TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Ex 10.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration Exercise 10.2

Question 1.
A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. [use π = 3.14] (AS4)
Solution:
Diameter of the base of the cone (d)= 6 cm
∴ Radius of the base (r) = d/2 = 3 cm
Radius of the hemisphere = 3 cm
Let the slant height of the cone be ‘l’
l2 = h2 + r2
= 42 + 32
= 16 + 9 = 25
∴ l = \(\sqrt{25}=\) = 5 cm.
TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 1
Surface area of the toy = curved surface area of the cone + curved surface area of the hemisphere
= πrl + 2πr2
= \(\frac{22}{7}\) × 3 × 5 + 2 × \(\frac{22}{7}\) × 3 × 3
= \(\frac{22}{7}\) [3 × 5 + 2 × 3 × 3]
= \(\frac{22}{7}\) [15 + 18] = \(\frac{22}{7}\) × 33 = \(\frac{726}{7}\)
= 103.71 cm2

TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 2.
A Solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm. and heights of the cylindrical and conical portions are 10 cm and 6cm respectively. Find the total surface area of the solid. [use π = 3.14](AS4)
Solution:
Total surface area = C.S.A of the cone + C.S.A of cylinder + C.S.A of the hemisphere
Cone : Radius (r) = 8 cm
Height (h) = 6 cm
Slant height
l = \(\sqrt{\mathrm{r}^2+\mathrm{h}^2}\)
= \(\sqrt{8^2+6^2}\)
= \(\sqrt{64+36}\)
= \(\sqrt{100}\)
= 10 cm
TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 2
C.S.A = πrl
= \(\frac{22}{7}\) × 8 × 10 = \(\frac{1760}{7}\) cm2
Cylinder : Radius (r) = 8 cm
Height(h) = 10 cm
C.S.A = 2πrh = 2 × \(\frac{22}{7}\) × 8 × 10
= \(\frac{3520}{7}\) cm
Hemisphere : Radius (r) = 8 cm
C.S.A = 2πr2 = 2 × \(\frac{22}{7}\) × 8 × 8
= \(\frac{2816}{7}\) cm2
∴ Total surface area of the given solid
= \(\frac{1760}{7}\) + \(\frac{3520}{7}\) + \(\frac{2816}{7}\)
T.S.A = \(\frac{8096}{7}\) = 1156.57 cm2

TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 3.
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5mm. Find its surface area. (AS4)
TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 3
Solution:
Surface area of the capsule = C.S.A of 2 hemispheres + C.S.A of the cylinder
Hemisphere : Radius (r) = \(\frac{\mathrm{d}}{2}\) = \(\frac{5}{2}\)
= 2.5 mm
C.S.A of two hemispheres = 2 × 2πr2
= 4 × \(\frac{22}{7}\) × 2.5 × 2.5 7
= \(\frac{550}{7}\) mm2 = 78.57 mm2
Cylinder : Radius (r) = \(\frac{\mathrm{d}}{2}\) = \(\frac{5}{2}\) = 2.5 mm
Height (h) = 14 mm
C.S.A = 2πrh
= 2 × \(\frac{22}{7}\) × 2.5 × 14 = 220 mm2
∴ Surface area of the capsule = 78.57 + 220 = 298.57mm2

Question 4.
Two cubes each of volume 64 cm3 are joined end to end together. Find the surface area of the resulting cuboid. (AS1) (Mar ’15 (A.P.))
Solution:
Given, volume of the cube, V = a3 = 64 cm3
∴ a3 = 4 × 4 × 4 = 43
Hence a = 4 cm
When two cubes are added, the length of cuboid = 2a
= 2 × 4 = 8 cm
breadth = a = 4 cm
height = a = 4 cm is formed
TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 4
∴ T.S.A. of the cuboid
= 2 (lb + bh + lh)
= 2(8 × 4 + 4 × 4 + 8 × 4)
= 2(32 + 16 + 32)
= 2(80) = 160 cm2.
∴ The surface area of resulting cuboid is 160 cm2.

TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 5.
A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 in and its length be 8 m. Find the cost of pointing it on the outside at rate of ₹ 20 per m2. (AS4)
Solution:
Total surface area of the tank = 2 × C.S.A of hemisphere + C.S.A of cylinder
TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 5
Hemisphere :
Radius (r) = \(\frac{\text { diameter }}{2}\)
= \(\frac{1.4}{2}\)
= 0.7m
C.S.A of hemisphere = 2πr2
= 2 × \(\frac{22}{7}\) × 0.7 × 0.7 7
= 3.08 m2
2 × C.S.A = 2 × 3.08 m2 = 6.16 m2
Cylinder : Radius (r) = \(\frac{\text { diameter }}{2}\)
= \(\frac{1.4}{2}\)
= 0.7 m
Height (h) = 8 m
C.S.A of the cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 0.7 × 8 = 35.2 m2
∴ Total surface area of the storage tank
= 35.2 + 6.16 = 41.36 m2
Cost of painting is surface area ₹ 20 per sq.m = 41.36 × 20 = ₹ 827.2

TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 6.
A sphere, a cylinder and a cone have the same radius and same height. Find the ratio of their volumes. (AS4)
[Hint: Diameter of the sphere is equal to heights of the cylinder and the cone].
Solution:
It is given that the three solids namely a sphere, a cylinder and a cone have the same radius. It is persumed that the three solids are of equal heights.
TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 6
Sphere :
Radius of sphere = r
∴ volume of the sphere = \(\frac{4}{3}\) πr3.

Cylinder :
Radius of cylinder = r
Height of cylinder h = 2r (diameter of sphere)
∴ volume of cylinder = πr2h
= πr2 × 2r
= 2πr × 2r
= 2πr3

Cone :
Radius of cone = r
Height of cone h = 2r
∴ Volume of cone = \(\frac{1}{3}\) πr2h
= \(\frac{1}{3}\) πr2 × 2r
= \(\frac{2}{3}\) πr3
∴ The ratios of the volumes
= sphere : cylinder : cone
= \(\frac{4}{3}\) πr3 : 2πr3 : \(\frac{2}{3}\) πr3
= 4 πr3 : 6πr3 : 2πr3
= 4 : 6 : 2
= 2 : 3 : 1.

TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 7.
A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube. Determine the surface area of the remaining solid. (AS4)
Solution:
T.S.A. of the remaining solid = area of each surface of cube + Area of hemisphere – Area of cutting part of hemisphere.
TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 7
Square surface : Side = a units.
Area a = a × a = a2 sq. units.
6 × square surface area = 6a2 sq. units.
Hemisphere : Diameter = a, Radius = \(\frac{\mathrm{a}}{2}\)
Area of hemisphere = 2πr2
= 2 × π × \(\left(\frac{\mathrm{a}}{2}\right)^2\)
= 2 × π × \(\frac{\mathrm{a}}{4}\) = \(\frac{\pi \mathrm{a}^2}{2}\)
Area of cutting part of hemisphere
= 2πr2 = πr2
= \(\frac{\pi \mathrm{a}^2}{2}\) – π\(\left(\frac{a}{2}\right)^2\)
= \(\frac{\pi \mathrm{a}^2}{2}\) – \(\frac{\pi \mathrm{a}^2}{4}\) = \(\frac{\pi \mathrm{a}^2}{4}\)

TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2

Question 8.
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its radius is of 3.5 cm, find the total surface are of the article. (AS4)
Solution:
Surface area of the given solid.
= C.S.A. of the cylinder + 2 × C.S.A. of hemisphre.
TS 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.2 8
= 2πrh + 2 × 2πr2
= 2πrh + 4πr2
= 2 × \(\frac{22}{7}\) × 3.5 × 10 + 4 × \(\frac{22}{7}\) × 3.5 × 3.5
= 220 + 2(77)
= 220 + 154 = 374 cm2

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