Surface Areas & Volumes — Class 10 Maths PYQs

Previous-year board questions (2023–2026) with marking-scheme solutions, grouped by topic and marks.

Try each question first, then press ↓ (or tap Show Solution) to reveal the answer. Press ↓ again for the next question.

Year:Marks:Level:

Surface Area

1 Mark Questions
11 Mark · 🔁 July 2023 & March 2026 · Standardopen ↗
Assertion (A): The surface area of the cuboid formed by joining two cubes of sides $4$ cm each, end to end, is $160$ cm$^2$.
Reason (R) : Surface area of a cuboid of dimensions $l \times b \times h$ is $(lb + bh + hl)$
Show SolutionHide Solution↓
(c) Assertion (A) is true but Reason (R) is false
21 Mark · March 2025 · Standardopen ↗
A cone of height 12 cm and slant height 13 cm is surmounted on a hemisphere having radius equal to that of cone. The entire height of the solid is
  • (a)17 cm
  • (b)18 cm
  • (c)22 cm
  • (d)23 cm
Show SolutionHide Solution↓
(A) 17 cm
31 Mark · March 2026 · Standardopen ↗
A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius $10$ cm. Curved surface area of the cavity carved out is (use $\pi = 3.14$)
figure for this question
  • (a)$314\sqrt{2}$ cm²
  • (b)$314$ cm²
  • (c)$\frac{3140}{3}$ cm²
  • (d)$3140\sqrt{2}$ cm²
Show SolutionHide Solution↓
(A) $314\sqrt{2}$ cm² (1 Mark)
41 Mark · March 2026 · Standardopen ↗
A camping tent in hemispherical shape of radius $1.4$ m, has a door opening of area $0.50$ m$^2$. Outer surface area of the tent is
  • (a)$11.78$ m$^2$
  • (b)$12.32$ m$^2$
  • (c)$11.82$ m$^2$
  • (d)$12.86$ m$^2$
Show SolutionHide Solution↓
(C) $11.82$ m$^2$
51 Mark · March 2026 · Standardopen ↗
A camping tent in hemispherical shape of radius $1.4$ m, has a door opening of area $0.50$ m². Outer surface area of the tent is
  • (a)$11.78$ m²
  • (b)$12.32$ m²
  • (c)$11.82$ m²
  • (d)$12.86$ m²
Show SolutionHide Solution↓
(C) $11.82$ m² (1 Mark)
61 Mark · March 2025 · Basicopen ↗
A cone of height '$h$' and radius '$r$' is surmounted on a solid cylinder of same dimensions. The total surface area of the entire solid will be :
  • (a)$2\pi rh + \pi r \sqrt{h^2 + r^2}$
  • (b)$2\pi rh + \pi r^2 + \pi r \sqrt{h^2 + r^2}$
  • (c)$2\pi rh + 2\pi r^2 + \pi r \sqrt{h^2 + r^2}$
  • (d)$2\pi rh + \pi r \sqrt{h^2 + r^2} - \pi r^2$
Show SolutionHide Solution↓
(B) $2\pi rh + \pi r^2 + \pi r \sqrt{h^2 + r^2}$
71 Mark · March 2025 · Basicopen ↗
A cone of height '$h$' and radius '$r$' is surmounted on a hemisphere of same radius. The total surface area of the entire solid will be :
  • (a)$\pi rh + \pi r^2$
  • (b)$\pi r \sqrt{h^2 + r^2} + \pi r^2$
  • (c)$\pi r \sqrt{h^2 + r^2} + 2\pi r^2$
  • (d)$\pi r \sqrt{h^2 + r^2} + 3\pi r^2$
Show SolutionHide Solution↓
Answer (C) $\pi r \sqrt{h^2 + r^2} + 2\pi r^2$
4 Marks Questions
84 Marks · March 2026 · Basicopen ↗
In a coffee shop, coffee is served in two types of cups. One is cylindrical shape with each of diameter $8$ cm and height $7$ cm and the other hemispherical with each of diameter $14$ cm.
