CBSE Class 10 · Maths
1. Calculate the volume of the hemispherical dome if the height of the dome is 21 m.
(A) $19404$ cu. m (B) $2000$ cu. m (C) $15000$ cu. m (D) $19000$ cu. m
2. The formula to find the volume of a sphere is:
(A) $\dfrac{2}{3}\pi r^{3}$ (B) $\dfrac{4}{3}\pi r^{3}$ (C) $4\pi r^{2}$ (D) $2\pi r^{2}$
3. The cloth required to cover the hemispherical dome if the radius of its base is 14 m is:
(A) 1222 sq. m (B) $1232$ sq. m (C) $1200$ sq. m (D) $1400$ sq. m
4. The total surface area of the combined figure — hemispherical dome (radius 14 m) and cuboidal top (8 m × 6 m × 4 m) — is:
(A) 1200 sq. m (B) 1232 sq. m (C) $1392$ sq. m (D) 1932 sq. m
5. The volume of the cuboidal shaped top with dimensions $8$ m × $6$ m × $4$ m is:
(A) $182.45 \mathrm{~m}^{3}$ (B) $282.45 \mathrm{~m}^{3}$ (C) $292 \mathrm{~m}^{3}$ (D) $192 \mathrm{~m}^{3}$
1. What is the outer surface area of its wall?
2. What is the total surface area (in m²) of the lid?
3. The lid is made of cement. What is the volume of the cement used?
CBSE Class 10 Maths Surface Areas and Volumes Previous Year Questions
Help your child ace CBSE Class 10 Maths Surface Areas and Volumes Previous Year Questions with this carefully curated collection drawn from real board papers spanning 2015–2023. Every question comes with a detailed step-by-step solution, giving your child the practice needed to confidently handle cones, cylinders, spheres, and combination solids — topics that carry significant marks in the board exam.
CBSE Class 10 Maths Surface Areas and Volumes — Questions with Solutions
The volume of a right circular cone whose area of the base is $156 \mathrm{~cm}^{2}$ and the vertical height is 8 cm, is:
Solution
Answer: Option (D) is correct.
Explanation: Volume of cone $= \dfrac{1}{3} \pi r^{2} h$
Since the area of the base $= \pi r^{2} = 156 \mathrm{~cm}^{2}$, we substitute directly:
$$= \frac{1}{3} \times 156 \times 8 = 416 \mathrm{~cm}^{3}$$
Explanation: Volume of cone $= \dfrac{1}{3} \pi r^{2} h$
Since the area of the base $= \pi r^{2} = 156 \mathrm{~cm}^{2}$, we substitute directly:
$$= \frac{1}{3} \times 156 \times 8 = 416 \mathrm{~cm}^{3}$$
A cuboid of base area $P$ sq units is filled with water upto a height of $Q$ units. A sphere of volume $R$ cu units is dropped into the cuboid such that it is completely submerged. Which of these represents the increase in the height of water?
Solution
Answer: Option (B) is correct.
Explanation: Let the increase in height of water be $h_1$.
Volume of water rise = Volume of sphere
$$P \times h_1 = R \Rightarrow h_1 = \frac{R}{P} \text{ units}$$
Explanation: Let the increase in height of water be $h_1$.
Volume of water rise = Volume of sphere
$$P \times h_1 = R \Rightarrow h_1 = \frac{R}{P} \text{ units}$$
Two cubes each with 6 cm edge are joined end to end. The surface area of the resulting cuboid is:
Solution
Answer: Option (B) is correct.
Explanation: For the resulting cuboid: Length $l = 12$ cm, Breadth $b = 6$ cm, Height $h = 6$ cm.
Surface area $= 2(lb + bh + lh)$
$$= 2[(12)(6) + (6)(6) + (12)(6)] = 2[72 + 36 + 72] = 360 \mathrm{~cm}^{2}$$
Explanation: For the resulting cuboid: Length $l = 12$ cm, Breadth $b = 6$ cm, Height $h = 6$ cm.
Surface area $= 2(lb + bh + lh)$
$$= 2[(12)(6) + (6)(6) + (12)(6)] = 2[72 + 36 + 72] = 360 \mathrm{~cm}^{2}$$
A sphere of maximum volume is cut out from a solid hemisphere of radius 7 cm. Then the ratio of the volume of the original hemisphere to that of the cutout sphere is:
Solution
Answer: Option (D) is correct.
Explanation: Radius of hemisphere $r = 7$ cm. Volume of hemisphere $= \dfrac{2}{3}\pi r^{3}$.
Radius of the largest sphere carved out $R = \dfrac{r}{2} = 3.5$ cm.
Volume of sphere $= \dfrac{4}{3}\pi R^{3} = \dfrac{4}{3}\pi \left(\dfrac{r}{2}\right)^{3} = \dfrac{2}{3}\pi\dfrac{r^{3}}{4}$
Required ratio $= \dfrac{\frac{2}{3}\pi r^{3}}{\frac{2}{3}\pi \frac{r^{3}}{4}} = 4:1$
Explanation: Radius of hemisphere $r = 7$ cm. Volume of hemisphere $= \dfrac{2}{3}\pi r^{3}$.
Radius of the largest sphere carved out $R = \dfrac{r}{2} = 3.5$ cm.
Volume of sphere $= \dfrac{4}{3}\pi R^{3} = \dfrac{4}{3}\pi \left(\dfrac{r}{2}\right)^{3} = \dfrac{2}{3}\pi\dfrac{r^{3}}{4}$
Required ratio $= \dfrac{\frac{2}{3}\pi r^{3}}{\frac{2}{3}\pi \frac{r^{3}}{4}} = 4:1$
Savita has a lamp placed at the centre of her square yard, each side measuring 20 m. The light of the lamp covers a circle of radius 10 m on the yard. What area of the yard is NOT lit by the lamp?
Solution
Answer: Option (D) is correct.
