Class 10 Surface Areas and Volumes Practice Questions

Q: Do I need to show the formula before substituting?

Yes. Writing the combined formula first, such as pi r l + 2 pi r to the power 2, shows the examiner which faces you counted, and earns method marks even if the arithmetic slips.

This set practises surface area and volume of solids made by joining cubes, cuboids, cylinders, cones and hemispheres. It has 14 original questions, from quick two-mark questions to full exam-length problems with costs, water levels and litres. Every solution is worked step by step, with the unit given in the answer.

Quick recap

Solid	Curved surface area	Total surface area	Volume
Cube (edge $a$ )	$4a^2$	$6a^2$	$a^3$
Cuboid	$2h(l + b)$	$2(lb + bh + hl)$	$lbh$
Cylinder	$2\pi r h$	$2\pi r(r + h)$	$\pi r^2 h$
Cone	$\pi r l$	$\pi r(l + r)$	$\displaystyle \frac{1}{3}\pi r^2 h$
Hemisphere	$2\pi r^2$	$3\pi r^2$	$\displaystyle \frac{2}{3}\pi r^3$
Sphere	$4\pi r^2$	$4\pi r^2$	$\displaystyle \frac{4}{3}\pi r^3$

Slant height of a cone: $l = \sqrt{h^2 + r^2}$ .
Surface area of a combined solid: add only the faces still on the outside; joined faces are hidden.
Volume of a combined solid: add the volumes of the pieces; subtract any piece cut out.
$1000 \text{ cm}^3 = 1$ litre and $1 \text{ m}^3 = 1000$ litres.
Rise in level in a cylindrical container of radius $R$ : $\displaystyle x = \frac{V_{\text{solid}}}{\pi R^2}$ .

Short answer

Question 1

Two cubes, each of volume 216 $\text{cm}^3$ , are placed side by side and glued along one face. Find the surface area and the volume of the cuboid formed.

Edge of each cube: $a = \sqrt[3]{216} = 6$ cm. The cuboid measures $12 \times 6 \times 6$ cm.

$\begin{aligned} \text{Surface area} &= 2(lb + bh + hl) = 2(12 \times 6 + 6 \times 6 + 6 \times 12) \\ &= 2(72 + 36 + 72) = 360 \text{ cm}^2 \\ \text{Volume} &= 12 \times 6 \times 6 = 432 \text{ cm}^3 \end{aligned}$

Why this works: the two joined faces (36 $\text{cm}^2$ each) are hidden, so the area is $2 \times 216 - 72 = 360$ , while the volume is simply $216 + 216$ .

Answer: 360 $\text{cm}^2$ and 432 $\text{cm}^3$ .

Question 2

A solid is a cone of radius 5 cm and height 12 cm with a hemisphere of radius 5 cm joined to its base. Find, in terms of $\pi$ , its total surface area and its volume.

Slant height: $l = \sqrt{12^2 + 5^2} = \sqrt{169} = 13$ cm.

$\displaystyle \begin{aligned} S &= \pi r l + 2\pi r^2 = 65\pi + 50\pi = 115\pi \text{ cm}^2 \\ V &= \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi(25)(12) + \frac{2}{3}\pi(125) \\ &= 100\pi + \frac{250}{3}\pi = \frac{550}{3}\pi \text{ cm}^3 \end{aligned}$

Answer: $115\pi$ $\text{cm}^2$ and $\displaystyle \frac{550\pi}{3}$ $\text{cm}^3$ .

Question 3

A capsule is a cylinder with two hemispherical ends. Its total length is 12 mm and its diameter is 4 mm. Find its volume, correct to two decimal places, using $\displaystyle \pi = \frac{22}{7}$ .

$r = 2$ mm, and the cylinder is $12 - 4 = 8$ mm long. The two ends form one sphere.

$\displaystyle \begin{aligned} V &= \pi r^2 h + \frac{4}{3}\pi r^3 = \pi(4 \times 8) + \frac{4}{3}\pi(8) \\ &= 32\pi + \frac{32}{3}\pi = \frac{128}{3}\pi \\ &= \frac{128}{3} \times \frac{22}{7} = \frac{2816}{21} \approx 134.10 \text{ mm}^3 \end{aligned}$

Answer: about 134.10 $\text{mm}^3$ .