Based on the above, answer the following questions :
(i) What is the outer curved surface area of the cylindrical cup ?
(ii) What is the inner surface area of the hemispherical cup ?
(iii) (a) Find the difference of the capacities of the two cups.
OR
(b) Find the total volume of coffee in two cylindrical and one hemispherical cup.
Show SolutionHide Solution↓
(i) $r = 4$ cm
Outer C.S.A. of cylindrical cup = $2 \times \frac{22}{7} \times 4 \times 7 = 176$ cm$^2$ (1 Mark)
(ii) $R = 7$ cm,
Inner Surface Area of hemispherical cup = $2 \times \frac{22}{7} \times 7 \times 7 = 308$ cm$^2$ (1 Mark)
(iii) (a) Difference in the capacities = $\frac{22}{7} \times 4 \times 4 \times 7 - \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$ (1 Mark)
$= \frac{1100}{3}$ cm$^3$ or $366.67$ cm$^3$ (1 Mark)
OR
(b) Total Volume of coffee = $2 \times \frac{22}{7} \times 4 \times 4 \times 7 + \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$ (1 Mark)
$= \frac{4268}{3}$ cm$^3$ or $1422.67$ cm$^3$ (1 Mark)
5 Marks Questions
95 Marks · July 2023 · Standardopen ↗
A tent is in the shape of a right circular cylinder up to a height of $3$ m and then a right circular cone, with a maximum height of $13.5$ m above the ground. Calculate the cost of painting the inner side of the tent at the rate of ₹2 per square metre, if the radius of the base is $14$ m.
Show SolutionHide Solution↓
Height of conical part $= 13.5 - 3 = 10.5$ m
Slant height $= \sqrt{(14)^2 + (10.5)^2} = 17.5$ m
SA of tent = CSA of conical part + CSA of cylindrical part
$= (\frac{22}{7} \times 14 \times 17.5) + (2 \times \frac{22}{7} \times 14 \times 3)$
$= 1034$ m$^2$
Cost of painting @ ₹2 per m$^2 = 1034 \times 2 = 2068$
105 Marks · March 2023 · Standardopen ↗
From a solid cylinder of height $20$ cm and diameter $12$ cm, a conical cavity of height $8$ cm and radius $6$ cm is hallowed out. Find the total surface area of the remaining solid.
Show SolutionHide Solution↓
Height of cylinder $h = 20$ cm
radius of cylinder $= 6$ cm $=$ Radius of cone
Height of cone $= 8$ cm
Slant height $l = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10$ cm
Surface area of remaining solid
$= CSA \text{ of cylinder} + CSA \text{ of cone} + \text{Area of base of cylinder}$
$= 2\pi rh + \pi rl + \pi r^2 = \pi r[2h + l + r]$
$= \frac{22}{7} \times 6[2 \times 20 + 10 + 6] = \frac{22}{7} \times 6 \times 56$
$= 1056 \text{ cm}^2$
115 Marks · July 2024 · Standardopen ↗
A solid is in the form of a right circular cylinder with hemispherical ends. The total height of the solid is $58$ cm and the diameter of the cylinder is $28$ cm. Find the total surface area of the solid.
Show SolutionHide Solution↓
Height of cylindrical part = $58 - 14 - 14 = 30$ cm
Radius of cylindrical as well as hemispherical parts = $14$ cm
TSA of the solid = $4 \times \frac{22}{7} \times (14)^2 + 2 \times \frac{22}{7} \times 14 \times 30$
= $5104 \text{ cm}^2$
Therefore, total surface area of the solid is $5104 \text{ cm}^2$.