Explanation:
Area of square yard $= (20)^{2} = 400 \mathrm{~m}^{2}$
Area of circle (lit region) $= \pi r^{2} = \pi(10)^{2} = 100\pi \mathrm{~m}^{2}$
Area not lit $=$ Area of square $-$ Area of circle $= (400 – 100\pi) \mathrm{~m}^{2}$
Explanation:
Area of square yard $= (20)^{2} = 400 \mathrm{~m}^{2}$
Area of circle (lit region) $= \pi r^{2} = \pi(10)^{2} = 100\pi \mathrm{~m}^{2}$
Area not lit $=$ Area of square $-$ Area of circle $= (400 – 100\pi) \mathrm{~m}^{2}$
A vessel having $30 \mathrm{~m}^{3}$ of water is emptied through two openings. Water flows out through the smaller opening at the rate of $U \mathrm{~m}^{3}/\mathrm{h}$ and through the larger one at the rate of $V \mathrm{~m}^{3}/\mathrm{h}$. Given that $3U + 2V = 70$ and that the vessel gets fully emptied in 1 hour, what is $V$?
Solution
Answer: Option (B) is correct.
Explanation: Since the vessel empties in 1 hour: $U + V = 30$ …(i)
Given: $3U + 2V = 70$ …(ii)
Multiply (i) by 3: $3U + 3V = 90$
Subtract (ii): $V = 20$
Therefore $V = 20 \mathrm{~m}^{3}/\mathrm{h}$.
Explanation: Since the vessel empties in 1 hour: $U + V = 30$ …(i)
Given: $3U + 2V = 70$ …(ii)
Multiply (i) by 3: $3U + 3V = 90$
Subtract (ii): $V = 20$
Therefore $V = 20 \mathrm{~m}^{3}/\mathrm{h}$.
The radius of a sphere (in cm) whose volume is $12\pi \mathrm{~cm}^{3}$, is:
Solution
Answer: Option (C) is correct.
Explanation: Volume of sphere $= \dfrac{4}{3}\pi r^{3} = 12\pi$
$$r^{3} = \frac{12\pi \times 3}{4\pi} = 9 \Rightarrow r = 9^{1/3} = (3^{2})^{1/3} = 3^{2/3}$$
Explanation: Volume of sphere $= \dfrac{4}{3}\pi r^{3} = 12\pi$
$$r^{3} = \frac{12\pi \times 3}{4\pi} = 9 \Rightarrow r = 9^{1/3} = (3^{2})^{1/3} = 3^{2/3}$$
A cylindrical pencil sharpened at one edge is the combination of:
Solution
Answer: Option (A) is correct.
Explanation: The sharpened tip of the pencil forms a cone, while the remaining unsharpened part is a cylinder. Together they make a combination of a cone and a cylinder.
Explanation: The sharpened tip of the pencil forms a cone, while the remaining unsharpened part is a cylinder. Together they make a combination of a cone and a cylinder.
A plumbline (Sahul) is the combination of:
Solution
Answer: Option (B) is correct.
Explanation: A plumbline is an instrument used to check the verticality of an object. Its shape is a combination of a hemisphere (the rounded top) and a cone (the pointed bottom).
Explanation: A plumbline is an instrument used to check the verticality of an object. Its shape is a combination of a hemisphere (the rounded top) and a cone (the pointed bottom).
The shape of a gilli, in the gilli-danda game, is a combination of:
Solution
Answer: Option (C) is correct.
Explanation: The shape of the gilli in the gilli-danda game is a combination of two cones (tapered at both ends) and a cylinder (in the middle), similar in shape to a medicine capsule.
Explanation: The shape of the gilli in the gilli-danda game is a combination of two cones (tapered at both ends) and a cylinder (in the middle), similar in shape to a medicine capsule.
The sum of the length, breadth and height of a cuboid is $6\sqrt{3}$ cm and the length of its diagonal is $2\sqrt{3}$ cm. The total surface area of the cuboid is:
Solution
Answer: Option (C) is correct.
Explanation: Given: $l + b + h = 6\sqrt{3}$ …(i) and diagonal $= 2\sqrt{3}$, so $l^{2} + b^{2} + h^{2} = 12$ …(ii)
Squaring (i): $(l+b+h)^{2} = 108$
$$l^{2}+b^{2}+h^{2}+2(lb+bh+hl) = 108$$ $$12 + 2(lb+bh+hl) = 108 \Rightarrow 2(lb+bh+hl) = 96 \mathrm{~cm}^{2}$$
Explanation: Given: $l + b + h = 6\sqrt{3}$ …(i) and diagonal $= 2\sqrt{3}$, so $l^{2} + b^{2} + h^{2} = 12$ …(ii)
Squaring (i): $(l+b+h)^{2} = 108$
$$l^{2}+b^{2}+h^{2}+2(lb+bh+hl) = 108$$ $$12 + 2(lb+bh+hl) = 108 \Rightarrow 2(lb+bh+hl) = 96 \mathrm{~cm}^{2}$$
A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that $\dfrac{1}{8}$ space of the cube remains unfilled. Then the number of marbles that the cube can accommodate is:
Solution
Answer: Option (A) is correct.
Explanation: Radius of marble $r = 0.25$ cm. Side of cube $l = 22$ cm.
Let $n$ be the number of marbles. Volume of $n$ marbles $= \dfrac{7}{8}$ of cube volume:
$$n \cdot \frac{4}{3}\pi r^{3} = \frac{7}{8} \times l^{3}$$ $$n = \frac{7 \times 3 \times 22 \times 22 \times 22 \times 7}{8 \times 4 \times 22 \times 0.25 \times 0.25 \times 0.25} = 142296$$
Explanation: Radius of marble $r = 0.25$ cm. Side of cube $l = 22$ cm.
Let $n$ be the number of marbles. Volume of $n$ marbles $= \dfrac{7}{8}$ of cube volume:
$$n \cdot \frac{4}{3}\pi r^{3} = \frac{7}{8} \times l^{3}$$ $$n = \frac{7 \times 3 \times 22 \times 22 \times 22 \times 7}{8 \times 4 \times 22 \times 0.25 \times 0.25 \times 0.25} = 142296$$
A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. The capacity of the capsule is:
Solution
Answer: Option (A) is correct.
Explanation: $r = 0.25$ cm. Total length $= 2r + h \Rightarrow h = 2 – 0.5 = 1.5$ cm.
