NCERT Class 10 Mathematics — Surface Areas and Volumes, Exercise 12.2. All 8 questions solved.

Exercise 12.1 asked which surfaces survive when solids are joined. This exercise asks how much space they take up, and the answer is much simpler: the volume of a combined solid is the sum of the volumes of its parts, and the volume of a solid with a cavity is the whole minus the cavity. The formulae, for radius $r$ and height $h$:

$$\text{cylinder} = \pi r^2 h \qquad \text{cone} = \tfrac{1}{3}\pi r^2 h \qquad \text{sphere} = \tfrac{4}{3}\pi r^3 \qquad \text{hemisphere} = \tfrac{2}{3}\pi r^3$$

and $l \times b \times h$ for a cuboid.

Key insight. Nothing vanishes at a joint. Volumes add, and cavities subtract, with no correction. So the real work in every question is reading the dimensions of each part correctly — above all the length of a cylinder once cones or hemispheres have taken up part of the total length.

Unless stated otherwise, take $\pi = \dfrac{22}{7}$.

Question 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 $\pi$.

Solution. Here $r = 1$ cm and the cone’s height is also $h = 1$ cm. The solid is the hemisphere plus the cone:

$$V = \tfrac{2}{3}\pi r^3 + \tfrac{1}{3}\pi r^2 h = \tfrac{2}{3}\pi(1)^3 + \tfrac{1}{3}\pi(1)^2(1) = \tfrac{2}{3}\pi + \tfrac{1}{3}\pi = \pi\ \text{cm}^3$$

“In terms of $\pi$” means leave $\pi$ as a symbol rather than substituting $\frac{22}{7}$.

Question 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.)

Solution. The radius of the cones and the cylinder is $r = \frac{3}{2}$ cm. Each cone has height $h_1 = 2$ cm, and the two cones use up $4$ cm of the $12$ cm length, so the cylinder has length $h_2 = 12 – 4 = 8$ cm.

Because the sheet is thin, the air inside fills the whole shape:

$$V = \tfrac{1}{3}\pi r^2 h_1 + \pi r^2 h_2 + \tfrac{1}{3}\pi r^2 h_1 = \tfrac{1}{3}\pi r^2(h_1 + 3h_2 + h_1)$$

$$= \frac{1}{3} \times \frac{22}{7} \times \frac{9}{4} \times (2 + 24 + 2) = \frac{1}{3} \times \frac{22}{7} \times \frac{9}{4} \times 28 = 66\ \text{cm}^3$$

Question 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 (see Fig. 12.15).

Solution. Fig. 12.15 shows a bowl of gulab jamuns, each the capsule shape drawn above. The radius is $r = 1.4$ cm. The two hemispherical ends take up $2 \times 1.4 = 2.8$ cm of the length, so the cylindrical middle is $h = 5 – 2.8 = 2.2$ cm long. The two hemispheres together make one sphere:

$$V_{\text{one}} = \pi r^2 h + \tfrac{4}{3}\pi r^3 = \pi r^2\left(h + \tfrac{4}{3}r\right)$$

$$= \frac{22}{7} \times 1.96 \times \left(2.2 + \frac{5.6}{3}\right) = 6.16 \times \frac{12.2}{3} \approx 25.05\ \text{cm}^3$$

For $45$ of them, and then $30\%$ of that:

$$45 \times 6.16 \times \frac{12.2}{3} = 1127.28\ \text{cm}^3, \qquad \text{syrup} = 0.3 \times 1127.28 = 338.184\ \text{cm}^3$$

The question asks for an approximate amount, so about $338\ \text{cm}^3$.

Question 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 (see Fig. 12.16).

Solution. Fig. 12.16 shows the block with four holes in its top face and pens standing in them. The wood is the cuboid minus the four conical holes:

$$\text{cuboid} = 15 \times 10 \times 3.5 = 525\ \text{cm}^3$$

$$\text{one depression} = \tfrac{1}{3} \times \frac{22}{7} \times 0.5^2 \times 1.4 = \frac{11}{30}\ \text{cm}^3, \qquad \text{four} = \frac{22}{15} \approx 1.47\ \text{cm}^3$$

$$\text{wood} = 525 – \frac{22}{15} = 523.53\ \text{cm}^3$$

Question 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. The vessel was full, so every lead shot that goes in pushes out its own volume of water. The water that flows out therefore equals the total volume of the shots.

