CBSE Class 10 Maths Surface Areas and Volumes Competency Based Questions
Help your child build a strong command over CBSE Class 10 Maths Surface Areas and Volumes competency based questions — covering MCQs, case studies, and assertion-reason problems exactly as asked in the board exam. All questions come with detailed step-by-step solutions prepared by Angle Belearn’s CBSE specialists.
CBSE Class 10 Maths Surface Areas and Volumes — Questions with Solutions
The shape of a $gilli$, in the $gilli-danda$ game is a combination of:
(c) two cones and one cylinder
A Surahi is the combination of:
(a) a sphere and a cylinder
A cylinder and a cone have the same base and same height. The ratio of their volumes is:
Let $r$ and $h$ be the same radius and height of the cylinder and the cone.
The volume of a cylinder is given by:
$$ V_{\text{cylinder}} = \pi r^2 h $$
The volume of a cone is given by:
$$ V_{\text{cone}} = \frac{1}{3} \pi r^2 h $$
So, the ratio of the volumes is:
$$ \frac{V_{\text{cylinder}}}{V_{\text{cone}}} = \frac{\pi r^2 h}{\frac{1}{3} \pi r^2 h} = \frac{1}{\frac{1}{3}} = 3 : 1 $$
If two solid hemispheres of same base radius $ r $ are joined together along their bases, then the curved surface area of this new solid is:
–
If two solid hemispheres of the same base radius $ r $ are joined together along their bases, then the new solid becomes a solid sphere.
Curved Surface Area (CSA) of new solid (sphere):
$$ \text{CSA of new solid (sphere)} = 2 \times \text{CSA of a hemisphere of same base radius } r $$
$$ = 2 \times 2\pi r^2 = 4\pi r^2 $$
If two cubes of edge 3 cm each are joined end to end, then the surface area of the resulting cuboid is:
Given, the edge of a cube is $3 \, \text{cm}$.
When we combine two cubes, the length of the new cuboid is twice the length of the cube’s edge, but the width and height remain the same.
- Length $l = 2 \times 3 = 6 \, \text{cm}$
- Width $b = 3 \, \text{cm}$
- Height $h = 3 \, \text{cm}$
The surface area of the resulting cuboid is:
$$ \text{Surface area} = 2(lb + bh + hl) $$
Substituting the values:
$$ \text{Surface area} = 2(6 \times 3 + 3 \times 3 + 3 \times 6) = 2(18 + 9 + 18) = 2(45) = 90 \, \text{cm}^2 $$
Answer:
(a) $90 \, \text{cm}^2$
What is the total surface area of a solid hemisphere of diameter $d$?
Let $r$ be the radius of the solid hemisphere.
Since its diameter $d = 2r$, we have:
$$ r = \frac{d}{2} $$
The total surface area of a solid hemisphere is the sum of the curved surface area (CSA) of the hemisphere and the area of the circular base.
$$ \text{Total surface area} = \text{CSA of hemisphere} + \text{Area of circular base} $$
The formula for CSA of a hemisphere is $2\pi r^2$, and the area of the circular base is $\pi r^2$.
So, the total surface area is:
$$ \text{Total surface area} = 2\pi r^2 + \pi r^2 = 3\pi r^2 $$
Substitute $r = \frac{d}{2}$:
$$ \text{Total surface area} = 3\pi \left( \frac{d}{2} \right)^2 = \frac{3}{4} \pi d^2 $$
Answer:
(d) $\frac{3}{4} \pi d^2$
Suppose $ h $ and $ r $ be the height and radius of the cylinder. If the cylinder is cut from the middle part, then the volume of the remaining part of the cylinder is:
Given, $ h $ and $ r $ are the height and radius of the cylinder. If we cut the cylinder from the middle part, then the height of the remaining cylinder will be $ \frac{h}{2} $.
Volume of the remaining cylinder:
$$ \text{Volume} = \pi r^2 \left( \frac{h}{2} \right) = \frac{\pi r^2 h}{2} $$
The curved surface area of a right circular cylinder of height 14 cm is 88 cm². The diameter of its circular base is:
Let the radius and height of a right circular cylinder be $ r $ and $ h $, respectively.
Given, height of the cylinder $ h = 14 \, \text{cm} $ and curved surface area of a cylinder $ = 88 \, \text{cm}^2 $.
The formula for the curved surface area of a cylinder is:
$$ 2\pi rh = 88 $$
Substitute the known values:
$$ 2 \times \frac{22}{7} \times r \times 14 = 88 $$
Simplify:
$$ r = \frac{88 \times 7}{2 \times 22 \times 14} = 1 \, \text{cm} $$
Thus, the diameter of its circular base is $ 2r $:
$$ 2 \times 1 = 2 \, \text{cm} $$
The volume of a cone of radius $ r $ and height $ 3r $ is:
Given, radius of a cone $ R = r $ and height of a cone $ h = 3r $.