Volume of capsule $= \dfrac{4}{3}\pi r^{3} + \pi r^{2}h = \pi r^{2}\left(\dfrac{4}{3}r + h\right)$
$$= \frac{22}{7} \times 0.0625 \times \left(\frac{1}{3} + 1.5\right) = \frac{22}{7} \times 0.0625 \times \frac{11}{6} \approx 0.36 \mathrm{~cm}^{3}$$
Explanation: $r = 0.25$ cm. Total length $= 2r + h \Rightarrow h = 2 – 0.5 = 1.5$ cm.
Volume of capsule $= \dfrac{4}{3}\pi r^{3} + \pi r^{2}h = \pi r^{2}\left(\dfrac{4}{3}r + h\right)$
$$= \frac{22}{7} \times 0.0625 \times \left(\frac{1}{3} + 1.5\right) = \frac{22}{7} \times 0.0625 \times \frac{11}{6} \approx 0.36 \mathrm{~cm}^{3}$$
Anishka melted 11 chocolate cubes in a cylindrical cup. If the length of the side of each cube is $k$ cm and the radius of the cup is $r$ cm, which of these represents the height of the melted chocolate in the cup? (Take $\pi = \dfrac{22}{7}$)
Solution
Answer: Option (B) is correct.
Explanation: Volume of 11 cubes = Volume of cylinder
$$11k^{3} = \frac{22}{7} \times r^{2} \times h \Rightarrow h = \frac{11k^{3} \times 7}{22r^{2}} = \frac{7k^{3}}{2r^{2}} \mathrm{~cm}$$
Explanation: Volume of 11 cubes = Volume of cylinder
$$11k^{3} = \frac{22}{7} \times r^{2} \times h \Rightarrow h = \frac{11k^{3} \times 7}{22r^{2}} = \frac{7k^{3}}{2r^{2}} \mathrm{~cm}$$
For which of the following solids is the lateral/curved surface area and total surface area the same?
Solution
Answer: Option (D) is correct.
Explanation: For a cube and cuboid, the lateral surface area excludes the top and bottom faces, so it differs from the total surface area. For a hemisphere, the curved surface area does not include the circular base. Only for a sphere is the entire surface curved — it has no flat faces — so the lateral/curved surface area equals the total surface area.
Explanation: For a cube and cuboid, the lateral surface area excludes the top and bottom faces, so it differs from the total surface area. For a hemisphere, the curved surface area does not include the circular base. Only for a sphere is the entire surface curved — it has no flat faces — so the lateral/curved surface area equals the total surface area.
Two cones have their heights in the ratio $1:3$ and radii in the ratio $3:1$. What is the ratio of their volumes?
Answer
Let heights be $h_1, h_2$ and radii be $r_1, r_2$. Given $\dfrac{h_1}{h_2} = \dfrac{1}{3}$ and $\dfrac{r_1}{r_2} = \dfrac{3}{1}$.
$$\frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^{2} h_1}{\frac{1}{3}\pi r_2^{2} h_2} = \left(\frac{r_1}{r_2}\right)^{2} \times \frac{h_1}{h_2} = \left(\frac{3}{1}\right)^{2} \times \frac{1}{3} = \frac{9}{3} = 3$$ The ratio of their volumes is $3:1$.
$$\frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^{2} h_1}{\frac{1}{3}\pi r_2^{2} h_2} = \left(\frac{r_1}{r_2}\right)^{2} \times \frac{h_1}{h_2} = \left(\frac{3}{1}\right)^{2} \times \frac{1}{3} = \frac{9}{3} = 3$$ The ratio of their volumes is $3:1$.
A cylindrical glass tube with radius 10 cm has water up to a height of 9 cm. A metal cube of 8 cm edge is immersed completely. By how much will the water level rise in the glass tube?
Answer
Let the height of water rise be $h$ cm.
Volume of water displaced in cylinder $= \pi(10)^{2} h$
Volume of cube $= 8 \times 8 \times 8 = 512 \mathrm{~cm}^{3}$
$$\pi(100)h = 512 \Rightarrow h = \frac{512}{100\pi} \approx 1.629 \mathrm{~cm}$$ The water level will rise by approximately $1.629$ cm.
Volume of water displaced in cylinder $= \pi(10)^{2} h$
Volume of cube $= 8 \times 8 \times 8 = 512 \mathrm{~cm}^{3}$
$$\pi(100)h = 512 \Rightarrow h = \frac{512}{100\pi} \approx 1.629 \mathrm{~cm}$$ The water level will rise by approximately $1.629$ cm.
Shown below is a right circular cone of volume $13{,}600 \mathrm{~cm}^{3}$ with base radius 20 cm. Find the angle which the slant height makes with the base radius. (Take $\pi = 3$, $\sqrt{2} = 1.4$, $\sqrt{3} = 1.7$)
Answer
Using the volume formula to find height $h$:
$$\frac{1}{3} \times 3 \times 20 \times 20 \times h = 13600 \Rightarrow h = 34 \mathrm{~cm}$$
The angle $\theta$ which the slant height makes with the base radius:
$$\tan\theta = \frac{h}{r} = \frac{34}{20} = 1.7 = \tan 60^{\circ} \Rightarrow \theta = 60^{\circ}$$ The slant height makes an angle of $60°$ with the base radius.
$$\frac{1}{3} \times 3 \times 20 \times 20 \times h = 13600 \Rightarrow h = 34 \mathrm{~cm}$$
The angle $\theta$ which the slant height makes with the base radius:
$$\tan\theta = \frac{h}{r} = \frac{34}{20} = 1.7 = \tan 60^{\circ} \Rightarrow \theta = 60^{\circ}$$ The slant height makes an angle of $60°$ with the base radius.
From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of the same height and same base is removed. Find the volume of the remaining solid.
