lower than cumulative frequency, upper limits
Answer:
A) more than cumulative frequency, lower limits

Question 36.
Observe the following tables. (T.S. Mar. ’15 )

For finding Arithmetic Mean by Direct method, the suggested frequency distribution table is ……………..
A) Only (1) is true
B) Only (2) is true
C) (1) and (2) are true
D) None of the above.
Answer:
A) Only (1) is true

Question 37.
If \(\overline{\mathrm{x}}\), is the mean of x₁, x₂, x₃, ………….. x_n (n items) then \(\sum_{i=1}^n\)(x₁ – \(\overline{\mathrm{x}}\)) = ……………….. (T.S. Mar. ’16 )
A) 0
B) n\(\overline{\mathrm{x}}\)
C) \(\frac{\overline{\mathrm{x}}}{\mathrm{n}}\)
D) \(\frac{2 \overline{\mathrm{x}}}{\mathrm{n}}\)
Answer:
A) 0

Question 38.
Mode can be calculated by = l + \(\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)\) × h here f₁ represents ………………. (T.S. Mar. ’16)
A) frequency of the modal class
B) frequency of class preceding modal class
C) frequency of class succeeding in the modal class
D) cumulative frequency of the class preceding the modal class
Answer:
A) frequency of the modal class

Question 39.
If the mode of (X > 0) is 5 then X = \(\frac{\mathrm{x}}{4}\), x, \(\frac{\mathrm{x}}{5}\), \(\frac{\mathrm{x}}{6}\), \(\frac{\mathrm{x}}{4}\)
A) 20
B) 10
C) 15
D) 8
Answer:
A) 20

Question 40.
If 20 is removed from the data 20, 24, 25, 26, 27, 28, 29, 30 then the median is increased by
A) 1
B) 1.5
C) 0.5
D) 2
Answer:
C) 0.5

Question 41.
Which of the following measure of central tendency is mostly effected by the extreme?
A) Mean
B) Median
C) Mode
D) Range
Answer:
B) Median

Question 42.
Match the following
1) Mean of first 10 natural numbers  ( )   p) 4.5
2) Median of first 10 whole numbers ( )  q) 5.5
3) Mode of first 10 natural numbers  ( )   r) does not exist

1) 1 → r, 2 → p, 3 → q
2) 1 → q, 2 → p, 3 → r
3) 1 → p, 2 → r, 3 → q
4) 1 → q, 2 → r, 3 → p
Answer:
2) 1 → q, 2 → p, 3 → r

Question 43.
The middle most value of a data is called
A) Mean
B) Mode
C) Median
D) Both B and C
Answer:
B) Mode

Question 44.
If the mean of 4, X, 6,9, Y, 13 is 8 then the relation between X and Y is
A) X + Y = 16
B) X – Y = 16
C) XY = 16
D) 2X – 3Y = 16
Answer:
A) X + Y = 16

Question 45.
From the adjoining ‘ogive Curve, the value 15 represents

A) Mean
B) Mode
C) Median
D) Range
Answer:
C) Median

Question 46.
The symbol for implies is
A) ⇒
B) ⇔
C) ∀
D) ∃
Answer:
A) ⇒

Question 47.
Range of the data 16, 17, 15, 11, 18, 23, 10, 9, 10
A) 14
B) 13
C) 7
D) 13
Answer:
A) 14

Question 48.
The class marks of a class interval is …………….
A) Upper boundary + lower boundary
B) Upper boundary – Lower boundary
C) \(\frac{\text { Upper boundary }+\text { Lower boundary }}{2}\)
B) \(\frac{\text { Upper boundary }-\text { Lower boundary }}{2}\)
Answer:
C) \(\frac{\text { Upper boundary }+\text { Lower boundary }}{2}\)

Question 49.
If mode of the following data is 7, then the value of 6, 3, 5, 6, 7, 5, 8, 7, 6, 2k + 1, 9 7, 13 is
A) \(\frac{5}{2}\)
B) 3
C) 7
D) 5
Answer:
B) 3

Question 50.
Mode of the following distribution is

A) 50
B) 56
C) 52
D) 54
Answer:
C) 52

Question 51.
The model class for the following distribution is

A) 40 – 50
B) 30 – 40
C) 50 – 60
D) 0 – 10
Answer:
B) 30 – 40

Question 52.
The lower limit of the modal class of the following data is

A) 15
B) 25
C) 10
D) 5
Answer:
C) 10

Question 53.
Which of the following is not a measure of central tendency ?
A) Mean
B) Median
C) Range
D) Mode
Answer:
C) Range

Question 54.
For a symmetrical distribution, which is correct ?
A) Mean < Mode < Median
B) Mean > Mode > Median
C) Mode = \(\frac{\text { Mean }+\text { Median }}{2}\)
D) Mean = Median = Mode
Answer:
D) Mean = Median = Mode

Question 55.
The upper limit of median class of the following distribution is

A) 17
B) 17.5
C) 18
D) 18.5
Answer:
C) 18

Question 56.
The measure of central tendency which take into account all data terms is
A) Mode
B) Mean
C) Median
D) None of these
Answer:
B) Mean

Question 57.
A data arrange in descending order has 25 observations. Which observation represents the median ?
A) 12^th
B) 13^th
C) 14^th
D) 15^th
Answer:
B) 13^th

Question 58.
Construction of cumulative frequency table is useful in determining the
A) Mean
B) Median
C) Mode
D) All the above
Answer:
B) Median

Question 59.
For a given data with 60 observations, the less than ogive’ and the more than ogive intersect at (66.5, 30). The median of the data is ……………..
A) 30
B) 66.5
C) 60
D) 36.5
Answer:
B) 66.5

Question 60.
For a given data with 50 observations the less than ogive’ and the more than ogive intersect at (15.5, 20). The median of the data is …………
A) 10.5
B) 4.5
C) 20
D) 15.5
Answer:
D) 15.5

Question 61.
The abscissa of the point of intersection of the less than type and ‘more than type’ cumulative frequency curves of a grouped data gives its ……………
A) Mean
B) Mode
C) Mode
D) None
Answer:
C) Mode

Question 62.
In the figure, the value of median of the data using the graph of less than ogive and more than ogive is
A) 5
B) 40
C) 80
D) 20

Answer:
D) 20

Question 63.
For a distribution with odd number (n) of observations, the median is ………… th observation.
A) \(\frac{\mathrm{n}}{2}\)
B) \(\frac{\mathrm{n}-1}{2}\)
C) \(\frac{\mathrm{n}+1}{2}\)
D) \(\frac{\mathrm{n}}{2}\) – 1
Answer:
C) \(\frac{\mathrm{n}+1}{2}\)

Question 64.
For a distribution with even number (n) of observations, the median is …………… terms
A) \(\frac{1}{2}\left[\frac{\mathrm{n}^{\text {th }}}{2}+\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }}\right]\)
B) \(\frac{1}{2}\left[\frac{\mathrm{n}}{2}-\frac{\mathrm{n}-1}{2}\right]\)
C) \(\frac{\mathrm{n}}{2}\) + \(\frac{\mathrm{n}+1}{2}\)
D) None
Answer:
A) \(\frac{1}{2}\left[\frac{\mathrm{n}^{\text {th }}}{2}+\left(\frac{\mathrm{n}}{2}+1\right)^{\text {th }}\right]\)

Question 65.
For a continuous grouped frequency distri¬bution, the median is given by
A) l + \(\left(\frac{\frac{n}{2}-f}{c}\right)\)h
B) l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
C) l² + \(\left(\frac{\frac{n}{2}+c f}{h}\right)\)
D) None
Answer:
B) l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h

Question 66.
Class Marks of a class x – y is …………..
A) \(\frac{\mathrm{x}}{2}\)
B) \(\frac{\mathrm{x}}{2}\)
C) xy
D) \(\frac{x+y}{2}\)
Answer:
D) \(\frac{x+y}{2}\)

Question 67.
Mean of n-observations x₁, x₂, …., x_n repeated f₁, f₂, f₃, f_n times is ………….

Answer:
A

Question 68.
Mode of a continuous grouped distribution is …………..
A) l + \(\frac{\mathrm{f}_1-\mathrm{f}_0}{\left(\mathrm{f}_1-\mathrm{f}_0\right)+\left(\mathrm{f}_1-\mathrm{f}_2\right)}\) × h
B) l + \(\frac{\mathrm{f}_1-\mathrm{f}_0}{\mathrm{f}_1-\mathrm{f}_0}\) × h
C) l² + \(\frac{\mathrm{f}_1-\mathrm{f}_0}{\mathrm{f}_1-\mathrm{f}_0+\mathrm{f}_2}\) × h
D) None
Answer:
A) l + \(\frac{\mathrm{f}_1-\mathrm{f}_0}{\left(\mathrm{f}_1-\mathrm{f}_0\right)+\left(\mathrm{f}_1-\mathrm{f}_2\right)}\) × h

Question 69.
If assumed mean is ‘a’ then mean = …………….
A) a² + Σf_id_i
B) a + Σf_id_i
C) a – Σf_i
D) a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)
Answer:
D) a + \(\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}\)

Question 70.
Mode is the value of variate which occurs …………… number of times
A) 2
B) Maximum
C) minimum
D) None
Answer:
B) Maximum

Question 71.
If each observation of a data is increased by ‘a’ then mean is increases by …………
A) a²
B) a
C) \(\frac{a}{2}\)
D) a + 1
Answer:
B) a

Question 72.
If the mean of x₁, x₂, …………, x_n is \(\overline{\mathrm{x}}\) then the mean of \(\frac{x_1}{a}\), \(\frac{x_2}{a}\), ……….., \(\frac{x_n}{a}\) is …………
A) \(\frac{\overline{\mathrm{x}}}{a}\)
B) \(\frac{\mathrm{x}}{a}\) – 1
C) x – a
D) xa
Answer:
A) \(\frac{\overline{\mathrm{x}}}{a}\)

Question 73.
\(\frac{\text { Sum of observations }}{\text { Number of observations }}\) = ………
A) mode
B) median
C) scale
D) mean
Answer:
D) mean

Question 74.
Mean of 1, 2, 3., ………., n is …………….
A) \(\frac{\mathrm{n}}{2}\) – 1
B) \(\frac{\mathrm{n}}{2}\)
C) \(\frac{\mathrm{n}+1}{2}\)
D) none
Answer:
C) \(\frac{\mathrm{n}+1}{2}\)

Question 75.
A.M. of 23, 24, 24, 22, 10 is ……………
A) 21.6
B) 22.6
C) 12.6
D) 81.6
Answer:
B) 22.6

Question 76.
Mode of 1, 2, 3, ……….10, 10 is …………
A) 1
B) 0
C) no mode
D) 10
Answer:
D) 10

Question 77.
Mode of 5, 6, 9, 10, 6, 12, 3, 6, 11, 10, 4, 6, 7 is …………….
A) 8
B) 7
C) 6
D) 5
Answer:
C) 6

Question 78.
Mode of 20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 2, 3 is …………..
A) 3, 7
B) 7, 10
C) 13, 3
D) none
Answer:
A) 3, 7

Question 79.
Mode of 0, 1, 2, 3, 3, 3, 7, is ………………..
A) 3
B) 0
C) 1
D) 9
Answer:
A) 3

Question 80.
Representing the data with the help of pictures is called …………..
A) data
B) pictograph
C) bar graph
D) none
Answer:
B) pictograph

Question 81.
Mid value of the class 10 – 20 is ……………
A) 13
B) 12
C) 10
D) 15
Answer:
D) 15

Question 82.
Histogram consists of ………….
A) rectangles
B) circles
C) triangles
D) none
Answer:
A) rectangles

Question 83.
Pie diagram consists of …………..
A) circles
B) sectors
C) rectangles
D) none
Answer:
B) sectors

Question 84.
Data having two modes is called ………… data.
A) unimodal
B) bimodal
C) trimodal
D) none
Answer:
B) bimodal

Question 85.
Mid values are used to calculate …………..
A) mean
B) mode
C) median
D) none
Answer:
A) mean

Question 86.
1 – 8, 9 – 16, 17 – 24, then C.I. = …………..
A) 12
B) 10
C) 9
D) 8
Answer:
D) 8

Question 87.
Range of 1, 2, 3, ………, 10 is …………..
A) 13
B) 12
C) 8
D) 9
Answer:
D) 9

Question 88.
Mean of 7, 6, 5, 0, 7, 8, 9 is ……………
A) 6
B) 8
C) 9
D) none
Answer:
A) 6

Question 89.
The mean of the following data is …………

A) 16.2
B) 15.1
C) 15.9
D) none
Answer:
B) 15.1

Question 90.
If mode of a distribution is 8 and its mean is then median is
A) 6.1
B) 18.2
C) 9
D) 8
Answer:
D) 8

Question 91.
If the mean of 10, 12, 18, 13, p and 17 is 15 then p = ………….
A) 20
B) 10
C) 30
D) 12
Answer:
A) 20

Question 92.
The mean of first five prime numbers is …………….
A) 8.1
B) 7.3
C) 6.5
D) 5.6
Answer:
D) 5.6

Question 93.
The median of the data 5, 3, 10, 7, 2, 9, 11, 2, 6 is ……………
A) 6
B) 2
C) 1
D) none
Answer:
A) 6

Question 94.
Mode of first ‘n’ natural numbers is ……………..
A) n – 1
B) n²
C) n + 1
D) no mode
Answer:
D) no mode

Question 95.
……………… is effected by extreme values.
A) Mean
B) Mode
C) Median
D) None
Answer:
A) Mean

Question 96.
Mean of -8, -4 and 4, 8 is …………….
A) -4
B) 8
C) 0
D) 7
Answer:
C) 0

Question 97.
Range of first 5 natural numbers is …………….
A) 9
B) 0
C) 5
D) 4
Answer:
D) 4

Question 98.
Empirical relation among mean, median and mode is …………….
A) mode = 3 median – 2 mean
B) mode = 2 median – mean
C) mode = 3 median – mean
D) all the above
Answer:
A) mode = 3 median – 2 mean

Question 99.
AM of 1², 2², 3²,4², ……….., 20² = ………….
A) 40
B) 50
C) 60
D) none
Answer:
D) none

Question 100.
Which of the following is not a measure of central tendency ?
A) mean
B) range
C) median
D) none
Answer:
B) range

Question 101.
A data has 13 observations arranged in descending order which observation represents the median of data ?
A) 17^th
B) 6^th
C) 7^th
D) none
Answer:
C) 7^th

Question 102.
The modal class in the following frequency distribution is ……………….

A) 30 – 40
B) 40 – 50
C) 50 – 60
D) none
Answer:
A) 30 – 40

Question 103.
In the formula of mode in the ground data represents.
A) upper boundary
B) lower boundary
C) limit
D) lower limit of the class with highest frequency
Answer:
D) lower limit of the class with highest frequency

Question 104.
In an arranged series of an even number 2n terms the median is ……………
A) \(\frac{1}{2}\)(n + 1)^th
B) \(\frac{1}{2}\)(n^th and (n + 1)^th term)
C) \(\frac{1}{2}\)(n^th)
D) none
Answer:
B) \(\frac{1}{2}\)(n^th and (n + 1)^th term)

Question 105.
AM of 1, 2, x, 3 is 0 then x = ……………
A) -6
B) 6
C) 7
D) none
Answer:
A) -6

Question 106.
AM of first n odd numbers is …………..
A) 2n
B) n²
C) n/2
D) n
Answer:
D) n

Question 107.
Mean of 6, -4, \(\frac{2}{3}\), \(\frac{5}{4}\), \(\frac{7}{6}\) is …………
A) \(\frac{12}{7}\)
B) \(\frac{11}{4}\)
C) \(\frac{11}{20}\)
D) none
Answer:
C) \(\frac{11}{20}\)

Question 108.
The AM of 10 consecutive numbers starting with n + 1 is ……………
A) x + 5
B) x + 5.5
C) x – 5
D) none
Answer:
B) x + 5.5

Question 109.
x = 2p + q, M = p + 2q then Z = ……………
A) 4q – p
B) q – 4p
C) 4q + p
D) p – q
Answer:
A) 4q – p

Question 110.
The information collected is called ………….
A) median
B) mean
C) mode
D) data
Answer:
D) data

Question 111.
In a data mean = 72.5 and median = 73.9 then mode is …………..
A) 70.7
B) 69.1
C) 60.2
D) none
Answer:
D) none

Question 112.
Mode of any 3 consecutive numbers is ………….
A) x + 1
B) 4
C) 3
D) no mode
Answer:
D) no mode

Question 113.
……………. is based on all observations.
A) Mean
B) Median
C) Mode
D) None
Answer:
A) Mean

Question 114.
The mean of first 5 odd multiples of 5 is ………….
A) 25
B) 20
C) 35
D) 15
Answer:
A) 25

Question 115.
Median = 52.5, mean = 54, mode = ……………..
A) 48.5
B) 60.1
C) 49.5
D) 40.5
Answer:
C) 49.5

Question 116.
The mean of the following data is …………….

A) 9.4
B) 3.8
C) 6.3
D) 8.1
Answer:
D) 8.1

Question 117.
Mode = 24.5, mean = 29.75 then median …………….
A) 28
B) 16
C) 82
D) 20
Answer:
A) 28

Question 118.
Find the sum of lower limit of median class and upper limit of modal class is ……………

A) 60
B) 40
C) 50
D) 90
Answer:
D) 90

Question 119.
…………… of all bars is same in bar graph.
A) width
B) length
C) circle
D) none
Answer:
A) width

Question 120.
The median class of the following distribution is ……………

A) 60 – 70
B) 50 – 60
C) 40 – 50
D) 30 – 40
Answer:
D) 30 – 40

Question 121.
For a given data with 120 observations, the ‘less than ogive’ and the ‘more than ogive’ intersect at (42.5, 60) the median of the data is …………….
A) 42.5
B) 32.7
C) 90.2
D) none
Answer:
A) 42.5

Question 122.
Consider the following frequency distribution.

The number of families having income range from ₹ 16000 to ₹ 19000 is …………….
A) 19
B) 10
C) 20
D) none
Answer:
A) 19

We are offering TS 10th Class Maths Notes Chapter 3 Polynomials to learn maths more effectively.

TS 10th Class Maths Notes Chapter 3 Polynomials

→ Polynomials are algebraic expressions constructed using constants and variables.

→ If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x).

→ Zero polynomial: A polynomial of degree zero is called zero polynomial.
E.g.: f(x) = 8,
g(x) = \(\frac{-5}{8}\)

→ Linear polynomial: A polynomial of degree ‘1’ is called linear polynomial.
E.g.: f(x) = 3x + 5, g(y) = 7y – 1, h(z) = 5z – 3

→ Quadratic polynomial: A polynomial of degree ‘2’ is called a quadratic polynomial.
E.g.: f(x) = 5x², f(x) = 7x² – 5x, f(x) = 6x² – 7x + 5

→ Cubic polynomial: A polynomial of degree ‘3’ is called a cubic polynomial.
E.g.: f(x) = 5x³ + 4x² – 3x + 6

→ Polynomial of degree ‘n’ in Standard Form : A polynomial in one variable Y of degree ‘n’ is an expression of the form.
F(x) = a₀xⁿ + a₁x^n-1 + a₂x^n-2 + ………… + a_n-1x + a_n, where a₀, a₁ , an are real coefficients and a ^ 0.

→ A real number k is said to be a zero of a polynomial p(x), if p(k) = 0.

→ The zero of the linear polynomial ax + b is \(\frac{-b}{a}\).

→ If a and P are the zeroes of the quadratic polynomial ax² + bx + c then

• α + β = \(\frac{-b}{a}=\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^2}\)
• αβ = \(\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

→ Division Algorithm : Dividend = Divisor × Quotient + Remainder.

→ Value of a polynomial at a given point: If p(x) is a polynomial in x and a is real number then the value obtained by putting x = a in p(a) is called the value of p(x) at x = a.
E.g.: Let p(x) = 5x² – 4x + 2, then its value at x = 2 is given by
p(2) = 5(2)² – 4(2) + 2 = 5(4) – 8 + 2 = 20 – 8 + 2 = 14
Thus, the value of p(x) at x = 2 is 14.

→ Graph of a polynomial: In algebraic or in set theory language, the graph of a polynomial f(a) is the collection (or set) of all points (a, y) where y = f(x)

• Graph of a linear polynomial ax + b is a straight line.
• The graph of quadratic polynomial (ax² + bx + c) is U – shaped, called Parabola.
  • If a > 0 in ax² + bx + c, the shape of parabola is opening upwards ‘U’.
  • If a < 0 in ax² + bx + c, the shape of parabola is opening downwards ‘n’.

→ How to make a quadratic polynomial with the given zeroes : Let the zeroes of a quadratic polynomial be α & β.
x = α, x = β
Then, obviously the quadratic polynomial is (x – α)(x – β) i.e., x² – (α + β) x + aβ i.e., x² – (sum of the zeroes) x + product of the zeroes.

→ Some useful relations :

• α² + β² = (α + β)² – 2αβ
• (α – β)² = (α + β)² – 4αβ
• α² – β² = (α + β)(α – β) = (α + β) \(\sqrt{(\alpha+\beta)^2-4 \alpha \beta}\)
• α³ + β³ = (α + β)³ – 3αβ (α + β)
• α³ – β³ = (α – β)³ + 3αβ (α – β)

→ If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d = 0 then

• α + β + γ = \(\frac{-b}{a}\)
• αβ + βγ + γα= \(\frac{c}{a}\)
• αβγ = \(\frac{-d}{a}\)

Important Formula:

• α + β = \(\frac{-b}{a}=\frac{-\text { coefficient of } x}{\text { coefficient of } x^2}\)
• αβ = \(\frac{c}{a}=\frac{\text { constant term }}{\text { coefficient of } x^2}\)
• Dividend = Divisor × Quotient + Remainder
• α + β + γ = \(\frac{-b}{a}\)
• αβ + βγ + γα = \(\frac{c}{a}\)
• αβγ = \(\frac{-d}{a}\)

Flow Chart

Ex.: (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.
(x + y)³ = x³ + y³ + 3xy (x + y)
(x – y)³ = x³ – y³ – 3xy (x – y)
(x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)

Pavuluri Mailana (11th Century):

• Pavuluri Mailana of 11th century A.D. was a mathematician – poet of repute, who wrote his magnum opus in Telugu (Prosodical form) and named it Sarasangraha Ganitham.
• Sarasangraha Ganitham Mailana, it seems wrote 10 chapters, but only three chapters are available. These contains mostly arithmetic and some elementary algebra dealing with linear and quadratic equations.

Telangana SCERT 10th Class Physics Study Material Telangana 5th Lesson Human Eye and Colourful World Textbook Questions and Answers.

TS 10th Class Physical Science 5th Lesson Questions and Answers Human Eye and Colourful World

Improve Your Learning
I. Reflections on concepts

Question 1.
How do you correct the eye detect Myopia?
(OR)
Explain the Myopia using the diagram.
Answer:
1. Some people cannot see objects at long distances but can see nearby objects clearly. This type of defect In vision is called ‘Myopia’ or ‘near sightedness’.

2. Myopia is corrected by using a concave lens of focal length equal to the distance of the far point F from the eye.

3. This lens diverges the parallel rays from distant objects as If they are coming from the far point.
4. Finally the eye lens forms a clear image at the retina.
5. Here object distance (u) is infinity and image distance (v) is equal to the far point, u = ∞, v distance of far point = – D, f = focal length of bi-concave lens.
We know = \( \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{f}=-\frac{1}{d}\)
⇒ f = -D
Here ‘f’ is negative showing that it is a concave lens.

Question 2.
Explain the correction of the eye detect Hypermetropia.
(OR)
Explain the Hypermetropia with the help of diagrams.
Answer:
1. A person with hypermetropia can see distant objects clearly but cannot see objects at near distances. This is also known as ‘far-sightedness’.