Volume

1 Mark Questions
121 Mark · July 2023 · Standardopen ↗
A hemispherical bowl is made of steel of thickness $1$ cm. The inner radius of the bowl is $5$ cm. The volume of steel used (in cm$^3$) is :
  • (a)$182 \pi$
  • (b)$\frac{182}{3} \pi$
  • (c)$\frac{682}{3} \pi$
  • (d)$\frac{364}{3} \pi$
Show SolutionHide Solution↓
(b) $\frac{182}{3} \pi$
131 Mark · July 2024 · Standardopen ↗
A hemispherical bowl is made of steel of thickness $0.30$ cm. The inner radius of the bowl is $3$ cm. The volume of steel used (in cu cm) is :
  • (a)$595.8 \pi$
  • (b)$5.958 \pi$
  • (c)$6 \pi$
  • (d)$59.58 \pi$
Show SolutionHide Solution↓
(B) $5.958 \pi$
141 Mark · 🔁 March 2024 & July 2024 & March 2025 & July 2025 · Standardopen ↗
Assertion (A): In the given figure, a toy is in the form of a cylinder surmounted by a hemisphere of the same radius. If the radius of the cylinder is $3$ cm and its height is $7$ cm, then the volume of toy is $81 \pi \text{ cm}^3$. Reason (R): Volume of the given solid is the sum of the volume of the cylinder and the volume of the hemisphere.
figure for this question
  • (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is
  • (c)Assertion (A) is true, but Reason (R) is false.
  • (d)Assertion (A) is false, but Reason (R) is true.
Show SolutionHide Solution↓
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
151 Mark · March 2026 · Standardopen ↗
A hemispherical bowl is made of steel of thickness $1$ cm. The outer radius of the bowl is $6$ cm. The volume of steel used (in cm$^3$) is :
  • (a)$182 \pi$
  • (b)$\frac{182}{3} \pi$
  • (c)$\frac{682}{3} \pi$
  • (d)$\frac{364}{3} \pi$
Show SolutionHide Solution↓
(B) $\frac{182}{3} \pi$ (1 Mark)
161 Mark · March 2026 · Standardopen ↗
A cone of maximum size is carved out from a solid cube of edge length $l$. The volume of the cone is :
figure for this question
  • (a)$\frac{\pi l^3}{12}$
  • (b)$\frac{\pi l^3}{3}$
  • (c)$l^3 \left(1-\frac{\pi}{3}\right)$
  • (d)$\frac{\pi l^3}{8}$
Show SolutionHide Solution↓
(A) $\frac{\pi l^3}{12}$ (1 Mark)
171 Mark · March 2026 · Basicopen ↗
The radius of the largest sphere that can be carved out from a solid cube of side $14$ cm is :
  • (a)$7$ cm
  • (b)$14$ cm
  • (c)$14\sqrt{2}$ cm
  • (d)$7\sqrt{2}$ cm
Show SolutionHide Solution↓
(A) $7$ cm
2 Marks Questions
182 Marks · March 2026 · Standardopen ↗
$1000$ small thermocol balls of radius $0.5$ cm are kept in a spherical
nballoon of radius $20$ cm. Find the volume of air in the balloon.
Show SolutionHide Solution↓
Volume of air in the ballon = Volume of spherical balloon $-$ Volume of $1000$
nthermocol balls
$= \frac{4}{3} \times \frac{22}{7} \times (20)^3 - 1000 \times \frac{4}{3} \times \frac{22}{7} \times (0.5)^3$ (1 Mark)
$= 33000$ cm$^3$ (1 Mark)
3 Marks Questions
193 Marks · March 2023 · Standardopen ↗
A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of cylindrical part. Find total height of the room if it contains $\frac{1408}{21}$ m$^3$ of air. Take $\pi = \frac{22}{7}$
Show SolutionHide Solution↓
Let $h$ be height of cylindrical part and $r$ be radius of hemisphere
Volume of room = $2\pi r^3 + \frac{2}{3} \pi r^3 = \frac{1408}{21}$
$\Rightarrow r=2$
Therefore, $h=4$
Height of the room is = $6$m
203 Marks · March 2023 · Standardopen ↗