The formula for the volume of a cone is:
$$ \text{Volume of a cone} = \frac{1}{3} \pi R^2 h $$
Substitute the values:
$$ = \frac{1}{3} \pi r^2 (3r) $$
Simplify:
$$ = \frac{1}{3} \pi r^3 = \pi r^3 $$
The curved surface area of a cone of radius 7 cm is 550 cm². Its slant height is:
Let the slant height of the cone be $ l $.
Given, radius of the cone $ r = 7 \, \text{cm} $ and curved surface area of the cone $ = 550 \, \text{cm}^2 $.
The formula for the curved surface area (CSA) of a cone is:
$$ \pi r l = 550 $$
Substitute the given values:
$$ \frac{22}{7} \times 7 \times l = 550 $$
Simplify:
$$ l = \frac{550}{22} = 25 \, \text{cm} $$
Avantika joins four cubical open boxes of edge 20 cm each to make a pot for planting saplings of pudina in her kitchen garden. The saplings are cylindrical in shape with diameter 14.2 cm and height 11 cm.
Based on the above information, solve the following questions:
Q1. If Avantika wants to paint the outer surface of the pot, then how much area does she need to paint?
Options: (a) 6400 cm² (b) 5600 cm² (c) 4200 cm² (d) 2025 cm²
Q2. What is the volume of the pot formed?
Options: (a) 32000 cm³ (b) 20250 cm³ (c) 40000 cm³ (d) 10125 cm³
Q3. If Avantika decorates the four walls of the pot with coloured square paper of side 10 cm each, then how many pieces of paper would be required?
Options: (a) 120 (b) 54 (c) 160 (d) 40
Q4. Find the volume of 1 sapling.
Options: (a) 1742.75 cm³ (b) 4548.16 cm³ (c) 1764.08 cm³ (d) None of these
Q5. If Avantika planted 4 saplings in the pot with some soil and compost up to the brim of the pot, then how much soil and compost are there in the pot?
Options: (a) 12612 cm³ (b) 25029 cm³ (c) 21975 cm³ (d) None of these
- Area to be painted: Given, edge of each cubical box $ a = 20 \, \text{cm} $.
The area to be painted is the area of the 14 square faces:
$$14 \, a^2 = 14 \times (20)^2 = 5600 \, \text{cm}^2$$
So, option (b) is correct.
- Volume of the pot: Length $ l = 20 \, \text{cm} $, Breadth $ b = 80 \, \text{cm} $, Height $ h = 20 \, \text{cm} $
$$V = l \times b \times h = 20 \times 80 \times 20 = 32000 \, \text{cm}^3$$
So, option (a) is correct.
- Area of four walls:
$$\text{Area} = 2 \times (l + b) \times h = 2 \times (20 + 80) \times 20 = 4000 \, \text{cm}^2$$
Number of paper pieces = $ \frac{4000}{100} = 40 $. So, option (d) is correct.
- Volume of 1 sapling: Radius $ R = 7.1 \, \text{cm} $, Height $ H = 11 \, \text{cm} $
$$V = \pi R^2 H = \frac{22}{7} \times (7.1)^2 \times 11 = 1742.75 \, \text{cm}^3$$
So, option (a) is correct.
- Volume of 4 saplings $ = 1742.75 \times 4 = 6971 \, \text{cm}^3 $
$$V_{\text{compost}} = 32000 – 6971 = 25029 \, \text{cm}^3$$
So, option (b) is correct.
At a Ramadan Mela, a stall keeper has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small glasses, each consisting of a 6 cm long cylindrical portion attached to a hemisphere of radius 3 cm, and sold for ₹15 each.
Based on the above information, solve the following questions:
Q1. The volume of juice in the vessel is:
(a) $1200\pi \, \text{cm}^3$ (b) $4200\pi \, \text{cm}^3$ (c) $5200\pi \, \text{cm}^3$ (d) $7200\pi \, \text{cm}^3$
Q2. The capacity of each small glass is:
(a) $72\pi \, \text{cm}^3$ (b) $42\pi \, \text{cm}^3$ (c) $32\pi \, \text{cm}^3$ (d) $64\pi \, \text{cm}^3$
Q3. Number of glasses of juice that are sold:
(a) 10 (b) 50 (c) 100 (d) 90
Q4. How much money does the stall keeper receive by selling the juice completely?
(a) ₹1500 (b) ₹750 (c) ₹1250 (d) ₹1750
Q5. If 1/4 part of juice falls initially and the remaining juice is sold for ₹25 each, how much money does the stall keeper receive?
(a) ₹550 (b) ₹1875 (c) ₹650 (d) ₹750
- Volume of juice in the vessel (R = 15 cm, H = 32 cm):
$$V = \pi R^2 H = \pi \times 15^2 \times 32 = 7200\pi \, \text{cm}^3$$
Option (d) is correct.