2. Eye lens can form a clear Image on the retina when any object Is placed beyond near point.
3. To correct the defect of hypermetropia, we need to use a lens which forms an image of an object beyond near point at H when the object is between H and L. This is possible only when a double convex lens is used.
4. The image acts like an object for the eye lens. Hence final Image to eye is formed at retina.
Here object distance (u) = – 25 cm
Image distance (v) distance of near point = – d.
f is the focal length of bi-convex lens.
we \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{1}{-d}-\frac{1}{-25} \Rightarrow \frac{1}{f}=\frac{1}{-d}+\frac{1}{25} \Rightarrow \frac{1}{f}=\frac{d-25}{25 d} \)
5) If d > 25, then f becomes positive, It indicates that we need to use biconvex lens.

Question 3.
How do you find experimentally the refractive index of material of a prism?
Answer:
Aim: To find the refractive index of material of e-prism.
Material required: Prism, piece of white chart, pencil, pens, scale and protractor.

Procedure:
1. Take a prism and place it on the white chart in such a way that the triangular base of the prism is on the chart.
2. Draw a line around the prism base using pencil. Remove the prism and name the vertices of the triangle so formed as P, Q and R.
3. Measure the angle between PQ and PR. This Is the angle of the prism (A).
4. Mark M on the side PQ of PQR and also draw a perpendicular to PQ at M.
5. Draw a line with 300 to the normal at M. This line denotes the incident ray.
Note this value n the table.

6. Place the prism In Its position (Δle) again.
7. Now fix two pins vertically on the line at the points A and B.
8. Look for the images of pins through the prism from the other side (PR) and fix another two pins at points C and D in such a way that all the four pins appear to lie along the same straight line.
9. Now remove the prism and take out pins.
10. Draw a line joining C and D and extend it to meet PR at N. This is the emerging ray.
11. Draw a normal to PR at N.

12. The angle between the normal at N and emergent ray Is the angle of emergence. Measure this angle and note its value in the above table.
13. Now join the points M and N by a straight line. The line passing through A, B, M, N, C and D represents the path of light when it suffers refraction through the prism.
14. Extend both incident arid emergence rays till they meet at ‘O’.
15. Measure angle between MO and ON, This is the angle of deviation, denoted by ‘d’. Note this value ¡n the table.
16. Repeat the process for different angles of incidence and measure corresponding angles of deviation.
17. Take angle of incidence along X-axis and the angle of deviation along Y-axis, draw a graph. D
18. We obtain a curve.
19. Draw a tangent line to the curve, parallel to X – axis, at the lowest point of the graph. The point where this line cuts.
Y – axis gives the angle of minimum deviation (D)

20. The refractive Index of prism
21. Using this formula, we can measure the refractive Index of the material of the prism.

Question 4.
Explain the formation of rainbow.
(OR)
Explain the formation of rainbow with the help of water drop diagram.
Answer:

1. Rainbow: The rainbow is an arch of seven colours visible in the sky which is produced by the dispersion of sun’s light by raindrops in the atmosphere.
2. Observe the figure. The rays of sunlight enter the drop near its top surface.
3. At this first refraction, the white light Is dispersed into its spectrum of colours, violet being deviated the most and red the least.
4. Reaching the opposite side of the drop, each colour is reflected back into the drop because of total internal reflection.
5. Arriving at the surface of the drop, each colour is again refracted into air.
6. At the second refraction, the angle between red and violet rays further increases when compared to the angle between those at first refraction.
7. We observe bright rainbow when the angle between incoming and outgoing rays is near the maximum angle of 42°.

Question 5.
Explain two activities for the formation of artificial rainbow.
Answer:
ActivIty – 1:

1. Take a metal tray and fill It with water.
2. Place a mirror in the water such that it makes an angle to the water surface.
3. Now focus white light on the mirror through the water as shown ¡n the figure.
4. Try to obtain colours on a white card board sheet kept above the water surface.
5. We can observe the seven colours of VISGYOR on the cardboard sheet.
   

Activity – 2:

1. Select a white-coloured wall on which the sun rays fall.
2. Stand in front of a wall in such a way that the sun rays fall on your back.
3. Hold a tube through which water is flowing.
4. Place your finger in the tube to obstruct the flow of water.
5. Water comes Out through the small gaps between the tube and your finger like a fountain.
6. Observe the changes on the wall while the water shower is maintained.
7. We observe different colours on the wall, which are similar colours of VIBGYOR.

Question 6.
Light of wavelength λ₁ enters a medium with refractive Index n₂ from a medium with refractive index n₁ What is the wavelength of light In second medium?
Answer:
Speed of light v = nλ
if light can travel from one medium to second medium v₁ = v₂
n₁λ₁ = n₂λ₂
λ₂ = \(\frac{n_1 \lambda_1}{n_1} \)

Question 7.
Why does the sky sometimes appear white?
Answer:
Our atmosphere contains atoms and molecules of different sizes. According to their sizes, they are able to scatter different wavelengths of light. For example, the size of the water molecules is greater than the size of the N, or 02 in air. It acts as a scattering centre for other frequencies which are lower than the frequency cf blue light.

On a “hot day due to rise in the temperature, water vapour enters into atmosphere which leads to abundant presence of water molecules in atmosphere. These water molecules scatter the colours of other frequencies. All such colours of other frequencies reaches our eye and mix together to give white colours.

Question 8.
A person is viewing an extended object. If a converging lens Is placed in front of his eye, will he feel that the size of object has increased? Why?
Answer:

1. The person feels that the size of object has increased.
2. He used the converging lens, i.e., the convex lens and the image s an extended object.
3. This image is formed when the object is in between foci (f) and lens centre (p) of the lens.
4. Hence the image size seems to be increased.

Question 9.
Explain briefly the reason for the blue colour of the sky.
Answer:

1. Scattering of sunlight through molecules of atmosphere is the reason for the blue of the sky.
2. Our atmosphere contains atoms and molecules of different sizes.
3. According to their sizes they are able to scatter different wavelengths of light.
4. For example, the size of the water molecule is greater than the size of the N₂ or O₂.
5. It acts as scattering centre for other frequencies which are lower than the frequency of blue light.

Question 10.
Derive an expression for the refractive index of the material of a prism.
(OR)
Derive the formula for refractive index of a prsim.
Answer:
Observe the ray diagram in the figure.
[d is the exterior angle of ΔOMN. Exterior angle is equal to the sum of interior opposite angles.]
From ΔOMN, we get
d = i₁ – r₁ + i₂ – r₂ ………………………… (1)
From ΔPMN, We have
A+(90°-r₁)+(90°-r₂)= 180°
A+ 180°- (r₁ + r₂) = 180°
A=r₁+r₂ ……………………………… (2)
From (1) and (2) d= i₁ + i₂ — A
⇒ A+d=i₁+i₂ ……………………….. (3)
From Sneil’s law we know that n₁ sin i = n₂ sin r.
Using Snell’s law at M, with refractive index of air.
n₁ = 1; i = i₁; n₂= n; r= r₁

∴ sin i₁ = n sin r₁ ………………………. (4)
Similarly at N, with n₁ = n; i = r₂; n₂ = 1; r = i₂
∴ n sin r₂ = sin i₂ ………………………… (5)
We know that at the angle of minimum deviation (D), i₁ = i₂ ⇒ MN || QR
When i₁ = i₂ angle of deviation (d) becomes angle of minimum deviation (D).
∴ (3) ⇒ A +D = 2i₁ ⇒ i₁ = \(\frac{A+D}{2} \)
When i₁ = i₂ then it is clear that r₁ = r₂
∴ (2) ⇒ 2r₁ ⇒ ⇒ r₁ = \(\frac{A}{2} \)

Substitute and r₁ in (4), we get sin\(\left(\frac{A+D}{2}\right) \) = nsin\(\left(\frac{A}{2}\right) \)
where ‘r s the refractive index of the material of the prism.

Application of Concepts

Question 1.
indent ray on one of the face (AB) of a prism and emergent ray from the tace AC are given In figure. Complete the ray diagram.

Answer:

Question 2.
Glass is known to be transparent material. But ground glass is opaque and white in colour. Why?
Answer:

• Glass is generally a transparent material because it transmits most of the light incident on it.
• When glass Is at ground it’s surface becomes rough due to microscopic unevenness.
• When light is incident on such a rough surface, it is reflected in many (different) directions.
• This type of reflection is known as defeat reflection. Due to this ground glass is opaque (does not transmit high) and white in colour.

Question 3.
A light ray falls on one of the faces of a prism at an angle 40° so that It suffers angle of minimum deviation of 30°. Find the angle of prism and angle of refraction at the given surface.
Answer:
i = 40⁰, D = 30°
At angle of minimum deviation i₁ = i₂ = i; r₁ = r₂ = r
i₁ + i₂= A+D
hence 2i =A+D
2(40°) = A + 30°
80-30 = A

The angle of the prism A = 50°
r₁+ r₂ = A
2r =A
r = A/2
∴ angle of refraction r= \(\frac{50^{\circ}}{2}\)
∴ r = 25°

Question 4.
The focal length of a lens suggested to a person with Hypermetropia is 100cm. Find the distance of near point and power of the lens.
Answer:
1. Distance of near point:
If ‘f’ is the local length and ‘d’ is the distance of near point then
f = \(\frac{25 d}{d-25} \)
(Recall f should be In cm)

Here, focal length f = 100cm
∴ 100 = \(\frac{25 d}{d-25} \Rightarrow d-25=\frac{25 d}{100}=\frac{d}{4} \)
That is, 4d – 100 = d
⇒ 4d-d =100 or 3d = 100
∴ d = \(\frac{100}{3}=33.33\) (nearly)
So, distance of near point = 33.33 cm

Power of lens is measured as reciprocal of focal length in metre.
Power of lens P = \(\frac{1}{f(\text { in } \mathrm{mt})}=\frac{100}{f(\text { incm })}\)
Power of the lens = 100/f in cm
100/100 = 1 dioptre.

Question 5.
How do you appreciate the role of molecules ¡n the atmosphere for the blue colour of the sky?
Answer:

1. The sky appear blue due to atmospheric refraction and scattering of light through molecules.
2. Molecules are scattering centres.
3. The reason to blue sky is due to the molecules N2 and 02.
4. The sizes of these molecules are comparable to the wavelength of blue light.
5. In the absence of these molecules there will be no scattering of sunlight and the sky will appear dark.
6. We should appreciate the molecules which are scattering centres.

Question 6.
How do you appreciate the working of Clliary muscles in the eye?
Answer:

1. The ciliary muscles to which eye lens is attached help the eye lens to change its focal length by changing the radii of curvature of the eye lens.
2. When the eye is focused on a distant object, the ciliary muscles are relaxed so that the focal length of eye lens has its maximum value.
3. The parallel rays coming into the eye are then focussed on to the retina and we see the object clearly.
4. When the eye Is focused on a closer object, the ciliary muscles are strained and focal length of eye lens decreases.
5. The ciliary muscles adjust the focal length in such a way that the Image is formed on retina and we see the object clearly. This process of adjusting focal length of eye lens Is called ‘accommodation’.
6. Really this ‘accommodation’ is a wonderful phenomenon through which we are able to see the distant and near objects.

Multiple choice questions

Question 1.
The size of an object as perceived by an eye depends mainly on ……………………… . ( )
(A) actual size of the object
(B) distance of the object from the eye
(C) aperture of the pupil
(D) size of the image formed on the retina
Answer:
(B) distance of the object from the eye

Question 2.
When objects at different distances are seen by the eye which of the following remain constant? ( )
(A) focal length of eye-lens
(B) object distance from eye lens
(C) the radii of curvature of eye lens
(D) image distance from eye lens
Answer:
(D) image distance from eye lens

Question 3.
During refraction ………………………… will not change. ( )
(A) wavelength
(B) frequency
(C) speed of light
(D) all the above
Answer:
(B) frequency

Question 4.
A ray of light falls on one of the lateral surfaces of an equilateral glass prism placed on the horizontal surface of a table as shown in the figure.
2. For minimum deviation of ray, which of the following is true?

(A) PQ is horizontal Q
(B) QR is horizontal
(C) RS is horizontal
(D) either PQ or RS is horizontal
Answer:

Question 5.
Far point of a person s 5m. In order that he has normal vision what kind of spectacles should he use? ( )
(A) concave lens with focal length 5 m.
(B) concave lens with focal length 10 m.
(C) convex lens with focal length 5 m.
(D) convex lens with focal length 2.5 m.
Answer:

Question 6.
The process of re-emission of absorbed light in all directions with different intensities by the atom or molecule is called ( )
(A) Scattering of light
(B) dispersion of light
(C) reflection of light
(D) refraction of light
Answer:

Suggested Experiments

Question 1.
Conduct an experiment to produce a rainbow in your classroom and explain the procedure.
Answer:
What you’ll need:
A glass of water (about three-quarters full) White paper, A sunny day
Instructions:

1. Take the glass of water and the white paper to a part of the room with bright sunlight (near a window is good).
2. Hold the glass of water (being careful not to spill it) above the paper and watch as sunlight passes through the glass of water, refracts (bends) and forms a rainbow of colors on your sheet of paper.
3. Try holding the glass of water at different heights and angles to see if it has a different effect.

What’s happening?
While you normally see a rainbow as an arc of color in the sky, they can also form In other situations. You may have seen a rainbow in a water fountain or In the mist of a waterfall and you can even make your own such as you did in this experiment.

Rainbows form in the sky when sunlight refracts (bends) as it passes through raindrops. It acts in the same way when It passes through your glass of water. The sunlight refracts, separating It into the colors red, orange, yellow, green, blue, indigo and violet.

Question 2.
conduct an experiment to find the refractive Index of a prism.
Answer:
Aim: Finding the refractive index of a prism.
Material required: Prism, piece of white chart of size 20 x 20 cm, Pencil, Pins, Scale and Protractor.

Procedure:

1. Take a prism and place it on the white chart in such a way that the rectangular base of the prism is on the chart.
2. Draw a line around the prism (boundary) using a pencil. Remove the prism.
3. You will get a triangle and name Its vertices as P, Q and R.
4. The angle between the surfaces PQ and QR Is called angle of the prism (A).
5. Make M’ on the side of triangle PQ and also draw a perpendicular to PQ at “M’
6. Draw a line AB making an angle of 30° with the normal which represents incident ray and this angle is called angle of Incidence.
7. Place the prism in its position again. Now fix two pins vertically on the line at ‘A’ and ‘B’ as shown in fig.
8. Look for the images of pins through the prism from the other side (PR) and another two pins ‘C’ and ‘D’ in such a way that all the four pins appear lie along the same straight line.
9. Now remove the pnsm and take out pins. Now join the two pin-holes formed by the pins to meet surface PR at N.
10. This Is called emergent ray and the angle between the normal at N and emergent ray is the angle of emergence (i₂). Measure this angle and note down In the table.
11. Extend both incident and emergent ray till they meet at a point ‘O Measure the angle between these two rays. This Is the angle of deviation (d).
12. Repeat this procedure for various angles of incident such as 40°, 5O° 60° etc. Find the corresponding angles of deviation and angles of emergence and note them in the following table.

l-d Graph:
1) Take the angle of Incidence along X – axis and the angle of deviation along Y – axis. You will get a U – shaped graph.
2) From the graph identify the angle of incidence (i₁) for which the angle of deviation is minimum. Note this point as ‘D’ on the Y – axis which Is called angle of minimum deviation (D).

Now refractive index of the prism can be calculated using formula
n = \(\frac{\operatorname{Sin}\left(\frac{A+D}{2}\right)}{\operatorname{Sin} A / 2}\)

Question 3.
Conduct an experiment to demonstrate the scattering of light.
Answer:
Required material: beaker, sodium-thiol-sulphate (hypo), sulphuric acid, water.

1. Take a solution of sodium-thio-sulphate (hypo) and sulphuric acid in a glass beaker.
2. Place the beaker in an open place where abundant sunlight is available.
3. Watch the formation of grains of sulphur and observe changes in the beaker.
4. Sulphur precipitates as the reaction Is in progress.
5. At the beginning, the grains of sulphur are smaller In size and appear blue in colour. As the reaction progresses, the size of grains increases and slowly their colour becomes white.
6. At the beginning, the size of grains is small and almost comparable to the wavelength of blue light. Hence they appear blue.
7. As the size of grains Increases, their size becomes comparable to the wavelengths of other colours and these grains act as scattering centres for other colours.
8. The combinations of all these colours appears as white.

Suggested Projects

Question 1.
Prisms are used in binoculars. Collect Information why prisms are used In binoculars.
Answer:

1. Binoculars are two identical telescopes placed side by side to each other.
2. At front side of each telescope Is a lens, called the objective lens. Its role is to gather light from whatever you are looking at and bring It to a focus in the eyepiece, where the light is formed into a visible image and magnified to take up a large portion of the retina.
3. The image produced by this telescope will be upside down and backwards.
4. This Is why binoculars use corrective elements between the objective and the eyepiece, called prisms.
5. Prisms used in binoculars are blocks of glass that function as mirrors but without a mirror’s reflective backing.
6. Their role is to bring the light beams from the objective closer together by means of internal reflection and also turn the image right-side up and orient the view properly left to right.
7. The diagram shows the path of the light that enters the objectives, passes through a set of prisms that turn the image right side up and finally leave the eyepiece to enter the observer’s eye. This applies to all binoculars.

Question 2.
Collect the information about different types of eye defets from your nearest eye specialist or optical shop and write a report.
Answer:
There are many eye related problems and defects of the eye which are cadue to the loss of ability of accommodate of eye lens. Some of these discussed below :
Defects of the eye:

1. Eye-Related Problems:

Question 3.
Collect the different types of lenses used for correcting the eye defects and write a report.
Answer:

Non – a – days so many people are using contact lenses in the place of above-mentioned lenses. There are two types of contact lenses in use. They are

1. The rigid, gas-permeable lens.
2. The soft, water-absorbing lens.

Question 4.
Collect the information about the dispersion phenomenon occurs in the daily life.
Answer:
In the year 1665, Newton discovered by his experiments with glass prisms that white light (like sunlight) consists of a mixture of seven colours. Newton found that if a beam of white light is passed through a triangular glass prism, the white light spits to form a band of seven colours on a white screen.

Dispersion: The splitting up of white light into seven colours on passing through a transparent medium like a glass prism is called dispersion of light.

Examples for dispersion phenomenon in daily life:
(a) Formation of Ralnblow:

1. One of the most beautiful examples of spectrum formed by the dispersion of sunlight Is provided by nature in the form of rainbow.
2. The rainbow is actually a nature spectrum of sunlight in the sky.
3. The rainbow is formed in the sky when the sun is shining and it is raining at the same time.
4. We can see the rainbow if we stand with our back towards the sun and rain Infront of us. A rainbow is always formed is a direction opposite to that of the sun.
5. A rainbow is produced by the dispersion of white sunlight by raindrops in the atmosphere.
6. The raindrops in the atmosphere act like many small prisms.
7. As white light enters and leaves these raindrops, the various coloured rays present in white light are refracted by different amounts due to which an arch of seven colours called rainbow is formed in the sky.

(b) Another example for dispersion of light In daily life are:

• CDs and DVDs disperse the light and produce seven colours.
• The diesel and petrol surface layers which fall on the road also disperse white light and produce seven colours.

TS 10th Class Physical Science Human Eye and Colourful World Intext Questions

Page 83

Question 1.
Why do the values of least distance of distinct vision and angle of vision change with person and age?
Answer:

1. The ciliary muscle which attached with eye lens helps the eye lens to change its focal length by changing radii of curvature of eye lens.
2. When the eye Is focussed on a distant object, the ciliary muscles are relaxed so that the focal length of eye lens has its maximum value which is equal to its distance from the retina.
3. The working of ciliary muscle n eye changes from person to person.
4. So, the values of least distance of distinct vision and angle of vision change with person and age.

Page 84

Question 2.
How can we get same image distance for various positions of objects?
Answer:
For different positions of object, the image distance remains constant only when focal length of lens changes.

Question 3.
Can you answer above question using concepts of refraction through lenses?
Answer:
The focal length of a lens depends on the material by which it has been made and radii of curvatures of surface’s lens. We need to change focal length of lens to get same image distance for various positions of object.

Page 85

Question 4.
How does eye lens changes its focal length?
Answer:
The ciliary muscles to which eye lens is attached help the eye lens to change its focal length by changing radii of curvature of eye lens.

Question 5.
How does this change takes place in eyeball?
Answer:
When the eye is focused on a distant object the ciliary muscles are relaxed and so the focal length of eye lens has its maximum value which ¡s equal to its distance from the retina. The parallel rays coming into the eye are focused on the retina and we see the object clearly.

Question 6.
How does the image formed on retina help us to perceive the object without change in Its shape, size and colour?
Answer:
The eye lens forms a real and inverted image on retina. This retina s a delicate membrane, which contains about 126 million receptors called rods’ and ‘cones. They receive the light signals and identify the colour and the intensity of light. These signals are transmitted to brain through optic-nerve fibres. The brain interprets these signals and finally processes the information so that we perceive the object in terms of its shape, size and colour.

Question 7.
Does eye tens form a real image or virtual image?
Answer:
Eye lens forms a real and inverted image.

Question 8.
Is there any limit to change of focal length of eye lens?
Answer:
Yes, when the object is at Infinity, the parallel rays from the object falling on the eye lens are refracted and they form a point-sized Image on retina. In this situation, eye lens has a maximum focal length.

Question 9.
What are the maximum and minimum focal lengths of the eye lens?
Answer:
Maximum focal length is 2.5 cm and minimum focal length is 2.27 cm.

Page 85

Question 10.
What happens ¡f the eye lens Is not able to adjust its focal length?
Answer:
In this case the person cannot see the object clearly and comfortably.

Question 11.
What happens It the focal length of eye lens is beyond the range of 2.5 cm to 2.27 cm?
Answer:
The vision (image) becomes blurred due to defects of eye lens.

Page 87

Question 12.
What can we do to correct myopia?
Answer:
To correct myopia, we use concave lens In spectacles.

Page 88

Question 13.
How can you decide the focal length of the lens to be used to correct myopia?
Answer:
Let the object distance (u) is infinity and image distance (v) is equal to distance of far point.
u = -∞’; v= distance of far point = – D
Let ‘f be the focal length of bi-concave lens.
Using the formula : \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{f}=\frac{1}{-D} \Rightarrow f=-D \)
Here ‘f’ is negative for a concave lens.

Question 14.
What happens when the eye has a minimum focal length greater than 2.27 cm?
Answer:
In this case, the rays coming from the nearby object after refraction at eye lens, form image beyond the retina.

Page 89

Question 15.
How can you decide the focal length of convex lens to be used?
Answer:
Here u = – 25
Image distance v = d (distance of near point)
Let ‘f’ be the focal length of bi- convex lens.
Using the formula :
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(\frac{1}{f}=\frac{1}{-d}-\frac{1}{(-25)} \Rightarrow \frac{1}{f}=\frac{(d-25)}{25 d} \Rightarrow f=\frac{25 d}{(d-25)} \) (f in centimetres)
If d > 25 cm, ⇒ ‘f’ becomes positive then we use biconvex lens to correct hypermetropia.

Page 90

Question 16.
Have you ever observed details in the prescription?
Answer:
A prescription contains some information regarding type of lens to be used to correct vision.

Question 17.
You might have heard people saying “my sight Is Increased or decreased”. What does It mean?
Answer:
Usually, doctors after testing the defects of vision prescribe correcting lenses indicating their power which determines the type of lens to be used and its focal length.

Question 18.
What do you mean by power of lens?
Answer:
The reciprocal of focal length is called power of lens.

Question 19.
Doctor advised to use 2D lens. What is the focal length of it?
Answer:
Given power of lens P = 2D
We know, P = \(\frac{100}{f(\text { incm })} \Rightarrow 2=\frac{100}{f} \Rightarrow f=\frac{100}{2}=50 \mathrm{~cm} \)
Focal length of lens (f) = 50 cm.

Page 91

Question 20.
How could the white light of the sun give us various colours of the rainbow?
Answer:
Due to reflection, refraction and dispersion of sunlight.

Question 21.
What happens to a light ray when ¡t passes through a transparent medium bounded by praise surfaces which are inclined to each other?
Answer:
when light incident on one of the plane surfaces, it emerges from the other.