$$\text{volume of cone} = \tfrac{1}{3} \times \frac{22}{7} \times 5^2 \times 8 = \frac{4400}{21}\ \text{cm}^3$$

$$\text{water out} = \tfrac{1}{4} \times \frac{4400}{21} = \frac{1100}{21}\ \text{cm}^3$$

$$\text{one shot} = \tfrac{4}{3} \times \frac{22}{7} \times 0.5^3 = \frac{11}{21}\ \text{cm}^3$$

$$\text{number of shots} = \frac{1100}{21} \div \frac{11}{21} = 100$$

Question 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 $1\ \text{cm}^3$ of iron has approximately $8$ g mass. (Use $\pi = 3.14$)

Solution. The lower cylinder has radius $12$ cm (half the diameter) and height $220$ cm; the upper one has radius $8$ cm and height $60$ cm.

$$V = \pi(12^2 \times 220 + 8^2 \times 60) = 3.14 \times (31680 + 3840) = 3.14 \times 35520 = 111532.8\ \text{cm}^3$$

At $8$ g per $\text{cm}^3$:

$$\text{mass} = 111532.8 \times 8 = 892262.4\ \text{g} \approx 892.26\ \text{kg}$$

Question 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. The solid is $60 + 120 = 180$ cm tall, exactly the height of the cylinder, so it is completely under water and pushes out its whole volume. The water left is the cylinder’s volume minus the solid’s:

$$\text{cylinder} = \pi \times 60^2 \times 180 = 648000\pi\ \text{cm}^3$$

$$\text{solid} = \tfrac{1}{3}\pi \times 60^2 \times 120 + \tfrac{2}{3}\pi \times 60^3 = 144000\pi + 144000\pi = 288000\pi\ \text{cm}^3$$

$$\text{water left} = 648000\pi – 288000\pi = 360000\pi = 360000 \times \frac{22}{7} \approx 1131428.57\ \text{cm}^3$$

Since $1\ \text{m}^3 = 100^3\ \text{cm}^3 = 10^6\ \text{cm}^3$, this is about $1.131\ \text{m}^3$.

Question 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 $345\ \text{cm}^3$. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.

Solution. The vessel holds a sphere of radius $\frac{8.5}{2} = 4.25$ cm and a cylinder of radius $1$ cm and length $8$ cm:

$$\text{sphere} = \tfrac{4}{3} \times 3.14 \times 4.25^3 = \tfrac{4}{3} \times 3.14 \times 76.765625 \approx 321.39\ \text{cm}^3$$

$$\text{neck} = 3.14 \times 1^2 \times 8 = 25.12\ \text{cm}^3$$

$$\text{total} \approx 321.39 + 25.12 = 346.51\ \text{cm}^3$$

This is more than $345\ \text{cm}^3$, so her measurement is not correct, though it is close. As in the textbook’s own answer, the neck’s $8$ cm is measured from the sphere, and the small overlap where the neck meets the curved surface is neglected.

Common mistakes

• Question 2, using $12$ cm for the cylinder. The $12$ cm is the whole model, tip to tip. The cones take $2$ cm at each end, leaving $8$ cm of cylinder. Using $12$ gives about $84.86\ \text{cm}^3$ for the cylinder alone, already more than the whole model holds.
• Question 3, forgetting one of the steps. There are three: the volume of one gulab jamun, then $45$ of them, then $30\%$ of that. Stopping at $1127.28\ \text{cm}^3$ gives the volume of the sweets, not the syrup.
• Question 4, adding the depressions. The holes are empty space, so they are subtracted from the cuboid. Adding them would make a block with holes heavier than a solid one.
• Question 5, using the whole cone. Only one-fourth of the water flows out, so only one-fourth of the cone’s volume is displaced. Using the whole cone gives $400$ shots.
• Question 6, using $24$ cm as a radius. The lower cylinder’s $24$ cm is a diameter, but the upper one’s $8$ cm is a radius. The question gives the two different ways on purpose.
• Question 7, converting to $\text{m}^3$ wrongly. One metre is $100$ cm, so one cubic metre is $100^3 = 10^6$ cubic centimetres, not $100$.
• Question 8, using $8.5$ cm as the radius. $8.5$ cm is the diameter of the spherical part. The volume depends on $r^3$, so doubling the radius makes the sphere’s volume $2^3 = 8$ times too large.

Practise next

• Exercise 12.1 — surface areas of the same kinds of solids, where, unlike here, the faces at a joint have to be taken away.
• Chapter 11, Exercise 11.1 — sectors, segments and circle areas, the plane-figure work that these solid formulae extend.