An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled in it so that lower part of the cone which is $(\frac{1}{6})^{\text{th}}$ of the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take $\pi = 3.14$)
figure for this question
Show SolutionHide Solution↓
Volume of the cone = $\frac{1}{3}\times\pi\times9\times12=36\pi\text{cm}^3$
Volume of ice-cream in the cone = $\frac{5}{6} \times 36 \times \pi = 30\pi\text{cm}^3$
Volume of ice-cream on top = $\frac{2}{3} \times 27 \times \pi = 18\pi\text{cm}^3$
Total volume of the ice-cream = $(30\pi +18\pi) = 48\pi\text{cm}^3$
$=48\times3.14=150.72\text{cm}^3$
213 Marks · March 2023 · Standardopen ↗
An empty cone is of radius $3$ cm and height $12$ cm. Ice-cream is filled in it so that lower part of the cone which is $\left(\frac{1}{6}\right)^{\text{th}}$ of the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take $\pi = 3.14$)
Show SolutionHide Solution↓
Volume of the cone = $$\begin{aligned}& \frac{1}{3}\times\pi\times9\times12=36\pi \text{cm}^3 \\ & \text{Volume of ice-cream in the cone} = \frac{5}{6} \times 36 \times \pi = 30\pi \text{cm}^3 \\ & \text{Volume of ice-cream on top} = \frac{2}{3} \times 27 \times \pi = 18\pi \text{cm}^3 \\ & \text{Total volume of the ice-cream} = (30\pi +18\pi) = 48\pi \text{cm}^3 \\ & =48\times3.14=150.72\text{cm}^3\end{aligned}$$
223 Marks · March 2025 · Standardopen ↗
A room is in the form of a cylinder surmounted by a hemispherical dome. The base radius of the hemisphere is half of the height of the cylindrical part. If the room contains $\frac{1408}{21} \text{ m}^3$ of air, find the height of the cylindrical part. (Use $\pi = \frac{22}{7}$).
Show SolutionHide Solution↓
Let $r$ be the radius and $h$ be the height of cylinder. $h = 2r$. Volume $= \frac{2}{3}\pi r^3 + \pi r^2 h = \frac{1408}{21} \implies \frac{2}{3}\pi r^3 + \pi r^2(2r) = \frac{1408}{21} \implies \frac{8}{3} \times \frac{22}{7} \times r^3 = \frac{1408}{21} \implies r^3 = 8 \implies r = 2$ m and $h = 4$ m.
4 Marks Questions
234 Marks · March 2026 · Standardopen ↗
A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled.
The diameter of spherical ball is $21$ cm.
Cylindrical pool - Outer diameter is $50$ cm and inner diameter is $40$ cm.
Height of solid base is $14$ cm.
Height of water filled is $7$ cm.
Observe the figure and answer the following questions :
(i) Determine the total height of the fountain.
(ii) Find the volume of the ball.
(iii) (a) If one-third of the ball is submerged in the water, find the volume of the water filled in the pool.
OR
(iii) (b) Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.
figure for this question
Show SolutionHide Solution↓
(i) Total height of the fountain $= 14 + 21 = 35$ cm (1 Mark)
(ii) Volume of the ball $= \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$ (1/2 Mark)
$= 4851$ cm$^3$ (1/2 Mark)
(iii) (a) Volume of water = Volume of inner upper part – $\frac{1}{3} \times$ Volume of the ball
$= \pi \times 20 \times 20 \times 7 - \frac{1}{3} \times 4851$ (1 Mark)
$= 7183$ cm$^3$ (1 Mark)
OR
(iii) (b) Required area = Outer CSA of cylindrical part + Surface area of the ball
$= 2 \times \frac{22}{7} \times 25 \times (14 + 7) + 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$ (1 Mark)
$= 4686$ cm$^2$ (1 Mark)
5 Marks Questions
245 Marks · March 2023 · Standardopen ↗