Question 22.
What is a prism?
Answer:
A prism is a transparent medium separated from the surrounding medium by consisting two retracting plane surfaces which are Inclined.

Page 92

Question 13.
What is the shape of the outline drawn for a prism?
Answer:
A triangular shape.

Page 93

Question 24.
How do you find the angle of deviation?
Answer:
The angle between the extended Incident and emergent rays is called angle of deviation.

Question 25.
What do you notice from the angle of deviation?
Answer:
The angle of deviation decreases first and then increases with increase of the angle of incidence.

Question 26.
Can you draw a graph between angle of incidence and angle of deviation?
Answer:
Yes, we can draw the graph between angle of incidence and angle of deviation.

Question 27.
From the graph, can you find the minimum of the angles of deviation?
Answer:
Yes we can. Draw a tangent line to the curve, parallel to X – axis, at the lowest point of the graph. The point where the line cuts y-axis cuves the angle of minimum deviation.

Question 28.
Is there any relation between the angle of incidence (i) and angle of emergence (r) and angle of deviation (d)?
Answer:
(i₁+i₂) = A+D
i+r = A + D

Page 94

Question 29.
can we derive n = from the equation 5?
Answer:
Yes.
From equAtion (3) we get
A + D = 2i₂ [since i₁ = i₂ at angle of minimum deviation (D)]
i₂ = \( \frac{A+D}{2}\)
From equation (2) we get
2r₂ = A (since r₁ = r₂ when i₁ = i₂)
r₂ = \(\frac{A}{2} \)
From equation (5)
n sinr₂ = sin i₂
Substitute i₂ = r₂ values in (5) we get

Page 96

Question 30.
In activity-3, we noticed that tight has chosen different paths. does this mean that the refractive index of the prism varies from colour to colour?
Answer:
Yes, refractive index of the prism varies from colour to colour.

Question 31.
Is the speed of light of each colour different?
Answer:
In vacuum – Speed of each colour is constant.
In medium – Speed Is different for different colours.

Question 32.
Can you guess now, why light splits into different colours when it passes through a prism?
Answer:
Due to dispersion of light and different wavelengths of colours in medium.

Page 97

Question 33.
Does It split into more colours? Why?
Answer:
We know the frequency of light is the property of the source and It is equal to number of waves leaving the source per second This cannot be changed by any medium. Hence frequency doesn’t change due to refraction. The coloured light passing through any transparent medium retains its colour.

Question 34.
Can you give an example ¡n nature, where you observe colours as seen In activity 3?
Answer:
Yes, in rainbow. It is a good example of dispersion of light.

Question 35.
When do you see a rainbow ¡n the sky?
Answer:
Due to the refraction, reflection and dispersion of sunlight. When the sunlight passes through the raindrops then we can see the rainbow in the sky.

Question 36.
Can we create a rainbow artificially?
Answer:
Yes, we can create a rainbow artificially.

Page 99

Question 37.
Why is the sky blue?
Answer:
A clear cloudless day-time sky is blue because molecules in the air scatter blue light from the sun more than they scatter red light.

Question 38.
What is scattering?
Answer:
Atoms or molecules which are exposed to absorb light energy and emit some part of the light energy In different directions and it Is called scattering of light.

Page 101

Question 39.
Why is that the sky appears white sometimes when you view it in certain direction on hot days?
Answer:
In a hot day due to raise of temperature, water vapour enters into atmosphere which leads to abundant presence of water molecules in atmosphere. These water molecules scatter the colours of other frequencies (other than blue). All such colours of other frequencies reaches our eye and white colour is appeared to us.

Page 102

Question 40.
Do you know the reasons for appearance the red colour of sun during sunrise and at sunset?
(Or)
What causes the red colour of sun during sunrise and at sunset?
Answer:
The light rays from the sun travel more distance In atmosphere to reach our eye in morning and evening times. During sunrise and sunset except red colour all colours scatter more and vanishes before reach us. Since scattering of red light is very less, so It reaches us. As a result sun appears reddish during sunrise and sunset.

Question 41.
Can you guess the reason why sun does not appear red during noon hours?
Answer:
During noon hours, the distance to be travelled by the sun rays in atmosphere is less than when compared to morning and evening hours. Therefore all colours reach our eye without scattering. Hence light appears white in noon hours.

Think And Discuss

Question 1.
Can you Imagine the shape of rainbow when observed during travel in an airplane? Discuss with your friends and collect information.
Answer:
When you look at a rainbow from a height high enough so that the sun shines on water particles below you, you will see a full-circle rainbow instead of a horseshoe shape.

TS 10th Class Physical Science Human Eye and Colourful World Activities

Activity 1

Question 1.
How do you find least distance of distinct vision?
Answer:

1. Hold the textbook at certain distance with our hands.
2. Try to read the contents on the page.
3. Gradually move the books towards eye, till it reaches very close to your eyes.
4. You may see that printed letters on the page appear blurred or you will feel strain to read.
5. Now move the book backwards to a position where you can see clear printed letters without strain.
6. Ask your friend to measure distance between your eye and textbook at this position.
7. Note down its value.
8. Repeat this activity with other friends and note down the distances for distinct vision in each case.
9. Find the average of all these distances of clear vision.
10. We notice that to see an object comfortably and distinctly, keep it at a distance about 25 cm from your eyes.
11. This 25cm distance is called least distance of distinct vision.
12. This value vanes from person to person and with age.

Activity 2

Question 2.
How do you measure the angle of vision?
Answer:
1. Collect a few wooden sticks or PVC pipes and make pieces of 20 cm, 30 cm, 35 cm, 40 cm, 50 cm from them.
2. Place a retort stand on a table and stand near the table such that your head is beside the vertical stand.
3. Adjust the clamp on the horizontal rod and fix it at a distance of 25 cm from your eyes.
4. Ask one of your friends to fix a wooden stick of 30 cm height to the damp in a vertical position as shown in the figure.
5. Now keeping your vision parallel to horizontal rod of the stand, try to see the top and bottom of wooden stick kept in vertical position.
6. II you are not able to see both ends of the stick at this instance, adjust the vertical stick on the horizontal rod till you are able to see both ends of the stick, from smallest possible distance from your eye. Fix the vertical stick at this position with the help of the clamp.
7. Without changing the position of the clamp on the horizontal rod, replace this stick of 30 cm length with other sticks of various lengths one by one and try to see the top and bottom of the simultaneously without any change in the position of eye either upwards, downwards or sideways.

8. From the given figure, you notice that you will be able to see only part EF of the object A and B’ because the rays coming from E and F enter your eye.
9. These rays form an angle at the eye. ¡f this angle is below 60°, we can see the whole object. If this angle is above 60°, then we can see only the part of the object.
10. The maximum angle at which we are able to see the whole object is called angle of vision.

Lab Activity

Aim: Finding to refractIve Index of a prism.
Answer:

Aim: Finding the refractive index of a prism.
Material required: Prism, piece of white chart of size 20 x 20 cm, Pencil, Pins, Scale and Protractor.

Procedure:

1. Take a prism and place it on the white chart in such a way that the rectangular base of the prism is on the chart.
2. Draw a line around the prism (boundary) using a pencil. Remove the prism.
3. You will get a triangle and name Its vertices as P, Q and R.
4. The angle between the surfaces PQ and QR Is called angle of the prism (A).
5. Make M’ on the side of triangle PQ and also draw a perpendicular to PQ at “M’
6. Draw a line AB making an angle of 30° with the normal which represents incident ray and this angle is called angle of Incidence.
7. Place the prism in its position again. Now fix two pins vertically on the line at ‘A’ and ‘B’ as shown in fig.
8. Look for the images of pins through the prism from the other side (PR) and another two pins ‘C’ and ‘D’ in such a way that all the four pins appear lie along the same straight line.
9. Now remove the pnsm and take out pins. Now join the two pin-holes formed by the pins to meet surface PR at N.
10. This Is called emergent ray and the angle between the normal at N and emergent ray is the angle of emergence (i₂). Measure this angle and note down In the table.
11. Extend both incident and emergent ray till they meet at a point ‘O Measure the angle between these two rays. This Is the angle of deviation (d).
12. Repeat this procedure for various angles of incident such as 40°, 5O° 60° etc. Find the corresponding angles of deviation and angles of emergence and note them in the following table.

l-d Graph:
1) Take the angle of Incidence along X – axis and the angle of deviation along Y – axis. You will get a U – shaped graph.
2) From the graph identify the angle of incidence (i₁) for which the angle of deviation is minimum. Note this point as ‘D’ on the Y – axis which Is called angle of minimum deviation (D).

Now refractive index of the prism can be calculated using formula
n = \(\frac{\operatorname{Sin}\left(\frac{A+D}{2}\right)}{\operatorname{Sin} A / 2}\)

Activity 3

Question 3.
Describe an activity for dispersion of light.
Answer:

1. Do this experiment In the dark room.
2. Take a prism and place It on a table near a vertical white wall.
3. Take a thin wooden plank.
4. Make a small hole in it and fix it vertically on the table.
5. Place the prism between the wooden plank and wall.
6. Place a white light source behind the hole of the wooden plank.
7. Switch on the light.
8. The rays coming out of the hole of plank become a narrow beam of light.
9. Adjust the height of the prism such that the light falls on one of the lateral surfaces.
10. Observe the changes in emerged rays of the prism.
11. Adjust the prism by slightly rotating ¡t till you get an image on the wall.
12. We observe a coloured image on the wall.
13. The white light s splits into colours because of dispersion.
14. We see seven different colours i.e., Violet. Indigo, Blue, Green, Yellow, Orange and Red. (VIBGYOR).

Activity 4

Question 4.
Suggest an experiment to produce a rainbow In your classroom and explain the procedure.
Answer:

1. Take a metal try and fill it with water.
2. Place a mirror in the water such that it makes
3. No: focus white light on the mirror through the water as shown In figure.
4. Keep a white cardboard sheet above the water surface. Mirror
5. We can observe the colours VIBGYOR on the board.
6. The splitting of white light into different colours (VIBGYOR) Is called dispersion.
7. So consider a white light Is a collection of waves with different wavelengths.
8. Violet has shortest wavelength and red has longest wavelength.

Activity 5

Question 5.
Explain two activities for the formation of artificial rainbow?
Answer:
Activity – I:

1. Select a white-coated wall on which the sun rays fall.
2. Stand in front of a wall in such a way that the sun rays fall on your back.
3. Hold a tube through which water is flowing.
4. Place your finger to obstruct the flow of water.
5. Water comes from small gaps between the tube and figures like a fountain.
6. Observe the changes on wall while showering the water.

Activity – II :

1. Take a metal try and fill it with water.
2. Place a mirror in the water such that it makes an angle with water surface.
3. Now focus white light on the mirror through the water.
4. Keep a white cardboard sheet above the water surface as shown In the figure.
5. We may observe the colours VIBGYOR on the board.
6. The splitting of white light into different colours (VIBGYOR) is called dispersion.
7. So consider a white light is a collection of waves with different wavelengths.
8. Violet has shortest wavelength and red has longest wavelength.

Activity 6

Question 6.
Describe an experiment for scattering of light.
Answer:

1. Take a solution of sodium-thin-sulphate (hypo) and sulphuric acid in a glass beaker.
2. Place the beaker in an open place where abundant sunlight is available.
3. Watch the formation of grains of sulphur and observe changes in the beaker.
4. You will notice that sulphur precipitates as the reaction is in progress.
5. At the beginning, the grains of sulphur are smaller in size and as the reacton progresses, their size increases due to precipitation.
6. Sulphur grains appear blue in colour at the beginning and slowly their colour becomes white as their size increases. Dispensed light
7. The reason for this is scattering of light.
8. At the beginning, the size of grains is small and almost comparable to the wavelength of blue light.
9. Hence they appear blue in the beginning.
10. As the size of grains increases, their size becomes comparable to wavelengths of other colours.
11. 11. As a result of this, they act as scattering centres for other colours.
12. 12. The combination of all these colours appears as white.

We are offering TS 10th Class Maths Notes Chapter 13 Probability to learn maths more effectively.

TS 10th Class Maths Notes Chapter 13 Probability

→ Probability :
Probability means the number of occasions that a particular statement (Or) event is likely to occur in a large population of events.

→ Random experiment:
Random experiments (or) tail is an act (or) process that leads to a result, that cannot be predicted in advance.
(or)
An experiment is said to be a random experiment if its outcome cannot be predicted. That is the outcome of an experiment does not obey any rule.

Ex:
(i) If a coin is tossed, we can’t say whether head or tail will appear. So tossing a coin is a random
experiment.
(ii) If we throw a die numbered 1,2,3,4,5,6 on its six faces, it is impossible to say which numbered face will appear. There is no rule to know it. So throwing a die is a random experiment.

→ Equally likely events :
Two or more events are said to be equally likely if each one of them has an equal chance of occurrence.

For example

• When a coin is tossed, the two possible outcomes, head and tail, are equally likely.
• When a die is thrown, the six possible outcomes, 1, 2, 3,4, 5 and 6 are equally likely.

→ Mutually exclusive events :
Two (or) more events are mutually exclusive if the occurrance of each event prevents the every other event.

→ Complementary events :
Consider an event has few outcomes. Event of all other outcomes in the sample survey which are not in the favourable event is called complementary event.

→ Exhaustive events :
All the events are exhaustive if their union is the sample space.

→ Sure events:
The sample space of a random experiments is called sure(or) certain event as any one of its elements will surely occur in any trail of experiment.

→ Impossible event:
An event which will not occur on any account is called an impossible event.

→ Theoretical event:
The theoretical (classical) probability of an event E, written as P(E), is defined as No. of trails in which the events happened
P(E) = \(\frac{\text { No. of trails in which the events happened }}{\text { Total number of trials }}\)
Where we assume that the outcomes of the experiment are equally likely.

→ Elementary event: An event having only one outcome is called an elementary event.
The sum of the probabilities of all the elementary events of an experiment is 1.
The probability of a sure (or certain event) is 1.
The probability of an impossible event is 0.
The probability of an event E is a number P(E) such that 0 < P(E) < 1.
For an event E, P(E) + P(Ē) = 1 where E stands for ‘not E’.
E and Ē are called complementary events. In general, it is true that for an event E,
P(Ē) = 1 – P(E)

→ Sample space : The set of all possible outcomes of an experiment is called a sample space (or) probability space.
Suppose we throw a die once. As it has six faces, every face has equal chance to appear.
Therefore, sample space (s) = {1, 2, 3, 4, 5, 6}
Number of events n(s) = 6
If a coin is tossed, either head (or) tail may appear.
Hence, sample space (s) = {H, T}
Number of events n (s) = 2

→ A cubic dice is a six faced cube, the six faces are marked as 1, 2, 3,4, 5, 6. When such a dice is thrown, any one of the faces come upwards. The number on this face is the outcome of the experiment.

→ Remember the following points :
(a) There are 52 cards in a pack of cards.
(b) Out of these, 26 are red and 26 are black.
(c) Out of 26 red cards, 13 are hearts and 13 are diamonds.
(d) Out of 26 black cards, 13 are spades and 13 are clubs.
(e) Each of four varities (hearts, diamonds, spades, clubs) has an ace (i.e.,) a pack of 52 cards has 4 aces.
Similarly these are 4 kings, 4 queens and 4 jacks.

Flow Chat:

Pierre Simon Laplace:

• The definition of probability was given by Pierre Simon Laplace in 1795.
• Probability theory had its origin in the 16th Century, when an Italian physician and mathematician J.Cardan wrote the first book on the subject, “The Book on Games of Chance”.
• James Bernoulli (1654 – 1705), A. De Moivre(1667-1754) and Pierre Simon Laplace (1749-1827) are among those who made significant contributions to this field.
• In recent years, probability has been used extensively in many areas such as Biology, Economics, Genetics, Physics, Sociology etc.

Telangana SCERT 10th Class Telugu Guide Telangana 3rd Lesson వీర తెలంగాణ Textbook Questions and Answers.

TS 10th Class Telugu 3rd Lesson Questions and Answers Telangana వీర తెలంగాణ

చదవండి – ఆలోచించి చెప్పండి (T.B. P.No. 26)

తెలుగు గుండెల బిగువు
తెలిపింది తెలగాణ
తెలుగు జోదుల తెగువ
చూపించి తెలగాణ

దేశానికే ముందు నిల్చిందిరా తెలగాణ
దేశానికే పేరు తెచ్చిందిరా !
పరాన్నభుక్కులకు
పక్కలో బల్లెమై
దేశద్రోహుల కింక
తావులేదని చాటి

ఢంకా బజాయించెరా తెలగాణ
దౌర్జన్య మెదిరించెరా !
జయభేరి మోగించెరా తెలగాణ
జయము రైతుల కందెరా !
– (కొత్తపల్లి రంగారావు తెలంగాణ పోరాట పాటలు)

ప్రశ్నలు – జవాబులు

ప్రశ్న 1.
ఈ గేయాన్ని రాసిందెవరు ?
జవాబు:
ఈ గేయాన్ని కొత్తపల్లి రంగారావుగారు రాశారు.

ప్రశ్న 2.
ఈ గేయం దేని గురించి తెలియజేస్తున్నది ?
జవాబు:
ఈ గేయం తెలంగాణ పోరాటం గురించి తెలియజేస్తున్నది.

ప్రశ్న 3.
ఈ గేయాన్ని ఎందుకు రాసి ఉంటాడు ?
జవాబు:
ఈ గేయాన్ని తెలంగాణ రాష్ట్ర సాధన కోసం రాసిఉంటాడు.

ప్రశ్న 4.
ఇట్లా తెలంగాణ గురించి రాసిన కవులు, రచయితలు ఎవరు ? ఎందుకు రచనలు చేసి ఉంటారు ?
జవాబు:
తెలంగాణ గురించి రాసిన కవులు, రచయితలు కాళోజి, డా॥ సినారె, సామల సదాశివ, అలిశెట్టి ప్రభాకర్, డా॥ దాశరథి కృష్ణమాచార్య మొదలగు వారు. వీరంతా తెలంగాణ సాధన కోసం రచనలు చేసారు.

ఆలోచించండి – చెప్పండి (T.B. P.No. 28)

ప్రశ్న 1.
ఈ భూమండలమంతా ఎందుకు ప్రతిధ్వనించింది ?
జవాబు :
తెలంగాణలో రజాకార్ల దుర్మార్గాలు మితిమీరి పోయాయి. మొత్తం తెలంగాణ ప్రజానీకమంతా ఏకమైంది. తన శత్రువులైన రజాకార్లపై ఎదురు తిరిగింది. పొలికేకలు పెట్టింది. ఆ బొబ్బలు ఎలా ఉన్నాయంటే తెలంగాణ మొత్తం ఒకేసారి యుద్ధ శంఖం పూరించినట్లుగా ఉంది. ఈ భూమండలంలో ప్రతిచోటా తెలంగాణ వాళ్ళు ఉన్నారు. ఎక్కడి వాళ్ళు అక్కడి నుండే సమరశంఖం పూరించారు. అందుకే అది భూమండలమంతా ధ్వనించిందని కవి చెప్పాడు.

ప్రశ్న 2.
బ్రతుకుతోవ చూపే కాలం రావడం అంటే ఏమిటి ?
జవాబు :
రజాకార్లు తెలంగాణను అభివృద్ధి చెందనివ్వలేదు. తర తరాల నుండి తెలంగాణ దోపిడీకి గురయ్యింది. తెలంగాణ ప్రజలు జీవనోపాధులను కోల్పోయారు. దెయ్యాలు, పిశాచాల వంటి దోపిడీదారుల ఉక్కుపాదాల కింద నలిగిపోయారు.

ఇప్పుడు క్రొత్త బ్రతుకు తోవ ఏర్పడే కాలం వచ్చింది. అంటే స్వేచ్ఛగా తమ బ్రతుకు నిర్ణయాలు తామే తీసుకొనే మంచి సమయం వచ్చిందని అర్థం.

ఆలోచించండి – చెప్పండి (T.B. P.No. 29)

ప్రశ్న 1.
“తెలంగాణ నేలలో ఎంత బలం ఉన్నదో కదా !” అని కవి ఎందుకన్నాడు ?
జవాబు:
కోటిమంది తెలుగు పిల్లలను తెలంగాణ ఒడిలో పెంచింది. కత్తులనిచ్చి, వజ్రాయుధమంతటి కఠినమైన భుజ పరాక్రమాలను చూపేటట్లు రాజుతో తలపడు తుంది. కావున ఈ తెలుగునేలలో ఎంత బలం (కాంతి) ఉన్నదని అర్థం.

ప్రశ్న 2.
‘గడ్డి పోచకూడా కత్తిలా మారటం’ అంటే ఏమిటి ?
జవాబు:
సాధారణంగా గడ్డి మనం నడిస్తే పాదాల కింద పడినలిగి తలవంచి నిలబడుతుంది. కానీ కొన్నిసార్లు ఆ గడ్డి పోచలే బిరుసెక్కి అరికాళ్ళలో ముళ్ళలా గుచ్చుకుంటున్నప్పుడు ఒక్క అడుగు కూడా సరిగా వేయలేం. అట్లాగే పిల్లలుగా పుట్టిన తెలంగాణ బిడ్డలు యుక్తవయసు రాగానే కత్తి చేపట్టి నిర్దయుడైన రాజుతో యుద్ధం చేయటానికి సిద్ధమవటాన్ని కవి అలా పోల్చాడు.

ప్రశ్న 3.
నవోదయం రావడమంటే మీరేమని అనుకొంటున్నారు ?
జవాబు:
ఉదయం అనేది రోజుకు మొదలు. ఉదయంతో రోజు ప్రారంభం అవుతుంది. అంటే ఉదయం అనేది రోజు మొత్తానికి తొలి అడుగు.

ప్రతిరోజూ ఉదయం జరుగుతూనే ఉంటుంది. రోజులు గడుస్తూనే ఉంటాయి. కాని, నవోదయం అంటే క్రొత్త ఉదయం. అంతవరకూ జరిగిన దానికి భిన్నంగా రోజు ప్రారంభమవడం. పాత రోజులలో ఉండే కష్టాలు, బాధలు లేకుండా క్రొత్త ఉత్సాహం, అభివృద్ధి చెందే ఆలోచనలకు అవకాశం కల్పించేది నవోదయం. అటువంటి అభివృద్ధే నవోదయం రావడమంటే అని అనుకొంటున్నాము.

ఆలోచించండి – చెప్పండి

ప్రశ్న 1.
తెలంగాణ వీరుల ప్రత్యేకత ఏమిటి ? (T.B. P.No. 29)
జవాబు:
తెలంగాణా వీరులు భూమండలాన్నంతా సవరించి ఉజ్జ్వలమైన, కాంతిమంతమైన సూర్యుడిని పిలిచి దేశమంతా కొత్తకాంతి సముద్రాలు నింపారు. వీరులు, యోధులు మరియు న్యాయం తెలిసిన పరోపకారులైన తెలుగు వీరులు తెలంగాణ వీరులు.

ప్రశ్న 2.
బ్రతుకు ఎప్పుడు దుర్భరం అవుతుంది ?
జవాబు:
మత పిశాచం విజృంభిస్తే బ్రతుకులు దుర్భరం అవుతాయి. స్వేచ్ఛ లేకపోతే బ్రతుకు భరించలేము మతకల్లోలాలు జరిగితే చాలామంది బ్రతుకులు నాశనం అవుతాయి. ఏ దిక్కూ మొక్కూ లేకపోయినా బ్రతకడం కష్టం. నీతి, న్యాయం, ధర్మం, విద్య లేని సమాజంలో బ్రతుకు దుర్భరం.

ప్రశ్న 3.
ఆకాశాన జెండాలు రెపరెపలాడటం దేనికి సంకేతం ?
జవాబు:
రుద్రమదేవి పరాక్రమించినపుడు తెలుగు జెండాలు ఆకాశాన రెపరెపలాడాయి. ఇది రుద్రమదేవి పరా క్రమానికి నిదర్శనం. తెలంగాణ విజయానికి సంకేతం.

ఇవి చేయండి

I. అవగాహన – ప్రతిస్పంద – ప్రతిస్పందన

ప్రశ్న 1.
కింది అంశాలను గూర్చి చర్చించండి.