Answer
Volume of remaining solid = Volume of cylinder $-$ Volume of cone
$$= \pi r^{2}h – \frac{1}{3}\pi r^{2}h = \frac{2}{3}\pi r^{2}h$$ $$= \frac{2}{3} \times \frac{22}{7} \times 6 \times 6 \times 14 = \mathbf{1056 \mathrm{~cm}^{3}}$$
$$= \pi r^{2}h – \frac{1}{3}\pi r^{2}h = \frac{2}{3}\pi r^{2}h$$ $$= \frac{2}{3} \times \frac{22}{7} \times 6 \times 6 \times 14 = \mathbf{1056 \mathrm{~cm}^{3}}$$
Isha and her father visited a juice shop. The juice seller served juice in two types of glasses — both with inner radius 3 cm and height 10 cm. The first type had a hemispherical raised bottom; the second had a conical raised bottom of height 1.5 cm. Isha chose the first type and her father chose the second. Who got more juice and by how much?
Answer
First glass (hemispherical bottom):
Juice volume $= \pi r^{2}H – \dfrac{2}{3}\pi r^{3} = \pi(3)^{2}\left[10 – \dfrac{2}{3}\times 3\right] = 9\pi \times 8 = 72\pi \approx 226.28 \mathrm{~cm}^{3}$
Second glass (conical bottom):
Conical volume $= \dfrac{1}{3}\pi r^{2}h = \dfrac{1}{3}\pi(3)^{2}(1.5) = 4.5\pi$
Juice volume $= \pi(3)^{2}(10) – 4.5\pi = 85.5\pi \approx 268.71 \mathrm{~cm}^{3}$
Suresh (father, second glass) got more juice — by approximately $42.43 \mathrm{~cm}^{3}$.
Juice volume $= \pi r^{2}H – \dfrac{2}{3}\pi r^{3} = \pi(3)^{2}\left[10 – \dfrac{2}{3}\times 3\right] = 9\pi \times 8 = 72\pi \approx 226.28 \mathrm{~cm}^{3}$
Second glass (conical bottom):
Conical volume $= \dfrac{1}{3}\pi r^{2}h = \dfrac{1}{3}\pi(3)^{2}(1.5) = 4.5\pi$
Juice volume $= \pi(3)^{2}(10) – 4.5\pi = 85.5\pi \approx 268.71 \mathrm{~cm}^{3}$
Suresh (father, second glass) got more juice — by approximately $42.43 \mathrm{~cm}^{3}$.
A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio $8:5$, then find the ratio of the radius of their bases to their heights.
Answer
Curved surface area of cylinder $= 2\pi rh$; Curved surface area of cone $= \pi r l$
$$\frac{2\pi rh}{\pi rl} = \frac{8}{5} \Rightarrow \frac{2h}{l} = \frac{8}{5} \Rightarrow \frac{h}{l} = \frac{4}{5}$$ where $l = \sqrt{h^{2}+r^{2}}$
$$\frac{h^{2}}{h^{2}+r^{2}} = \frac{16}{25} \Rightarrow 25h^{2} = 16h^{2}+16r^{2} \Rightarrow 9h^{2} = 16r^{2}$$ $$\frac{r^{2}}{h^{2}} = \frac{9}{16} \Rightarrow \frac{r}{h} = \frac{3}{4}$$ The ratio of radius to height is $3:4$.
$$\frac{2\pi rh}{\pi rl} = \frac{8}{5} \Rightarrow \frac{2h}{l} = \frac{8}{5} \Rightarrow \frac{h}{l} = \frac{4}{5}$$ where $l = \sqrt{h^{2}+r^{2}}$
$$\frac{h^{2}}{h^{2}+r^{2}} = \frac{16}{25} \Rightarrow 25h^{2} = 16h^{2}+16r^{2} \Rightarrow 9h^{2} = 16r^{2}$$ $$\frac{r^{2}}{h^{2}} = \frac{9}{16} \Rightarrow \frac{r}{h} = \frac{3}{4}$$ The ratio of radius to height is $3:4$.
A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use $\pi = \dfrac{22}{7}$)
Answer
Radius $r = 3.5$ cm. Height of cylindrical part $= 20 – 7 = 13$ cm (subtracting both hemispherical ends of radius 3.5 cm each).
Total volume $= \pi r^{2} h_{\text{cyl}} + \dfrac{4}{3}\pi r^{3}$
$$= \frac{22}{7} \times \frac{49}{4} \times 13 + \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8}$$ $$= \frac{77 \times 53}{6} \approx \mathbf{680.17 \mathrm{~cm}^{3}}$$
Total volume $= \pi r^{2} h_{\text{cyl}} + \dfrac{4}{3}\pi r^{3}$
$$= \frac{22}{7} \times \frac{49}{4} \times 13 + \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8}$$ $$= \frac{77 \times 53}{6} \approx \mathbf{680.17 \mathrm{~cm}^{3}}$$
A cylindrical tank of radius 40 cm is filled up to a height of 3.15 m by a cylindrical pipe at the rate of 2.52 km/h in $\dfrac{1}{2}$ hour. Calculate the diameter of the cylindrical pipe.
Answer
Volume of water in tank $= \pi(0.4)^{2} \times 3.15 = 0.504\pi \mathrm{~m}^{3}$
Distance covered by water in pipe in $\dfrac{1}{2}$ hour $= 2520 \times 0.5 = 1260$ m
Let pipe radius $= r_p$. Volume delivered $= \pi r_p^{2} \times 1260 = 0.504\pi$
$$r_p^{2} = \frac{0.504}{1260} = 0.0004 \Rightarrow r_p = 0.02 \mathrm{~m} = 2 \mathrm{~cm}$$ Diameter of the cylindrical pipe $= 4$ cm.
Distance covered by water in pipe in $\dfrac{1}{2}$ hour $= 2520 \times 0.5 = 1260$ m
Let pipe radius $= r_p$. Volume delivered $= \pi r_p^{2} \times 1260 = 0.504\pi$
$$r_p^{2} = \frac{0.504}{1260} = 0.0004 \Rightarrow r_p = 0.02 \mathrm{~m} = 2 \mathrm{~cm}$$ Diameter of the cylindrical pipe $= 4$ cm.
A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?
Answer
Radius $r = 12$ m, height $h = 3.5$ m.