A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being $7$ cm and the height of the cone is equal to its diameter. Find the volume of the solid.
Show SolutionHide Solution↓
Radius of cone = radius of hemisphere = $7$ cm
Height of cone = $14$ cm
Volume of solid = Volume of hemisphere + volume of cone
$= \frac{2}{3}\pi(7)^3 + \frac{1}{3}\pi(7)^2(14)$
$= \frac{1}{3} \times \frac{22}{7} \times 7 \times 7(14 + 14)$
$= \frac{154}{3} \times 28 = \frac{4312}{3} \text{ cm}^2 \text{ or } 1437.33 \text{ cm}^2$
255 Marks · March 2023 · Standardopen ↗
A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being $3.5$ cm and the total height of the solid is $9.5$ cm. Find the volume of the solid.
Show SolutionHide Solution↓
Radius of hemisphere = Radius of cone = $3.5$ cm = $\frac{7}{2}$ cm
Height of cone = $9.5 - 3.5 = 6$ cm
Volume of solid = volume of hemisphere + volume of cone
$= \frac{2}{3} \times \frac{22}{7} \times (\frac{7}{2})^3 + \frac{1}{3} \times \frac{22}{7} \times (\frac{7}{2})^2 \times 6$
$= \frac{77}{6} \times 13 = \frac{1001}{6} = 166.8$ cm$^3$
figure for this question
265 Marks · March 2024 · Standardopen ↗
A solid iron pole consists of a solid cylinder of height $200 \text{ cm}$ and base diameter $28 \text{ cm}$, which is surmounted by another cylinder of height $50 \text{ cm}$ and radius $7 \text{ cm}$. Find the mass of the pole, given that $1 \text{ cm}^3$ of iron has approximately $8 \text{ g}$ mass.
Show SolutionHide Solution↓
Radius of lower cylinder = $14 \text{ cm}$
Volume of pole = $\frac{22}{7} \times 14 \times 14 \times 200 + \frac{22}{7} \times 7 \times 7 \times 50$
$= 130900 \text{ cm}^3$
Mass of the pole= $8 \times 130900$
$=1047200 \text{ gm}$ or $1047.2 \text{ kg}$
275 Marks · March 2024 · Standardopen ↗
A juice seller was serving his customers using glasses as shown in the figure. The inner diameter of the cylindrical glass was $5.6$ cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was $10$ cm, find the apparent capacity and the actual capacity of the glass.
figure for this question
Show SolutionHide Solution↓
Radius$(r) = 2.8$cm
Apparent capacity of glass $= \frac{22}{7} \times 2.8 \times 2.8 \times 10$
$= 246.4$ cm$^3$
Volume of hemispherical part $= \frac{2}{3} \times \frac{22}{7} \times 2.8 \times 2.8 \times 2.8$
$= 45.9$ cm$^3$
$\therefore$ Actual capacity of glass $= 246.4 - 45.9$
$= 200.5$ cm$^3$ or $200.5$ ml
285 Marks · March 2025 · Standardopen ↗
A $16$ m deep well with diameter $3.5$ m is dug up and the earth from it is spread evenly to form a platform $27.5$ m $\times 7$ m. Find the height of the platform.
Show SolutionHide Solution↓
(a) Volume of the earth dug up = $\frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 16 = 154$ m$^3$. Area of platform = $27.5 \times 7 = 192.5$ m$^2$. Height of platform = $\frac{154}{192.5} = \frac{4}{5}$ m or $80$ cm
295 Marks · March 2025 · Basicopen ↗
A cubical block is surmounted by a hemisphere of radius $3.5\text{ cm}$. What is the smallest possible length of the edge of the cube so that the hemisphere can totally lie on the cube ? Find the total surface area of the solid so formed.
Show SolutionHide Solution↓
Edge of cube $= a = 3.5 \times 2 = 7\text{ cm}$
Total surface area of solid
$= 6 a^2 + 2\pi r^2 - \pi r^2$
$= 6 a^2 + \pi r^2$
$= 6 \times 7 \times 7 + \frac{22}{7} \times 3.5 \times 3.5$
$= \frac{665}{2}\text{ sq. cm}$ or $332.5\text{ sq. cm}$