అ) ‘వీర తెలంగాణ’ అనే పాఠం పేరు వినగానే మీకు ఎటువంటి అనుభూతి కలిగింది ? దాశరథి తెలంగాణను వీర తెలంగాణ అనడాన్ని తగిన ఉదాహరణలతో సమర్థించండి.
జవాబు:
‘వీర తెలంగాణ’ పేరు వినగానే గొప్ప పోరాటాలు కనుల ముందు మెదిలి వెంట్రుకలు నిక్కబొడుచుకున్నాయి. గడచిన బాధల మబ్బులు ఒక్కసారిగా తొలిగి వెలుతురు వచ్చినట్లు అయింది. తెలంగాణ పేరు ముందు ఉన్న ‘వీర’ అను పదం తెలంగాణలో ఉన్న పోరాటాల చరిత్రను గుర్తు చేసి ఆలోచింపచేసింది.

దాశరథి రాసిన ‘రుద్రవీణ’ అనే పేరుతోనే ఆవేశము, పోరాటము, ఉద్యమము వినిపిస్తాయి. అందులోనూ ఈ పాఠంలో ప్రతి పద్యపాదంలోని పదాలు చరిత్ర చెబుతూ స్ఫూర్తిని ఇస్తాయి. తెలంగాణ పెదవుల మీది నుంచి వచ్చిన శంఖపు చప్పుడు నాలుగు చెరగులా ప్రతిధ్వనించింది అనటం వల్ల తెలంగాణ పోరాటం ఎంత ఉద్వేగభరితంగా జరిగిందో చెప్పవచ్చు.

కోటిమంది తెలుగు పిల్లలు తెలంగాణ యోధులై కత్తులు దూసే వజ్రాయుధమంతటి భుజ పరాక్రమం కలిగి నిజాం రాజుతో తలపడమని తెలంగాణ నేల చెప్పినదన్నప్పుడు తెలంగాణ వీరుల పరాక్రమం అక్షరాల్లో కనిపించి ‘వీర తెలంగాణ’ అనటం సరియైనదే అనిపిస్తుంది.

తెలంగాణలో మొలచిన గడ్డిపోచ కూడా కత్తిపడుతుందని, తెలంగాణ స్వాతంత్ర్య పోరాటం సముద్రంలాగా ఉప్పొంగుతున్నదని, నవాబుల ఆజ్ఞలకు కాలం చెల్లించిందనీ అనటం కూడా ఒక ఉదాహరణగా చెప్పవచ్చు.

కాకతీయుల పోరాటాలు గుర్తుకు తెస్తూ రుద్రమనూ, కాపయ నాయకుడినీ గుర్తు చేసినపుడు, దొంగ దెబ్బలకు భయపడకుండా శత్రురాజుల గుండెలు ఆగేలా ఉరిమిన తెలంగాణ మేఘపు గర్జనలను గుర్తుచేసినపుడు, దాశరథి ఈ పాఠాన్ని ‘వీర తెలంగాణ’ అనటం సబబే అనిపించి తీరుతుంది.

ప్రశ్న 2.
కింది అపరిచిత కవితను చదువండి. ప్రశ్నలకు జవాబులు రాయండి.

తోటమాలి బలిదానం చేస్తేనే
పువ్వులు పరిమళాల నీనగలవు
మానవుడు కలవాలి మానవుణ్ణి
తిడితే ఏం లాభం కనిపించని దేవుణ్ణి
ఆకాశానికి శోభ చందమామ
మిణుగురుతో విద్యుత్కాంతులు ప్రసరించవు
మారాలి నేటి నాటువ్యక్తి
కాకుంటే లే దెన్నటికి విముక్తి
మానవునికి మానవుడే ధ్యేయం
మానవత్వమే మానవజాతికి శ్రేయం
చరిత్రలు మన ఉనికికి కావు ప్రమాణం
ధరిత్రిని వెనక్కి నెట్టి చేయాలి ప్రయాణం
(కవిరాజ మూర్తి)

ప్రశ్నలు

అ) పూలు ఎప్పుడు తమ పరిమళాలను వెదజల్లగల్గుతాయి ?
జవాబు:
తోటమాలి బలిదానం చేస్తే పువ్వులు పరిమళాలను వెదజల్లగల్గుతాయి.

ఆ) ఎవరిని తిట్టగూడదు ?
జవాబు:
దేవుడ్ని (కనిపించని) తిట్టకూడదు.

ఇ) ఎవరు మారాలి ? ఎందుకు మారాలి ?
జవాబు:
నేటి నాటువ్యక్తి (సామాన్యుడు) మారాలి. ఎందుకంటే మారకపోతే మానవునకు అజ్ఞానం నుండి, బాధల నుండి విముక్తి కలగదు.

ఈ) మానవుడు ఏవిధంగా ప్రయాణం చేయాలి ?
జవాబు:
ఈ భూమిని (ధరిత్రిని) వెనక్కి నెట్టి ప్రయాణం చేయాలి. అంటే భూమికంటే వేగంగా అభివృద్ధి మార్గంలో మానవుడు ప్రయాణించాలి.

ఉ) పై కవితకు శీర్షిక నిర్ణయించండి.
జవాబు:
పై కవితకు శీర్షిక “మానవత్వం”.

ఊ) పై కవితను రాసింది ఎవరు ?
జవాబు:
కవిరాజ మూర్తిగారు రచించారు.

ప్రశ్న 3.
రెండో పద్యానికి ప్రతిపదార్థం క్రింద ఉంది. ఇదే విధంగా 3, 4, 6 సంఖ్య గల పద్యాలకు ప్రతిపదార్థాలు రాయండి.
2వ పద్యం ప్రతిపదార్ధము
తల్లీ = అమ్మా !
నీ = నీ యొక్క
ప్రతిభా విశేషములు = ప్రజ్ఞా విశేషాలు
కొన్ని తరాలదాక = కొన్ని తరాల వరకు
భూతప్రేత = చెడు శక్తుల (భూతప్రేతాల)
హస్తమ్ములన్ = చేతులలో
డుల్లెన్ = పడిపోయినవి (చిక్కుకున్నవి)
ఇపుడు = ఇప్పుడు
అడ్డుల్ + పోయెన్ = అడ్డంకులు తొలగిపోయాయి
సౌదామనీవల్లీ = మెరుపుతీగల
ఫుల్ల = విచ్చుకున్న
విభా + ఆవళుల్ = కాంతులవరుసలు
బ్రతుకుత్రోవల్ = బ్రతుకు దారులను
చూపు = చూపే
కాలమ్ములున్ = సమయములు
మళ్ళెన్ = తిరిగివచ్చినవి (అదిగో)
స్వచ్ఛతర = అత్యంత స్వచ్ఛమైన
ఉజ్జల = ప్రకాశవంతమైన
ప్రథమ సంధ్యా = తొలి పొద్దు
భానువు = సూర్యుడు
ఏతెంచెడిన్ = వస్తున్నాడు (ఉదయిస్తున్నాడు)

జవాబు:
ప్రతిపదార్థ తాత్పర్యాలలో 3, 4, 6 పద్యాలు చూడుము.

II. వ్యక్తీకరణ – సృజనాత్మకత

ప్రశ్న 1.
క్రింది ప్రశ్నలకు ఐదేసి వాక్యాల్లో జవాబులు రాయండి.

అ) “తెలంగాణ గొప్పతనపు విశేషాలు కొన్ని తరాలవరకు దుర్మార్గుల చేతిలో చిక్కుకొన్నాయి” అన్న కవి మాటలను మీరెట్లా సమర్థిస్తారు ?
జవాబు:
తెలంగాణ ఎంతో సాంస్కృతిక వికాసం కలది. భాషా సంస్కృతులు ఎంతో గొప్పవి. కానీ ఇవన్నీ చరిత్రలో ఇనుప పద ఘట్టనల కింద నలిగిపోయాయి. తెలంగాణ నేలమీద విముక్తి ఉద్యమాలు, సాయుధ పోరాటాలు, ప్రత్యేక రాష్ట్ర మహోద్యమాలు సముద్రంలో అలల మాదిరిగా పొంగాయి.

తెలంగాణ గొప్పతన విశేషాలు అన్నీ కొన్ని తరాల వరకు దుర్మార్గుల చేతిలో చిక్కుకున్నాయి. ఇప్పుడు ఆ రోజులు పోయినాయి. అడ్డంకులు తొలగి పోయాయి అన్నాడు దాశరథి. కవి మాటలను నేను సమర్థిస్తున్నాను. “నేను నా తెలంగాణ నిగళాలు తెగద్రొచ్చి ఆకాశమంత ఎత్తు అరచినాను” అన్నాడు కవి. ఈ విధంగా ఎందరో వీరుల త్యాగఫలమే నేటి తెలంగాణ. ఎన్నో పోరాటాల ఫలితంగా 2014 జూన్ రెండవ తేదీన పరిపూర్ణ స్వాతంత్య్రం పొందింది. తెలంగాణ రాష్ట్రం ఏర్పడింది. దానితో తెలంగాణ ఆశలు నెరవేరాయి.

ఆ) “తెలగాణమ్మున గడ్డిపోచయును సంధించెన్ కృపాణమ్ము” అని దాశరథి ఎందుకన్నాడు ?
జవాబు:
ఈ తెలంగాణలో గడ్డిపోచ కూడా కత్తి బట్టి ఎదిరించింది. గొప్ప రాజుగా పేరొందినవాని గర్వాన్ని అణచేటట్లుగా యుద్ధం సాగించింది. ఏమి జరుగు తుందో తెలియక జగమంతా భయపడిపోయింది. దిగంతాలన్నీ ఆకాశంలో ఇంద్రధనుస్సుల పరంపరలచేత సయ్యాటలాడాయి.

కాబట్టి “ఇక్కడ పుట్టిన చిగురు కొమ్మైనా చేవగలది”. “తెలంగాణలో చాలా చిన్నదయిన గడ్డి పోచకు కూడా కత్తిబట్టి యుద్ధం చేయగల సత్తా ఉందని” కవి చెప్పాడు.

ఇ) తెలంగాణలో సంధ్యాభానువు మొదటిసారి ఉదయించాడని కవి ఎందుకన్నాడు ?
జవాబు:
అమ్మా తెలంగాణమా ! నీ గొప్పతనపు విశేషాలు కొన్ని తరాల వరకు దుర్మార్గుల చేతులలో చిక్కుకొన్నాయి. ఇప్పుడు ఆ రోజులు గతించాయి. అడ్డంకులు తొలిగాయి. విచ్చుకున్న మెరుపు తీగల కాంతిరేఖలు బతుకుతోవ చూపే కాలం వచ్చింది. స్వచ్ఛమైన కాంతవంతమైన సంధ్యాసూర్యుడు మొదటి సారి ఉదయించాడు.
కాబట్టి కాంతి జ్ఞానానికి సంకేతం – విముక్తికి సంకేతం. కాంతివంతము, స్వచ్ఛమయిన సూర్యకాంతి తెలంగాణకు వచ్చిందని కవి భావన.

ప్రశ్న 2.
క్రింది ప్రశ్నకు పది వాక్యాలలో జవాబు రాయండి.

అ) వీర తెలంగాణ పాఠ్యభాగ సారాంశాన్ని సొంత మాటలలో రాయండి.
(లేదా)
వీర తెలంగాణ పాఠం ఆధారంగా తెలంగాణ వీరుల ఘనతను వర్ణించండి. (June ’16)
జవాబు:
ఓ తెలంగాణమా ! నీ పెదవులతో ఊదిన శంఖధ్వనులు ఈ భూమండలమంతా ఒక్కమారుగా బొబ్బలు పెట్టినట్లుగా ప్రతిధ్వనించాయి. ఆహా ! ఉదయించిన సూర్యుని కిరణాలచేత ప్రీతిపొందిన పద్మాలచే, చలించిన ఆకాశగంగా తరంగాలు అన్ని దిక్కులను తెలవారేటట్లు చేశాయి. అమ్మా తెలంగాణమా ! నీ గొప్పతనపు విశేషాలు కొన్ని తరాల వరకు దుర్మార్గుల చేతులలో చిక్కుకొన్నాయి. ఇప్పుడు ఆ రోజులు గతించాయి. అడ్డంకులు తొలిగాయి. విచ్చుకున్న మెరుపు తీగల కాంతిరేఖలు బతుకుతోవ చూపే కాలం వచ్చింది. స్వచ్ఛమైన, కాంతిమంతమైన సంధ్యా సూర్యుడు
మొదటిసారి ఉదయించాడు.

అమ్మా ! కోటిమంది తెలుగు పిల్లలను నీ ఒడిలో పెంచావు. వారికి వయసురాగానే చేతులకు కత్తులనిచ్చి, వజ్రాయుధమంతటి కఠినమైన భుజపరాక్రమాలను చూపేటట్లు రాజుతో తలపడమన్నావు. అమ్మా ! ఈ తెలుగు నేలలో ఎంత బలం ఉన్నదో కదా ! ఈ తెలంగాణలో గడ్డిపోచకూడా కత్తి బట్టి ఎదిరించింది. గొప్ప రాజుగా పేరొందినవాని గర్వాన్ని అణచేటట్లుగా యుద్ధం సాగించింది. ఏమి జరుగుతుందో తెలియక జగమంతా భయపడిపోయింది. దిగంతాలన్నీ ఆకాశంలో ఇంద్ర ధనుస్సుల పరంపరలచేత సయ్యాట లాడాయి.

తెలంగాణా స్వాతంత్ర్య పోరాటం సముద్రం మాదిరిగా ఉప్పొంగుతున్నది. నాలుగువైపుల నుండి సముద్రానికి గండికొట్టి తెలంగాణ నేలనంతా స్వాతంత్ర్యపు నీటితో తడుపుతున్నారు. ఉద్రిక్తత కలిగించిన నవాబుల ఆజ్ఞలకు కాలం చెల్లిపోయింది. అమ్మా తెలంగాణా ! నీ పిల్లలలో ప్రకాశించే విప్ల వాత్మకమైన కదలిక ఊరికే పోలేదు. భూమండ లాన్నంతా సవరించి ఉజ్జ్వలమైన కాంతిమంతమైన సూర్యుడిని పిలిచి దేశమంతా కొత్త కాంతి సముద్రాలు నింపారు. వారంతా వీరులు, యోధులేకాదు. న్యాయం తెలిసిన పరోపకారులైన తెలుగువీరులు సుమా !

అమ్మా ! మతం అనే పిశాచి తన క్రూరమైన కోరలతో మా నేలను ఆక్రమించి మా గొంతులు కోస్తున్నప్పుడు, ఏ దిక్కూ తోచనప్పుడు, బ్రతకడమే భారమైనప్పుడు తెలుగుదనాన్ని కోల్పోలేదు. యుద్ధ రంగంలో రుద్రాదులు మెచ్చేటట్లు చివరికి విజయాన్ని సాధించాం. ఇక్కడ కాకతీయ రాజుల కంచుగంట మ్రోగినప్పుడు దుర్మార్గులైన శత్రురాజులు కలవర పడ్డారు. రుద్రమదేవి పరాక్రమించినప్పుడు తెలుగు జెండాలు ఆకాశాన రెపరెపలాడాయి. కాపయ్య నాయకుడు తన విజృంభణం చూపినప్పుడు శత్రు రాజులకు గుండెలు ఆగిపోయాయి.

చాళుక్య రాజులు పశ్చిమ దిక్కున పరిపాలన చేసేటప్పుడు మంగళకరమైన జయధ్వనులు మోగాయి. నాటి నుండి నేటి వరకు తెలంగాణం శత్రువుల దొంగ దెబ్బలకు ఓడిపోలేదు. శ్రావణ మాసంలోని మేఘం మాదిరిగా గంభీరమైన గర్జనలు అలరారుతుండగా నా తెలంగాణం ముందుకు సాగుతూనే ఉన్నది.

ప్రశ్న 3.
కింది అంశాన్ని గురించి సృజనాత్మకంగా / ప్రశంసిస్తూ రాయండి.

అ) ‘తెలంగాణ తల్లి’ తన గొప్పదనాన్ని వివరిస్తున్నట్లుగా ఏకపాత్రా భినయం రాసి ప్రదర్శించండి. లేదా ఆత్మకథ రాయండి.
జవాబు:
ఏకపాత్ర
ఏంది అట్టా సూత్తుండ్రి ? నేనే మీ యమ్మను. అదే బిడ్డా ! మీ తెలంగాణా తల్లిని. ఏందట్లో నవ్వుతుండావే ఎన్నేళ్ళయింది బిడ్డల్లారా ! మీ మొగాన నగవు చూసి, మీరు నా కోసం ఎన్ని బాదలు పడ్డారో ? ఎంతమంది నా బిడ్డల ఉసురు కోల్పోయేరో ? వాళ్ళందర్నీ తల్చుకొంటే కడుపు చెరువయ్యేను. ఒకటే పట్టు, మొండి పట్టు, జనిగె పట్టు పట్టారు, సాధించారు.

ఇదిగో ఇయ్యేల నాకు పూలకాలం తియ్యటి తీపులదినం. ఏమనుకున్నవో ? ఓ బిడ్డా ! ఇయ్యేల నా తెలంగాణాలో అదేనో మనింట్లో ఎక్కడ జూసిన కమ్మటి వాసన. ఇను సొంపైన రొద. అదో ! అక్కడి దిక్కు జెర నా చెవు వారిచ్చి ఇను, ఎడవాసిన ఎదలను ఏకంజేయ పేరుపెట్టి పిలుస్తున్నట్లు ఆ కోయిల కూతలు.

మంగళార్తులు ఈయడానికి ముత్తైదలొత్తన్నారు. జర ఆనందంగా ఉండండి. ఇగ మనకు కష్టాలు లేవు. సల్లగా నవ్వాలి నా వోళ్ళంతా.

ఆత్మకథ

ఎన్నో కష్టాలు పడ్డాను. నా కళ్ళ ముందే నా బిడ్డలు పిట్టల్లా రాలిపోతుంటే వెక్కి వెక్కి ఏడ్చాను. పరాయి పాలనలో ఎన్ని బాధలు పడ్డానో ? తలుచు కొంటే ఒళ్ళు జలదరిస్తుంది. పాల్కురికి సోమన, బమ్మెర పోతన వంటి భక్త కవులను తలచుకొంటే భక్తి భావంతో పరవశించి పోతాను. యాదగిరిగుట్టలో లక్ష్మీనరసింహస్వామిని చూసినప్పుడల్లా అనుకొంటాను కష్టాలు అశాశ్వతమని. ఆయన హిరణ్య కశిపుడిని చంపాక ప్రహ్లాదుడిని కాపాడాడు కదా !

అలాగే ఏదో రోజు పరాయి పాలన అనే హిరణ్య కశిపుడు అంతమౌతాడు. ప్రహ్లాదుడు లాంటి నా తెలంగాణా బిడ్డలు రక్షించబడతారు అని నా నమ్మకం. అది నెరవేరింది.

దాశరథిలాంటి విప్లవ కవులు పుట్టడం నా అదృష్టం. తెలంగాణ పోరాటాలలో నా ‘కవి కుమారుల’ గొప్పతనాన్ని మరిచిపోగలనా? ఎంతోమంది కళా కారులను కన్నాను. మీకెవ్వరికీ అన్నానికి లోటులేదు. ఇది తెలుగు మాగాణి. నేనెప్పుడూ పదిమందికి పెట్టాను. అందుకే మీకూ నా లక్షణాలే వచ్చాయి.

ఇది మన ఇల్లు. మనం అభివృద్ధి చేసుకోవాలి. మీరంతా బాగా చదువండి. నా పేరు ప్రఖ్యాతులు పెంచండి. తెలంగాణ తల్లి దీవెనలు మీ అందరికీ ఎప్పుడూ ఉంటాయి. ఇంకోసారి కలిసినప్పుడు ఇంకా మాట్లాడుకొందాం.

III. భాషాంశాలు

పదజాలం

ప్రశ్న 1.
క్రింది వాక్యాలు చదువండి. గీత గీసిన పదాల అర్థాలను ఉపయోగించి సొంతవాక్యాలు రాయండి.

అ) గాలికి ఊగుతున్న పువ్వులు చిగురుటాకులతో సయ్యాటలాడుతున్నాయి.
అర్థం = పరిహాసాలాడు
జవాబు:
సొంతవాక్యం : రవి తన స్నేహితులతో పరిహాసాలాడాడు.

ఆ) స్వాతంత్ర్యోద్యమం బ్రిటిష్ వారి గుండెల్లో కల్లోలం రేపింది.
అర్థం = అలజడి
జవాబు:
సొంతవాక్యం : పరీక్షా ఫలితాల రోజున మా విద్యార్థుల మనస్సులో అలజడి రేగుతుంది.

ఇ) వీరులెప్పుడూ ప్రాణాలను అర్పించడానికి వెనుకాడరు.
అర్థం = ఆగిపోరు
జవాబు:
సొంతవాక్యం : దేశభక్తులైన భారతీయ సైనికులు యుద్ధంలో శత్రువుల బలాన్ని చూసి ఆగిపోరు.

ఈ) దిక్కు తోచనప్పుడు అయోమయంలో పడుతాం.
అర్థం = ఏమీ పాలుపోనప్పుడు
జవాబు:
సొంతవాక్యం : విశాఖలో ప్రజలకు హుదూద్ తుఫాను వచ్చినప్పుడు ఏమీ పాలుపోలేదు.

ప్రశ్న 2.
క్రింది పదాలకు నానార్థాలు రాయండి.

అ) ఉదయము = పుట్టుక, తూర్పుకొండ, ప్రాతకాలం, వడ్డి, సృష్టి, సూర్యోదయం
ఆ) ఆశ = కోరిక, దిక్కు
ఇ) అభ్రము = ఆశ్చర్యము, అపురూపం, అచ్చెరువు, మేఘం, స్వర్గం

ప్రశ్న 3.
క్రింది వాక్యాల్లో గీతగీసిన పదాలకు పర్యాయపదాలు రాయండి.

అ) మురళీరవము మానసిక ఆహ్లాదాన్నిస్తుంది.
జవాబు:
ధ్వని, కంఠధ్వని, శబ్దం, రవళి

ఆ) రుద్రమదేవి కృపాణముతో శత్రువులను చెండాడింది.
జవాబు:
ఖడ్గము, కత్తి, అసి

ఇ) జలధి అనేక జీవరాశులకు నిలయం.
జవాబు:
వార్థి, సముద్రం, అంబుధి, పయోధి

ఈ) జాతీయ జెండాను గౌరవించాలి.
జవాబు:
పతాకం, టెక్కెం, కేతనం, పతాక

ఉ) హనుమంతుడు సముద్రాన్ని లంఘించాడు.
జవాబు:
దూకుడు, దాటుట, తరించాడు, పరిస్థుతం

వ్యాకరణాంశాలు

1. కింది పదాలను విడదీసి రాసి, సంధిపేరు రాయండి. 

అ) జగమెల్ల = జగము + ಎల్ల – ఉకార సంధి
ఆ) సయ్యాటలాడెన్ = సయ్యాటలు + ఆడెన్ – ఉకార సంధి
ఇ) దారినిచ్చిరి = దారిని + ఇచ్చిరి – ఇకార సంధి
ఈ) ధరాతలమెల్ల = ధరాతలము + ఎల్లన్ – ఉత్వసంధి
ఉ) దిశాంచలము = దిశ + అచలము – సవర్ణదీర్ఘ సంధి
ఊ) శ్రావణాభ్రము = శ్రావణ + అభ్రము – సవర్ణదీర్ఘ సంధి
ఋ) మేనత్త = మేన + అత్త – అకార సంధి

2. క్రింది పదాలకు విగ్రహవాక్యాలు రాసి సమాసం పేరు వ్రాయండి. 

