NCERT Solutions Class 10th Maths Chapter – 12 Surface Areas and Volumes Exercise – 12.2

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Solution: Here r = 1 cm and h = 1 cm.
The diagram is as follows.

Now, Volume of solid = Volume of conical part + Volume of hemispherical part
We know the volume of cone = ⅓ πr²hAnd,
The volume of hemisphere = ⅔πr³
So, volume of solid will be

= π cm³

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution:

Given, Height of cylinder = 12–4 = 8 cm
Radius = 1.5 cm
Height of cone = 2 cm
Now, the total volume of the air contained will be = Volume of cylinder+2×(Volume of cone)
∴ Total volume = πr²h+[2×(⅓ πr²h )]
= 18 π+2(1.5 π)
= 66 cm³.

5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution: For the cone,
Radius = 5 cm,
Height = 8 cm
Also, Radius of sphere = 0.5 cm
The diagram will be like

It is known that,
Volume of cone = volume of water in the cone
= ⅓πr²h = (200/3)π cm³
Now, Total volume of water overflown= (¼)×(200/3) π =(50/3)π
Volume of lead shot
= (4/3)πr³
= (1/6) π
Now, The number of lead shots = Total Volume of Water over flown/ Volume of Lead shot
= (50/3)π/(⅙)π
= (50/3)×6 = 100

6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass.

Solution:

7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution:

Here, the volume of water left will be = Volume of cylinder – Volume of solid
Given, Radius of cone = 60 cm,
Height of cone = 120 cm
Radius of cylinder = 60 cm
Height of cylinder = 180 cm
Radius of hemisphere = 60 cm
Now, Total volume of solid = Volume of Cone + Volume of hemisphere
Volume of cone= 1/3πr²h = 1/3 × π×60²×120cm³ = 144×10³π cm³
Volume of hemisphere  = (⅔)×π×60³ cm³ = 144×10³π cm³
So, total volume of solid =  144×10³π cm³ + 144×10³π cm³ = 288 ×10³π cm³
Volume of cylinder = π×60²×180 = 648000 = 648×10³ π cm³
Now, volume of water left will be = Volume of cylinder – Volume of solid
= (648-288) × 10³×π = 1.131 m³

8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution: Given, For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm
For the spherical part, Radius (r) = (8.5/2) = 4.25 cm

Now, volume of this vessel = Volume of cylinder + Volume of sphere
= π×(1)²×8+(4/3)π(4.25)³
= 346.51 cm³