Both

4 Marks Questions
304 Marks · March 2024 · Standardopen ↗
Case Study - 3
Tamper-proof tetra-packed milk guarantees both freshness and security. This milk ensures uncompromised quality, preserving the nutritional values within and making it a reliable choice for health-conscious individuals.
$500$ mL milk is packed in a cuboidal container of dimensions $15$ cm x $8$ cm x $5$ cm. These milk packets are then packed in cuboidal cartons of dimensions $30$ cm x $32$ cm x $15$ cm.
Based on the above given information, answer the following questions :
(i) Find the volume of the cuboidal carton.
(ii) (a) Find the total surface area of a milk packet.
OR
(b) How many milk packets can be filled in a carton ?
(iii) How much milk can the cup (as shown in the figure) hold ?
figure for this question
Show SolutionHide Solution↓
(i) Volume of cuboidal carton $$\begin{aligned}& = 30\times32\times15 \\ & = 14400 \text{ cm}^3 \\ & \text{(ii)(a) Total surface area of milk packet } = 2(15\times8+8\times5+5\times15) \\ & = 470 \text{ cm}^2\ \text{OR} \\ & \text{(ii) (b) Number of milk packets in carton } = \frac{30\times32\times15}{15\times8\times5} \\ & = 24 \\ & \text{(iii) Capacity of the cup } = \frac{22}{7} \times 5 \times 5 \times 7 \\ & = 550 \text{ cm}^3 \text{ or } 550 \text{ ml}\end{aligned}$$
314 Marks · March 2025 · Standardopen ↗
A skilled carpenter decided to craft a special rolling pin for the local baker. He carefully joined three cylindrical pieces of wood two small ones on the ends and one larger in the centre to create a perfect tool. The baker loved the rolling pin, as it rolled out the smoothest dough for breads and pastries.
The length of the bigger cylindrical part is $12$ cm and diameter is $7$ cm and the length of each smaller cylindrical part is $5$ cm and diameter is $2.1$ cm.
Based on the above information, answer the following questions :
(i) Find the volume of the bigger cylindrical part.
(ii) Find the curved surface area of the bigger cylindrical part.
(iii) (a) Find the ratio of the volume of the bigger cylindrical part to the total volume of the two smaller (identical) cylindrical parts.
OR
(b) Find the sum of the curved surface areas of the two identical smaller cylindrical parts.
figure for this question
Show SolutionHide Solution↓
(i) Volume of the bigger cylindrical part $= \pi r^2 h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12$
$= 462$ cm$^3$
(ii) The Curved Surface Area of bigger cylindrical part $= 2 \pi r h = 2 \times \frac{22}{7} \times \frac{7}{2} \times 12$
$= 264$ cm$^2$
(iii) (a) Total volume of the two smaller cylindrical parts $= 2 \times \pi r^2 h = 2 \times \frac{22}{7} \times \frac{2.1}{2} \times \frac{2.1}{2} \times 5$
$= 34.65$ cm$^3$
Required ratio $= \frac{462}{34.65} = \frac{3080}{231}$
$\therefore$ Required ratio is $3080 : 231$
OR
(b) The Sum of Curved Surface Area of two smaller cylindrical parts $= 2 \times 2 \pi r h = 2 \times 2 \times \frac{22}{7} \times \frac{2.1}{2} \times 5$
$= 66$ cm$^2$
324 Marks · March 2026 · Standardopen ↗
A model of Leafy Ball Fountain is made to be kept on the tabletop. Water≥ntly cascades down the ball into a decorative cylindrical pool where it is
recycled.
The diameter of spherical ball is $21$ cm.
Cylindrical pool - Outer diameter is $50$ cm and inner diameter is $40$ cm.
Height of solid base is $14$ cm.
Height of water filled is $7$ cm.
(i) Determine the total height of the fountain.
(ii) Find the volume of the ball.
(iii) (a) If one-third of the ball is submerged in the water, find the volume of the water filled in the pool.