సమాసపదం – విగ్రహవాక్యం – సమాసం పేరు

అ) కాకతీయుల కంచుగంట – కాకతీయుల యొక్క కంచుగంట – షష్ఠీ తత్పురుష సమాసము
ఆ) కళ్యాణ ఘంటలు – కళ్యాణ ప్రదమైన ఘంటలు – విశేషణ పూర్వపద కర్మధారయం
ఇ) బ్రతుకుత్రోవ – బ్రతుకు యొక్క త్రోవ – షష్ఠీ తత్పురుష సమాసము
ఈ) మహారవము – మహాయైన (గొప్పదైన) రవము – విశేషణ పూర్వపద కర్మధారయం
ఉ) వికారదంష్ట్రలు – వికారమైన దంష్ట్రలు – విశేషణ పూర్వపద కర్మధారయం
ఊ) కాంతివార్డులు – కాంతి అనెడి వార్థులు – రూపకం
ఋ) తెలంగాణ రాష్ట్రం – తెలంగాణ అను పేరుగల రాష్ట్రం – సంభావనాపూర్వపద కర్మధారయం
ౠ) మతపిశాచి – మతమనెడు పిశాచి – రూపక సమాసం

చీకానుప్రాసాలంకారం :

ఈ వాక్యాన్ని పరిశీలించండి.

నీటిలో పడిన తేలు తేలుతదా !
ఈ వాక్యంలో తేలు, తేలు అనే పదాలకు అర్థాలకు వేరు వేరుగా ఉన్నాయి. ఆ పదాలు వెంట వెంటనే ప్రయోగించబడ్డాయి. హల్లుల జంట అర్థభేదంతో వెంటవెంటనే వాడబడితే దానిని ‘ఛేకానుప్రాసాలంకారం’ అంటారు.
మరికొన్ని వాక్యాలు చూడండి. సమన్వయం చేయండి.

హల్లుల జంట అర్థభేదంతో
అవ్యవధానంగా వస్తే
ఛేకానుప్రాసం.

అ) అరటితొక్క తొక్కరాదు.
ఆ) నిప్పులో పడితే కాలు కాలుతుంది.
ఇ) తమ్మునికి చెప్పు ! చెప్పు తెగిపోకుండా నడువుమని.

ఇట్లాంటివి పాఠంలో వెతకండి. కొన్ని సొంతంగా తయారు చేయండి.

ప్రాజెక్టు పని

తెలంగాణా పోరాట నేపథ్యంలో వచ్చిన ఏవైనా రెండు మూడు పాటలు సేకరించండి. వాటిని పాడి వినిపించండి.
(లేదా)
దాశరథి రాసిన ఏదైనా ఒక పుస్తకం / పాట / కవిత చదువండి. దాని ఆధారంగా నివేదిక రాసి చదివి వినిపించండి.
జవాబు:
గద్దర్ తం
1. బండెనక బండి కట్టి
పదహారు బండ్లు కట్టి
ఏ బండ్లో పోతావ్ కొడకో
నైజాము సర్కరోడా
నాజీల మించినవురో
నైజాము సర్కారోడా
పోలీసు మిల్టీ రెండు
బలవంతులా అనుకోని
నువ్వు పల్లెలు దోస్తివి కొడుకో
నువ్వు పల్లెలు దోస్తివి కొడుకో
మా పల్లెలు దోస్తివి కొడుకో
నైజాము సర్కారోడా

2. పల్లె పల్లెను లేపి
గుండె గుండెను ఊపి
నిండుశక్తిని జూపి, నింగినేలను దాచే
దుంకో దుంకర దుంకో ॥ దుర్గ దుంకిన
దుంకో !
కాశ్మీర్ చూడరో ! కథ మారిపోయరా
అస్సామీ నాడురో ! నెత్తురు మడుగాయరా
ఖలిస్తాన్ మాటరో ! కడుపున చిచ్చాయరో
ముడుచుకు కూర్చుంటెరో ! ముక్కలేను
దేశమ్మురో || పల్లె ||

3. పహరాహుషార్ పహరా హుషార్ పహరా ॥ హుషార్ ॥
నీవు లేచిఉండాలిరా కాపు కాచి ఉండాలిరా || హుషార్ ॥
స్వార్థబుద్ధి రాజ్యమేల ప్రగతి శూన్య మాయరా!
అన్యాయం అక్రమాలు పెచ్చుమీరిపోయరా ! ॥ పహరా ॥
నీతికి న్యాయానికీ నేతవునీవై
జాతికి నవశక్తి దాతవు కాగా ॥ నీవులేచి ॥
దాశరథి కవిత :
ఆ చల్లని సముద్రగర్భం
దాచిన బడబానల మెంతో ?
ఆ నల్లని ఆకాశంలో
కానరాని భాస్కరు లెందరో ?
భూగోళం పుట్టుకకోసం
కూలిన సురగోళాలెన్నో ?
ఈ మానవరూపంకోసం
జరిగిన పరిణామాలెన్నో ?

ఒక రాజును గెలిపించుటలో
ఒరిగిన నరకంఠాలెన్నో ?
శ్రమజీవులు పచ్చినెత్తురులు
త్రాగని ధనవంతులెందరో ?

అన్నార్తులు అనాథలుండని
ఆ నవయుగ మదెంత దూరమో ?
కరువంటూ కాటకమంటూ
కనుపించని కాలాలెపుడో ?

అణగారి అగ్నిపర్వతం
కని పెంచిన “లావా” యెంతో ?
ఆకలితో చచ్చే పేదల
శోకంలో కోపం యెంతో ?

పసిపాపల నిదుర కనులలో
ముసిరిన భవితవ్యం యెంతో ?
గాయపడిన కవి గుండెల్లో
రాయబడని కావ్యాలెన్నో ?

కులమతాల సుడిగండాలకు
బలియైన పవిత్రులెందరో ?
భరతావని బలపరాక్రమం
చెర వీడే దింకెన్నాళ్ళకో ?

నివేదిక

ప్రశ్నల రూపంలో కొనసాగుతూ అసంఖ్యాకమైన ఆలోచనలను రేకెత్తించే గేయం ఇది. ఇరవై నాలుగు చరణాల్లో సమస్త మానవ ప్రపంచాన్ని, విశ్వవిజ్ఞానశాస్త్ర విషయాల సారాన్ని కవిత్వంలో సంక్షిప్తీకరించాడు. మొదటి చరణం సముద్రం, ఖగోళశాస్త్రాల సమ్మేళనం. రెండో చరణం భూమి, మనిషి పుట్టుకల తీరుని వివరిస్తుంది. ఇందులో విశ్వ ఆవిర్భావం, మానవ పరిణామక్రమం కనిపిస్తుంది.

మిగిలిన చరణాలు కవి లోకానుభవంనుండి వచ్చిన చారిత్రక వాస్తవాలు. కవి కలలు, అందమైన ఊహలు, మరో కొత్త ప్రపంచపు ఆశలు, ఆశయాలు, ఆవేదనలు, ఆగ్రహాలతో ఉద్వేగంగా నడుస్తుంది. కవి సున్నితమైన భావాలు మనల్ని ఆనందానికి, ఆగ్రహానికి గురిచేస్తాయి. కవి అన్నీ ప్రశ్నలే వేశాడు. ఇవి మానవాళికి సంబంధించిన ఆత్మవిమర్శనాత్మక అస్త్రాలు. ప్రపంచాన్ని చుట్టుముట్టిన ఈ ప్రశ్నలకు సమాధానం, సమస్యలకు పరిష్కారం మనలోనే, మన దగ్గరే ఉంది. కవి వేసే ప్రశ్న మనకు గుచ్చుకుంటుంది. ఇలాంటి ప్రశ్నలకు చోటులేని ‘నవయుగం’ కోసం కవి చేసిన అక్షరయాగమే ఈ గేయం.

మీకు తెలుసా ?…

1. దాశరథి 1952లో డా॥ సి. నారాయణ రెడ్డి, వట్టికోట ఆళ్వారు స్వామి, డా॥ బిరుదురాజు రామరాజు మొదలగువారితో కలిసి తెలంగాణా రచయితల సంఘాన్ని స్థాపించాడు. ఈయన వట్టికోట ఆళ్వారుస్వామితో కలిసి నిజామాబాద్లో జైలుశిక్ష అనుభవించాడు. తెలంగాణ ఉద్యమ కావ్యాలలోకెల్ల అగ్రగణ్యమైనది దాశరథి ‘అగ్నిధార’. ఇది పీడిత ప్రజల మనోభావాలకు ప్రతీక. ‘అగ్నిధార’ను నిజామాబాద్లో జైలు సహవాసి వట్టికోట అళ్వారుస్వామికి అంకితమిచ్చాడు. అట్లే ‘రుద్రవీణ’ అనే కవితాసంపుటిని తెలంగాణ ప్రజానీకానికి అంకితమిచ్చాడు.

విశేషాంశాలు :

1. కాపయనాయకుడు: క్రీ॥శ॥ 1328 నుండి 1369 వరకు ఓరుగల్లు కోటను కేంద్రంగా చేసుకొని స్వాతంత్య్ర పోరాటం చేసినవాడు. మహా బలవంతుడైన ఢిల్లీ చక్రవర్తి మహమ్మద్ బీన్ తుగ్లక్ అధికారమును ధిక్కరించి స్వాతంత్య్రోద్యమమును లేవదీసిన ఘనత కాపయప్రోలయ నాయకులది. కాపానీడు లేక కాపయనాయకుడు ముసు నూరి నాయకుల వంశమునకు చెందినవాడు. కాపయనాయకుడు తెలంగాణను మ్లేచ్ఛ పాలన నుండి విముక్తి చేశాడు. కాకతీయుల తరువాత ఓరుగల్లును రాజధానిగా చేసుకొని పరిపాలించాడు. ముసునూరి యుగం చరిత్రలో సువర్ణ ఘట్టం.

సూక్తి : సంకల్పమే సకల విజయాలకు మూలం సాధించాలనే తపనే విజయంవైపు వేసే తొలి అడుగు

ప్రతిపదార్థ తాత్పర్యాలు

I

1. ఉ.
ఓ తెలంగాణ ! నీ పెదవులొత్తిన శంఖ మహారవమ్ములీ
భూతలమెల్ల నొక్కమొగి బొబ్బలు పెట్టినయట్లు తోచె, ఓ
హో ! తెలవార్చివేసినవి ఒక్కొక దిక్కు నవోదయార్క రుక్
ప్రీత జలేజ సూన తరళీకృత దేవనదీతరంగముల్

కవి పరిచయం
ఈ పద్యం డా॥ దాశరథి కృష్ణమాచార్య రచించిన దాశరథి సాహిత్యం ఒకటవ సంపుటి “రుద్రవీణ” లోనిది.

ప్రతిపదార్థము (June 2017)

ఓ తెలంగాణ = ఓ తెలంగాణమా !
నీ పెదవుల = నీ పెదవులతో
ఒత్తిన = ఊదిన
శంఖమహారవమ్ములు = శంఖధ్వనులు
ఈ భూతలము = ఈ భూమండలము
ఎల్లన్ = అంతా
ఒక్కమొగి = ఒక్కమారుగా
బొబ్బలు = బొబ్బలు
పెట్టిన + అట్లు = పెట్టినట్లుగా
తోచే = అనిపించింది.
ఓహో ! = ఆహా !
తెల్లవార్చి = తెల్లవారేటట్లు
వేసినవి = చేశాయి
ఒక్కొక్క = అన్ని
దిక్కు = దిక్కులను
నవోదయ + అర్క = ఉదయించిన సూర్యుని
రుక్ = కిరణాలచేత
ప్రీత = ప్రీతి పొందిన
జలేజసూన = పద్మాలచే
తరళీకృత = చలించిన
దేవనదీతరంగముల్ = ఆకాశగంగా తరంగాలు

తాత్పర్యము:
ఓ తెలంగాణమా ! నీ పెదవులతో ఊదిన శంఖ ధ్వనులు ఈ భూమండలమంతా ఒక్కమారుగా బొబ్బలు పెట్టినట్లుగా అనిపించింది. ఆహా ! ఉదయించిన సూర్యుని కిరణాలచేత ప్రీతిపొందిన పద్మాలతో, చలించిన ఆకాశ గంగా తరంగాలు అన్ని దిక్కులను తెలవారేటట్లు చేశాయి.

2. శా.
తల్లీ ! నీ ప్రతిభా విశేషములు భూతప్రేత హస్తమ్ములన్
డుల్లెన్ కొన్ని తరాలదాక ! ఇపుడడ్డుల్ వోయె; సౌదామనీ
వల్లీ పుల్లవిభావళుల్ బ్రతుకుత్రోవల్ జూపు కాలమ్ములున్
మళ్ళెన్ ! స్వచ్ఛతరోజ్జ్వల ప్రథమ సంధ్యాభానువేతెంచెడిన్

కవి పరిచయం

ఈ పద్యం డా॥ దాశరథి కృష్ణమాచార్య రచించిన దాశరథి సాహిత్యం ఒకటవ సంపుటి “రుద్రవీణ” లోనిది.

ప్రతిపదార్ధము (June 2017)

తల్లీ = అమ్మా (తెలంగాణమా)
నీ = నీ యొక్క
ప్రతిభా = గొప్పతనపు
విశేషములు = విశేషాలు
భూతప్రేత = దుర్మార్గుల (చెడుశక్తుల)
హస్తమ్ములన్ = చేతులలో
డుల్లెన్ = చిక్కుకున్నాయి
కొన్ని = కొన్ని
తరాలదాక = తరాలవరకు
ఇపుడు = ఇప్పుడు (ఆ రోజులు గతించాయి)
అడ్డుల్ = అడ్డంకులు
పోయెన్ = తొలిగాయి (తొలగిపోయాయి)
సౌదామనీ వల్లీ = మెరుపు తీగల
ఫుల్ల = విచ్చుకున్న
విభా + ఆవళుల్
(విభావళుల్) = (కాంతి రేఖల) కాంతుల వరుసలు
బ్రతుకుత్రోవల్ = బతుకు దారులను
చూపు = చూపే
కాలమ్ములన్ + మళ్ళెన్ = కాలం వచ్చింది (సమయం వచ్చింది)
స్వచ్ఛతర = స్వచ్ఛమైన
ఉజ్జ్వల = కాంతివంతమైన
ప్రథమసంధ్యా = తొలిపొద్ద
భానువు = సూర్యుడు
ఏతెంచిడిన్ = ఉదయించాడు (వస్తున్నాడు)

తాత్పర్యము
అమ్మా తెలంగాణమా ! నీ గొప్పతనపు విశేషాలు కొన్ని తరాల వరకు దుర్మార్గుల చేతులలో చిక్కుకొన్నాయి. ఇప్పుడు అడ్డంకులు తొలిగాయి. విచ్చుకున్న మెరుపు తీగల కాంతి రేఖలు బతుకుతోవ చూపే కాలం వచ్చింది. స్వచ్ఛమైన కాంతిమంతమైన సంధ్యాసూర్యుడు ఉదయించాడు.

II

3. ఉ.
నీ యొడిలోన పెంచితివి నిండుగ కోటి తెలుంగు కుర్రలన్ !
ప్రాయము వచ్చినంతనె కృపాణములిచ్చితి, యుద్ధమాడి వా
శ్రేయ భుజాబలమ్ము దరిసింప జగమ్ము, నవాబుతో సవాల్
చేయుమటంటి; వీ తెలుగు రేగడిలో జిగి మెండు మాతరో !

కవి పరిచయం
ఈ పద్యం డా॥ దాశరథి కృష్ణమాచార్య రచించిన దాశరథి సాహిత్యం ఒకటవ సంపుటి “రుద్రవీణ” లోనిది.

ప్రతిపదార్థము

మాతరో = అమ్మా !
కోటి = కోటి మంది
తెలుంగు కుర్రలన్ = తెలుగు పిల్లలను
నీ = నీ
ఒడిలోన = ఒడిలో
పెంచితివి = పోషించావు
ప్రాయము వచ్చినంతనే = వయసురాగానే
కృపాణములు ఇచ్చితి = చేతులకు కత్తులనిచ్చి
వాత్రేయ = వజ్రాయుధము అంతటి కఠినమైన
భుజాబలమ్ము = భుజపరాక్రమాలను
దరిసింప = చూపేటట్లు
జగమ్ము = లోకం
నవాబుతో = నిజాం నవాబుతో
సవాల్ =ఎదురొడ్డి ప్రశ్నించుట
చేయుము + అట + అంటివి = చేయమన్నావు
యుద్ధము + ఆడి = తలపడమన్నావు
ఈ తెలుగు రేగడిలో = సారవంతమైన ఈ తెలంగాణ భూమిలో
జిగి = ఎంత బలం
మెండు = ఉన్నదో కదా !

తాత్పర్యము
అమ్మా ! కోటిమంది తెలుగు పిల్లలను నీ ఒడిలో పెంచావు. వారికి వయసురాగానే చేతులకు కత్తులనిచ్చి, వజ్ర సమానమైన భుజపరాక్రమాలను లోకం చూసేటట్లు నిజాం రాజుతో తలపడమన్నావు. ఈ తెలుగు నేలలో ఎంత బలం ఉన్నదో కదా !

4. మ.
తెలగాణమ్మున గడ్డిపోచయును సంధించెన్ కృపాణమ్ము ! రా
జలలాముం డనువాని పీచమడచన్ సాగించె యుద్ధమ్ము ! భీ
తిలిపోయెన్ జగమెల్ల యేమియగునో తెల్యంగరాకన్ ! దిశాం
చలముల్ శక్రధనుఃపరంపరలతో సయ్యాటలాడెన్ దివిన్

కవి పరిచయం
ఈ పద్యం డా॥ దాశరథి కృష్ణమాచార్య రచించిన దాశరథి సాహిత్యం ఒకటవ సంపుటి “రుద్రవీణ” లోనిది.

ప్రతిపదార్ధము
తెలగాణమ్మున గడ్డిపోచయును కృపాణమ్ము సంధించెన్

తెలగాణమ్మున = తెలంగాణలో
గడ్డిపోచయును = గడ్డిపోచకూడా
కృపాణమ్ము = కత్తి బట్టి
సంధించెన్ = ఎదిరించింది
రాజలలాముండు = గొప్ప రాజుగా
అనువాని = పేరొందినవాని
పీచమున్ = గర్వాన్ని
అడచన్ = అణచేటట్లుగా
యుద్ధమ్ము = యుద్ధం
సాగించె = సాగించింది
ఏమి + అగునో = ఏమి జరుగుతుందో
తెల్యంగరాకన్ = తెలియక
జగము + ఎల్లన్ = జగమంతా
భీతిలిపోయెన్ = భయపడిపోయింది
దిశ + అంచలమున్ = దిగంతాలన్నీ
దివిన్ = ఆకాశంలో
శక్తధనుః = ఇంద్రధనుస్సుల
పరంపరలతో = పరంపరలచేత
సయ్యాటలు + ఆడెన్ = సయ్యాటలాడాయి

తాత్పర్యము
ఈ తెలంగాణలో గడ్డిపోచకూడా కత్తి బట్టి ఎదిరించింది. గొప్ప రాజునని అనుకొనేవాని గర్వాన్ని అణచేటట్లుగా యుద్ధం సాగించింది. ఏమి జరుగుతుందో తెలియక జగమంతా భయపడిపోయింది. దిగంతాలన్నీ ఆకాశంలో ఇంద్రధనుస్సుల వరుసలతో సయ్యాటలాడాయి.

5. ఉ.
నాలుగు వైపులన్ జలధి నాల్కలు సాచుచు కూరుచుండె ! క
ల్లోలము రేపినారు భువిలో ! నలుదిక్కుల గండికొట్టి సం
డ్రాలకు దారినిచ్చిరి ! ధరాతలమెల్ల స్వతంత్ర వారి ధా
రాలులితమ్ము కాదొడగె, రాజు రివాజులు బూజు పట్టగన్

కవి పరిచయం
ఈ పద్యం డా॥ దాశరథి కృష్ణమాచార్య రచించిన దాశరథి సాహిత్యం ఒకటవ సంపుటి “రుద్రవీణ” లోనిది.

ప్రతిపదార్థము

నాలుగువైపులన్ = నాలుగు వైపుల నుండి
జలధి = సముద్రం
నాల్కలుసాచుచు = నోరు తెరచినట్లుగా
కూరుచుండె = ఉప్పొంగుతున్నది
కల్లోలం = అలజడి
భువిలో = భూమిలో
రేపినారు = కలిగించినారు
నలుదిక్కులు = నాలుగు దిక్కులు
గండికొట్టి = గండికొట్టి
సంద్రాలకు = సముద్రాలకు
దారినిచ్చిరి = దారిని ఇచ్చారు.
స్వతంత్ర వారి ధారా = స్వాతంత్ర్యపు నీటిలో
లులితమ్ము = తడుపుతున్నారు.
రాజు = నవాబుల
రివాజులు = ఆజ్ఞలకు
బూజుపట్టగన్ = కాలం చెల్లిపోయింది

తాత్పర్యము
తెలంగాణా స్వాతంత్య్ర పోరాటం సముద్రం మాదిరిగా ఉప్పొంగుతున్నది. నాలుగు వైపుల నుండి సముద్రానికి ‘గండికొట్టి తెలంగాణ నేలనంతా స్వాతంత్య్రపు నీటితో తడుపుతున్నారు. ఉద్రిక్తత కలిగించిన నవాబుల ఆజ్ఞలకు కాలం చెల్లిపోయింది.

III

6. మ.
తెలగాణా ! భవదీయ పుత్రకులలో తీండ్రించు వైప్లవ్య సం
చలనమ్మూరక పోవలేదు ! వసుధా చక్రమ్ము సారించి ఉ
జ్జ్వల వైభాతిక భానునిన్ పిలిచి దేశంబంతటన్ కాంతి వా
ర్ధులు నిండించిరి, వీరు వీరులు పరార్థుల్ తెల్గుజోదుల్ బళా !

కవి పరిచయం
ఈ పద్యం డా॥ దాశరథి కృష్ణమాచార్య రచించిన దాశరథి సాహిత్యం ఒకటవ సంపుటి “రుద్రవీణ” లోనిది.

ప్రతిపదార్దము

తెలంగాణా = అమ్మా తెలంగాణ !
భవదీయ + పుత్రకులలో = నీ పిల్లలలో
తీండ్రించు = ప్రకాశించే
వైప్లవ్య = విప్లవాత్మకమైన
సంచలనమ్ము = కదలిక
ఊరకపోవలేదు = ఊరికే పోలేదు
వసుధా చక్రము = భూమండలాన్నంతా
సారించి = సవరించి
ఉజ్జ్వల = ఉజ్జ్వలమైన
వైభౌతిక = కాంతివంతమైన
భానునిన్ = సూర్యుడిని
పిలిచి = పిలిచి
దేశంబు + అంతటన్ = దేశమంతా
కాంతి = క్రొత్తకాంతి
వార్ధులు = సముద్రాలు
నిండించిరి = నింపారు
వీరు = వారంతా
వీరులు = యోధులేకాదు
పరార్థుల్ = న్యాయం తెలిసిన పరోపకారులైన
తెల్గుజోదుల్ = తెలుగువీరులు
బళా ! = సుమా !

అమ్మా తెలంగాణా ! నీ పిల్లలలో ప్రకాశించే విప్లవాత్మకమైన కదలిక ఊరికే పోలేదు. వీరు భూ మండలాన్నంతా సవరించి ఉజ్జ్వలమైన కాంతివంతమైన సూర్యుడిని పిలిచి దేశమంతా కొత్త కాంతి సముద్రాలు నింపారు. వారంతా వీరులు, యోధులేకాదు. పరోపకారులు కూడా.

7. మ.
మతపైశాచి వికార దంష్ఠికలతో మా భూమి లంఘించి మా
కుతుకల్ గోసెడి వేళ గూడ, యెటు దిక్కున్ తోచకున్నప్పుడున్
బ్రతుకే దుర్భరమైన యప్పుడును ఆంధ్రత్వమ్ము పోనాడ లే
దు, తుదిన్ గెల్చితిమమ్మ యుద్ధమున రుద్రుల్ మెచ్చనాంధ్రాంబికా!

కవి పరిచయం
ఈ పద్యం డా॥ దాశరథి కృష్ణమాచార్య రచించిన దాశరథి సాహిత్యం ఒకటవ సంపుటి “రుద్రవీణ” లోనిది.