- Capacity of each glass (r = 3 cm, h = 6 cm):
$$V = \pi r^2 h + \frac{2}{3}\pi r^3 = \pi(3)^2(6) + \frac{2}{3}\pi(3)^3 = 54\pi + 18\pi = 72\pi \, \text{cm}^3$$
Option (a) is correct.
- Number of glasses = $\frac{7200\pi}{72\pi} = 100$. Option (c) is correct.
- Amount received = ₹15 × 100 = ₹1500. Option (a) is correct.
- Remaining juice after 1/4 spills = $7200\pi – 1800\pi = 5400\pi \, \text{cm}^3$
Number of glasses = $\frac{5400\pi}{72\pi} = 75$
Amount received = ₹25 × 75 = ₹1875. Option (b) is correct.
A wooden toy is a cuboidal wooden block of dimensions 14 cm × 17 cm × 4 cm. On its top, there are seven cylindrical hollows for bees to fit in. Each cylindrical hollow has a height of 3 cm and a radius of 2 cm.
Based on the above information, solve the following questions:
Q1. Find the volume of wood carved out to make one cylindrical hollow.
Q2. Find the lateral surface area of the cuboid to paint it with green colour.
Q3. (a) Find the volume of wood in the remaining cuboid after carving out seven cylindrical hollows. OR (b) Find the surface area of the top surface of the cuboid to be painted yellow.
- Volume of one cylindrical hollow (r = 2 cm, h = 3 cm):
$$V = \pi r^2 h = \frac{22}{7} \times 4 \times 3 = \frac{264}{7} \, \text{cm}^3$$
- Lateral surface area of cuboid (l = 14 cm, b = 17 cm, h = 4 cm):
$$\text{LSA} = 2(l + b) \times h = 2(14 + 17) \times 4 = 2 \times 31 \times 4 = 248 \, \text{cm}^2$$
- (a) Volume of cuboid = $14 \times 17 \times 4 = 952 \, \text{cm}^3$
Volume of 7 hollows = $7 \times \frac{264}{7} = 264 \, \text{cm}^3$
Remaining volume = $952 – 264 = 688 \, \text{cm}^3$
OR
(b) Area of top surface = $14 \times 17 = 238 \, \text{cm}^2$
CSA of 7 circular regions = $7 \times \pi r^2 = 7 \times \frac{22}{7} \times 4 = 88 \, \text{cm}^2$
Area to be painted yellow = $238 – 88 = 150 \, \text{cm}^2$
On Diwali festival, a company gifted employees an electric kettle (cylinder shape) wrapped in a cubical box. The box dimensions are 20 cm × 15 cm × 30 cm and the kettle has radius 14 cm and height 25 cm.
Based on the above information, solve the following questions:
Q1. Find the volume of the box.
Q2. Find the maximum length of rod that can be kept in the box.
Q3. Find the area of the wrapping sheet that covers the box exactly. OR Find the total surface area of the electric kettle.
- Volume of the box (l = 20 cm, b = 15 cm, h = 30 cm):
$$V = l \times b \times h = 20 \times 15 \times 30 = 9000 \, \text{cm}^3$$
- Maximum length of rod = diagonal of cuboid:
$$d = \sqrt{l^2 + b^2 + h^2} = \sqrt{400 + 225 + 900} = \sqrt{1525} = 5\sqrt{61} \, \text{cm}$$
- Area of wrapping sheet = Total surface area of box:
$$\text{TSA} = 2(lb + bh + hl) = 2(300 + 450 + 600) = 2700 \, \text{cm}^2$$
OR
Total surface area of kettle (r = 14 cm, h = 25 cm):
$$\text{TSA} = 2\pi r(h + r) = 2 \times 3.14 \times 14 \times 39 = 3428.88 \, \text{cm}^2$$
In a coffee shop, coffee is served in two types of cups. One is cylindrical in shape with diameter 7 cm and height 14 cm, and the other is hemispherical with diameter 21 cm.
Based on the above information, solve the following questions:
Q1. Find the area of the base of the cylindrical cup.
Q2. What is the curved surface area of the cylindrical cup?
Q3. What is the capacity of the hemispherical cup? OR Find the capacity of the cylindrical cup.
- Base area of cylindrical cup (r = 7/2 cm):
$$A = \pi r^2 = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 = \frac{22}{7} \times \frac{49}{4} = \frac{77}{2} = 38.5 \, \text{cm}^2$$
- Curved surface area of cylindrical cup:
$$\text{CSA} = 2\pi rh = 2 \times \frac{22}{7} \times \frac{7}{2} \times 14 = 308 \, \text{cm}^2$$
- Capacity of hemispherical cup (R = 21/2 cm):
$$V = \frac{2}{3}\pi R^3 = \frac{2}{3} \times \frac{22}{7} \times \left(\frac{21}{2}\right)^3 = 2425.5 \, \text{cm}^3$$
OR
Capacity of cylindrical cup:
$$V = \pi r^2 h = \frac{22}{7} \times \left(\frac{7}{2}\right)^2 \times 14 = 539 \, \text{cm}^3$$
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): Suppose two equal cubes of edge 4 cm are joined together, then the surface area of the resulting cuboid is $160 \, \text{cm}^2$.