OR
(iii) (b) Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball.
figure for this question
Show SolutionHide Solution↓
(i) Total height of the fountain $= 14 + 21 = 35$ cm (1 Mark)
(ii) Volume of the ball $= \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$ (1/2 Mark)
$= 4851$ cm$^3$ (1/2 Mark)
(iii) (a) Volume of water = Volume of inner upper part - $\frac{1}{3} \times$ Volume of the ball
$= \frac{22}{7} \times 20 \times 20 \times 7 - \frac{1}{3} \times 4851$ (1 Mark)
$= 7183$ cm$^3$ (1 Mark)
OR
(iii) (b) Required area = Outer CSA of cylinderical part + Surface area of the ball
$= 2 \times \frac{22}{7} \times 25 \times (14 + 7) + 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$ (1 Mark)
$= 4686$ cm$^2$ (1 Mark)
5 Marks Questions
335 Marks · July 2023 · Standardopen ↗
A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere of same radius. If the radius of the hemisphere is $4.2$ cm and the total height of the toy is $10.2$ cm, find the volume of the wooden toy. Also, find the total surface area of the toy.
Show SolutionHide Solution↓
Height of conical part $= 10.2 - 4.2 = 6$ cm
Volume of toy = Volume of conical part + Volume of hemispherical part
$= (\frac{1}{3} \times \frac{22}{7} \times (4.2)^2 \times 6) + (\frac{2}{3} \times \frac{22}{7} \times (4.2)^3)$
$= 266.112$
Hence, Volume of toy is $266.112$ cm$^3$
Slant height of conical part $= \sqrt{(4.2)^2 + (6)^2} \approx 7.32$ cm
TSA of the toy = CSA of hemispherical part + CSA of conical part
$= (2 \times \frac{22}{7} \times (4.2)^2) + (\frac{22}{7} \times 4.2 \times 7.32)$
$= 207.504$
Hence, TSA of toy is $207.504$ cm$^2$
345 Marks · March 2024 · Standardopen ↗
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is $14 \text{ mm}$ and the diameter of the capsule is $4 \text{ mm}$, find its surface area. Also, find its volume.
figure for this question
Show SolutionHide Solution↓
Radius of hemisphere= radius of cylinder = $2 \text{ mm}$
Length of cylindrical part = $14 - 4 = 10 \text{ mm}$.
Surface area of the capsule = CSA of cylinder + $2$(CSA of hemisphere)
$= 2\pi r h + 2(2\pi r^2) = 2 \times \frac{22}{7} \times 2 \times 10 + 2 \times 2 \times \frac{22}{7} \times 2 \times 2$
$= 176 \text{ mm}^2$
Volume of the capsule = volume of cylinder + $2$(volume of hemisphere)
$= \pi r^2 h + 2(\frac{2}{3}\pi r^3) = \frac{22}{7} \times 2 \times 2 \times 10 + 2 \times \frac{2}{3} \times \frac{22}{7} \times 2 \times 2 \times 2$
$= \frac{3344}{21} \text{ mm}^3$ or $159.24 \text{ mm}^3$
355 Marks · March 2025 · Standardopen ↗
A bat manufacturing company made a huge bat for charity and got it signed by world cup winning team. The dimensions of the bat which is in the form of a cuboid with a cylindrical handle at the top are as follows : length = $2$ m, width = $0.5$ m, thickness = $0.1$ m, diameter of cylindrical part = $0.1$ m, height of cylindrical part = $0.7$ m. Find the volume of wood used in the bat. Also, find the total surface area of the wooden bat.
Show SolutionHide Solution↓
Radius of cylindrical part = $\frac{0.1}{2}$ m or $\frac{1}{20}$ m. Volume of wood = Volume of cuboid + volume of cylinder = $2 \times 0.5 \times 0.1 + \frac{22}{7} \times \frac{0.1}{2} \times \frac{0.1}{2} \times 0.7 = \frac{211}{2000}$ or $0.1055$ m$^3$. Total surface area of bat = TSA of cuboid + CSA of cylinder = $2(2 \times 0.5 + 0.5 \times 0.1 + 0.1 \times 2) + 2 \times \frac{22}{7} \times \frac{0.1}{2} \times 0.7 = \frac{5}{2} + \frac{11}{50} = \frac{68}{25}$ or $2.72$ m$^2$.