Volume of rice $= \dfrac{1}{3} \times \dfrac{22}{7} \times 12 \times 12 \times 3.5 = \mathbf{528 \mathrm{~m}^{3}}$
Slant height $l = \sqrt{12^{2} + 3.5^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5$ m
Canvas required $= \pi r l = \dfrac{22}{7} \times 12 \times 12.5 = \mathbf{471.43 \mathrm{~m}^{2}}$
Volume of rice $= \dfrac{1}{3} \times \dfrac{22}{7} \times 12 \times 12 \times 3.5 = \mathbf{528 \mathrm{~m}^{3}}$
Slant height $l = \sqrt{12^{2} + 3.5^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5$ m
Canvas required $= \pi r l = \dfrac{22}{7} \times 12 \times 12.5 = \mathbf{471.43 \mathrm{~m}^{2}}$
A sphere of diameter 6 cm is dropped in a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Answer
Volume of sphere $= \dfrac{4}{3}\pi(3)^{3} = 36\pi \mathrm{~cm}^{3}$
Volume of water rise in cylinder $=$ Volume of sphere
$$\pi(6)^{2} \times h = 36\pi \Rightarrow 36h = 36 \Rightarrow h = 1 \mathrm{~cm}$$ The water level will rise by $1$ cm.
Volume of water rise in cylinder $=$ Volume of sphere
$$\pi(6)^{2} \times h = 36\pi \Rightarrow 36h = 36 \Rightarrow h = 1 \mathrm{~cm}$$ The water level will rise by $1$ cm.
A paint roller is 26 cm long with an outer diameter of 7 cm. Find the maximum area of the surface that gets painted when the roller makes 6 complete rotations vertically. (Take $\pi = \dfrac{22}{7}$)
Answer
Radius $r = 3.5$ cm, length $h = 26$ cm.
Curved surface area of roller $= 2\pi r h = 2 \times \dfrac{22}{7} \times 3.5 \times 26 = 572 \mathrm{~cm}^{2}$
Area painted in 6 rotations $= 6 \times 572 = \mathbf{3432 \mathrm{~cm}^{2}}$
Curved surface area of roller $= 2\pi r h = 2 \times \dfrac{22}{7} \times 3.5 \times 26 = 572 \mathrm{~cm}^{2}$
Area painted in 6 rotations $= 6 \times 572 = \mathbf{3432 \mathrm{~cm}^{2}}$
Dinesh is building a greenhouse with a circular base of diameter 12 m and a hemispherical dome on top. The cylindrical wall has height 2 m. How much will it cost to cover the walls and top with transparent plastic at ₹77 per sq m? (Take $\pi = \dfrac{22}{7}$)
Answer
Radius $r = 6$ m.
Curved surface area of cylindrical wall $= 2\pi rh = 2\pi \times 6 \times 2 = 24\pi \mathrm{~m}^{2}$
Curved surface area of hemispherical dome $= 2\pi r^{2} = 2\pi \times 36 = 72\pi \mathrm{~m}^{2}$
Total area $= 96\pi \mathrm{~m}^{2}$
Cost $= 96 \times \dfrac{22}{7} \times 77 = 96 \times 22 \times 11 = \mathbf{₹23{,}232}$
Curved surface area of cylindrical wall $= 2\pi rh = 2\pi \times 6 \times 2 = 24\pi \mathrm{~m}^{2}$
Curved surface area of hemispherical dome $= 2\pi r^{2} = 2\pi \times 36 = 72\pi \mathrm{~m}^{2}$
Total area $= 96\pi \mathrm{~m}^{2}$
Cost $= 96 \times \dfrac{22}{7} \times 77 = 96 \times 22 \times 11 = \mathbf{₹23{,}232}$
Subodh is baking a cylindrical cake 21 cm tall with radius 15 cm. He removes a cylindrical portion of height 21 cm from the centre. The cake weighs 0.5 g per cubic cm, and the remaining cake weighs 6600 g. Find the radius of the central portion that is cut. (Take $\pi = \dfrac{22}{7}$)
Answer
Volume of full cake $= \dfrac{22}{7} \times 15^{2} \times 21 = 14850 \mathrm{~cm}^{3}$
Weight of full cake $= 14850 \times 0.5 = 7425$ g
Weight of removed portion $= 7425 – 6600 = 825$ g
Volume of removed portion $= 825 \div 0.5 = 1650 \mathrm{~cm}^{3}$
$$\frac{22}{7} \times r^{2} \times 21 = 1650 \Rightarrow r^{2} = \frac{1650}{66} = 25 \Rightarrow r = 5 \mathrm{~cm}$$ The radius of the central portion cut is $5$ cm.
Weight of full cake $= 14850 \times 0.5 = 7425$ g
Weight of removed portion $= 7425 – 6600 = 825$ g
Volume of removed portion $= 825 \div 0.5 = 1650 \mathrm{~cm}^{3}$
$$\frac{22}{7} \times r^{2} \times 21 = 1650 \Rightarrow r^{2} = \frac{1650}{66} = 25 \Rightarrow r = 5 \mathrm{~cm}$$ The radius of the central portion cut is $5$ cm.
A semi-circular waffle sheet of radius 5 cm is folded into an ice-cream cone. Due to overlap while folding, the radius of the base of the cone is 80% of what it would be without overlap. Find the approximate volume of the cone. (Take $\pi = \dfrac{22}{7}$)
Answer
The slant height of the cone $= 5$ cm (radius of the semicircle).