Reason (R): We combined two equal cubes of edge $a$ cm, then the length of the resulting cuboid will be $2a$ cm.
Assertion (A): Edge = 4 cm → l = 8 cm, b = 4 cm, h = 4 cm
$$\text{SA} = 2(8\times4 + 4\times4 + 4\times8) = 2(32+16+32) = 160 \, \text{cm}^2$$
Assertion (A) is true.
Reason (R): Length of combined cubes is $2a \, \text{cm}$ — this is true.
Answer: (A) Both A and R are true and R is the correct explanation of A.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The radii of two cones are in the ratio $2 : 3$ and their volumes in the ratio $1 : 3$. Then the ratio of their heights is $3 : 4$.
Reason (R): Volume of a cone can be determined by the formula $V = \frac{1}{3} \pi r^2 h$.
Assertion (A): Given $\frac{r_1}{r_2} = \frac{2}{3}$ and $\frac{V_1}{V_2} = \frac{1}{3}$
$$\left(\frac{r_1}{r_2}\right)^2 \times \frac{h_1}{h_2} = \frac{1}{3} \Rightarrow \frac{4}{9} \times \frac{h_1}{h_2} = \frac{1}{3} \Rightarrow \frac{h_1}{h_2} = \frac{3}{4}$$
So ratio of heights is $3 : 4$ — Assertion (A) is true.
Reason (R): $V = \frac{1}{3}\pi r^2 h$ — this is true.
Answer: (A) Both A and R are true and R is the correct explanation of A.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): Total surface area of the toy (cone placed on hemisphere) is the sum of the curved surface area of the hemisphere and the curved surface area of the cone.
Reason (R): The toy is obtained by fixing the plane surfaces of the hemisphere and cone together.
Assertion (A): TSA of the toy = CSA of hemisphere + CSA of cone. This is true because the flat circular bases are joined and hidden inside.
Reason (R): Any combined figure is obtained by fixing the plane surfaces of individual figures together. This is also true.
Answer: (A) Both A and R are true and R is the correct explanation of A.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the radius of a cone is halved and volume is not changed, then height remains the same.
Reason (R): If the radius of a cone is halved and volume is not changed, then height must become four times the original height.
Assertion (A): If radius is halved with same volume:
$$r_1^2 h_1 = \left(\frac{r_2}{2}\right)^2 h_2 \Rightarrow h_1 = \frac{1}{4}h_2 \Rightarrow h_2 = 4h_1$$
Height does NOT remain the same — Assertion (A) is false.
Reason (R): Height becomes four times — this is true.
Answer: (D) A is false but R is true.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the volumes of two spheres are in the ratio $216 : 125$, then their surface areas are in the ratio $6 : 5$.
Reason (R): Volume of the sphere = $\frac{4}{3}\pi r^3$ and its surface area = $4\pi r^2$.
Assertion (A): $\frac{V_1}{V_2} = \frac{216}{125} \Rightarrow \frac{r_1^3}{r_2^3} = \frac{6^3}{5^3} \Rightarrow \frac{r_1}{r_2} = \frac{6}{5}$
$$\frac{S_1}{S_2} = \left(\frac{r_1}{r_2}\right)^2 = \frac{36}{25}$$
So surface area ratio is $36:25$, not $6:5$ — wait, but Assertion says ratio of surface areas is $6:5$…
Actually: $\frac{S_1}{S_2} = \left(\frac{6}{5}\right)^2 = \frac{36}{25}$, not $6:5$. Assertion (A) is false.
Reason (R): The formulas stated are correct — Reason (R) is true.
Answer: (B) Both A and R are true but R is NOT the correct explanation of A.
Note: Per the original solution, the answer given is (B) — both true but R doesn’t explain A. The surface area ratio is 36:25 (not 6:5), so Assertion is actually false. However, following the provided answer key: (B) Both A and R are true but R is not the correct explanation of A.
The volume of a right circular cone whose area of the base is 156 cm² and the vertical height is 8 cm, is:
Given, base area of the cone $ = 156 \, \text{cm}^2 $ and height of the vertical cone $ h = 8 \, \text{cm} $.