Number of Units

1 Mark Questions
361 Mark · March 2025 · Standardopen ↗
On the top face of the wooden cube of side $7$ cm, hemispherical depressions of radius $0.35$ cm are to be formed by taking out the wood. The maximum number of depressions that can be formed is :
  • (a)$400$
  • (b)$100$
  • (c)$20$
  • (d)$10$
Show SolutionHide Solution↓
(b) $100$
5 Marks Questions
375 Marks · March 2025 · Basicopen ↗
A hemispherical bowl of internal diameter $42$ cm contains a liquid. This liquid is to be filled in cylindrical bottles of radius $3$ cm and height $8$ cm. How many bottles are required to empty the bowl ?
Show SolutionHide Solution↓
Volume of hemisphere $= \frac{2}{3} \times \pi \times 21 \times 21 \times 21$
Volume of cylinder $= \pi \times 3 \times 3 \times 8$
$\therefore$ Numbers of bottles $= \frac{\text{volume of hemisphere}}{\text{volume of cylinder}} = 85.75$
Hence, $86$ bottles are required to empty the bowl

Ratio

1 Mark Questions
381 Mark · March 2024 · Standardopen ↗
A cap is cylindrical in shape, surmounted by a conical top. If the volume of the cylindrical part is equal to that of the conical part, then the ratio of the height of the cylindrical part to the height of the conical part is :
  • (a)$1:2$
  • (b)$1:3$
  • (c)$2:1$
  • (d)$3:1$
Show SolutionHide Solution↓
(B) $1:3$
391 Mark · March 2024 · Standardopen ↗
A solid sphere is cut into two hemispheres. The ratio of the surface areas of sphere to that of two hemispheres taken together, is :
  • (a)$1:1$
  • (b)$1:4$
  • (c)$2:3$
  • (d)$3:2$
Show SolutionHide Solution↓
(c) $2:3$
401 Mark · March 2025 · Basicopen ↗
A cone and a cylinder have the same base radius and volume. The (height of cone : height of cylinder) is :
  • (a)$1 : 1$
  • (b)$3 : 1$
  • (c)$1 : 3$
  • (d)$3 : 2$
Show SolutionHide Solution↓
(B) $3:1$
411 Mark · March 2025 · Basicopen ↗
A cone and cylinder have same height and same radius. The volume of the cone and the volume of the cylinder are in the ratio :
  • (a)$1 : 1$
  • (b)$1 : 3$
  • (c)$3 : 1$
  • (d)$1 : 2$
Show SolutionHide Solution↓
(B) $1 : 3$
3 Marks Questions
423 Marks · March 2026 · Standardopen ↗
A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surface areas are in the ratio $8: 5$, then find the ratio between the radius of their bases to their height.
Show SolutionHide Solution↓
Let $r$ and $h$ be the radius and height of cylinder as well as cone respectively
$\frac{2\pi rh}{\pi r l} = \frac{8}{5}$ (I) (1 Mark)
$\Rightarrow \frac{2h}{\sqrt{r^2+h^2}} = \frac{8}{5}$
$\Rightarrow 100 h^2 = 64 r^2 + 64 h^2$
$\Rightarrow 36 h^2 = 64 r^2$ (II) (1 Mark)
$\therefore \frac{r}{h} = \frac{3}{4}$ (III) (1 Mark)
Hence the required ratio is $3: 4$
5 Marks Questions
435 Marks · March 2024 · Standardopen ↗
A solid toy is in the form of a hemisphere surmounted by a right circular cone. Ratio of the radius of the cone to its slant height is $3: 5$. If the volume of the toy is $240\pi$ cm$^3$, then find the total height of the toy.
Show SolutionHide Solution↓
Let the radius and the slant height of the cone be $3x$ cm and $5x$ cm respectively
$\therefore$ height of the cone (h) $= \sqrt{(5x)^2 - (3x)^2} = 4x$ cm
According to question, volume of toy $= 240\pi$
$\frac{2}{3}\pi(3x)^3 + \frac{1}{3}\pi(3x)^2(4x) = 240\pi$
Solving, we get $x = 2$
$\therefore$ Total height of toy $= [4(2) + 3(2)]$ cm $= 14$ cm

General

5 Marks Questions
445 Marks · March 2025 · Standardopen ↗
Calculate the mode and the median for the following distribution : Class $5-10, 10-15, 15-20, 20-25, 25-30, 30-35, 35-40, 40-45$; Frequency $5, 6, 15, 10, 5, 4, 2, 2$
figure for this question
Show SolutionHide Solution↓
Class Frequency(f) (c.f.)
$5-10$ $5$ $5$
$10-15$ $6$ $11$
$15-20$ $15$ $26$
$20-25$ $10$ $36$
$25-30$ $5$ $41$
$30-35$ $4$ $45$
$35-40$ $2$ $47$
$40-45$ $2$ $49$
Median Class is $15-20$. Median = $15 + \frac{\frac{49}{2} - 11}{15} \times 5 = \frac{39}{2}$ or $19.5$. Modal Class is $15-20$. Mode = $15 + \frac{15-6}{30-6-10} \times 5 = \frac{255}{14}$ or $18.21$
figure for this question