Without overlap: circumference of base $= \pi r_1 \Rightarrow 2\pi r_2 = \pi \times 5 \Rightarrow r_2 = \dfrac{5}{2}$ cm
With 80% overlap: actual radius $= 0.80 \times \dfrac{5}{2} = 2$ cm
Height of cone: $h = \sqrt{l^{2} – r^{2}} = \sqrt{25 – 4} = \sqrt{21} \approx 5$ cm (using $\sqrt{21} \approx 5$ for approximation as per the solution)
Volume $= \dfrac{1}{3} \times \dfrac{22}{7} \times 4 \times 5 = \dfrac{440}{21} \approx \mathbf{20 \mathrm{~cm}^{3}}$
Without overlap: circumference of base $= \pi r_1 \Rightarrow 2\pi r_2 = \pi \times 5 \Rightarrow r_2 = \dfrac{5}{2}$ cm
With 80% overlap: actual radius $= 0.80 \times \dfrac{5}{2} = 2$ cm
Height of cone: $h = \sqrt{l^{2} – r^{2}} = \sqrt{25 – 4} = \sqrt{21} \approx 5$ cm (using $\sqrt{21} \approx 5$ for approximation as per the solution)
Volume $= \dfrac{1}{3} \times \dfrac{22}{7} \times 4 \times 5 = \dfrac{440}{21} \approx \mathbf{20 \mathrm{~cm}^{3}}$
A hemispherical glass cloche has a radius of 13 cm. Kanan wants to cover a cylindrical cake of volume $3168 \mathrm{~cm}^{3}$. Find one set of values of radius and height of the cake such that the cloche does not touch the cake. (Take $\pi = \dfrac{22}{7}$)
Answer
Volume of cake: $\pi r^{2}h = 3168 \Rightarrow r^{2}h = 1008$
Since the cloche radius is 13 cm, the cake radius must be less than 13 cm. Taking $r = 12$ cm:
$$h = \frac{1008}{12 \times 12} = \frac{1008}{144} = 7 \mathrm{~cm}$$ One valid set: radius $= 12$ cm and height $= 7$ cm. (The diagonal of this cylinder $= \sqrt{12^{2}+7^{2}} = \sqrt{193} \approx 13.9$ cm, so the cloche fits around it.)
Since the cloche radius is 13 cm, the cake radius must be less than 13 cm. Taking $r = 12$ cm:
$$h = \frac{1008}{12 \times 12} = \frac{1008}{144} = 7 \mathrm{~cm}$$ One valid set: radius $= 12$ cm and height $= 7$ cm. (The diagonal of this cylinder $= \sqrt{12^{2}+7^{2}} = \sqrt{193} \approx 13.9$ cm, so the cloche fits around it.)
A student made a model shaped like a cylinder with two cones attached to its ends using thin aluminium sheet. The diameter of the model is 3 cm and its total length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model.
Answer
Radius $r = 1.5$ cm. Height of cylindrical part $= 12 – (2+2) = 8$ cm.
Volume of model $= $ Volume of cylinder $+ 2 \times$ Volume of cone
$$= \pi r^{2} h_{\text{cyl}} + 2 \times \frac{1}{3}\pi r^{2} h_{\text{cone}}$$ $$= \pi(1.5)^{2}(8) + \frac{2}{3}\pi(1.5)^{2}(2)$$ $$= 18\pi + 3\pi = 21\pi = 21 \times \frac{22}{7} = \mathbf{66 \mathrm{~cm}^{3}}$$
Volume of model $= $ Volume of cylinder $+ 2 \times$ Volume of cone
$$= \pi r^{2} h_{\text{cyl}} + 2 \times \frac{1}{3}\pi r^{2} h_{\text{cone}}$$ $$= \pi(1.5)^{2}(8) + \frac{2}{3}\pi(1.5)^{2}(2)$$ $$= 18\pi + 3\pi = 21\pi = 21 \times \frac{22}{7} = \mathbf{66 \mathrm{~cm}^{3}}$$
A solid is in the shape of a right-circular cone surmounted on a hemisphere. The radius of each is 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid.
Answer
Radius $r = 7$ cm. Height of cone $= 2r = 14$ cm.
Volume of solid $=$ Volume of cone $+$ Volume of hemisphere
$$= \frac{1}{3}\pi r^{2}h + \frac{2}{3}\pi r^{3} = \frac{1}{3}\pi r^{2}(h + 2r)$$ $$= \frac{1}{3} \times \frac{22}{7} \times 49 \times (14+14) = \frac{154}{3} \times 28 = \frac{4312}{3} \approx \mathbf{1437.33 \mathrm{~cm}^{3}}$$
Volume of solid $=$ Volume of cone $+$ Volume of hemisphere
$$= \frac{1}{3}\pi r^{2}h + \frac{2}{3}\pi r^{3} = \frac{1}{3}\pi r^{2}(h + 2r)$$ $$= \frac{1}{3} \times \frac{22}{7} \times 49 \times (14+14) = \frac{154}{3} \times 28 = \frac{4312}{3} \approx \mathbf{1437.33 \mathrm{~cm}^{3}}$$
Shown below is a cuboid with $480 \mathrm{~cm}^{3}$ of water in two different orientations. If the height of water in orientation II is half of that in orientation I, find the heights of water in both orientations.
Answer
Let the vertical dimension of the cuboid in orientation I be $h$ cm, and the height of water be $(h – 4)$ cm.
Height of water in orientation II $= \dfrac{1}{2}(h-4)$
Setting up the volume equation (base area $= 5 \times h$):
$$5 \times h \times \frac{1}{2}(h-4) = 480 \Rightarrow h^{2} – 4h – 192 = 0$$ $$(h-16)(h+12) = 0 \Rightarrow h = 16 \text{ cm (rejecting } h = -12\text{)}$$ Height of water in orientation I $= 16 – 4 = 12$ cm
Height of water in orientation II $= \dfrac{1}{2} \times 12 = 6$ cm
Height of water in orientation II $= \dfrac{1}{2}(h-4)$
Setting up the volume equation (base area $= 5 \times h$):
$$5 \times h \times \frac{1}{2}(h-4) = 480 \Rightarrow h^{2} – 4h – 192 = 0$$ $$(h-16)(h+12) = 0 \Rightarrow h = 16 \text{ cm (rejecting } h = -12\text{)}$$ Height of water in orientation I $= 16 – 4 = 12$ cm
Height of water in orientation II $= \dfrac{1}{2} \times 12 = 6$ cm
Two rectangular sheets of dimensions 45 cm × 155 cm are folded to make hollow right circular cylinders — Sheet 1 is folded along its length and Sheet 2 along its width, each with 1 cm overlap. Both cylinders are closed on both ends.
(i) Find the difference in the curved surface areas of the two cylinders.
(ii) Find the ratio of the volumes of the two cylinders. (Use $\pi = \dfrac{22}{7}$)
(i) Find the difference in the curved surface areas of the two cylinders.
