CBSE Class 10 Mathematics Surface Areas And Volumes Worksheet Set 02

Question. A cylindrical pencil sharpened at one edge is the combination of
(a) a cone and a cylinder
(b) frustum of a cone and a cylinder
(c) a hemisphere and a cylinder
(d) two cylinders
Answer : A

Question. A surahi is the combination of
(a) a sphere and a cylinder
(b) a hemisphere and a cylinder
(c) two hemispheres
(d) a cylinder and a cone
Answer : A

Question. In a right circular cone, the cross-section made by a plane parallel to the base is a
(a) Triangle
(b) Circle
(c) Square
(d) None of these
Answer : B

Question. If two solid hemispheres of same base radius r are joined together along their bases, then curved surface area of this new solid is
(a) 4πr²
(b) 3πr²
(c) 2πr²
(d) πr²
Answer : A

Assertion-Reason Type Questions

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion (A): The number of coins 1.75 cm in diameter and 2 mm thick is formed from a melted cuboid 10 cm × 5.5 cm × 3.5 cm is 400.
Reason (R): Volume of a cylinder = πr2h cubic units and volume of cuboid = (l × b × h) cubic units.
Answer : A

Question. Assertion (A): Number of spherical balls that can be made out of a solid cube of 1ead whose edge is 44 cm, each ball being 4 cm in diameter is 2541.
Reason (R): Number of balls = Volume of one ball/Volume of lead
Answer : C

Answer the following.

Question. What is the capacity of cylindrical vessel with the hemispherical bottom portion raised upwards?

Answer : πr²/3 [3h - 2r]

Question. A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas of two parts are equal, then what is the ratio of its radius and the slant height of the conical part?
Answer : 1 : 2
 

Case Study Based Questions

I. The Great Stupa at Sanchi is one of the oldest stone structures in India, and an important monument of Indian Architecture. It was originally commissioned by the emperor Ashoka in the 3rd century BCE. Its nucleus is a simple hemispherical brick structure built over the relics of the Buddha. It is a perfect example of combination of solid figures. A big hemispherical dome with a cuboidal structure mounted on it. (Take = 22/7)

Question. The volume of the hemispherical dome if the height of the dome is 21 m, is
(a) 19404 cu. m
(b) 2000 cu. m
(c) 15000 cu. m
(d) 19000 cu. m
Answer : A

Question. The formula to find the volume of sphere is
(a) (2/3)πr³
(b) (4/3)πr³
(c) 4 πr²
(d) 2 πr²
Answer : B

Question. 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
Answer : B

Question. The total surface area of the combined figure, i.e. hemispherical dome with radius 14 m and cuboidal shaped top with dimensions 8 m × 6 m × 4 m is
(a) 1200 sq. m
(b) 1232 sq. m
(c) 1392 sq.m
(d) 1932 sq. m
Answer : C

Question. The volume of the cuboidal shaped top with dimensions mentioned in question 4, is
(a) 182.45 m³
(b) 282.45 m³
(c) 292 m³
(d) 192 m³
Answer : D

The following questions consist of two statements—Assertion(A) and Reason(R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A) : If the volumes of two spheres are in the ratio 27:8. Then their surface areas are in the ratio 3:2.
Reason (R) : Volume of the sphere \( = \frac{4}{3} \pi r^3 \) and its surface area \( = 4 \pi r^2 \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.
Solution : We have, \( \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{27}{8} \)

\( \implies \frac{r_1^3}{r_2^3} = \frac{27}{8} \)

\( \implies \frac{r_1}{r_2} = \frac{3}{2} \)
\( \therefore \) Ratio of surface area \( = \frac{4 \pi r_1^2}{4 \pi r_2^2} = \frac{r_1^2}{r_2^2} = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \)
So, A is false but R is true.
Hence, option (d) is correct.

Question. Assertion (A) : If the radii of two cones are in the ratio 2:3 and their volumes are the ratio 1:3 then the ratio of their heights is 3:2.
Reason (R) : Volume of the cone \( = \frac{1}{3} \pi r^2 h \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.
Solution : Let radii of two cones are \( 2x \) and \( 3x \).
We have, ratio of volumes \( = \frac{\frac{1}{3} \pi (2x)^2 \times h_1}{\frac{1}{3} \pi (3x)^2 \times h_2} \)

\( \implies \frac{1}{3} = \frac{4}{9} \times \frac{h_1}{h_2} \)

\( \implies \frac{h_1}{h_2} = \frac{3}{4} \)
\( \therefore h_1 : h_2 = 3 : 4 \)
So, A is false but R is true.
Hence, option (d) is correct.

Question. Assertion (A) : If the volume of two hemispheres are in the ratio 27 : 8 then the ratio of their surface area is \( \frac{9}{4} \).
Reason (R) : The ratio of surface areas of two hemispheres is equal to the ratio of square of their respective radii.
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation for A.
Solution : Ratio of volumes of hemispheres \( = \frac{\frac{2}{3} \pi r_1^3}{\frac{2}{3} \pi r_2^3} \)

\( \implies \left( \frac{r_1}{r_2} \right)^3 = \frac{27}{8} \)

\( \implies \frac{r_1}{r_2} = \frac{3}{2} \)
Now ratio of surface areas of hemispheres \( = \frac{2 \pi r_1^2}{2 \pi r_2^2} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \)
Hence, option (a) is correct.

Question. Assertion (A) : If the height of cone is equal to radius then its volume is \( \frac{1}{4} \) of the volume of sphere.
Reason (R) : Volume of sphere \( = \frac{4}{3} \pi r^3 \) and Volume of cone \( = \frac{1}{3} \pi r^2 h \).
(a) Both A and R are true and R is the correct explanation for A.
(b) Both A and R are true but R is not the correct explanation for A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.
Solution : From both the formulae it is clear that \( \frac{4}{3} \pi r^3 = 4 \left( \frac{1}{3} \pi r^2 \times r \right) \quad [\because h = r] \)

\( \implies \text{Volume of sphere} = 4 (\text{volume of cone}) \)
So, volume of cone \( = \frac{1}{4} \) volume of sphere
Thus assertion is false.
But formulae given in reason are true.
Hence, option (d) is correct.

Please click on below link to download CBSE Class 10 Mathematics Surface Areas And Volumes Worksheet Set B