ప్రతిపదార్థము

ఆంధ్రాంబికా ! = అమ్మా !
మతం = మతం అనే
పైశాచి = పిశాచి
వికార = తన క్రూరమైన
దంష్ట్రికలతో = కోరలతో
మా భూమి = మా నేలను
లంఘించి = ఆక్రమించి
మాకుతుకల్ = మా గొంతులు
గోసెడివేళ కూడా = కోస్తున్నప్పుడు కూడా
యెటుదిక్కున్ = ఏ దిక్కూ
తోచకున్నప్పుడున్ = తోచనప్పుడు
బ్రతుకే = బ్రతకడమే
దుర్భరమైన + అప్పుడును = భారమైనప్పుడు
ఆంధ్రత్వమ్ము = తెలుగుదనాన్ని
పోనాడలేదు = కోల్పోలేదు
యుద్ధమున = యుద్ధరంగంలో
రుద్రుల్ = రుద్రాదులు
మెచ్చున్ = మెచ్చేటట్లు
తుదిన్ = చివరికి
గెల్చితిమమ్మ = విజయాన్ని సాధించాం

తాత్పర్యము
అమ్మా ! మతం అనే పిశాచి తన క్రూరమైన కోరలతో మా నేలను ఆక్రమించి మా గొంతులు కోస్తున్నప్పుడు, ఏ దిక్కూ తోచనప్పుడు, బ్రతకడమే భారమైనప్పుడు కూడా తెలుగుదనాన్ని కోల్పోలేదు. రుద్రాదులు మెచ్చేటట్లు చివరికి విజయాన్ని సాధించాం.

8. సీ.
కాకతీయుల కంచు గంట మ్రోగిననాడు
కరకు రాజులకు తత్తరలు పుట్టె
వీర రుద్రమదేవి విక్రమించిన నాడు
తెలుగు జెండాలు నర్తించే మింట
కాపయ్య నాయకుండేపు సూపిన నాడు
పరరాజులకు గుండె పట్టుకొనియె
చాళుక్య పశ్చిమాశా పాలనమ్మున
కళ్యాణ ఘంటలు గణగణమనె

తే.గీ॥ నాడు నేడును తెలగాణ మోడలేదు.
శత్రువుల దొంగదాడికి; శ్రావణాభ్ర
మటుల గంభీర గర్ణాట్టహాసమలర
నా తెలంగాణ పోవుచున్నది పథాన

కవి పరిచయం
ఈ పద్యం డా॥ దాశరథి కృష్ణమాచార్య రచించిన దాశరథి సాహిత్యం ఒకటవ సంపుటి “రుద్రవీణ” లోనిది.

ప్రతిపదార్ధము

కాకతీయులు = కాకతీయరాజుల
కంచు గంట = కంచుగంట
మ్రోగిననాడు = మ్రోగినప్పుడు
కరకురాజులకు = దుర్మార్గులైన శత్రురాజులు
తత్తరలు + పుట్టె = కలవర పడ్డారు
వీరరుద్రమదేవి = రుద్రమదేవి
విక్రమించిననాడు = పరాక్రమించినప్పుడు
తెలుగు జెండాలు = తెలుగు జెండాలు
మింటన్ = ఆకాశాన
నర్తించె = రెపరెపలాడాయి
కాపయ్యనాయకుడు = కాపయ్య నాయకుడు
ఏపు + సూపిననాడు = విజృంభణం చూపినప్పుడు
పరరాజులకు = శత్రురాజులకు
గుండెపట్టుకొనియె = గుండెలు ఆగిపోయాయి
చాళుక్య = చాళుక్య రాజులు
పశ్చిమ + ఆశా = పశ్చిమ దిక్కున
పాలనమ్మున్ = పరిపాలన చేసేటప్పుడు
కళ్యాణఘంటలు = మంగళకరమైన జయధ్వనులు
గణగణమనె = గణగణమంటూ మోగాయి
నాడు = నాటినుండి
నేడును = నేటివరకు
తెలగాణము = తెలంగాణం
శత్రువుల = శత్రువుల
దొంగదాడికి = దొంగదెబ్బలకు
ఓడలేదు = ఓడిపోలేదు
శ్రావణాభ్రము = శ్రావణ మాసంలోని మేఘం
అటుల = మాదిరిగా
గంభీర = గంభీరమైన
గర్జాట్టహాస = గర్జనలు
అలర = అలరారుతుండగా
నా తెలంగాణ = నా తెలంగాణం
పథాన = ముందుకు
పోవుచున్నది = సాగుతూనే ఉన్నది

తాత్పర్యము

ఇక్కడ కాకతీయ రాజుల కంచుగంట మ్రోగినప్పుడు దుర్మార్గులైన శత్రురాజులు కలవరపడ్డారు. రుద్రమదేవి పరాక్రమించినప్పుడు తెలుగు జెండాలు ఆకాశాన రెపరెప లాడాయి. కాపయ్య నాయకుడు తన విజృంభణం చూపినప్పుడు శత్రురాజులకు గుండెలు ఆగిపోయాయి. చాళుక్య రాజులు పశ్చిమ దిక్కున పరిపాలన చేసేటప్పుడు మంగళకరమైన జయధ్వనులు గణగణమంటు మోగాయి. నాటి నుండి నేటి వరకు శత్రువుల దొంగ దెబ్బలకు తెలంగాణం ఓడిపోలేదు. శ్రావణ మాసంలోని మేఘం మాదిరిగా గంభీరమైన గర్జనలు అలరారుతుండగా నా తెలంగాణం ముందుకు సాగుతూనే ఉన్నది.

పాఠం నేపథ్యం / ఉద్దేశ్యం

తెలంగాణ వీరుల పురిటిగడ్డ. ఎందరో యోధులు తెలంగాణ విముక్తి కోసం తుదిశ్వాస వరకు పోరాడారు. దుర్మార్గులైన రజాకార్ల అరాచకత్వాన్ని ఎదిరించిన రణక్షేత్రం తెలంగాణ. అటువంటి నేల అస్తిత్వ పరిరక్షణ కోసం తెలంగాణ ప్రజాసమూహం తమదైన పద్ధతుల్లో ధిక్కారస్వరం వినిపించింది. ఆయుధం ధరించి పోరాడినవారు కొందరైతే, అక్షరాయుధంతో పోరాడినవారు మరికొందరు. సాహితీ యోధుడు డా॥ దాశరథి కృష్ణమాచార్య ప్రత్యక్షంగా పోరాటంలో మమైకమౌతూనే ధైర్య సాహసాలను పద్యాలలో ప్రశంసించాడు.

వీరుల త్యాగాలను స్మరించడం, తెలంగాణ తల్లి ఔన్నత్యాన్ని కీర్తించడం, స్వస్థానాభిమానాన్ని ప్రేరేపించడమే ఈ పాఠ్యభాగ ఉద్దేశం.

పాఠ్యభాగ వివరాలు

ఈ పాఠం పద్య ప్రక్రియకు చెందినది. చారిత్రక అంశాలను వస్తువుగా తీసుకొని రాసిన పద్యాలివి. డా॥ దాశరథి కృష్ణమాచార్య రచించిన “దాశరథి సాహిత్యం” ఒకటవ సంపుటి ‘రుద్రవీణ’ లోనిది.

కవి పరిచయం

కవి : డా॥ దాశరథి కృష్ణమాచార్య
జననం : 22.7.1925
మరణం : 05.11.1987
జన్మస్థలం : మహబూబాబాద్ జిల్లా, చిన్నగూడూరు.

ఇతర రచనలు :

1. అగ్నిధార,
2. రుద్రవీణ,
3. మహాంధ్రోదయం,
4. పునర్నవం
5. కవితాపుష్పకం,
6. తిమిరంతో సమరం,
7. అమృతాభిషేకం
8. ఆలోచనాలోచనాలు మొదలగు నవి. నవమి (నాటికలు), యాత్రాస్మృతి (స్వీయచరిత్ర) వంటి పలు గ్రంథాలను రచించారు. సినిమాలకు చక్కని పాటలు రాశారు. గాలిబ్ గజళ్ళను అనువదించారు.

కవితారీతి :

నాటి పాలకులపై వ్యతిరేకంగా ప్రజా పోరాటాలలో పాల్గొన్నారు. ఆచరణాత్మక వైఖరి ప్రదర్శించారు. తన కవిత్వంతో ప్రజలను చైతన్యపరిచారు. దాశరథి ఉద్యమ కవి. ‘నా గీతావళి ఎంత దూరము ప్రయాణం బౌనొ అందాక ఈ భూగోళంబున కగ్గి పెట్టెద’నన్నారు.

నిజాంకు వ్యతిరేకంగా పద్యాలను జైలు గోడలపై రాసిన ధీరుడు. ఆయన కవిత లలో అక్షరాలకు ఆవేశం ఎక్కువ. ఆయన కవిత్వం అభ్యుదయ మార్గంలో నడు స్తుంది. సున్నితమైన భావుకతతో ఉంటుంది. ప్రాచీన పద్యశైలిలో ఉంటుంది. ప్రజల హృదయాలలో శాశ్వతంగా నిలిచే కవితా రీతుల ప్రదర్శనలో తలపండిన మహాకవి దాశరథి. ఆయన ప్రజాకవి, మహాకవి.

సన్మానాలు-సత్కారాలు : తెలుగులో గజల్ ప్రక్రియకు ప్రాణం పోసినవాడు దాశరథి. 1961లో గాలిబ్ గజళ్ళను అనువదించాడు. ప్రసిద్ధులైన ఉర్దూ కవుల కవిత్వాన్ని అనువదించాడు. విమర్శకుల ప్రశంసలను అందుకొన్నాడు.

పురస్కారాలు :

1. ఆంధ్రప్రదేశ్ సాహిత్య అకాడమీ అవార్డు (1967)
2. కేంద్ర సాహిత్య అకాడమీ అవార్డు (1974)

శైలి : సున్నితమైన భావుకత, ప్రాచీన పద్య శైలి.
విశేషాంశములు : సినీగేయ కవి, ఆంధ్రప్రదేశ్ రాష్ట్ర చివరి ఆస్థాన కవి.

ప్రవేశిక

సముద్రం ఉప్పొంగుతుండగా చూసేవారు అరుదుగా ఉంటారు. సముద్రం చెలియలికట్ట దాటడం ఎవరూ ఊహించలేరు. కాని తెలంగాణ నేల ఈ అరుదైన పరిణామా లను, అద్భుతాలను ప్రపంచానికి చూపెట్టింది. తెలంగాణ నేలమీద జరిగిన విముక్తి ఉద్యమంలో తెలంగాణ ప్రజలు సముద్రంలో అలల మాదిరిగా ఉవ్వెత్తున ఉప్పొంగి ఎగిశారు. ఆ హోరును, తెలంగాణ వీరుల తిరుగుబాటు జోరును, మహోన్నత త్యాగాల తీరును దాశరథి పద్యాల్లో విని ఉత్తేజితులం కావడానికి… ఈ పాఠంలోకి పయనిద్దాం.

విద్యార్థులకు సూచనలు

• పాఠం ప్రారంభంలోని ప్రవేశిక చదువండి. పాఠంలోని విషయాన్ని ఊహించండి.
• పాఠం చదువండి. అర్థంకాని పదాల కింద గీత గీయండి.
• వాటి అర్థాలను పుస్తకం చివర ఉన్న ‘పదవిజ్ఞానం’ పట్టికలో చూసి లేదా నిఘంటువులో చూసి తెలుసుకోండి.

ప్రక్రియ – ఆధునిక పద్యం

ఈ పాఠం పద్య ప్రక్రియకు చెందినది. ఆధునిక తెలుగు సాహిత్య ప్రక్రియల్లో ఆధునిక పద్యం ఒకటి. చారిత్రక, వాస్తవిక అంశాలను వస్తువుగా తీసుకొని ఆధునిక భావ వ్యక్తీకరణతో రాసిన పద్యాలివి. గ్రాంథిక భాషకు దూరంగా ఉంటాయి. ఆధునిక భాష పద్యాలలో ఎక్కువగా ఉంటుంది. సమాజాభ్యుదయం ప్రధానాంశంగా ఉంటుంది.

Telangana SCERT 10th Class Physics Study Material Telangana 10th Lesson Electromagnetism Textbook Questions and Answers.

TS 10th Class Physical Science Solutions 10th Lesson Electromagnetism

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I. Reflections on concepts

Question 1.
Are the magnetic field lines closed? Explain.
Answer:
1. The field lines appear to be closed loops but you can’t conclude that lines are closed or open loops by looking at the picture of field lines because we do not know about alignment of lines that are passing through the bar magnet.

2. The direction of a magnetic line at any point gives the direction of the magnetic force on a north pole placed at that point.

3. Since the direction of magnetic field line is the direction of force on a north pole, the magnetic lines always begin from the ‘N’ pole of the magnet and end on the ‘S’ pole of the magnet.

4. Inside the magnet however the direction of magnetic lines is from S-pole of the magnet to the N-pole of the magnet.

5. Thus the magnetic field lines are closed.

Question 2.
See figure, magnetIc lines are shown. What is the direction of the current flowing through the wire?

Answer:
1) The magnetic field lines are in the anti-clockwise direction.
2) The direction of current is vertically upwards. This can be demonstrated with right-hand thumb rule.

Question 3.
A bar magnet with North pole facing towards a coil moves as shown In figure given below. What happens to the magnetic flux passing through the coil?

Answer:
The magnetic flux passing through the coil induces current in the coil. This current is called induced EMF. This induced N EMF is equal to the rate of change of magnetic flux passing through it.

Question 4.
A coil is kept perpendicular to the page. At P, current flows into the page and at Q it comes out of the shown in figure. What is the page as direction of magnetic field due to the coil?

Answer:
To know the direction of magnetic field, we use right-hand rule i. e., ‘when you curl your right-hand fingers ¡n the direction of current, thumb gives the direction of magnetic field’. According to this the direction of magnetic field is as shown in the following figure.

Application of Concepts

Question 1.
The direction of current flowing in a coil Is shown in figure. What type of magnetic pole Is formed at the face that has flow of current as shown in fig.

Answer:
The top surface of the coil shown in the fig. behaves as North pole and direction of magnetic field points towards us.

Question 2.
Why does the picture appear distorted when a bar magnet is brought close to the screen of a television? Explain.
Answer:
Picture on a television screen Is due to motion of the electrons reaching the screen. These electrons are affected by magnetic field of bar magnet. This is due to the fact that the magnetic field exerts a force on the moving charges. This force is called magnetic force. Due to this magnetic force, the picture is distorted when you remove the bar magnet away from the screen, the motion of electron is not affected by the magnetic force and the picture will be normal.

Question 3.
Symbol ‘X’ indicates the direction of a magnetic field into the page. A straight long wire carrying current along its length is kept perpendicular to the magnetic field. What A is the magnitude of force experienced by the wire? In what direction does it act?

Answer:
From the figure, a straight wire carrying current which Is kept perpendicular to a uniform magnetic field B. This ‘B’ is directed Into the page. Let the field be confined to the length L. We know that the electric current means charges in motion. Hence they move with a certain velocity called drift velocity V.

The magnetic force on a single charge is given by F₀ = qv B.
Let total charge inside the magnetic field be Q. So magnetic force on the current carrying wire s given by F=QvB ……………………………. (1)
The time taken by the charge (Q) to cross the field be
t = \(\frac{L}{v} \Rightarrow v=\frac{L}{t} \) ……………………………. (2)
∴ (1) ⇒ F = \(\frac{\mathrm{QLB}}{\mathrm{t}} \Rightarrow \frac{\mathrm{Q}}{\mathrm{t}}(\mathrm{LB}) \) …………………………. (3)
We know that \( \frac{\mathrm{Q}}{\mathrm{t}}\) Is equal to the electhc current in the wire.
I = \(\frac{Q}{t} \)
∴ (3) ⇒ FILB.

The direction of the force: The direction of force can be find by using right-hand rule. Fore finger points towards the velocity of current, middle finger points to the direction of magnetic field (B), then the thumb gives direction of force when the three fingers are stretched in such away that they are perpendicular to each other.

Question 4.
An 8N force acts on a rectangular conductor 20 cm long placed perpendicular to a magnetic field. Determine the magnetic field induction if the current in the conductor is 40 A.
Answer:
Formula, F = ILB (or) B = F/IL
Where F = magnetic force = 8N
I = Electrical current in the conductor = 40 A
L, length of conductor = 20 cm. = \(\frac{20}{100}\) = 0.2 mt
∴ B =?
B= \(\frac{8}{40 \times 0.2}=\frac{80}{40 \times 2} \) = 1
Magnetic field induction = 1 tesla.

Question 5.
As shown in figure both coil and bar magnet move in the same direction. Your friend is arguing that there is no change in flux. Do you agree with his statement’ If not what doubts do you have? Frame questions about the doubts you have regarding change in flux.

Answer:
Agree: Yes. I will agree.

1. The induced EMF will not produce when the coil and magnet are moving in the same direction with same velocity.
2. Hence my friend’s argument is correct.

Disagree:

1. What happens if both magnet and coil move in same direction?
2. What happens if both magnet and coil move in opposite direction?
3. What is the direction of the current in the coil?
4. If both move in same direction, is there any linkage of flux with the coil?
5. When N pole is moved towards the coil what is the direction of current?
6. If magnet is reversed, what is the direction of current in the coil?

Question 6.
Give a few applications of Faraday’s law of induction in daily life.
Answer:
The daily life applications of Faraday’s law of induction are :

1. Generation of electricity.
2. Transmission of electricity.
3. Metal detectors in security checking.
4. The tape recorder.
5. Use of ATM cards
6. Induction stoves
7. Transformers
8. Induction coils (spark plugs in automobiles)
9. Break system in railway wheels
10. AC and DC generators
11. Windmills etc.

Question 7.
Explain the working of an electric motor with a neat diagram.
Answer:
1. Consider a rectangular coil kept in uniform magnetic field as shown in figure.
2. Switch on the circuit so that the current flows through the coil. The direction of current is shown in the figure.
3. The sides AB and CD of the coil are always at rigi angles to the magnetic field.
4. According to right-hand rule, at AB the magnetic force acts inward perpendicular to the field of magnet ai
on CD, it acts outward.
5. The top view of coil is shown in the figure.
6. The force on the sides BC and DA varies because they make different angles at different positions of the coil in the field. At BC, magnetic force pulls the coil up and at DA magnetic force pulls it down.
7. The net force acting on AB and on CD is zero because they carry equal currents in the opposite direction.
Similarly, the sum of the forces on sides BC and DA is also zero. So, net force is zero on the coil.
8. But the rectangular coil comes into rotation in clockwise direction because equal and opposite pair of forces acting on the twos sides of the coil.


9. If the direction of current in the coil is unchanged, it rotates, upto half rotation In one direction and the next half In the direction opposite to previous like to and fr0 motion.
10. It the direction of current in the coil is changed the coil will rotate continuously in one and the same direction.
11. To achieve this, brushes B, and B₂ are used.
12. These brushes are connected to the battery. The ends of the coil are connected to slip rings C₁ and C₂ which rotate along with the coil.
13. Initially C₁ is In contact with B₁ and C₂ Is in contact with B₂
14. After half rotation, the brushes come into contact with the other slip rings in such a way that the direction of current through the coil Is reversed. This happens every half-rotation.
15. Thus the direction of rotation of the coil remains the same. This is the principle used in electric motor.
16. In electric motors electrical energy is converted Into mechanical energy.

Question 8.
Explain the working of AC electric generator with a neat diagram.
Answer:
(i) Electric Generator or Dynamo: It is a device which converts mechanical energy into electrical energy.
(ii) Principle: It works on the principle of the electromagnetic induction.
(iii) Construction:
It consists of
(i) Armature coil,
(ii) Brushes,
(iii) Slip rings,
(iv) Strong magnet and
(v) Rotating mechanism (or) motor.
The two A and B of the coil ABCD are connected to the slip rings.

Working:

• When the coil is at rest ¡n vertical position, with side (A) of coil at top position and side (B) at bottom position, no current will be induced in it. Thus current in the coil is zero at this position.
• When the coil is rotated in clockwise direction, current will be induced in it and it flows from A to B, in this position the current increases from zero to a maximum.
• If we continue the rotation of coil current decreases during the second quarter of the rotation and once again becomes zero when coil comes to vertical position with side B at top side and side A at bottom position.
• During the second half of the rotation, current generated follows the same patterns as that in the first half, except that the direction of current is reversed.
• Thus, after every rotation of the current ¡n the respective arm changes, there by generating an alternating current. This device is called A.C. generator.

Question 9.
Explain the working of DC generator with a neat diagram.
Answer:

1. Consider a rectangular coil. Let it be held between the poles of curve-shaped permanent magnets as shown in figure.
2. As the coil rotates the magnetic flux passing through the coil changes.
3. According to the law of electromagnetic induction, an induced current is generated in the coil.
4. If two half-slip rings are connected to the ends of the coil as shown in figure, this generator works as DC generator to produce DC current.

Working:

1. When the coil is in vertical position the induced current generated during the first half rotation, rises from zero to maximum and then falls to zero again.
   

2. As the coil moves further from this position, the ends of the coil go to other slip rings.

3. Hence, during the second half rotation, the current is reversed in the coil itself, the current generated in the second half rotation of the coil is identical with that during the first half of the direct current (DC), for one revolution.

4. Hence, this current ¡s called direct current (DC).

Question 10.
How do you appreciate Faraday’s law, which s the consequence of conservation of energy?
Answer:

1. Faraday’s law of electromagnetic induction, which is the outcome of law of conservation of energy, is very much useful.
2. The law led to the conclusion that mechanical energy can be converted into electrical energy.
3. The fruits of electrical energy, namely, the electric lights, the stoves, electric heaters, electrical irons, the fridge, the electric motor the television, the electronic locomotives and all electric gadgets are serving mankind and making people more comfortable.
4. All these gadgets work on electricity and electrical energy that can be produced by generators on large scale where energy Is created by law of conservation of energy on a very large scale.
5. So, we must appreciate the utilïtanan value of Faraday’s law which is the consequence of conservation of energy.

Question 11.
The value of magnetic field Induction which is uniform is 2T. What is the flux passing through a surface of area 1.5m1 perpendicular to the field?
Answer:
Magnetic Induction of uniform field B = 2T
Flux Φ =?
Surface area A = 1.5 m²
We know B = \(\frac{\varphi}{A} \) ⇒ BA cosθ
Φ= BA
Where B ¡s perpendicular to area is
ΔΦ = 2x 1.5=3webers

Question 12.
Which of the various methods of current generation protects nature well? Give examples to support your answer. )
Answer:
The methods of current generation that protect the nature well:
1. Hydra power plant A power plant that produces electricity by using flowing of water to rotate a turbine Is called hydropower plant.

Advantages:

• It does not produce any environmental pollution.
• It will never get exhausted.
• The construction of dams on rivers helps In controlling floods and in irrigation.

2. Wind generators: A wind generator s used to generate electricity by using wind energy.

Advantages:

• It does not cause any pollution.
• It will never get exhausted.
• Wind energy is available free of cost.

3. Solar cell: Solar cell is a device which converts solar energy directly into electricity.

Advantages:

1. They require no maintenance.
2. They can be set up in remote inaccessible and very sparsely inhabited areas where the laying of usual power transmission is different and expensive.
3. As seawater flows in and out of the tidal barrage during high and low tides It turns the turbines to generate electricity.
4. The energy of moving sea waves can be used to generate electricity.
5. The energy available due to the difference In the temperature of water at the surface of the ocean and at deeper levels is called ocean thermal energy. The ocean thermal energy can be converted into a usable form of energy like electricity.
6. Geothermal energy is the heat energy from hot rocks present inside the earth. It is also used to produce electricity. It is a clean and environmental friendly source of energy.
7. The energy produced during nuclear fission reactions is used for generating electricity at nuclear power plants.