(ii) Find the ratio of the volumes of the two cylinders. (Use $\pi = \dfrac{22}{7}$)
Answer
(i) Difference in curved surface areas:
Sheet 1 cylinder height $= 155$ cm; overlap area $= 155 \times 1 = 155 \mathrm{~cm}^{2}$
Sheet 2 cylinder height $= 45$ cm; overlap area $= 45 \times 1 = 45 \mathrm{~cm}^{2}$
Since sheets are identical, difference in CSA $= 155 – 45 = \mathbf{110 \mathrm{~cm}^{2}}$
(ii) Ratio of volumes:
Sheet 1: circumference $= 45 – 1 = 44$ cm $\Rightarrow 2\pi r_1 = 44 \Rightarrow r_1 = 7$ cm
Sheet 2: circumference $= 155 – 1 = 154$ cm $\Rightarrow 2\pi r_2 = 154 \Rightarrow r_2 = \dfrac{49}{2}$ cm
$$\frac{V_1}{V_2} = \frac{\pi r_1^{2} h_1}{\pi r_2^{2} h_2} = \frac{7^{2} \times 155}{\left(\frac{49}{2}\right)^{2} \times 45} = \frac{49 \times 155}{\frac{2401}{4} \times 45} = \frac{4 \times 49 \times 155}{2401 \times 45} = \mathbf{\frac{124}{441}}$$
Sheet 1 cylinder height $= 155$ cm; overlap area $= 155 \times 1 = 155 \mathrm{~cm}^{2}$
Sheet 2 cylinder height $= 45$ cm; overlap area $= 45 \times 1 = 45 \mathrm{~cm}^{2}$
Since sheets are identical, difference in CSA $= 155 – 45 = \mathbf{110 \mathrm{~cm}^{2}}$
(ii) Ratio of volumes:
Sheet 1: circumference $= 45 – 1 = 44$ cm $\Rightarrow 2\pi r_1 = 44 \Rightarrow r_1 = 7$ cm
Sheet 2: circumference $= 155 – 1 = 154$ cm $\Rightarrow 2\pi r_2 = 154 \Rightarrow r_2 = \dfrac{49}{2}$ cm
$$\frac{V_1}{V_2} = \frac{\pi r_1^{2} h_1}{\pi r_2^{2} h_2} = \frac{7^{2} \times 155}{\left(\frac{49}{2}\right)^{2} \times 45} = \frac{49 \times 155}{\frac{2401}{4} \times 45} = \frac{4 \times 49 \times 155}{2401 \times 45} = \mathbf{\frac{124}{441}}$$
Case Study — The Great Stupa at Sanchi
The Great Stupa at Sanchi is one of the oldest stone structures in India. It features a big hemispherical dome with a cuboidal structure on top. (Take $\pi = \dfrac{22}{7}$)1. Calculate the volume of the hemispherical dome if the height of the dome is 21 m.
(A) $19404$ cu. m (B) $2000$ cu. m (C) $15000$ cu. m (D) $19000$ cu. m
2. The formula to find the volume of a sphere is:
(A) $\dfrac{2}{3}\pi r^{3}$ (B) $\dfrac{4}{3}\pi r^{3}$ (C) $4\pi r^{2}$ (D) $2\pi r^{2}$
3. The cloth required to cover the hemispherical dome if the radius of its base is 14 m is:
(A) 1222 sq. m (B) $1232$ sq. m (C) $1200$ sq. m (D) $1400$ sq. m
4. The total surface area of the combined figure — hemispherical dome (radius 14 m) and cuboidal top (8 m × 6 m × 4 m) — is:
(A) 1200 sq. m (B) 1232 sq. m (C) $1392$ sq. m (D) 1932 sq. m
5. The volume of the cuboidal shaped top with dimensions $8$ m × $6$ m × $4$ m is:
(A) $182.45 \mathrm{~m}^{3}$ (B) $282.45 \mathrm{~m}^{3}$ (C) $292 \mathrm{~m}^{3}$ (D) $192 \mathrm{~m}^{3}$
Answer
1. Answer: Option (A) — $19404$ cu. m
Height of dome = radius = 21 m.
Volume $= \dfrac{2}{3}\pi r^{3} = \dfrac{2}{3} \times \dfrac{22}{7} \times 21^{3} = 19404 \mathrm{~m}^{3}$
2. Answer: Option (B) — $\dfrac{4}{3}\pi r^{3}$
The standard formula for the volume of a sphere is $\dfrac{4}{3}\pi r^{3}$.
3. Answer: Option (B) — 1232 sq. m
Cloth required $= 2\pi R^{2} = 2 \times \dfrac{22}{7} \times 14^{2} = 1232$ sq. m
4. Answer: Option (C) — 1392 sq. m
Total surface area $= 2\pi r^{2} + 2(lb+bh+hl) – lb$
$= 1232 + 2(48+24+32) – 48 = 1232 + 208 – 48 = 1392 \mathrm{~m}^{2}$
5. Answer: Option (D) — $192 \mathrm{~m}^{3}$
Volume $= l \times b \times h = 8 \times 6 \times 4 = 192 \mathrm{~m}^{3}$
Height of dome = radius = 21 m.
Volume $= \dfrac{2}{3}\pi r^{3} = \dfrac{2}{3} \times \dfrac{22}{7} \times 21^{3} = 19404 \mathrm{~m}^{3}$
2. Answer: Option (B) — $\dfrac{4}{3}\pi r^{3}$
The standard formula for the volume of a sphere is $\dfrac{4}{3}\pi r^{3}$.
3. Answer: Option (B) — 1232 sq. m
Cloth required $= 2\pi R^{2} = 2 \times \dfrac{22}{7} \times 14^{2} = 1232$ sq. m
4. Answer: Option (C) — 1392 sq. m
Total surface area $= 2\pi r^{2} + 2(lb+bh+hl) – lb$
$= 1232 + 2(48+24+32) – 48 = 1232 + 208 – 48 = 1392 \mathrm{~m}^{2}$
5. Answer: Option (D) — $192 \mathrm{~m}^{3}$
Volume $= l \times b \times h = 8 \times 6 \times 4 = 192 \mathrm{~m}^{3}$
Case Study — Open Water Tank
An open water tank has walls of thickness 10 cm. The outer dimensions are: length $l = 1.6$ m, breadth $b = 2$ m, height $h = 1.8$ m. A lid of thickness 3 cm covers the tank. Answer the following:1. What is the outer surface area of its wall?