Question 4

A hemisphere of diameter 6 cm is placed on one face of a cube of edge 6 cm. Find the total surface area of the solid, using $\pi = 3.14$ .

The hemisphere hides a circle of radius 3 cm and adds a dome of area $2\pi r^2$ .

$\begin{aligned} S &= 6a^2 - \pi r^2 + 2\pi r^2 = 6a^2 + \pi r^2 \\ &= 216 + 3.14 \times 9 = 216 + 28.26 = 244.26 \text{ cm}^2 \end{aligned}$

Answer: 244.26 $\text{cm}^2$ .

Question 5

A cylindrical drum has radius 14 cm and height 30 cm. How many litres of water can it hold?

$\displaystyle V = \pi r^2 h = \frac{22}{7} \times 196 \times 30 = 18480 \text{ cm}^3$

Dividing by 1000: $18480 \div 1000 = 18.48$ litres.

Answer: 18.48 litres.

Long answer

Question 6

A tent is a cylinder of radius 10.5 m and height 3 m surmounted by a cone of the same radius and height 14 m. Find the area of canvas used (no floor) and its cost at ₹80 per $\text{m}^2$ .

Slant height: $l = \sqrt{14^2 + 10.5^2} = \sqrt{196 + 110.25} = \sqrt{306.25} = 17.5$ m.

$\displaystyle \begin{aligned} \text{Canvas} &= 2\pi r h + \pi r l = \pi r(2h + l) \\ &= \frac{22}{7} \times 10.5 \times (6 + 17.5) \\ &= 33 \times 23.5 = 775.5 \text{ m}^2 \end{aligned}$

Cost $= 775.5 \times 80 =$ ₹62,040.

Answer: 775.5 $\text{m}^2$ ; ₹62,040.

Question 7

A toy is a cone mounted on a hemisphere, both of radius 6 cm. The total height of the toy is 14 cm. Using $\pi = 3.14$ , find (a) its total surface area and (b) its volume.

Height of the cone $= 14 - 6 = 8$ cm, so $l = \sqrt{8^2 + 6^2} = 10$ cm.

$\displaystyle \begin{aligned} S &= \pi r l + 2\pi r^2 = \pi(60) + \pi(72) = 132\pi = 414.48 \text{ cm}^2 \\ V &= \frac{1}{3}\pi(36)(8) + \frac{2}{3}\pi(216) = 96\pi + 144\pi = 240\pi = 753.6 \text{ cm}^3 \end{aligned}$

Answer: (a) 414.48 $\text{cm}^2$ ; (b) 753.6 $\text{cm}^3$ .

Question 8

From a solid cylinder of height 12 cm and radius 3.5 cm, a hemisphere of radius 3.5 cm is scooped out of each end. Find (a) the total surface area and (b) the volume of the remaining solid.

(a) The curved surface of the cylinder remains. Each flat end is replaced by a hemispherical hollow.

$\displaystyle \begin{aligned} S &= 2\pi r h + 2 \times 2\pi r^2 = 2\pi r(h + 2r) \\ &= 2 \times \frac{22}{7} \times 3.5 \times (12 + 7) = 22 \times 19 = 418 \text{ cm}^2 \end{aligned}$

(b) The two hollows together make one sphere, which is removed.

$\displaystyle \begin{aligned} V &= \pi r^2 h - \frac{4}{3}\pi r^3 \\ &= \frac{22}{7} \times 12.25 \times 12 - \frac{4}{3} \times \frac{22}{7} \times 42.875 \\ &= 462 - \frac{539}{3} = \frac{847}{3} \approx 282.33 \text{ cm}^3 \end{aligned}$

Answer: (a) 418 $\text{cm}^2$ ; (b) about 282.33 $\text{cm}^3$ .

Question 9

A pedestal is a cuboid 20 cm long, 20 cm wide and 10 cm high, with a solid cylinder of radius 7 cm and height 14 cm fixed at the centre of its top. Find (a) the surface area to be painted (including the base), (b) the cost of painting at 50 paise per $\text{cm}^2$ , and (c) the volume of the pedestal.

