NCERT Solutions for Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2#
Exercise 12.2 covers the volume of combinations of solids — finding total volume when solids are combined together.
Key Volume Formulas:
- Cylinder: $V = \pi r^2 h$
- Cone: $V = \dfrac{1}{3}\pi r^2 h$
- Sphere: $V = \dfrac{4}{3}\pi r^3$
- Hemisphere: $V = \dfrac{2}{3}\pi r^3$
- Cuboid: $V = l \times b \times h$; Cube: $V = a^3$
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. #
Answer
Radius $r$ = 1 cm
Height of cone $h$ = 1 cm
Volume of solid = Volume of cone + Volume of hemisphere
So we find the volume:
$$= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$$$$= \frac{1}{3}\pi(1)^2(1) + \frac{2}{3}\pi(1)^3$$
$$= \frac{\pi}{3} + \frac{2\pi}{3}$$
$$= \pi$$
∴ The volume of the solid is $\mathbf{\pi}$ 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 aluminium 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.) #
Answer
Radius $r = \dfrac{3}{2} = 1.5$ cm
Height of each cone $h_1$ = 2 cm
Height of cylinder $h_2$ = 12 − 2 − 2 = 8 cm
Volume of air = Volume of cylinder + 2 × Volume of cone
So we find the volume:
$$= \pi r^2 h_2 + 2 \times \frac{1}{3}\pi r^2 h_1$$$$= \pi r^2\left(h_2 + \frac{2h_1}{3}\right)$$
$$= \frac{22}{7} \times 2.25 \times \left(8 + \frac{4}{3}\right)$$
$$= \frac{22}{7} \times 2.25 \times \frac{28}{3}$$
$$= 66$$
∴ The volume of air in the model is 66 cm³.
3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. #
Answer
Radius $r = \dfrac{2.8}{2} = 1.4$ cm
Length of cylinder $h$ = 5 − 2 × 1.4 = 2.2 cm
Volume of one gulab jamun = Volume of cylinder + 2 × Volume of hemisphere
So we find the volume of one gulab jamun:
$$= \pi r^2 h + \frac{4}{3}\pi r^3$$$$= \pi r^2\left(h + \frac{4r}{3}\right)$$
$$= \frac{22}{7} \times 1.96 \times \left(2.2 + \frac{5.6}{3}\right)$$
$$= 6.16 \times \frac{12.2}{3}$$
$$= 25.05 \text{ (about)}$$
Volume of 45 gulab jamuns = $45 \times 25.05$
Volume of 45 gulab jamuns = 1127.25 (about)
Volume of syrup = 30% of 1127.25
Volume of syrup = $\dfrac{30}{100} \times 1127.25$
Volume of syrup = 338.18 (about)
∴ The 45 gulab jamuns contain about 338 cm³ of syrup.
4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand. #
Answer
Volume of cuboid = $15 \times 10 \times 3.5 = 525$ cm³
Using Volume of cone $= \dfrac{1}{3}\pi r^2 h$, we find the volume of one depression:
$$= \frac{1}{3} \times \frac{22}{7} \times 0.5^2 \times 1.4$$$$= \frac{11}{30}$$
Volume of 4 depressions = $4 \times \dfrac{11}{30} = \dfrac{22}{15} = 1.47$ cm³ (about)
Volume of wood = 525 − 1.47
Volume of wood = 523.53
∴ The volume of wood in the stand is about 523.53 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. #
Answer
Using Volume of cone $= \dfrac{1}{3}\pi r^2 h$, we find the volume of water in the vessel:
$$= \frac{1}{3}\pi \times 5^2 \times 8$$$$= \frac{200\pi}{3}$$
Volume of water that flows out = $\dfrac{1}{4} \times \dfrac{200\pi}{3} = \dfrac{50\pi}{3}$ cm³
Using Volume of sphere $= \dfrac{4}{3}\pi r^3$, we find the volume of one lead shot:
$$= \frac{4}{3}\pi \times 0.5^3$$$$= \frac{\pi}{6}$$
The volume of the lead shots equals the volume of water that flows out, so:
$$n \times \frac{\pi}{6} = \frac{50\pi}{3}$$$$\Rightarrow n = \frac{50}{3} \times 6$$
$$\Rightarrow n = 100$$
∴ 100 lead shots were dropped in the vessel.
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. (Use π = 3.14) #
Answer
The radius of the lower cylinder is $\dfrac{24}{2} = 12$ cm.
Volume of lower cylinder = $3.14 \times 12^2 \times 220 = 99475.2$ cm³
Volume of upper cylinder = $3.14 \times 8^2 \times 60 = 12057.6$ cm³
Volume of pole = 99475.2 + 12057.6
Volume of pole = 111532.8 cm³
Mass of pole = $111532.8 \times 8$
Mass of pole = 892262.4 g
∴ The mass of the pole is about 892.26 kg.
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. #
Answer
Volume of cylinder = $\pi \times 60^2 \times 180 = 648000\pi$ cm³
Volume of cone = $\dfrac{1}{3}\pi \times 60^2 \times 120 = 144000\pi$ cm³
Volume of hemisphere = $\dfrac{2}{3}\pi \times 60^3 = 144000\pi$ cm³
Volume of solid = $144000\pi + 144000\pi$
Volume of solid = $288000\pi$ cm³
Volume of water left = Volume of cylinder − Volume of solid
So we find the volume of water left:
$$= 648000\pi - 288000\pi$$$$= 360000\pi$$
$$= 360000 \times \frac{22}{7}$$
$$= 1131428.57 \text{ (about)}$$
∴ The volume of water left is about 1131428.57 cm³, that is about 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. #
Answer
Radius of neck = $\dfrac{2}{2} = 1$ cm
Radius of spherical part = $\dfrac{8.5}{2} = 4.25$ cm
Volume of neck = $3.14 \times 1^2 \times 8 = 25.12$ cm³
Using Volume of sphere $= \dfrac{4}{3}\pi r^3$, we find the volume of the spherical part:
$$= \frac{4}{3} \times 3.14 \times 4.25^3$$$$= \frac{4}{3} \times 3.14 \times 76.77$$
$$= 321.39 \text{ (about)}$$
Volume of vessel = 25.12 + 321.39
Volume of vessel = 346.51 cm³
∴ The child is not correct. The correct volume is 346.51 cm³.