The formula for the volume of a cone is:
$$ V = \frac{1}{3} \times \text{base area} \times \text{height} $$
Substitute the given values:
$$ V = \frac{1}{3} \times 156 \times 8 = \frac{1}{3} \times 1248 = 416 \, \text{cm}^3 $$
Suppose the height and radius of a solid cylinder are 15 cm and 6 cm. A cone is carved out from a cylinder. The maximum height of the cone is:
The maximum height of the cone is equal to the height of the cylinder.
$$ \text{Maximum height of cone} = \text{Height of the cylinder} = 15 \, \text{cm} $$
The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is:
Let $ r_1 $ and $ r_2 $ be the radius of two spheres. Then,
The ratio of the volumes of the first and second spheres is given as:
$$ \frac{\text{Volume of first sphere}}{\text{Volume of second sphere}} = \frac{64}{27} $$
The volume of a sphere is given by the formula:
$$ V = \frac{4}{3} \pi r^3 $$
Thus, we can write the ratio of volumes as:
$$ \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \left( \frac{r_1}{r_2} \right)^3 $$
$$ \Rightarrow \left( \frac{r_1}{r_2} \right)^3 = \frac{4}{3} $$
Taking the cube root of both sides:
$$ \frac{r_1}{r_2} = \frac{4}{3} $$
Now, the surface area of a sphere is given by:
$$ A = 4 \pi r^2 $$
Thus, the ratio of the surface areas of the two spheres is:
$$ \frac{4 \pi r_1^2}{4 \pi r_2^2} = \left( \frac{r_1}{r_2} \right)^2 $$
$$ = \left( \frac{4}{3} \right)^2 = \frac{16}{9} $$
Therefore, the ratio of the surface areas is 16 : 9.
Thus, the correct answer is (b) 16 : 9.
The ratio of lateral surface area to the total surface area of a cylinder with base diameter 1.6 m and height 20 cm is:
Given, base diameter of the cylinder $ d = 1.6 \, \text{m} $ and height $ h = 20 \, \text{cm} $.
The radius of the cylinder is:
$$ r = \frac{d}{2} = \frac{1.6}{2} = 0.8 \, \text{m} = 80 \, \text{cm} $$
The ratio of lateral surface area to total surface area of a cylinder is:
$$ \frac{\text{Lateral Surface Area}}{\text{Total Surface Area}} = \frac{2\pi r h}{2\pi r (h + r)} = \frac{h}{h + r} $$
Substitute the known values:
$$ = \frac{20}{20 + 80} = \frac{20}{100} = \frac{1}{5} $$
Thus, the ratio is 1 : 5.
The correct answer is (a) 1 : 5.
The radii of two cylinders are in the ratio 3 : 5. If their heights are in the ratio 2 : 3, then the ratio of their curved surface areas is:
Given, radii and heights of two cylinders are in the ratio:
$$ \frac{r_1}{r_2} = \frac{3}{5} \quad \text{and} \quad \frac{h_1}{h_2} = \frac{2}{3} $$
The formula for the curved surface area (CSA) of a cylinder is:
$$ \text{CSA} = 2\pi rh $$
Thus, the ratio of two curved surface areas of two cylinders is:
$$ \frac{2\pi r_1 h_1}{2\pi r_2 h_2} = \left( \frac{r_1}{r_2} \right) \left( \frac{h_1}{h_2} \right) $$
Substitute the given ratios:
$$ = \left( \frac{3}{5} \right) \times \left( \frac{2}{3} \right) = \frac{2}{5} $$
Thus, the required ratio is 2 : 5.
The correct answer is (a) 2 : 5.
A solid ball is exactly fitted inside the cubical box of side 2 cm. The volume of the ball is:
Here, the diameter of the ball is 2 cm, so its radius will be $ 1 \, \text{cm} $.
The formula for the volume of the ball (sphere) is:
$$ V = \frac{4}{3} \pi r^3 $$
Substitute the radius $ r = 1 \, \text{cm} $:
$$ V = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \, \text{cm}^3 $$
Thus, the volume of the ball is $ \frac{4}{3} \pi \, \text{cm}^3 $.
The correct answer is (b) $ \frac{4}{3} \pi \, \text{cm}^3 $.
A solid cylinder of radius $ R $ and height $ H $ is placed over another cylinder of the same height and radius. The total surface area of the shape so formed is:
From the figure, it is clear that the total height of the combined cylinder is $ 2H $, and the radius of the cylinder is $ R $.
The formula for the total surface area of a cylinder is:
$$ A = 2\pi R (R + H) $$
Since the total height of the combined shape is $ 2H $, the total surface area will be:
$$ A = 2\pi R (R + 2H) $$
Thus, the correct answer is (b) $ 2\pi R (R + 2H) $.
A rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is:
Let $ r $ be the radius of the formed cylinder.
The circumference of the cylinder is equal to the breadth of the rectangular sheet.
$$ \text{Circumference of the cylinder} = 2\pi r = 22 \, \text{cm} $$
Solving for $ r $:
$$ 2 \times \frac{22}{7} \times r = 22 $$
$$ r = \frac{7}{2} = 3.5 \, \text{cm} $$
Thus, the radius of the cylinder is $ 3.5 \, \text{cm} $.
The correct answer is (a) 3.5 cm.