Cuboid 20 cm by 20 cm by 10 cm with a cylinder of radius 7 cm and height 14 cm standing on the centre of its top face — The cylinder hides a circle on the cuboid's top, but its own top circle has exactly the same area.

(a) The circle hidden under the cylinder is replaced by the equal circle on top of the cylinder, so

$\displaystyle \begin{aligned} S &= \text{TSA of cuboid} - \pi r^2 + 2\pi r h + \pi r^2 = \text{TSA of cuboid} + 2\pi r h \\ &= 2(400 + 200 + 200) + 2 \times \frac{22}{7} \times 7 \times 14 \\ &= 1600 + 616 = 2216 \text{ cm}^2 \end{aligned}$

(b) 50 paise is ₹0.50, so the cost is $2216 \times 0.5 =$ ₹1108.

(c)

$\displaystyle V = 20 \times 20 \times 10 + \frac{22}{7} \times 49 \times 14 = 4000 + 2156 = 6156 \text{ cm}^3$

Answer: (a) 2216 $\text{cm}^2$ ; (b) ₹1108; (c) 6156 $\text{cm}^3$ .

Question 10

A cylindrical jar of radius 10 cm contains water. A solid made of a cylinder (radius 3 cm, height 5 cm) with a hemisphere of radius 3 cm on one end is dropped in and is completely under water. Find the rise in the water level.

$\displaystyle V = \pi(9)(5) + \frac{2}{3}\pi(27) = 45\pi + 18\pi = 63\pi \text{ cm}^3$

If the level rises by $x$ cm, then $\pi(10^2)x = 63\pi$ , so $\displaystyle x = \frac{63}{100} = 0.63$ cm.

Why this works: the extra water column is a cylinder with the jar's base, and its volume equals the solid's volume.

Answer: 0.63 cm.

Question 11

A glass is a cylinder of inner radius 3 cm and height 12 cm, but its bottom is a raised hemisphere of radius 3 cm. Using $\pi = 3.14$ , find the apparent capacity and the actual capacity of the glass.

Cylindrical glass of radius 3 cm and height 12 cm with a raised hemispherical base of radius 3 cm shown dashed inside at the bottom — The raised dome takes up space that looks as if it holds water.

$\displaystyle \begin{aligned} \text{Apparent capacity} &= \pi r^2 h = \pi(9)(12) = 108\pi = 339.12 \text{ cm}^3 \\ \text{Hemisphere} &= \frac{2}{3}\pi(27) = 18\pi \\ \text{Actual capacity} &= 108\pi - 18\pi = 90\pi = 282.6 \text{ cm}^3 \end{aligned}$

Answer: apparent 339.12 $\text{cm}^3$ ; actual 282.6 $\text{cm}^3$ .

Question 12

From a solid cylinder of radius 6 cm and height 8 cm, a cone of the same radius and height is hollowed out from the top. Find, in terms of $\pi$ , the total surface area and the volume of the remaining solid. Are the two answers the same quantity?

Slant height: $l = \sqrt{8^2 + 6^2} = 10$ cm.

$\displaystyle \begin{aligned} S &= 2\pi r h + \pi r^2 + \pi r l = \pi(96 + 36 + 60) = 192\pi \text{ cm}^2 \\ V &= \pi r^2 h - \frac{1}{3}\pi r^2 h = \frac{2}{3}\pi(36)(8) = 192\pi \text{ cm}^3 \end{aligned}$

The numbers match, but one is an area in $\text{cm}^2$ and the other a volume in $\text{cm}^3$ , so they are different quantities. (With $\pi = 3.14$ , each number is 602.88.)

Answer: $192\pi$ $\text{cm}^2$ and $192\pi$ $\text{cm}^3$ ; no, they are different quantities.

Question 13

A grain silo is a cylinder of radius 7 m and height 20 m with a hemispherical roof of the same radius. (a) Find the cost of painting its outside (not the base) at ₹15 per $\text{m}^2$ . (b) Find its volume, correct to two decimal places.

Silo drawn as a cylinder of radius 7 m and height 20 m with a hemispherical roof; the painted area is 2πrh + 2πr² without the base — Only the curved wall and the dome are painted.