2. What is the total surface area (in m²) of the lid?
3. The lid is made of cement. What is the volume of the cement used?
Answer
1. Outer surface area of wall:
Lateral surface area $= 2(l+b) \times h = 2(1.6+2) \times 1.8 = 2 \times 3.6 \times 1.8 = \mathbf{12.96 \mathrm{~m}^{2}}$
2. Total surface area of the lid:
Lid dimensions: $l = 1.6$ m, $b = 2$ m, thickness $h = 0.03$ m
TSA $= 2(lb+bh+hl) = 2(3.2 + 0.06 + 0.048) = 2 \times 3.308 = \mathbf{6.616 \mathrm{~m}^{2}}$
3. Volume of cement in the lid:
Volume $= l \times b \times h = 1.6 \times 2 \times 0.03 = \mathbf{0.096 \mathrm{~m}^{3}}$
Lateral surface area $= 2(l+b) \times h = 2(1.6+2) \times 1.8 = 2 \times 3.6 \times 1.8 = \mathbf{12.96 \mathrm{~m}^{2}}$
2. Total surface area of the lid:
Lid dimensions: $l = 1.6$ m, $b = 2$ m, thickness $h = 0.03$ m
TSA $= 2(lb+bh+hl) = 2(3.2 + 0.06 + 0.048) = 2 \times 3.308 = \mathbf{6.616 \mathrm{~m}^{2}}$
3. Volume of cement in the lid:
Volume $= l \times b \times h = 1.6 \times 2 \times 0.03 = \mathbf{0.096 \mathrm{~m}^{3}}$
A cylindrical can is placed in a cubical container with side $2p$. The radius of the can is $\dfrac{p}{4}$ and its height is $p$.
(i) How many of these cans can be packed in the container such that no more cans can fit?
(ii) If the capacity of one can is 539 ml, find the internal volume of the cubical container. (Take $\pi = \dfrac{22}{7}$)
(i) How many of these cans can be packed in the container such that no more cans can fit?
(ii) If the capacity of one can is 539 ml, find the internal volume of the cubical container. (Take $\pi = \dfrac{22}{7}$)
Answer
(i) Number of cans:
Each can has diameter $\dfrac{p}{2}$ and height $p$. In each of the length and breadth directions: $2p \div \dfrac{p}{2} = 4$ cans. In height: $2p \div p = 2$ layers.
Total cans $= 4 \times 4 \times 2 = \mathbf{32}$
(ii) Internal volume of the cubical container:
Volume of one can $= \pi r^{2} h = \dfrac{22}{7} \times \dfrac{p^{2}}{16} \times p = 539$
$$\frac{22p^{3}}{112} = 539 \Rightarrow p^{3} = \frac{539 \times 112}{22} = 2744 \Rightarrow p = 14 \mathrm{~cm}$$ Side of cube $= 2p = 28$ cm
Internal volume $= 28^{3} = \mathbf{21952 \mathrm{~cm}^{3}}$
Each can has diameter $\dfrac{p}{2}$ and height $p$. In each of the length and breadth directions: $2p \div \dfrac{p}{2} = 4$ cans. In height: $2p \div p = 2$ layers.
Total cans $= 4 \times 4 \times 2 = \mathbf{32}$
(ii) Internal volume of the cubical container:
Volume of one can $= \pi r^{2} h = \dfrac{22}{7} \times \dfrac{p^{2}}{16} \times p = 539$
$$\frac{22p^{3}}{112} = 539 \Rightarrow p^{3} = \frac{539 \times 112}{22} = 2744 \Rightarrow p = 14 \mathrm{~cm}$$ Side of cube $= 2p = 28$ cm
Internal volume $= 28^{3} = \mathbf{21952 \mathrm{~cm}^{3}}$
Frequently Asked Questions
What does the Surface Areas and Volumes chapter cover in CBSE Class 10 Maths? ⌄
This chapter covers the surface areas and volumes of solid shapes including cuboids, cubes, right circular cylinders, cones, spheres, and hemispheres. It also covers combination solids — shapes formed by joining two or more basic solids — and problems involving conversion of one solid into another, which are very commonly tested in board exams.
How many marks does Surface Areas and Volumes carry in the CBSE Class 10 board exam? ⌄
Surface Areas and Volumes is part of the Mensuration unit, which carries approximately 10 marks in the CBSE Class 10 Mathematics board paper. Questions from this chapter regularly appear as 1-mark MCQs, 2-mark short answers, 3-mark problems, and 5-mark case study questions, making it one of the most marks-dense chapters in the syllabus.
What are the most important topics students should focus on in Class 10 Surface Areas and Volumes? ⌄
The most frequently tested topics are: volume and surface area of combination solids (cone + cylinder, hemisphere + cylinder, etc.), problems on conversion of one solid into another (like melting a sphere into smaller spheres), and water-level rise problems involving displacement. Case study questions in recent years have also featured real-life contexts like stupas, greenhouses, and food items.
What common mistakes do students make when solving Surface Areas and Volumes questions? ⌄
A very common mistake is confusing total surface area with curved/lateral surface area — especially in combination solids, where the joined face must be excluded. Students also frequently apply the wrong formula for slant height in cones or forget to convert units consistently (cm to m, etc.). Practising a variety of board paper questions helps your child recognise these traps before exam day.
How does Angle Belearn help students score well in Surface Areas and Volumes? ⌄
Angle Belearn’s CBSE specialists have curated a question bank drawn from real board papers spanning multiple years, paired with clear, step-by-step solutions. Students who practise with these questions develop a habit of structured working — showing all steps — which earns full marks in board exams. Regular practice builds the formula recall and problem-solving speed your child needs to tackle this chapter confidently.