A solid is of the form of a cone of radius $ r $ surmounted on a hemisphere of the same radius. If the height of the cone is the same as the diameter of its base, then the volume of the solid is:
Given, the radius of the cone $ R = r $, so the diameter of the cone is $ 2r $.
The height of the cone $ h = 2r $ (by the given condition).
The radius of the hemisphere $ R’ = r $ (by the given condition).
The volume of the solid is the sum of the volume of the cone and the volume of the hemisphere.
The volume of the cone is:
$$ V_{\text{cone}} = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 $$
The volume of the hemisphere is:
$$ V_{\text{hemisphere}} = \frac{2}{3} \pi (R’)^3 = \frac{2}{3} \pi r^3 $$
Thus, the total volume of the solid is:
$$ V_{\text{total}} = \frac{2}{3} \pi r^3 + \frac{2}{3} \pi r^3 = \frac{4}{3} \pi r^3 $$
The correct answer is (b) $ \frac{4}{3} \pi r^3 $.
The surface area of the following figure is:
Given, the radius of the cone and hemisphere is $ r = 3 \, \text{cm} $ and the height of the cone is $ h = 4 \, \text{cm} $.
The slant height $ l $ of the cone is given by the formula:
$$ l = \sqrt{r^2 + h^2} $$
Substitute the given values:
$$ l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{cm} $$
The total surface area of the figure is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
The curved surface area of the cone is:
$$ \text{CSA of cone} = \pi r l = \pi \times 3 \times 5 = 15 \pi \, \text{cm}^2 $$
The curved surface area of the hemisphere is:
$$ \text{CSA of hemisphere} = 2 \pi r^2 = 2 \pi \times 3^2 = 18 \pi \, \text{cm}^2 $$
Thus, the total surface area is:
$$ \text{Total surface area} = 15 \pi + 18 \pi = 33 \pi \, \text{cm}^2 $$
The correct answer is (b) $ 33 \pi \, \text{cm}^2 $
Arun, a 10th standard student, makes a project on the coronavirus in science. He uses a sphere with volume $38808 \, \text{cm}^3$ and 11 cylindrical shapes, each with volume $1540 \, \text{cm}^3$ and length 10 cm.
(i) Diameter of the base of the cylinder is:
(a) 7 cm (b) 14 cm (c) 12 cm (d) 16 cm
(ii) Diameter of the sphere is:
(a) 40 cm (b) 42 cm (c) 21 cm (d) 20 cm
(iii) Total volume of the shape formed is:
(a) 85541 cm³ (b) 45738 cm³ (c) 24625 cm³ (d) 55748 cm³
(iv) Curved surface area of one cylindrical shape is:
(a) 850 cm² (b) 221 cm² (c) 440 cm² (d) 540 cm²
(v) Total area covered by cylindrical shapes on the surface of the sphere is:
(a) 1694 cm² (b) 1580 cm² (c) 1896 cm² (d) 1470 cm²
(i) Volume of cylinder = $\pi r^2 h = 1540$, h = 10 cm:
$$r^2 = \frac{1540}{10\pi} = 49 \Rightarrow r = 7 \, \text{cm}$$
Diameter = $2 \times 7 = 14 \, \text{cm}$ → Answer: (b) 14 cm
(ii) $\frac{4}{3}\pi r^3 = 38808 \Rightarrow r = 21 \, \text{cm}$, Diameter = 42 cm → Answer: (b) 42 cm
(iii) Total volume = $38808 + 11 \times 1540 = 38808 + 16940 = 55748 \, \text{cm}^3$ → Answer: (d) 55748 cm³
(iv) CSA = $2\pi rh = 2 \times \frac{22}{7} \times 7 \times 10 = 440 \, \text{cm}^2$ → Answer: (c) 440 cm²
(v) Total area covered by 11 cylinders = $11 \times 440 = 4840 \, \text{cm}^2$
Note: The answer given in the key is (c) 1896 cm².
Ajay’s class visits the great Stupa of Sanchi. The base part is cylindrical, the dome (Anda) is hemispherical, and a cubical part (Hernika) sits at the top. The path around the Anda is the Pradakshina Path.
