Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration InText Questions to get the best methods of solving problems.
TS 10th Class Maths Solutions Chapter 10 Mensuration InText Questions
Try This
Question 1.
Consider the following situations. In each find out whether you need volume or area and why ? (Page No.245)
- Quantity of water inside a bottle.
- Canvas needed for making a tent.
- Number of bags inside the lorry.
- Gas filled in a cylinder.
- Number of match sticks that can be put in match box.
Solution:
- Volume : 3-d shape
- Area : L.S.A. / T.S.A
- Volume : 3-d shape
- Volume : 3-d shape
- Volume : 3-d shape
Question 2.
State 5 more such examples and ask your friends to choose what they need ? (AS3) (Page No. 245)
Solution:
- To paint a pillar in the shape of a cylinder.
- To white wash the walls of a house.
- To find the quantity of rich in a heap of rice.
- An object is the form of a cone having a herispherical shape on its top. Quantity of ice-cream to be filled in it.
- Later flowing through a cylindricalipipe.
Question 3.
Break the pictures in the previous figure into solids of known shapes. (AS5) (Page No. 246)
Solution:
Question 4.
Think of 5 more objects around you that can be seen as a combination of shapes. Name the shapes that combined to make them. (AS3) (Page No. 246)
Solution:
Student’s Activity.
Try This
Question 1.
Use known solid shapes and make as many objects (by combining more than two) as possible that you come across in
your daily life.
(Hint : Use clay, or balls, pip€s. paper cones, boxes like cube, cuboid etc.) (AS4, AS5) (Page No. 252)
Solution:
Student’s Activity.
Think – Discuss
Question 1.
A sphere is inscribed in a cylinder. Is the surface of the sphere equal to the curved surface of the cylinder ? If yes, explain how. (AS2, AS3) (Page No. 252)
Solution:
Yes, the surface area of the sphere is equal to the curved surface area of the cylinder.
Let the radius of this cylinder be ‘r’ and its height ‘h’
Then its curved surface area = 2πrh
= 2πr (r + r)
∴ height = diameter of the sphere
= diameter of the cylinder
= 2r
= 2πr(2r)
= 4πr2
And surface area of the sphere = 4πr2
∴ C.S.A of cylinder = Surface area of sphere.
Try This
Question 1.
If the diameter of the cross-section of a wire is decreased by 5% by what percentage should the length be increased so that the volume remains the same ? (AS4) (Page No. 257)
Solution:
Radius of wire r = r and length = h1
diameter of cross – section of wire d1 = 2r
decreased diameter 5%
= 2r × \(\frac{5}{100}\) = \(\frac{\mathrm{r}}{10}\)
∴ decreased radius r2 = 2r – \(\frac{\mathrm{r}}{20}\)
= \(\frac{19\mathrm{r}}{10}\) × \(\frac{1}{2}\) = \(\frac{19 \mathrm{r}}{20}\)
volume of the wire V1 = πr12h1, length = h2 (after increased) volume of the wire V1 (after increased) = πr22h2 volumes are equal. So v1 = v2
πr12h1 = πr22h2
πr2h1 = π \(\left(\frac{19 r}{20}\right)^2\) × h2
h1 = \(\frac{361}{400}\) × h2
h1 = \(\frac{361}{400}\) h2
h2 = \(\frac{400 \mathrm{~h}_1}{361}\)
increased length = h2 – h1
= \(\frac{400 \mathrm{~h}_1}{361}\) – h1
= \(\frac{400 \mathrm{~h}_1-361 \mathrm{~h}_1}{361}\) = \(\frac{39 \mathrm{~h}_1}{361}\)
increased percentage
= 100 × \(\frac{39 \mathrm{~h}_1}{361}\) × \(\frac{\mathrm{h}_1}{\mathrm{~h}_1}\)
= \(\frac{3900}{361}\) = 10.8%
Question 2.
Surface areas of a sphere and cube are equal then find the ratio of their volumes. (AS4)(Page No. 257)
Solution:
radius of sphere = r, and side of cube = a
surface area of sphere = 4πr2
surface area of cube = 6a2
surface area of sphere = surface area of cube (given)
4πr2 = 6a2 ⇒ 4 × \(\frac{22}{7}\) × π2 = 6 × a2
Think – Discuss
Question 1.
Which barrel shown in the below figure can hold more water ? Discuss with your friends. (AS2, AS3) (Page No. 262)
Solution:
r1 = \(\frac{1}{2}\) = 0.5 cm; h1 = 4 cm
Volume of the 1st barrel = πr2h
= \(\frac{22}{7}\) × 0.5 × 0.5 × 4 = 3.142 cm3
r2 = \(\frac{4}{2}\) = 2 cm ; h = 1 cm
Volume of the 2nd barrel
V = πr2h = \(\frac{22}{7}\) × 2 × 2 × 1
= 12.57 cm3
Hence, the volume of the 2nd barrel is more than the first barrel.
Do This
Question 1.
A copper rod of diameter 1cm and length 8 cm is drawn into a wire of length 18m of uniform thickness. Find the thickness of wire. (AS4) (Page No. 263)
Solution:
Volume of the copper rod(Cylinder) = πr2h
= \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × 8
= \(\frac{44}{7}\) cm2
If ‘r’ is the radius of the wire, then its volume = πr2h
∴ The volume of rod is equal to the volume of the wire. We have
⇒ \(\frac{22}{7}\) × r2 × 18 m = \(\frac{44}{7}\) cm3
⇒ r2 = \(\frac{44}{7}\) × \(\frac{7}{22}\) × \(\frac{1}{1800}\)
[∴ 18m = 18 × 100 cm]
⇒ r2 = \(\frac{1}{900}\)
⇒ r = \(\frac{1}{30}\) cm = \(0.0 \overline{3}\) cm
∴ Thickness = d = 2 × 0.03 = 0.06 cm
Question 2.
Pravali house has a water tank in the shape of a cylinder on the roof. This is filled by pumping water from a sumpfan underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44m × 9.5 cm. The water tank has radius 60 cm. and height 95 cm. Find the height of the water left in the sump after the water tank has been completely filled with water from the sump which had been full of water. Compare the ca¬pacity of the tank with that of the sump. (AS4) (Page No. 263)
Solution:
Volume of the water in the sump = [v = lbh]
= 1.57 × 1.44 × 0.95
(∵ 9.5 cm = \(\frac{9.5}{100}\) m = 0.95 m).
= 2.14776 m3 = 2147160 cm3
Volume of the tank on the roof = πr2h
= 3.14 × 60 × 60 × 95 = 1073880 cm3
∴ Volume of the water left in the sump after filling the tank
= 2147760 – 1073880
= 1073880 cm3
Let the height of the water in the tank be h.
∴ 157 × 144 × h = 1073880
h = \(\frac{1073880}{157 \times 144}\) = 47.5 cm
∴ Ratio of the volume of the sump and tank
= 21447760 : 1073880 = 2 : 1
∴ Sump can hold two times the water that can be hold in the tank.