Multiple choice questions

Question 1.
Which of the following converts electrical energy into mechanical energy [ ]
(a) motor
(b) battery
(c) generator
(d) switch
Answer:
(a) motor

Question 2.
Which of the following converts mechanical energy into electrical energy [ ]
(a) motor
(b) battery
(c) generator
(d) switch
Answer:
(c) generator

Question 3.
The magnetic force on a current-carrying wire placed in uniform magnetic field if the wire is oriented perpendicular to magnetic field is [ ]
(a) 0
(b) ILB
(c) 2ILB
(d) \(\frac{\text { ILB }}{2} \)
Answer:
(b) ILB

Question 4.
One Tesla = [ ]
(a) Newion/Coloumb
(b) Newton / ampere – meter
(c) Ampere I meter
(d) Newton / ampere second
Answer:
(b) Newton/ampere – meter

Question 5.
Magnetic flux [ ]
(a) dyne
(b) Oersted
(c) Guass
(d) Weber
Answer:
(d) Weber

Question 6.
No force works on the conductor carrying electric current when kept [ ]
(a) parallel to magnetic field
(b) perpendicular to magnetic field
(c) in the magnetic field
(d) away from magnetic field
Answer:
(a) parallel to magnetic field

Suggested Experiments

Question 1.
Explain with the help of two activities that current-carrying wire produces magnetic field.
Answer:
Activity – 1.

1. Take a wooden plank and make a hole as shown in adjacent figure. Table
2. Place the plank on the table.
3. Place a retort stand on the plank as shown in figure.
4. Pass 24 gauge copper wire through a hole of the plank and rubber knob of the retort stand In such a way that the wire be arranged In a vertical position and not touch the stand.
5. Connect the two ends of the wire to a battery via switch.
6. Place 6 to lo compass needles In a circular path around the hole so that its centre coincides with the hole. Use 3-volt battery In the circuit.
7. Switch on current.
8. We see the compass needles change the direction In such a way as tangents to the circle.
9. This activity helps us to prove current carrying wire produces magnetic field.

Activity -2
1. Take a thin wooden plank covered with white paper.
2. Make equidistant holes on its surface as shown in adjacent figure.
3. Pass copper wire through the holes. This forms a coil. Join the ends of the coil to a battery through a switch.
4. Switch on the circuit. Current passes through the coil.
5. Now sprinkle iron filings on the surface of the plank around the coil.
6. Give small jerk to It. An orderly pattern of iron filing is seen on the paper.
7. This activity proves that a current-carrying wire develops a magnetic field around it.

Question 2.
How do you verify experimentally that the current-carrying conductor experiences a force when it is kept in a magnetic field?
(Or)
List out the apparatus and experimental procedure for the experiment to observe a current-carrying wire experiences a magnetic force when it is kept in uniform magnetic field.
Answer:
Required apparatus:

• Horse-shoe magnet
• conducting wire
• battery, switch

1. Take a wooden plank.
2. Fix two long wooden sticks on it.
3. These wooden sticks are split at their top ends.
4. A copper wire is passed through splêts of wooden sticks.
5. Connect the wire to 3 volts battery.
6. Close the switch to make the circuit. Current passes through the wire.
7. Bring a horseshoe magnet near the wire.
8. Then a force is experienced on the wire, following right thumb rule.
9. Reverse the polarities of the magnet, then the direction of the force is also reversed.
10. The night hand rule helps the direction of magnetic force exerted by the magnetic field on the current carring wire.

Question 3.
Explain Faraday’s law of induction with the help of activity.
Answer:

1. Connect the terminals of a coil to a sensitive ammeter or galvanometer as shown in the adjacent figure.
2. Now push a bar magnet towards the coil, with its north pole facing the coil.
3. While the magnet is moving towards the coil the needle in galvanometer deflects, showing that a current has been set up in the coil.
4. The needle of galvanometer does not deflect If the magnet is at rest.
5. If the magnet Is moved away from the coil, the needle In the galvanometer again deflects, but ¡n the opposite direction.
6. It means that a current is set up in the coil in the opposite direction.
7. If we use the end of south pole of a magnet instead of north pole in this activity, the experiment works just as described but the reactions are exactly in opposite directions.
8. Further experimentation enables us to understand that the relative motion of the magnet and coil sets up a current in the coil. It makes no difference whether the magnet is moved towards the coil or the coil towards, the magnet.
9. This proves the Faraday’s law: “Whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated In the coil.

Question 4.
What experiment do you suggest to understand Faraday’s law? What Instruments are required? What suggestions do you give to get good results of the experiment? Give precautions also.
Answer:
Aim: To understand Faraday’s law of induction. Materials required: A coil of copper wire, a bar magnet Galvanometer etc.

Procedure:

1. Connect the terminals of a coil to a sensitive galvanometer as shown In the figure.
2. Normally we would not expect any deflections of needle in the galvanometer because there is to be no electromotive force In this circuit.
3. Now if we push a bar magnet towards the coil, with its north pole facing the coil, we observe the needle in the galvanometer deflects, showing that a current Is set up in the coil.
4. The galvanometer does not deflect if the magnet Is at rest.
5. If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil, in the opposite direction.
6. If we use end of south pole of a magnet instead of north pole In the above activity, the deflections are exactly reversed.
7. This experiment proves “whenever there is a continuous change of magnetic flux linked with a closed coil, a current Is generated in the coil.

Precautions:

• The coil should be kept on an insulating surface.
• Bar magnet should be of good pole strengths.
• The centre of the Galvanometer scale must be zero.
• The deflections in the galvanometer must be observed while Introducing the bar magnet Into the coil and also while withdrawing It.

Question 5.
How can you verify that a current-carrying wire produces a magnetic field with the help of an experiment?
Answer:
Aim: To demonstrate that a magnetic field is produced around a current-carrying wire.
Apparatus: Copper wire, slits, 3volt battery, key, and connecting wires, thermocol sheets.

Procedure:

• Take a hermocoI sheet and fix two wooden slits of 1 cm at the top of their ends.
• Arrange a copper wire which passes through slits and make a circuit with the elements as shown in the figure.
• Now, keep a magnetic compass below the wire.
• Bring a barmagnet dose to the compass.

Suggested Projects

Question 1.
Collect Information about generation of current by using Faraday’s law.
Answer:
Faraday’s law is useful In generation of current.

1. According to this law, the change in magnetic flux Induces EMF in the coil.
2. He also proposed electromagnetic induction.
3. Electromagnetic induction is a base for generator, which produces electric Current.
4. Transformer also works on the principle of electromagnetic induction, which is helpful In transmission of electricity.
5. Hence Faraday’s law is used in the generation and transmission of current.

Question 2.
Collect information about material required and procedure of making a simple electric motor from Internet and make a simple motor on your Own.
Answer:
Material required:

1. A plastic pipe of 3 or 4 cm. diameter and 5 cm long.
2. A copper wire 30 cm long.
3. A horseshoe magnet.
4. Two corks that exactly fit Into the bore of the plastic pipe.
5. A wooden plank used as base for the whole arrangement.
6. Cello tape and small nails.

Procedure:

1. Wind the copper wire around the plastic pipe.
2. Connect the ends of the wire to the two brushes.
3. Fit in the corks into the hollow of the pipe such that they close either end of the pipe.
4. By means of vertical pieces of wood and the nails arrange the plastic pipe such that it can rotate freely on the pieces of wood. This acts as the armature.
5. Arrange the horseshoe magnet on the wooden base such that the armature lies between both of its poles.
6. Now connect the ends of the copper wire to a sensitive galvanometer.
7. Rotate the armature freely.
8. You can observe reflection in the galvanometer.
9. Thus it is an Improvised electric motor.

Question 3.
Collect information of experiments done by Faraday.
Answer:
Experiment – 1

1. Connect the terminals of a coil to a sensitive galvanometer as shown n the figure.
2. Normally, we would not expect any deflection of needle In the galvanometer because there Is no EMF In the N – circuIt.
3. Now, If we push a bar magnet towards the coil, with N Its north pole facing the coil, the needle In the galvanometer deflects, showing that a current has been set up in the coil, the galvanometer does not deflect if the magnet is at rest.
4. If the magnet is moved away from the coil, the needle in the galvanometer again deflects, but in the opposite direction, which means that a current is set up in the coil in the opposite direction,
5. If we use the end of south pole of a magnet Instead of north pole, the results i.e., the deflections in galvanometer are exactly opposite to the previous one.
6. This activity proves that the change in magnetic flux linked with a closed coil, produces current.
7. From this Faradays law of induction can be stated as ‘whenever there is a continuous change of magnetic flux linked with a closed coil, a current is generated in the coil” This induced EMF is equal to the rate of change of magnetic flux passing through it.

Experiment – 2
1. Prepare a coil of copper wire C₁ and connect the two ends of the coil to a galvanometer.
2. Prepare another coil of copper wire C₂ similar to C₁ and connect the two ends of the coil to a battery via switch.
3. Now arrange the two coils C₁ and C₂ nearby as shown in the figure.
4. Now switch on the coil C₂ We observe a deflection In the galvanometer connected to the coil C₁.
5. The steady current in C₂ produces steady magnetic field. If coil C₂ is moved towards the coil C₁, galvanometer shows a deflection.
6. This indicates that electric current Is induced in coil C₁.

7. When C₂ is moved away, the galvanometer shows a deflection again, but this time in the opposite direction.
8. The deflection lasts as long as coil C₂ in motion.
9. When C₂ is fixed and C₁ is moved, the same effects are observed.
10. This shows the induced EMF due to relative motion between two coils.

TS 10th Class Physical Science Electromagnetism Intext Questions

Page 209

Question 1.
How do they (most of the appliances) work?
Answer:
The electronic motor, generators, electric cranes, rice cookers, electrical irons, washing machines, grinders, fans, and many more appliances work on electricity.

Question 2.
How do electromagnets work?
Answer:
By passing D.C. current through a conductor wound around a core, a magnetic field Is developed around the current-carrying conductor and work as a magnet.

Question 3.
Is there any relation between electricity and magnetism?
Answer:
Yes. A current-carrying conductor develops a magnetic field around it.

Question 4.
Can we produce magnetism from electricity?
Answer:
Yes. By passing electric current (D.C) through an insulated wire wound around an iron rod we can produce magnetism by electricity.

Page 210

Question 5.
Does the needle get deflected by the bar magnet?
Answer:
Yes, the needle gets deflected.

Question 6.
why does the needle get deflected by the magnet?
Answer:
When the N-pole of compass needle and N-pole of magnet come close to each other the needle gets deflected due to repulsion. (or) When the S-pole of compass needle and S-pole of magnet come close to each other (S-pole and S-pole are like poles they repel each other) and the needle gets deflected.

Question 7.
What do you notice?
Answer:
The needle is deflected, as there is magnetic field around the wire.

Question 8.
Is there any change in the position of the compass needle?
Answer:
Yes.

Question 9.
Which force Is responsible for the deflection of the compass needle?
Answer:
The magnetic force developed around the current-carrying wire is responsible for the deflection of compass needle.

Question 10.
Does the current-carrying wire apply a force on the needle?
Answer:
Yes. The current-carrying wire applies a repulsive force on the needle.

Question 11.
What do we call this force?
Answer:
This force is called magnetic force of attraction and repulsion.

Page 211

Question 12.
How was this fleid produced?
Answer:
This field was produced due to the magnetic field developed ¡n the wire by the electric current.\

Question 13.
Can we observe the field of a bar magnet?
Answer:
Yes.

Question 14.
Why does this happen?
Answer:
The needle of the magnetic compass is affected by the bar magnet without any physical contact.

Question 15.
Is there any change in the direction of the needle of the magnetic compass? Why?
Answer:
Yes. The change in direction of needle of magnetic compass is due to the magnetic force developed in the conductor.

Question 16.
What is the nature of force that acts on the needle?
Answer:
The force which acting on the needle from a distance s the magnetic field of the bar magnet.

Question 17.
What do you observe?
Answer:
Almost the needle of compass shows single direction pointing north at far places from bar magnet.

Page 212

Question 18.
What does It mean?
Answer:
It means that strength of field vanes with distance from the bar magnet.

Question 19.
How can we find the strength of the field and direction of the field?
Answer:
The direction of field can be determined by using the compass. To determine the strenath we have to perform activity 3.

Question 20.
What are these curves?
Answer:
These curves are technically called as magnetic field lines,

Page 213

Question 21.
Are these field lines closed loops or open loops?
Answer:
The field lines appear to be closed loops. But we can’t conclude that lines are closed or open loops by looking at the figure of field lines because we do not know about alignment of lines that are passing through the bar magnet.

Question 22.
Can we give certain values to magnitude of the field at every point in the magnetic field?
Answer:
Yes. The number of lines passing through the plane of area ‘A’ perpendicular to the field is called magnetic flux and is deoted by Φ (Phi).

Page 214

Question 23.
What is the flux through unit area perpendicular to the field?
Answer:
It Is equal to Φ/A. (A = Area)

Question 24.
Can we generalize the formula of flux for any orientation of the plane taken in the field?
Answer:
The magnetic flux density denoted by ‘B’ is given by, B = magnetic flux / effective area i.e.; B= Φ /A cos θ)
Where ‘θ’ Is the angle between magnetic field (B) and normal to the plane with area ‘A’.

Question 25.
What is the flux through the plane taken parallel to the field?
Answer:
Since θ is zero between parallel lines. Cos0° = 1 and So, Φ= BA.

Question 26.
What Is the use of Introducing the Ideas of magnetic flux and magnetic flux density?
Answer:
The ideas of magnetic flux and magnetic flux density help us to estimate the strength of the magnetic field.

Question 27.
Are there any sources of magnetic field other than magnets?
Answer:
Yes. The current-carrying conductor develops a magnetic field around It.

Question 28.
Do you know how the old electric calling bells work?
Answer:
Yes. They work on the principle of magnetic effect of electric currents.

Page 215

Question 29.
How do the directions of the compass needles change?
Answer:
They are directed along tangents to the circle.

Question 30.
What is the shape of the magnetic field line around wire?
Answer:
The shape of magnetic line around a current-carrying wire is almost a circle,

Question 31.
What is the direction of magnetic field Induction at any point on the field line?
Answer:
If the current flows vertically in upward direction from the page (plane of paper),the field lines are in anti-clockwise direction.

Page 216

Question 32.
Can you tell the direction of the magnetic field of the coil?
Answer:
Yes. The direction In which compass needle comes to rest Indicates the directton of the field due to the coil. So, the direction of field is perpendicular to the plane of the coil.

Question 33.
Why does the compass needle point In the direction of field?
Answer:
We know that south pole is attracted by the north pole. The needle Is oriented in such a way that Its south pole points towards the north pole of the coil.

Page 217

Question 34.
How do they adjust ¡n such an orderly pattern?
Answer:
A solenoid Is a long wire wound in a closed-packed helix. The magnetic field lines set up by solenoid resemble those of a bar magnet indicating that a solenoid behaves like a bar magnet.

Question 35.
What happens when a current-carrying wire is kept in a magnetic field?
Answer:
When a current-carrying wire is kept in a magnetic field it undergoes orientation.

Question 36.
Do you feel any sensation on your skin?
Answer:
Yes. The hair on my skin rises up when I stand near TV screen.

Question 37.
What could be the reason for that?
Answer:
It is due to the magnetic field produced by electric charges in motion.

Question 38.
Take a bar magnet and bring it near the TV screen. What do you observe?
Answer:
I observed that the picture on the TV screen is not clear. It is distorted.

Question 39.
Why does the picture get distorted?
Answer:
It is because the motion of electrons that form the picture is affected by the magnetic field of bar magnet.

Page 218

Question 40.
Is the motion of electrons reaching the screen affected by magnetic field of the bar magnet?
Answer:
Yes.

Question 41.
Can we calculate the force experienced by a charge moving In a magnetic field?
Answer:
Yes. If the force is F, it is given by the expression, F=qvB
Where, q = Amount of charge
y = Velocity of charged particle perpendicular to the magnetic field.
B = Magnetic field

Question 42.
Can we generalize the equation for magnetic force on charge when there Is an angle ‘q’ between the directions of field ‘B’ and velocity ‘V’?
Answer:
Then force F is given by the formula F = qvB Sin θ.

Question 43.
What Is the magnetic force on the charge moving parallel to a magnetic field?
Answer:
When the charge moves parallel to the magnetic field, the value of θ becomes zero. In the equation F=qvB sin θ. since θ=0, sinθ=0 the value of force F also becomes zero.

Question 44.
What is the direction of magnetic force acting on a moving charge?
Answer:
By applying the “right-hand’ rule we can guess the direction of magnetic force acting on a moving charge is direction of the thumb.

Page 219

Question 45.
What is the direction of force acting on a negative charge moving in a field?
Answer:

1. At first find the direction of force acting on a positive charge, by applying right-hand rule for positive charge.
2. Then reverse the direction and this new direction is the direction of force acting on a negative charge.

Question 46.
What happens when a current-carrying wire ¡s placed in a magnetic field?
Answer:
The current-carrying wire experiences magnetic force when it is kept in magnetic field.

Page 220

Question 47.
Can you determine the magnetic force on a current-carrying wire which is placed along a magnetic field?
Answer:
The magnetic force (F) on a current-carrying wire of length L in a uniform magnetic field ‘B’ where the current through the wire is lis :
It θ is the angle between direction of field and velocity of current and magnetic field, then
F = ILB Sin θ
If the current carrying wire is placed along direction of field θ = 0
∴ F = 0

Question 48.
What Is the force on the wire if its length makes an angle ‘θ’ with the magnetic field?
Answer:
If ‘θ’ is the angle between direction of current and magnetic field, then the force acting on the current-carrying wire is given by F = IL B Sin θ.

Page 221

Question 49.
how could you find Its direction?
Answer:
I could find the direction of this force by applying right-hand rule.

Question 50.
What happens to the wire?
Answer:
The wire is set into deflection.

Question 51.
In which way does lt deflect?
Answer:
The wire deflects upwards according to the right-thumb rule.

Question 52.
Is the direction of deflection observed experimentally same as that of the theoretically expected one?
Answer:
Yes. But it depends on polarities of the horseshoe magnet.

Question 53.
Does the right-hand rule give the explanation for the direction of magnetic force exerted by magnetic field on the wire?
Answer:
The right-hand rule does not help us to explain the reason for deflection of wire.

Question 54.
Can you give a reason for it?
Answer:

1. There exists magnetic field due to external source, ie; horseshoe magnet.
2. When there is current in the wire, it also produces field.
3. These fields overlap and give non-uniform field. This is the reason for it.

Page 222

Question 55.
Does this deflection fit with the direction of magnetic force found by right-hand rule?
Answer:
Yes.

Question 56.
What happens when a current-carrying coil Is placed In uniform magnetic field?
Answer:
The coil is set to motion.

Question 57.
Can we use this knowledge to construct an electric motor?
Answer:
Yes. This is the principle of electric motor which converts electrical energy into kinetic energy.

Question 58.
What is the angle made by AB and CD with magnetic field?
Answer:
AB and CD are at right angles to the magnetic field. (always 900)

Question 59.
Can you draw direction of magnetic force on sides AB and CD?
Answer:
Yes, the direction of magnetic force on sides AB and CD can be determined by applying right hand rule.

Page 223

Question 60.
What are the directions of forces on BC and DA?
Answer:
At BC magnetic force pulls the coil up and at DA magnetic force pulls it down.

Question 61.
What is the net force on the rectangular coil?
Answer:
Net force on the rectangular coil is zero.

Question 62.
Why does the coil rotate?
Answer:
The rectangular coil rotates in clockwise direction because the equal and opposite pair of parallel forces acting on the edges of the coil constitute a couple.

Question 63.
What happens to the rotation of the coil if the direction of current in the coil remains unchanged?
Answer:
If the direction of current in the coil is unchanged, it rotates upto a vertical position and then due to law of conservation of inertia, it rotates further in clockwise direction. But now the sides of the coil experience forces which are in the opposite direction to the previous case. So these forces tend to rotate the wire in anti clockwise direction. As a result, the coil comes, to halt and rotates In anti-clock wire direction.

Question 64.
How could you make the coil to rotate continuously?
Answer:
If the direction of current in coil, after the first half rotation is reversed, the coil continues to rotate in the same direction.

Question 65.
How can we achieve this?
Answer:
To achieve this, brushes B1 and B2 are used.

Page 224

Question 66.
What happens when a coil without current is made to rotate In magnetic field?
Answer:
It a coil without current is made to rotate in magnetic field nothing happens.

Question 67.
How is current produced?
Answer:
The current is produced from the battery to the coil.

Question 68.
What do you notice? (ASI.) (iMark)
Answer:
We notice that the metal ring Is levitated on the coil.

Question 69.
Why is there a difference in behaviour in these two cases?
Answer:
An The A.C. supply changes its direction a number of times in a second. But D.C. is unidirectional current. So there is a difference in the behaviour of the metal ring In these two cases.

Question 70.
What force supports the ring against gravity when it is being levitated?
Answer:
The magnetic force developed in the coil of copper wire supports the ring against gravity when It is being levitated.

Page 225

Question 72.
Could the ring be levitated if D.C. is used?
Answer:
The metal ring is levitated only when the net force on it is zero according to Newton’s second law. So it is not possible If D.C is used.

Question 72.
What is this unknown force acting on the metal ring?
Answer:
The change in polarities at certain intervals at the ends of the solenoid causes the unknown force acting on the metal ring.

Question 73.
What is responsible for the current in the metal ring?
Answer:
The field through the metal ring changes so that flux linked with the metal ring changes and this is responsible for the current in metal ring.

Question 74.
If D.C. is used, the metal ring lifts up and falls down immediately. Why?
Answer:
The flux linked with metal ring is zero when the switch is on. At that instant there should be a change in flux linked with ring. So the ring raises up and falls down. If the switch is off, the metal ring again raises up and falls down. There is no change in flux linked with ring when the switch is off.

Page 226

Question 75.
What could you conclude from the above analysis?
Answer:
The relative motion of the magnet and coil sets up a current in the coil.

Page 227

Question 76.
What is its direction?
Answer:
The direction of the induced current ¡s such that it opposes the change that produced It.

Question 77.
Can you apply conservation of energy for electromagnetic induction?
Answer:
Yes. In electromagnetic induction, mechanical energy is converted into electrical energy and there is neither gain or loss of energy. So the conservation of energy Is maintained.

Page 228

Question 78.
Can you guess what could be the direction of induced current in the coil in such case?
Answer:
Invariably the direction of induced current in the coil must be In anti clockwise direction. In simple terms, when flux Increases through coil, the coil opposes the increase In flux and flux decreases through coil, it opposes the decrease in flux.

Question 79.
Could we get Faraday’s law of induction from conservation of energy?
Answer:
Yes. The mechanical energy used to move crosswire to distance s’ in one second Is converted Into electric energy, such that electronic energy = Φ/Δt and leads to conservation of energy.

Page 229

Question 80.
Can you derive an expression for the force applied on cross wire by the field B?
Answer:
Yes. The force applied F = BIL

Page 231

Question 81.
How could we use the principle of electromagnetic Induction In the case of using ATM card when Its magnetic strip is swiped through a scanner?
Answer:
When the magnetic strip of ATM card coated with Iron oxide moves past the small coil of wire, the magnetic field changes which leads to generation of current and it leads to the operation of the card.

Question 82.
What happens when a coil is continuously rotated In a uniform magnetic field?
Answer:
An Induced current is generated in the coil.

Question 83.
Does it help us to generate the electric current?
Answer:
Yes.

Question 84.
Is the direction of current induced in the coil constant? Does It change?
Answer:
Yes It changes. When the coil is at rest In vertical position, with side (A) of coil at top position and side (B) at bottom position no current will be Induced in It.

Page 232

Question 85.
Can you guess the reason for variation of current from zero to maximum and vice-versa during the rotation of coil?
Answer:
The reason for variation of current from zero to maximum and vice-versa during the second part of the rotation is current generated follows the same pattern as that in the first half except that the direction of current is reversed.

Question 86.
Can we make use of this current? If so, how?
Answer:
Two carbon brushes are arranged in sucti a way that they press the slip rings to obtain current from the coil. When these brushes are connected to external devices like TV, radio, we can make them work with current supplied from ends of carbon brushes.