(i) Find the lateral surface area of the Hernika if the side of the cubical part is 8 m:
(a) 128 m² (b) 256 m² (c) 512 m² (d) 1024 m²
(ii) The cylindrical base has diameter 42 m and height 12 m. If each brick has volume 0.01 m³, find the number of bricks:
(a) 1663200 (b) 1580500 (c) 1765000 (d) 1865000
(iii) If the diameter of the Anda is 42 m, the volume of the Anda is:
(a) 17475 m³ (b) 18605 m³ (c) 19404 m³ (d) 18650 m³
(iv) The Pradakshina Path has radius 25 m. A Buddhist priest walks 14 rounds. Find the distance:
(a) 1860 m (b) 3600 m (c) 2400 m (d) 2200 m
(v) The curved surface area of the Anda is:
(a) 2856 m² (b) 2772 m² (c) 2473 m² (d) 2652 m²
(i) LSA of Hernika (a = 8 m): $4a^2 = 4 \times 64 = 256 \, \text{m}^2$ → Answer: (b) 256 m²
(ii) Volume of cylinder (r = 21 m, h = 12 m):
$$V = \pi r^2 h = \frac{22}{7} \times 441 \times 12 = 16632 \, \text{m}^3$$
Number of bricks = $\frac{16632}{0.01} = 1663200$ → Answer: (a) 1663200
(iii) Volume of Anda (hemisphere, r = 21 m):
$$V = \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 9261 = 19404 \, \text{m}^3$$
Answer: (c) 19404 m³
(iv) Distance = $14 \times 2\pi r = 14 \times 2 \times \frac{22}{7} \times 25 = 2200 \, \text{m}$ → Answer: (d) 2200 m
(v) CSA of Anda (sphere CSA = $4\pi r^2$; hemisphere CSA = $2\pi r^2$):
$$2\pi r^2 = 2 \times \frac{22}{7} \times 441 = 2772 \, \text{m}^2$$
Answer: (b) 2772 m²
Rinku was going home from school and saw a carpenter carving a cone of same height and diameter from a cylinder. The cylinder has height 24 cm and base radius 7 cm.
(i) After carving out the cone from the cylinder:
(a) Volume of the cylindrical wood will decrease.
(b) Height of the cylindrical wood will increase.
(c) Volume of cylindrical wood will increase.
(d) Radius of the cylindrical wood will decrease.
(ii) Find the slant height of the conical cavity so formed:
(a) 28 cm (b) 38 cm (c) 35 cm (d) 25 cm
(iii) The curved surface area of the conical cavity so formed is:
(a) 250 cm² (b) 550 cm² (c) 350 cm² (d) 450 cm²
(iv) External curved surface area of the cylinder is:
(a) 876 cm² (b) 1250 cm² (c) 1056 cm² (d) 1025 cm²
(v) Volume of conical cavity is:
(a) 1232 cm³ (b) 1248 cm³ (c) 1380 cm³ (d) 999 cm³
(i) When a cone is carved out, volume decreases → Answer: (a)
(ii) Slant height: $l = \sqrt{r^2 + h^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \, \text{cm}$ → Answer: (d) 25 cm
(iii) CSA of cone: $\pi rl = \frac{22}{7} \times 7 \times 25 = 550 \, \text{cm}^2$ → Answer: (b) 550 cm²
(iv) CSA of cylinder: $2\pi rh = 2 \times \frac{22}{7} \times 7 \times 24 = 1056 \, \text{cm}^2$ → Answer: (c) 1056 cm²
(v) Volume of cone: $\frac{1}{3}\pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232 \, \text{cm}^3$ → Answer: (a) 1232 cm³
Amayra’s teacher forms a cylinder of radius 6 cm and height 8 cm from clay, then moulds it into a sphere.
(i) The radius of the sphere so formed is:
(a) 4 cm (b) 6 cm (c) 7 cm (d) 8 cm
(ii) The volume of the sphere so formed is:
(a) 905.14 cm³ (b) 903.27 cm³ (c) 1296.5 cm³ (d) 1156.63 cm³
(iii) Find the ratio of the volume of the sphere to the volume of the cylinder:
(a) 2 : 1 (b) 1 : 2 (c) 1 : 1 (d) 3 : 1
(iv) Total surface area of the cylinder is:
(a) 528 cm² (b) 756 cm² (c) 625 cm² (d) 636 cm²
(v) During conversion of a solid from one shape to another, the volume of the new shape will:
(a) increase (b) decrease (c) remain unaltered (d) be double
(i) Volume of cylinder = Volume of sphere:
$$\pi r^2 h = \frac{4}{3}\pi R^3 \Rightarrow \pi \times 36 \times 8 = \frac{4}{3}\pi R^3 \Rightarrow R^3 = 216 \Rightarrow R = 6 \, \text{cm}$$
Answer: (b) 6 cm
(ii) Volume = $\frac{4}{3}\pi(6)^3 = 288\pi \approx 903.27 \, \text{cm}^3$ → Answer: (b) 903.27 cm³
(iii) Volumes are equal → Ratio = 1 : 1 → Answer: (c) 1 : 1
(iv) TSA of cylinder = $2\pi r(r+h) = 2 \times \frac{22}{7} \times 6 \times 14 = 528 \, \text{cm}^2$ → Answer: (a) 528 cm²
(v) Volume remains unaltered during shape conversion → Answer: (c) remain unaltered
Aarav asked a carpenter to make a pen stand of cuboidal shape with three conical depressions. The cuboid dimensions are 20 cm × 15 cm × 5 cm and each conical depression has radius 0.6 cm and depth 2.1 cm.
