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Class 10 Exam  >  Class 10 Tests  >  RD Sharma Test: Surface Area & Volumes - Class 10 MCQ

RD Sharma Test: Surface Area & Volumes - Class 10 MCQ


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25 Questions MCQ Test - RD Sharma Test: Surface Area & Volumes

RD Sharma Test: Surface Area & Volumes for Class 10 2024 is part of Class 10 preparation. The RD Sharma Test: Surface Area & Volumes questions and answers have been prepared according to the Class 10 exam syllabus.The RD Sharma Test: Surface Area & Volumes MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RD Sharma Test: Surface Area & Volumes below.
Solutions of RD Sharma Test: Surface Area & Volumes questions in English are available as part of our course for Class 10 & RD Sharma Test: Surface Area & Volumes solutions in Hindi for Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt RD Sharma Test: Surface Area & Volumes | 25 questions in 25 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study for Class 10 Exam | Download free PDF with solutions
RD Sharma Test: Surface Area & Volumes - Question 1
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The cost of painting a cubical box of side 3m at the rate of Rs.2 per sq.m is

Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 1

Given: Side of the cube (a) = 3 m
∴ Surface Area of Cube = 6a2 = 6 x 3 x 3 = 54 sq. cm
Now, Cost of painting the cubical box of 1 sq. m = Rs. 2 .
∴ Cost of painting the cubical box of 54 sq. m = 54 x 2 = Rs. 108 

RD Sharma Test: Surface Area & Volumes - Question 2
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The volume of the cuboid whose length, breadth and height is 12cm, 8cm and 6cm is

Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 2

Volume of cuboid = Length x Breadth x Height
⇒ V = 12 x 8 x 6 = 576 cu. cm

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RD Sharma Test: Surface Area & Volumes - Question 3
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The base area of the cylinder is 80 sq.cm. If its height is 5cm, then its volume is

Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 3

Base area of the cylinder = 80s.q cm

Now, Volume of the cylinder = 

RD Sharma Test: Surface Area & Volumes - Question 4
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A cubical block of side 7 cm is surmounted by a hemisphere. The greatest diameter of the hemisphere is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 4


It is clear that Maximum diameter of hemisphere can be the side of the cube. 
∴ Greatest diameter of the hemisphere = 7 cm

RD Sharma Test: Surface Area & Volumes - Question 5
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The plural form of ‘frustum’ is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 5

The plural form of ‘frustum’ is frusta.

RD Sharma Test: Surface Area & Volumes - Question 6
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A solid cylinder of radius ‘r’ and height ‘h’ is placed over other cylinder of same height and radius. The total surface area of the shape so formed is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 6


Here,
TSA of new shape = 2πrh + 2πrh+ πr2  
= 4πrh + 2πr2 

RD Sharma Test: Surface Area & Volumes - Question 7
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The edge of the cube whose volume is 1728cm is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 7

Given : Volume of cube = 1728 cm

RD Sharma Test: Surface Area & Volumes - Question 8
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The radius and height of a right circular cone and that of a right circular cylinder are respectively equal. If the volume of the cylinder is 300 cu.cm, then the volume of the cone is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 8

Let Radius and height of the cone be r and h respectively and Radius and height of the right circular cylinder be R and H.

RD Sharma Test: Surface Area & Volumes - Question 9
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A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1cm and the height of the cone is equal to its radius. The volume of the solid is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 9

Radii of cone and hemisphere = r = 3 cm
Height of cone (h) = 1 cm
Volume of solid = Volume of cone + Volume of hemisphere 

RD Sharma Test: Surface Area & Volumes - Question 10
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The length of the diagonal of a cuboid of dimensions 5cm by 4cm by 3cm is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 10

Length of the diagonal of cuboid 

RD Sharma Test: Surface Area & Volumes - Question 11
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If two solid hemispheres of same base radius ‘x’ cm are joined together along their bases, then the CSA of the new solid formed is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 11

If two solid hemispheres of same base radius ‘x’ cm are joined together along their bases, then the CSA of the new solid formed is 4πr2 = 4πx2.

RD Sharma Test: Surface Area & Volumes - Question 12
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The volumes of two spheres are in the ratio 125 : 64. The ratio of their surface areas is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 12

Let r1 and r2 be the radius of the two spheres respectively. Therefore, the ratio of their surface areas, 

RD Sharma Test: Surface Area & Volumes - Question 13
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If the surface area of a sphere is 36π sq. cm, then its volume is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 13

Given: Surface area of sphere = 36πr sq. cm

∴ Volume of sphere = 

RD Sharma Test: Surface Area & Volumes - Question 14
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A frustum of a cone is of height 12cm with radii of its circular ends as 2cm and 4cm. The volume of the frustum is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 14

RD Sharma Test: Surface Area & Volumes - Question 15
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The longest rod that can be placed inside the cube is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 15

The longest rod that can be placed inside the cube is √3 times the edge of the cube, i.e., √3 edge

RD Sharma Test: Surface Area & Volumes - Question 16
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If the surface area of the sphere is same as the CSA of a right circular cylinder whose height and diameter are 12cm each, then the radius of the sphere is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 16

Let the radius of solid sphere be rr cm and radius of solid cylinder be R cm. Then according to question,

RD Sharma Test: Surface Area & Volumes - Question 17
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If the volume of a cube is 343 cm , then its edge is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 17

RD Sharma Test: Surface Area & Volumes - Question 18
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The inner dimensions of a closed box are 12cm, 10cm and 8cm. If the thickness of the wood is 1cm, then the capacity f the box is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 18

Given: l = 12 cm, b = 10 cm and h = 8 cm
∴ Capacity of a closed box = lbh = 12 x 10 x 8 = 960 cm. cm
Capacity of box of thickness 1 cm = 960/1 = 960 cu. cm 

RD Sharma Test: Surface Area & Volumes - Question 19
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A funnel is the combination of
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 19


A funnel is the combination of a frustum of a cone and a cylinder.

RD Sharma Test: Surface Area & Volumes - Question 20
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A plumbline is combination of
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 20

A plumbline is combination of a hemisphere and a cone

RD Sharma Test: Surface Area & Volumes - Question 21
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A rectangular piece of paper is 44cm long and 18cm wide. If a cylinder is formed by rolling the paper along its length, then the radius of the base of the cylinder is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 21


Let r be the radius of the cylinder. Given: Circumference of cylinder = 44 cm 

RD Sharma Test: Surface Area & Volumes - Question 22
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The surface areas of two spheres are in the ratio 16 : 9. The ratio of their volumes is
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 22

Let r1 and r2 be the radius of the two spheres respectively. Therefore. the ratio of their volumes.

Now, ratio of thier volumes,

RD Sharma Test: Surface Area & Volumes - Question 23
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During conversion of a solid from one shape to another, the volume of the new shape will
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 23

During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

RD Sharma Test: Surface Area & Volumes - Question 24
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The shape of a glass tumbler is usually in the form of
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 24


The shape of a glass tumbler is usually in the form of frustum of a cone.

RD Sharma Test: Surface Area & Volumes - Question 25
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A cylindrical pencil sharpened at one edge is the combination of
Detailed Solution for RD Sharma Test: Surface Area & Volumes - Question 25

A cylindrical pencil sharpened at one edge is the combination of a cone and a cylinder.

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