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Class 10 Exam  >  Class 10 Tests  >  Mathematics (Maths) Class 10  >  RS Aggarwal Test: Surface Areas & Volumes - Class 10 MCQ

RS Aggarwal Test: Surface Areas & Volumes - Class 10 MCQ


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10 Questions MCQ Test Mathematics (Maths) Class 10 - RS Aggarwal Test: Surface Areas & Volumes

RS Aggarwal Test: Surface Areas & Volumes for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The RS Aggarwal Test: Surface Areas & Volumes questions and answers have been prepared according to the Class 10 exam syllabus.The RS Aggarwal Test: Surface Areas & Volumes MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RS Aggarwal Test: Surface Areas & Volumes below.
Solutions of RS Aggarwal Test: Surface Areas & Volumes questions in English are available as part of our Mathematics (Maths) Class 10 for Class 10 & RS Aggarwal Test: Surface Areas & Volumes solutions in Hindi for Mathematics (Maths) Class 10 course. Download more important topics, notes, lectures and mock test series for Class 10 Exam by signing up for free. Attempt RS Aggarwal Test: Surface Areas & Volumes | 10 questions in 10 minutes | Mock test for Class 10 preparation | Free important questions MCQ to study Mathematics (Maths) Class 10 for Class 10 Exam | Download free PDF with solutions
RS Aggarwal Test: Surface Areas & Volumes - Question 1
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If a marble of radius 2.1 cm is put into a cylindrical cup full of water of radius 5 cm and height 6 cm, then how much water flows out of the cylindricall cup?

Detailed Solution for RS Aggarwal Test: Surface Areas & Volumes - Question 1

Given, a marble of radius 2.1 cm is put into a cylindrical cup full of water of radius 5 cm and height 6 cm.

We have to find the quantity of water that flows out of the cylindrical cup.

We know that when the marble is put inside the cylindrical cup full of water the volume of water displaced is equal to the volume of the marble.

So, volume of water that flows out = volume of the marble

Volume of sphere = (4/3)πr3

Given, radius = 2.1 cm

So, volume of marble = (4/3)(22/7)(2.1)3

= 38.808 cm³

Therefore, the volume of the water that flows out is 38.8 cm³

RS Aggarwal Test: Surface Areas & Volumes - Question 2
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If we change the shape of an object from a sphere to a cylinder, then the volume of cylinder will

Detailed Solution for RS Aggarwal Test: Surface Areas & Volumes - Question 2

Answer: (c) Remains unchanged

Explanation: If we change the shape of a three-dimensional object, the volume of the new shape will be same.

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RS Aggarwal Test: Surface Areas & Volumes - Question 3
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The lateral surface area of a cone with radius 6 cm and slant height 10 cm is:

Detailed Solution for RS Aggarwal Test: Surface Areas & Volumes - Question 3

Answer: b) 60 π cm²
Lateral Surface Area = πrl
Radius r = 6 cm
Slant height l = 10 cm
LSA = π × 6 × 10 = 60π cm²

RS Aggarwal Test: Surface Areas & Volumes - Question 4
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The volume of a cone is 100π cm³ and its height is 10 cm. What is its radius?
Detailed Solution for RS Aggarwal Test: Surface Areas & Volumes - Question 4

RS Aggarwal Test: Surface Areas & Volumes - Question 5
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A decorative fountain is designed in the shape of a cylinder with a hemispherical bowl on top. The height of the cylinder is 2 m, and the radius of both the cylinder and the hemisphere is 0.5 m. Find the total surface area of the fountain. (Use π=22/7)
Detailed Solution for RS Aggarwal Test: Surface Areas & Volumes - Question 5

RS Aggarwal Test: Surface Areas & Volumes - Question 6
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A cylinder and a cone are of the same base radius and same height. Find the ratio of the volumes of the cylinder to that of the cone.
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RS Aggarwal Test: Surface Areas & Volumes - Question 7
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A mason constructs a wall of dimensions 270 cm x 300 cm x 350 cm with the bricks each of size 22.5 cm x 11.25 cm x 8.75 cm and it is assumed that 1/8 space is covered by the mortar.
Then the number of bricks used to construct the wall is
Detailed Solution for RS Aggarwal Test: Surface Areas & Volumes - Question 7

To find the number of bricks used to construct the wall, we will follow these steps:

Step 1: Calculate the Volume of the Wall
The volume V of a cuboid is given by the formula:
V=Length×Breadth×Height
For the wall, the dimensions are:
- Length = 270 cm
- Breadth = 300 cm
- Height = 350 cm

Calculating the volume:
Vwall=270cm×300cm×350cm
Vwall=28350000cm3

Step 2: Adjust for Mortar
Since 18 of the volume is covered by mortar, the volume occupied by bricks is:
Vbricks=Vwall×78
Calculating the volume occupied by bricks:
Vbricks=28350000×78=24806250cm3

 

Step 3: Calculate the Volume of One Brick
The dimensions of one brick are:
- Length = 22.5 cm
- Breadth = 11.25 cm
- Height = 8.75 cm

Using the volume formula for the brick:
Vbrick=22.5cm×11.25cm×8.75cm
Calculating the volume of one brick:
Vbrick=22.5×11.25×8.75=2214.84375cm3

 

Step 4: Calculate the Number of Bricks
To find the number of bricks used, divide the volume occupied by bricks by the volume of one brick:
N=VbricksVbrick
Calculating the number of bricks:
N=248062502214.84375≈11200

Final Answer
The number of bricks used to construct the wall is 11200.

RS Aggarwal Test: Surface Areas & Volumes - Question 8
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 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 entire capsule is 2 cm. The capacity of the capsule is
Detailed Solution for RS Aggarwal Test: Surface Areas & Volumes - Question 8

RS Aggarwal Test: Surface Areas & Volumes - Question 9
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Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is:
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RS Aggarwal Test: Surface Areas & Volumes - Question 10
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Two identical solid cubes of side a are joined end to end. Then the total surface area of the resulting cuboid is
Detailed Solution for RS Aggarwal Test: Surface Areas & Volumes - Question 10

Answer: (b) 10a2

Explanation:

The total surface area of a cube having side a = 6a2

If two identical faces of side a are joined together, then the total surface area of the cuboid so formed is 10a2.

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