(a)

$\displaystyle \begin{aligned} S &= 2\pi r h + 2\pi r^2 = 2\pi r(h + r) \\ &= 2 \times \frac{22}{7} \times 7 \times 27 = 44 \times 27 = 1188 \text{ m}^2 \end{aligned}$

Cost $= 1188 \times 15 =$ ₹17,820.

(b)

$\displaystyle \begin{aligned} V &= \pi r^2 h + \frac{2}{3}\pi r^3 = \frac{22}{7} \times 49 \times 20 + \frac{2}{3} \times \frac{22}{7} \times 343 \\ &= 3080 + \frac{2156}{3} = \frac{11396}{3} \approx 3798.67 \text{ m}^3 \end{aligned}$

Answer: (a) ₹17,820; (b) about 3798.67 $\text{m}^3$ .

Question 14

An overhead tank is a cylinder of radius 0.7 m and height 1.5 m, with a cone of the same radius and height 0.6 m forming its bottom. How many litres of water can it hold?

$\displaystyle \begin{aligned} V &= \pi r^2 h_1 + \frac{1}{3}\pi r^2 h_2 = \pi r^2\left(h_1 + \frac{h_2}{3}\right) \\ &= \frac{22}{7} \times 0.49 \times (1.5 + 0.2) \\ &= 1.54 \times 1.7 = 2.618 \text{ m}^3 \end{aligned}$

Since $1 \text{ m}^3 = 1000$ litres, the tank holds $2.618 \times 1000 = 2618$ litres.

Answer: 2618 litres.

Key terms

Combined solid: A solid formed by joining basic solids such as cones, cylinders and hemispheres.
Hidden face: A face along a join that is no longer on the outside and is left out of the surface area.
Curved surface area: The area of only the curved part of a solid.
Slant height: For a cone, $l = \sqrt{h^2 + r^2}$ ; used in surface area, not in volume.
Capacity: The volume a container can hold, often given in litres.
Apparent capacity: The capacity a container seems to have from its outer measurements, before any raised base is removed.
Displacement: The rise of a liquid caused by sinking a solid in it; the raised volume equals the solid's volume.

Common questions

How do I know whether a question wants surface area or volume?

Painting, covering, canvas, sheet metal and polishing need surface area. Material used, melting, holding water, capacity and weight need volume.

Why is the painted area of the pedestal the cuboid's full area plus only the cylinder's curved surface?

The cylinder hides a circle on the cuboid, but the cylinder's own top is an exposed circle of the same size, so the two cancel.

What value of $\pi$ should I use in the exam?

Use the value given in the question. If none is given and the radius is a multiple of 7 or 3.5, $\displaystyle \frac{22}{7}$ usually gives a clean answer.

Do I need to show the formula before substituting?

Yes. Writing the combined formula first, such as $\pi r l + 2\pi r^2$ , shows the examiner which faces you counted, and earns method marks even if the arithmetic slips.

How do I change paise to rupees in cost questions?

Divide by 100: 50 paise is ₹0.50. Then multiply the area by this rate.

References

National Council of Educational Research and Training. Mathematics: Textbook for Class X. NCERT, New Delhi.
Aggarwal, R. S. Secondary School Mathematics for Class 10. Bharati Bhawan.
Sharma, R. D. Mathematics for Class 10. Dhanpat Rai Publications.

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Class 10 Surface Areas and Volumes Practice Questions

Quick recap#

Short answer#

Ques­tion 1#

Ques­tion 2#

Ques­tion 3#

Ques­tion 4#

Ques­tion 5#

Long answer#

Ques­tion 6#

Ques­tion 7#

Ques­tion 8#

Ques­tion 9#

Ques­tion 10#

Ques­tion 11#

Ques­tion 12#

Ques­tion 13#

Ques­tion 14#

Key terms#

Com­mon ques­tions#

How do I know whether a ques­tion wants sur­face area or vol­ume?#

Why is the painted area of the pedestal the cuboid's full area plus only the cylin­der's curved sur­face?#

What value of π\piπ should I use in the exam?#

Do I need to show the for­mula before sub­sti­tut­ing?#

How do I change paise to rupees in cost ques­tions?#

Ref­er­ences#