Mechanics (Physics) (Part-4)

Total Questions: 38

1. With reference to gravity, what is G called? [S.S.C. Online CHSL (T-I) 23.01.2017 (Shift-III)]

Correct Answer: (a) Gravitational constant
Solution:With reference to gravity, G is called the gravitational constant, also known as the universal gravitational constant.

G=6.67430×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²

F=Gm₁m₂/r²Where:

F is the gravitational force between two masses,
m and m₂ are the two masses.
r is the distance between their centers, and G is the gravitational constant.

2. Weight of a person at the height, of 2R from the centre of the earth, where R is the radius of the earth________. [S.S.C. Online CHSL (T-I) 18.01.2017 (Shift-II)]

Correct Answer: (d) becomes one-fourth
Solution:

At the height of R—
W = mg
where g (R) = Gm / R² ......(i)
At the height of 2R—
g(2R) = Gm / (2R)² = Gm / 4R² ......(ii)
From Equations (i) and (ii)
g(R) / g(2R) = Gm/R² / Gm/4R²
g(R) / g(2R) = 4
g(2R) = 1/4 g(R)
W = This will also reduce the weight to one-fourth of its original value.

3. The value of acceleration due to gravity (g) at a distance of 2R from the surface of earth, where R is the radius of earth is________. [S.S.C. Online CHSL (T-I) 18.01.2017 (Shift-I)]

Correct Answer: (b) g/9
Solution:

To find the acceleration due to gravity (g') at a distance of 2R from the surface of the Earth (where R is Earth's radius), we follow this process:

Total distance from the center of Earth
= Radius of Earth + 2R
= R + 2R = 3R
Formula:
g′=g/(r/R)²Where:
g = acceleration due to gravity on Earth's surface
r = distance from the center of Earth = 3R
g′=g/(3)²=g/9
So, the value of acceleration due to gravity at a distance 2R from the surface is g/9.

4. If the distance between two objects increases three-fold, then the gravitational force of attraction between then________. [S.S.C. JE Mechanical Exam 22.03.2021 (Shift-II)]

Correct Answer: (d) Becomes one-ninth of the original
Solution:Let the initial distance between the two objects be Now, if the distance between the two objects increases threefold, the new distance is .
The new gravitational force of attraction between them, , is:

5. If the radius of the earth decreases and its mass remains the same, then the value of "acceleration due to gravity" will_______. [S.S.C. Online CHSL (T-I) 23.01.2017 (Shift-I)]

Correct Answer: (b) increase
Solution:

The new acceleration due to gravity, g', is:

6. As per Newton's Law of Gravitation, the force between two bodies is ________ . [S.S.C. Online CHSL (T-I) 22.01.2017 (Shift-I)]

Correct Answer: (a) directly proportional to the product of their masses.
Solution:As per Newton's Law of Gravitation, the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Mathematically, it's given by:
F=Gm₁m₂/r²
Where:

F is the gravitational force between the two bodies,
G is the gravitational constant (6.674×10⁻¹¹ Nm²/kg²)
m1 and m2 are the masses of the two bodies,
r is the distance between the centers of the two masses.

7. Who first determined the value of G (gravitational constant)? [S.S.C. Online CHSL (T-I) 30.01.2017 (Shift-II)]

Correct Answer: (a) Lord Cavendish
Solution:G is called the universal gravitational constant.

Its value is 6.67 x 10⁻¹¹ Newton m²/kg.
Lord Cavendish first found its value in 1798.
The value of G is the same in the whole universe.

8. What is the value of acceleration due to gravity at the centre of the earth? [S.S.C. Online CHSL (T-I) 24.01.2017 (Shift-II)]

Correct Answer: (b) 0
Solution:At the center of the Earth, the value of acceleration due to gravity is zero.
Here's why:
Gravity inside the Earth decreases as you go deeper.
According to the shell theorem, when you're inside a spherical mass like the Earth, only the mass at a smaller radius than your position contributes to the gravitational pull. At the very center, you're surrounded equally in all directions, and all the gravitational forces cancel out.
So:
g(center)=0 m/s²
No net gravitational force acts on a mass placed at the Earth's center—it's in perfect gravitational equilibrium.

9. _______ is the process of allowing particles in suspension in water to settle out of the suspension under the effect of gravity. [S.S.C. Online CHSL (T-I) 16.04.2021 (Shift-I)]

Correct Answer: (d) Sedimentation
Solution:

Sedimentation is the process of allowing particles in suspension in water to settle out of the suspension under the effect of gravity.

This is a common method used in water treatment to remove suspended solids before further purification steps like filtration or disinfection.

10. The force which makes a vehicle to stop when brake is applied is called _____ . [S.S.C. Online Graduation Level (T-I) 7.09.2016 (Shift-III)]

Correct Answer: (c) Frictional force
Solution:Friction is a force that opposes the relative tangential motion between two planes.

Therefore, the force that stops the vehicle when brakes are applied is called frictional force.