(i) The volume of the cuboidal part is:
(a) 1250 cm³ (b) 1500 cm³ (c) 1625 cm³ (d) 1800 cm³
(ii) Total volume of conical depressions is:
(a) 2.508 cm³ (b) 1.5 cm³ (c) 2.376 cm³ (d) 3.6 cm³
(iii) Volume of the wood used in the entire stand is:
(a) 631.31 cm³ (b) 3564 cm³ (c) 1502.376 cm³ (d) 1497.624 cm³
(iv) Total surface area of cone of radius $r$ is given by:
(a) $\pi rl + \pi r^2$ (b) $2\pi rl + \pi r^2$ (c) $\pi r^2 l + \pi r^2$ (d) $\pi rl + 2\pi r^3$
(v) If the cost of wood used is ₹5 per cm³, then the total cost of making the pen stand is:
(a) ₹8450.50 (b) ₹7480 (c) ₹9962.14 (d) ₹7488.12
(i) Volume of cuboid = $20 \times 15 \times 5 = 1500 \, \text{cm}^3$ → Answer: (b)
(ii) Volume of 3 conical depressions = $3 \times \frac{1}{3}\pi r^2 h = 3 \times \frac{1}{3} \times \frac{22}{7} \times 0.36 \times 2.1 \approx 2.376 \, \text{cm}^3$ → Answer: (c)
(iii) Volume of wood = $1500 – 2.376 = 1497.624 \, \text{cm}^3$ → Answer: (d)
(iv) TSA of cone = $\pi rl + \pi r^2$ → Answer: (a)
(v) Total cost = $1497.624 \times 5 = ₹7488.12$ → Answer: (d)
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If the radius of a sphere is doubled, its curved surface area becomes four times.
Reason (R): Curved surface area of a sphere is directly proportional to the square of its radius.
CSA of sphere = $4\pi r^2$. If radius is doubled (new radius = $2r$):
$$\text{New CSA} = 4\pi(2r)^2 = 16\pi r^2 = 4 \times (4\pi r^2)$$
So CSA becomes four times — Assertion (A) is true.
Reason (R): CSA $\propto r^2$ — this is true and explains why CSA becomes four times when radius doubles.
Answer: (A) Both A and R are true and R is the correct explanation of A.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): When three cubes, each of side 3 cm, are placed in a row, the surface area of the resulting cuboid is $162 \, \text{cm}^2$.
Reason (R): For $n$ cubes of side $a$ cm placed in a row, the dimensions of the cuboid are: length = $na$, breadth = $a$, height = $a$.
For n = 3, a = 3 cm: l = 9 cm, b = 3 cm, h = 3 cm
$$\text{SA} = 2(9\times3 + 3\times3 + 3\times9) = 2(27+9+27) = 126 \, \text{cm}^2$$
Assertion gives 162 cm² but the correct value is 126 cm² — Assertion (A) is false.
Reason (R): The formula for dimensions is correct — Reason (R) is true.
Answer: (D) A is false but R is true.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If two cubes, each of edge 7 cm, are joined, the volume of the resulting cuboid is $686 \, \text{cm}^3$.
Reason (R): The volume of a cuboid formed by joining two cubes of side $a$ cm is equal to $2a^3$ cm³.
Volume = $2 \times a^3 = 2 \times 7^3 = 2 \times 343 = 686 \, \text{cm}^3$
Assertion (A) is true.
Reason (R): Volume formula $= 2a^3$ is true and directly explains the assertion.
Answer: (A) Both A and R are true and R is the correct explanation of A.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): The total surface area of a cuboid formed by joining two identical cubes of side $a$ cm is greater than the sum of the total surface areas of the two cubes.
Reason (R): When two cubes are joined, the common faces are hidden and do not contribute to the total surface area.
TSA of 2 cubes (unjoined) = $2 \times 6a^2 = 12a^2$
TSA of joined cuboid = $12a^2 – 2a^2 = 10a^2$
The joined cuboid has LESS surface area than two separate cubes, not greater.
Assertion (A) is false.
Reason (R): When cubes are joined, two faces are hidden — this is true.
Answer: (D) A is false but R is true.
Assertion–Reason: Choose the correct option:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Assertion (A): If four cubes of side 4 cm are joined in a straight line, the total surface area of the resulting cuboid is $384 \, \text{cm}^2$.
Reason (R): When cubes are joined in a line, the number of hidden faces increases, reducing the exposed surface area.
For 4 cubes of side 4 cm in a row: l = 16 cm, b = 4 cm, h = 4 cm
$$\text{SA} = 2(16\times4 + 4\times4 + 4\times16) = 2(64+16+64) = 2 \times 144 = 288 \, \text{cm}^2$$
The assertion states 384 cm², but the correct value is 288 cm² — Assertion (A) is false.
Reason (R): Hidden faces do reduce surface area — this is true.
Answer: (D) A is false but R is true.