Page 233

Question 87.
How can we get D.C. current using a generator?
Answer:
By connecting two half slip rings, Instead of a slip ring commutator on either side of the ends of the coil, we can get D.C current.

Question 88.
What changes do we need to make in an A.C. generator to be conected into a D.C. generator?
Answer:
Instead of two slip rings, we use a half slip ring commutator to change A.C. generator Into a D.C. generator.

Think And Discuss

Question 1.
How does a tape recorder which we use to listen songs / speeches work?
Answer:

1. The tape recorder which we use to listen songs/speeches and to record the voices works on the principle of electromagnetic induction.
2. it consists of a piece of plastic tape coated with ferromagnetic substance, iron oxide which is magnetised more in some parts than in others due to varying currents produced by our voice.
3. When the tape moves past as a small coil of wire (it is called head of the tape recorder.), the magnetic field produced by the tape changes which leads to generation of current in the small coil of wire.
4. These varying currents are fed into a loud speaker where they are reproduced as sounds.

Question 2.
How could we use the principle of electro-magnetic induction in the case of using ATM card?
Answer:
When the magnetic strip of ATM card coated with iron oxide moves past as a small coil of wire, the magnetic field changes which leads to generation of current and it leads to the operation of the card.

TS 10th Class Physical Science Electromagnetism Activities

Activity 1

Question 1.
With the help of an activity show the current-carrying wire produces magnetic field.
(or)
How do you prove magnetic field develops around a current-carrying conductor?
Answer:

1. Take a thermocol sheet and fix two thin wooden sticks of height 1cm which have small slit at the top
   of their ends.
2. Arrange a copper wire of 24 gauge which passes through these slits and make a circuit in which the elements1 at 3 volt battery, key and copper; wire are connected in series.
3. Now keep a magnetic compass below the wire.
4. Bring a bar magnet close to the compass.
5. Take a bar magnet far away from the circuit and switch on the circuit. Observe the changes In compass.
6. The compass needle deflects.
7. This deflection Is due to the magnetic field produced by current-carrying wire.

Activity 2

Question 2.
Show that the magnetic field around a bar magnet is three dimentional and its strength and direction varies from place to place.
Answer:

1. Take a sheet of white paper and place it on the horizontal table.
2. Place a bar magnet in the middle of the sheet.
3. Place a magnetic compass near the magnet. It settles to a certain direction:
4. Use a pencil and put dots on the sheet on either side of the needle. Remove the compass. Draw a small line segment connecting the two dots Draw an arrow on it from South Pole of the, needle to North Pole of the needle.
5. Repeat the same by placing the compass needles at various positions on the paper. The compass needle settles in different directions at different positions.
6. This shows that the direction of magnetic field due to a bar magnet varies from place to place.
7. Now take the compass needle to places far away from magnet, on the sheet and observe the orientation of the compass needle in each case.
8. The compass needle shows almost the same direction along north and south at places far from the magnet.
9. This shows that the strength of the field varies with distance from the bar magnet.
10. Now hold the compass a little above the table and at the top of the bar magnet.
11. We observe the deflection in compass needle. Hence we can say that the magnetic field is three dimensional i.e., a magnetic field surrounds its source.
12. From the above activities we can generalize that a magnetic field exists in the region surrounding a bar magnet and is characterized by strength and direction.

Activity 3

Question 3.
Explain the process of tracing magnetic field lines.
Answer:

1. Place a white sheet of paper on a horizontal table.
2. Place a compass in the middle of it. Put two dots on either side of the compass needle. Take it out.
   Draw a line connecting the dots which shows the North and South of the earth.
3. Now place a bar magnet on the line drawn in such a way that its north pole points towards the Magnetic field lines geographic north.
4. Now place the compass at the north pole of the bar magnet. Put a dot at the north pole of the compass needle.
5. Now remove the compass and place it at the dot. It will point in other direction. Again put a dot at the north pole of the compass needle.
6. Repeat the process till you reach the south pole of the bar magnet.
7. Connect the dots from ‘N’ of the bar magnet to S’ of the bar magnet, with a’ free hand ctirve. You will get a curved line.
8. Now select another point from the north pole of the bar magnet. Repeat the process for many points taken near the north pole.
9. You will get different curves as shown in the figure.
10. These lines are called magnetic field lines. They are Imaginary lines.

Activity 4

Question 4.
Describe an activity to find magnetic field due to straight wire carrying current. What Is its direction?
Answer:
1. Take a wooden plank and make a hole as shown in figure, Place this plank on the table.
2. Now place a retort stand on the plank as shown In figure.
3. Pass a 24 gauge copper wire through hole of the plank and rubber knob of the retort stand in such a way that the wire be arranged In a vertical position and not touch the stand.
4. Connect the two ends of the wire to a battery via switch.
5. Place 6 to lo compass needles in a circular path around the hole so that its centre coincides with the hole.
6. Use a 9V battery in the c!rcuet. Swtch on current and it flows through the wire.
7. We notice that they are directed as tangents to the circle.
8. The magnetic field around the wire is circular ¡n shape. This can be verified by sprinkling iron filings around the wire when current flows in the wire.
9. The direction of magnetic field around the wire will be as shown in the following figure.

10. The direction of the magnetic field around the current-carrying wire can be determined by right-hand thumb rule ie., grab the current-carrying wire with your right hand in such a way that thumb is in the direction of current, then the curled fingers show the direction of magnetic field as shown in the figure.

Activity 5

Question 5.
Trace the magnetic field due to circular coil.
Answer:
1. Take a thin wooden plank covered with white paper and make two holes on its surface as shown in the figure.
2. Pass insulated copper wire (24 guage) through the holes and wind the wire 4 to 5 times through holes
such that it looks like a coil.
3. The ends of the wire are connected to terminals of the battery through a switch. Now switch on the
circuit.
4. Place a compass needle on the plank at the centre of the coil. Put dots on either side of the compass needle.
5. Again place compass at one of the dots put other dots further.
6. Do the same till you reach the edge of the plank.
7. Now repeat this for the other side of the coil from the centre. Then draw a line joining the dots. We will get a field line of the circular coil.
8. Do the same for the other points taken In the between the holes. Draw corresponding lines. We will get field lines of the circular coil.
9. The direction of the field due to coil is s determined by using right-hand rule, which states that, when you curl your right-hand fingers in the direction of current, thumb waves the direction of magnetic field.

Activity 6

Question 6.
Find the magnetic field due to a solenoid.
(or)
Write the experimental procedure and observations of the experiment that is to be performed to observe the magnetic field formed due to solenoid.
Answer:

1. Take a wooden plank covered with a white paper.
2. Make equidistant holes on its surface as shown in the figure.
3. Pass copper wire through the holes. This forms a coil.
4. Join the ends of the coil to a battery through a switch.
5. Switch on the circuit. Current passes through the coil.
6. Now sprinkle Iron filings on the surface of the plank around the coil. Give-a small jerk to it, An orderly pattern Of iron filings is seen on the paper.
7. The long coil is called solenoid. The field of solenoid is shown in the figure.
8. This activity proves that current-carrying solenoid forms magnetic field.

Activity 7

Question 7.
Explain magnetic force on moving charge and current carrying wire.
Answer:
(a) TV Screen Activity:

1. Take a bar magnet and bring it near the TV screen.
2. Then the picture on the screen is distorted.
3. Here the distortion is due to the motion of the electrons reacting the screen are affected by the magnetic field.
4. Now move the bar magnet away from the screen.
5. Then the picture on the screen stabilizes.
6. This must be due to the fact that the magnetic field exerts a torce on moving charges. This force Is called magnetic force.
7. The magnitude of the force is F = Bqv where B Is magnetic induction, q’ is the charge and y Is the velocity of the charged partide.

(b) Procedure:

1. Take a bar magnet near to the TV screen. We observe that electrons are affected by the field produced by the bar magnet. Then the picture Is disturbed.
2. Move the bar magnet away from screen. Then you will get a clear picture.
3. The force on the moving charge is given by F = Baqv.
4. Right hand rule is used when velocity and field are perpendicular to each other

Fore-finger —- Direction of current (V).
Middle finger —- Direcrtion of field
Thumb —- direction of force
This rule is applicable to positive charge.

Activity 8

Question 8.
Explain the result of magnetic force applied on a current-carrying wire by an experiment.
(or)
Why the current carrying straight wire which is kept In a uniform magnetic field, perpendicularly to the direction of the field bends aside P Explain this process with a diagram showing the direction of forces acting on the wire?
Answer:
1. Take a wooden plank. Fix two long wooden sticks on It. These wooden sticks are split at their top ends.
2. A copper wire is passed through these splits and the ends of the wire are connected to a battery
of 3V, through a switch.
3. Close the switch to make the circuit. Current passes through the wire.
4. Now bring a horseshoe magnet near the copper wire as shown in the figure. Observe the deflection of the wire.
5. change polarities of the horseshoe magnet. Again observe the deflection. Repeat this by changing the direction of current in the circuit.
6. When current passes through wire, it produces a magnetic field and this field overlaps with the field by horseshoe magnet and gives a non-uniform field.
7. The field in between north arid south poles of horseshoe magnet is shown in the figure.

8. Let us make a wire passing perpendicular to the paper. Let the current pass through It. It produces magnetic field as shown in the figure.

9. The resultant field will be as shown below.
10. We can see that the direction of the field lines due to wire in upper part (of circular lines) coincides
with the direction of field lines of horseshoe magnet.

11. The direction of field lines by wire in lower part (or circular lines) Is opposite to the direction of the
field lines of horseshoe magnet. The net field in upper part is strong and in lower part it is weak. Hence a non-uniform field is created around the wire.
12. Therefore the wire tries to move to the weaker field region.

Activity 9

Question 9.
Take wooden base. Fix a soft iron cylinder on the wooden base vertically. Wind copper wire around the soft Iron as shown in the figure. Now take a metal ring which is slightly greater in radius than the radius of the soft iron cylinder and insert it through the cylinder on the wooden base. Now explain the behaviour of metal ring when the ends of copper wire are connected to (i) AC source and (ii) DC source.
Answer:
When the ends of copper wire are connected to AC source:

• Connect the two ends of copper wire to AC source and switch on the current.
• The metal ring Levitates due to net force acting on it is zero according to Newton’s second law.

Reason:

1. AC changes both its direction and magnitude in regular intervals.
2. Due to the magnetic field produced by current in the coil, one end of the coil behaves like North pole and the other end behaves like south pole for certain time interval.
3. For the next interval, the coil changes its polarties.
4. Assume that the current flows in clockwise direction in the solenoid as viewed from the top. Then the upper end becomes a south pole.
5. An upward force Is applied on the ring only when the upper side of the ring becomes north pole.
6. It Is only possible when there exists anti clock wise current viewed from the top in the ring.
7. After certain intervals, solenoid changes its polarities, so that the ring should also change its polarities in the same intervals.
8. This is the reason why the metal ring is levitated.

When the ends of copper wire are connected to a DC source:

1. Now connect the ends of copper wire to a DC source.
2. When the current is allowed to flow through the solenoid, It behaves like bar magnet. So the flux is linked to the metal ring when the switch is on.
3. At that instant there is a change in flux linked with the ring. Hence the ring rises up.
4. Thereafter, there Is no change In flux linked with coil, hence it falls down.
5. If the switch is off, the metal ring again lifts up and falls down. In this case also there is change
   in flux linked with ring when the switch is off.

Solving these TS 10th Class Maths Bits with Answers Chapter 10 Mensuration Bits for 10th Class will help students to build their problem-solving skills.

Mensuration Bits for 10th Class

Question 1.
To find out quantity of water in the bottle, we measure
A) surface area
B) total surface area
C) volume
D) base area
Answer:
C) volume

Question 2.
Lateral surface area of a cube is given by
A) 2a²
B) 4a²
C) 6a²
D) a³
Answer:
B) 4a²

Question 3.
Total surface area of a regular circular cylinder is
A) 2πrh
B) πrl
C) 2πr(π + r)
D) 2πr(r + h)
Answer:
D) 2πr(r + h)

Question 4.
The ratio of volumes of a cone and a cylinder whose radii and height are equal is ………………….
A) 3 : 1
B) 1 : 3
C) 1 : 2
D) 1 : 1
Answer:
B) 1 : 3

Question 5.
The diagonal of a cube whose side is ‘a’ units is ……………..
A) a
B) \(\sqrt{2}\) a
C) \(\sqrt{3}\) a
D) 2a
Answer:
C) \(\sqrt{3}\) a

Question 6.
The volume of a sphere of radius ‘r’ is obtained by multiplying its surface area by
A) 4/3
B) r/3
C) 4r/3
D) 3r
Answer:
B) r/3

Question 7.
The total surface area of a solid hemisphere of radius 7cm is
A) 239 π cm²
B) 449 π cm²
C) 221 π cm²
D) 129 π cm²
Answer:
A) 239 π cm²

Question 8.
The curved surface area of a right circular cone of height 15cm and base diameter 16cm is.
A) 144 π cm²
B) 136 π cm²
C) 105 π cm²
D) 120 π cm²
Answer:
B) 136 π cm²

Question 9.
The surface areas of two spheres are in the ratio 1 : 4 then, ratio of their volumes is
A) 1 : 4
B) 2 : 8
C) 1 : 16
D) 1 : 64
Answer:
A) 1 : 4

Question 10.
The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is
A) 19.4 cm³
B) 74.6 cm³
C) 9.7 cm³
D) 8.4 cm³
Answer:
A) 19.4 cm³

Question 11.
The ratio of volume of cone and cylinder of equal diameter and height
A) 3 : 1
B) 1 : 2
C) 2 : 1
D) 1 : 3
Answer:
D) 1 : 3

Question 12.
An iron cylindrical rod has a height 4 times of its radius is melted and cast into spherical balls of the same radius. The number of balls cast is
A) 4
B) 3
C) 2
D) 1
Answer:
D) 1

Question 13.
A cone and a hemi-sphere have equal bases and equal volumes then the ratio of their heights
A) 2 : 1
B) 3 : 1
C) 4 : 1
D) 1 : 1
Answer:
A) 2 : 1

Question 14.
The volume of the greatest cylinder that can be cut form a solid wooden cube of length of edge 14cm is
A) 2156 cm³
B) 1078 cm³
C) 539 cm³
D) 428 cm³
Answer:
A) 2156 cm³

Question 15.
A shuttle cock is a combination of
A) cylinder, sphere
B) sphere, cone
C) cylinder, hemisphere
D) hemisphere, first term cone
Answer:
D) hemisphere, first term cone

Question 16.
T.S.A of a solid hemisphere whose radius is 7cm is …………….. cm².
A) 327 π
B) 144 π
C) 147 π
D) 189 π
Answer:
C) 147 π

Question 17.
If the radius of base of a cylinder is doubled and the height remains unchanged, it’s C.S.A becomes
A) double
B) times
C) half
D) no change
Answer:
A) double

Question 18.
The number of cubes of side 2cm which can be cut from a cube of side 6 cm is
A) 3
B) 18
C) 27
D) 9
Answer:
C) 27

Question 19.
If the diameter of a sphere is’d’ then its volume is
A) \(\frac{1}{6}\) πd³
B) \(\frac{4}{3}\) πd³
C) \(\frac{1}{24}\) πd³
D) \(\frac{1}{3}\) πd³
Answer:
A) \(\frac{1}{6}\) πd³

Question 20.
A cylindrical, a cone and a hemisphere are of equal base and have the same height, then the ratio of their volumes is
A) 3 : 1 : 1
B) 3 : 2 : 1
C) 1 : 2 : 3
D) 1 : 3 : 2
Answer:
A) 3 : 1 : 1

Question 21.
Total surface area of a cube is 216 cm² then its volume is ……………….. cm³.
A) 216
B) 196
C) 212
D) 144
Answer:
A) 216

Question 22.
The total surface area of a cube is 54 cm² then its side is ………………. cm. (A.P. Mar. ’15 )
A) 6
B) 9
C) 12
D) 3
Answer:
D) 3

Question 23.
Base area of a regular cylinder is 154 cm² then its radius is ……………….. (A.P. Mar. ’16, ’15)
A) 49 cm
B) 7 cm
C) 22 cm
D) 14 cm
Answer:
B) 7 cm

Question 24.
If the height and radius of a cone are 1.5 and 8 cm then its slant height = ………………. cm
A) 2.5 cm
B) 7.5 cm
C) 5 cm
D) 10 cm
Answer:
C) 5 cm

Question 25.
Curved surface area of a hemi-sphere = ………………. (A.P. Mar. ’15)
A) πr²
B) \(\frac{1}{3}\)πr²
C) 3πr²
D) 2πr²
Answer:
D) 2πr²

Question 26.
Volume of a cube having 1 cm side is ……………….. (A.P. Mar. ’16, June ’15)
A) 1 cm³
B) 3 cm³
C) 1 cm²
D) 3 cm²
Answer:
A) 1 cm³

Question 27.
Ratio of volumes of two spheres is 8 : 27 then ratio of their curved surface areas is …………….. (A.P. June ’15)
A) 2 : 3
B) 4 : 27
C) 8 : 9
D) 4 : 9
Answer:
C) 8 : 9

Question 28.
Football is in a model of …………………. (A.P. Mar. ’16)
A) circle
B) cylinder
C) Sphere
D) cone
Answer:
C) Sphere

Question 29.
Radius of a cone is ‘r’, height is ‘h’ and its slant height is 7 then which of the following is false ? (A.P. Mar. ’16)
A) always l > h
B) always l > r
C) always r > p
D) l² = r² + h²
Answer:
C) always r > p

Question 30.
Radius, height, slant height of a cone are| r, h, l, then ‘l’ value in terms of r and h is ……………… (T.S. Mar. ’15)
A) \(\sqrt{h^2-r^2}\)
B) \(\sqrt{r^2+h^2}\)
C) \(\sqrt{r^2-h^2}\)
D) \(\sqrt{4 r^2+h^2}\)
Answer:
B) \(\sqrt{r^2+h^2}\)

Question 31.
To calculate the quantity of milk inside a bottle, we need to find out
A) Area
B) Volume
C) Density
D) TSA
Answer:
B) Volume

Question 32.
Sphere, cylinder and cone have same heights and radii, then its ratios of curved surface areas.
A) 4 : 4 : \(\sqrt{5}\)
B) 1 : 1 : \(\sqrt{5}\)
C) \(\sqrt{5}\) : 4 : 4
D) 4 : \(\sqrt{5}\) : 4
Answer:
A) 4 : 4 : \(\sqrt{5}\)

Question 33.
Diagonal of a cuboid is …………….. units.
A) \(\sqrt{l^2+b^2+h^2}\)
B) \(1 \sqrt{\mathrm{b}^2+\mathrm{h}^2}\)
C) \(b \sqrt{h^2+r^2}\)
D) none
Answer:
A) \(\sqrt{l^2+b^2+h^2}\)

Question 34.
The radius of a conical tent is 3 meter and height is 4 meter then its slant height is …………………. meter.
A) 5
B) 725
C) A and B
D) none
Answer:
A) 5

Question 35.
The total surface area of a solid hemisphere of radius 1 unit is
A) 3πr²
B) 2πr²
C) 3π
D) 2π
Answer:
C) 3π

Question 36.
Volume of is cuboid.
A) 16
B) 10
C) 6
D) 12
Answer:
C) 6

Question 37.
The diameter of a metallic sphere is 6 cm and melted to draw a wire of diameter 2cm, then the length of the wire is
A) 48 cm
B) 12 cm
C) 36 cm
D) 24 cm
Answer:
C) 36 cm

Question 38.
A solid sphere of radius r melted and recast into the shape of a solid cone of height r, then radius of the base of the cone is (of equal volume)
A) 2r
B) r
C) 3r
D) 4r
Answer:
A) 2r

Question 39.
If the radii of circular ends of a frustum of a cone are 20 cm and 12 cm and its height is 6cm, then the slant height of the frustum is …………………. cm.
A) 10
B) 6
C) 9
D) 8
Answer:
A) 10

Question 40.
The number of balls, each of radius 1 cm that can be made from a solid sphere of radius 8 cm is
A) 64
B) 216
C) 16
D) 512
Answer:
D) 512

Question 41.
The ratio of volume of two cones is 4 : 5 and the ratio of the radii of their base is 2 : 3 then ratio of their vertical heights is
A) 4 : 5
B) 9 : 5
C) 3 : 5
D) 2 : 5
Answer:
B) 9 : 5

Question 42.
If the ratio of radii of two spheres is 2 : 3 then the ratio of their surface areas is
A) 3 : 2
B) 27 : 8
C) 8 : 27
D) 4 : 9
Answer:
D) 4 : 9

Question 43.
If a cone is cut into two parts by a horizontal plane passing through the mid point of the axis, the ratio of the volumes of the
upper part and the cone is
A) 1 : 2
B) 1 : 4
C) 1 : 6
D) 1 : 8
Answer:
D) 1 : 8

Question 44.
The height of a cylinder is doubled and radius is tripled then its curved surface area will become …………….. times.
A) 7
B) 6
C) 9
D) 12
Answer:
B) 6

Question 45.
Diameter of a sphere which can in-scribe a cube of edge x cm is ……………
A) \(\frac{x}{3}\)
B) \(\frac{x^2}{3}\)
C) \(\frac{x}{\sqrt{3}}\)
D) x\(\sqrt{3}\)
Answer:
D) x\(\sqrt{3}\)

Question 46.
Total surface area of hemisphere of radius r is ………….
A) πr²
B) 2πr²
C) 3πr²
D) none
Answer:
C) 3πr²

Question 47.
Volume of frustrum of a cone is
A) \(\frac{\pi h}{3}\) (R² + r²2 +R.r)
B) \(\frac{\mathrm{h}}{3}\)(R² + r²)
C) \(\frac{\pi h}{3}\)(R² +r²)
D) none
Answer:
A) \(\frac{\pi h}{3}\) (R² + r²2 +R.r)

Question 48.
If the length of each diagonal of a cube is doubled, then its volume become ………………. times.
A) 7
B) 8
C) 9
D) none
Answer:
B) 8

Question 49.
If a right angled triangle is revolved about its hypotenuse then it will form a ………………….
A) double cone
B) triple cone
C) only cone
D) none
Answer:
A) double cone

Question 50.
A solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then radius of small spherical balls is ……………… cm.
A) 10
B) 9
C) 6
D) 5
Answer:
D) 5

Question 51.
In a hollow cuboid box of size 4 × 3 × 2m, the number of solid iron spherical balls of radius 0.5 m that can be packed ……………
A) 71
B) 24
C) 22
D) 16
Answer:
B) 24

Question 52.
If the external and internal radii of a hollow hemispherical bowl are R and r, then its total surface area is ……………….
A) πr² + R²
B) πR² + r²
C) πR² + r
D) π(3R² + r²)
Answer:
D) π(3R² + r²)

Question 53.
Volume of cylinder is …………………. cu. units.
A) πr²h
B) πr²
C) \(\frac{\pi}{r}\)
D) none
Answer:
A) πr²h

Question 54.
Volume of cone is ……………… cu. units.
A) \(\frac{1}{7}\) πr²h
B) \(\frac{1}{2}\) πr²h
C) πr²h
D) \(\frac{1}{3}\) πr²h
Answer:
D) \(\frac{1}{3}\) πr²h

Question 55.
Volume of sphere is ………………… cu.units.
A) \(\frac{4}{3}\) πr²h
B) \(\frac{4}{3}\) πr³
C) \(\frac{1}{3}\) πr³
D) none
Answer:
B) \(\frac{4}{3}\) πr³

Question 56.
Volume of cuboid = …………………… cu.units
A) l²b
B) lbh²
C) lbh
D) none
Answer:
C) lbh

Question 57.
Total surface area of cone is …………………. sq.units.
A) πr² + πrl
B) πr² + πr
C) πr² + πl
D) none
Answer:
A) πr² + πrl

Question 58.
Total surface area of cylinder is ……………………. sq.units.
A) πrh + πr²
B) 2πr + π
C) 2πrh²
D) 2πrh + 2πr²
Answer:
D) 2πrh + 2πr²