Gravitation Test

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Gravitation Test - 9

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Question 1
1 / -0

The gravitational force with which the earth attracts the moon

A
is less than the force with which the moon attracts the earth

B
is equal to the force with which the moon attracts the earth

C
is greater than the force with which the moon attracts the earth

D
varies with the phases of the moon

Solution

Every force has an equal and opposite reaction. The gravitational force between any two objects is an attractive force, which pulls both objects closer to each other, which always acts on both the objects with the same magnitude. So, the force with which the earth attracts the moon towards itself, the moon attracts the earth towards itself with the same force.
Question 2
1 / -0

The value of gravitational constant in CGS system is 6.67 10^–8. Its value in MKS system will be

A
6.67 10^–5

B
6.67 10^–7

C
6.67 10^–9

D
6.67 10^–11

Solution
Question 3
1 / -0

The tidal waves in the sea are primarily due to

A
gravitational effect of the moon on the earth

B
gravitational effect of the sun on the earth

C
gravitational effect of the sun and moon on the earth

D
gravitational effect of the earth on the sun

Solution

Moon is the heavenly body which is closest to the earth and the distance between earth and sun is 400 times more than the distance between earth and moon. That is why the tides are mainly caused by the gravitational effect of the moon on the earth.
Question 4
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If different planets have the same density, but different radii, then the acceleration due to gravity (g) on the surface of the planet will depend on its radius (R) as

A
g

B
g

C
g R

D
g R²

Solution

M = d x
g =
g R
Question 5
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The acceleration due to gravity at Earth is 'g', and the Earth suddenly shrinks uniformly to half its present size without losing any mass. The value of 'g' at the same point (assuming that the distance of the point from the centre of the Earth does not change) will now be

A
g

B
g/20

C
g/5

D
320 g

Solution
Question 6
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The value of g will be 1% of its value on the surface of Earth at a height of (R_e = 6400 km)

A
6400 km

B
57,600 km

C
2560 km

D
64,000 km

Solution

or or R + h = 10 R

or h = 9 R = 9 × 6400 km = 57,600 km
Question 7
1 / -0

The weight of a body at the centre of earth is

A
zero

B
slightly more than that at the surface

C
slightly less than that at the surface

D
maximum

Solution

The weight of a body at the centre of earth is zero.
Question 8
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If the force of gravity suddenly disappears,

A
the mass of all bodies will become zero

B
the weight of all bodies will become zero

C
both mass and weight of all bodies will become zero

D
neither mass nor weight of all bodies will become zero

Solution

Gravity is a natural phenomenon due to which all bodies attract each other with a force proportional to their masses. So, if the force is omitted, the bodies will remain where there are in space and will occupy no weight.
Question 9
1 / -0

The escape velocity for a planet is v_e. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be

A
v_e

B

C

D
zero

Solution
Question 10
1 / -0

The escape velocity from a planet of mass M and radius R is given by

A

B

C

D

Solution

The escape velocity from a planet of mass M and radius R is given by
Question 11
1 / -0

If 'g' is the acceleration due to gravity at the earth's surface and 'r' is the radius of the earth, the escape velocity for a body to escape out of earth's gravitation field is

A

B

C
g/r

D
r/g

Solution

The minimum velocity required to just free a body from the binding of the gravitational attraction is called the escape velocity and is calculated by using the given formula:
Question 12
1 / -0

The time period of a simple pendulum of infinite length is

A
infinite

B
2

C
2

D

Solution

The time period of a simple pendulum can be calculated by:
T = 2* pi /sqrt [(1/R+1/L)g]
If the length of the pendulum is very small compared to the radius of earth, then the first term in the denominator is neglected and time period of a simple pendulum can be calculated. But when the length of the pendulum is very large compared to the radius of earth, (infinite) the second term, that is1/L is ignored and the formula becomes2π√R/g.
Question 13
1 / -0

The rotation of the earth about its axis speeds up such that a man on the equator becomes weightless. In such a situation, what would be the duration of one day?

A
2

B

C
2

D

Solution
Question 14
1 / -0

Two identical trains A and B move with equal speeds on parallel tracks along the equator. A moves from east to west; and B from west to east. Which train will exert greater force on the tracks?

A
A

B
B

C
They will exert equal force.

D
The mass and the speed of each train must be known to reach a conclusion.

Solution
Question 15
1 / -0

For a satellite to orbit around Earth, which of the following must be true?

A
It must be above the equator at sometime.

B
It cannot pass over the poles at any time.

C
Its height above the surface cannot exceed 36,000 km.

D
Its period of rotation must be 2.

Solution

The orbit ellipse of a satellite lies in a plane known as the orbital plane. The orbital plane always goes through the centre of Earth, but may be tilted any angle relative to the equator. Hence, a satellite must be above the equator at sometime. Thus, option (1) is correct.
Question 16
1 / -0

The distances of two satellites from the surface of the earth are R and 7R. Their time periods of rotation are in the ratio

A
1 : 7

B
1 : 8

C
1 : 49

D
1 : 7^3/2

Solution

Radius of the orbit is taken from the centre of the earth.
R₁ = 2R and R₂ = 8R
According to Kepler's laws, the time period of rotation of satellite is given by:
Question 17
1 / -0

The distances of Neptune and Saturn from the sun are 10¹³ m and 10¹² m, respectively. The ratio of time periods of the planets is

A
100:1

B
1: √10

C
√10:1

D
10√10:1

Solution

T_N = (10¹³)^3/2
T_S = (10¹²)^3/2

or We have

T_N:T_S :: 10√10:1
Question 18
1 / -0

Two planets are moving around the sun. The periodic times of the orbits are T₁, T₂and mean radii of the orbits are r₁, r₂. The ratio of T₁/T₂ is equal to

A
(r₁/r₂)

B
(r₁/r₂)^3/2

C
(2r₁/r₂)²

D
(r₁/2r₂)^3/2

Solution

T² is directly proportional to r³ (Kepler's third law)
T₁/T₂ = (r₁/r₂)^3/2
Question 19
1 / -0

Which of the following graphs represents the motion of a planet moving around the sun?

A

B

C

D

Solution

T² is directly proportional to R³ (Kepler's third law).
Hence, option (1) is correct.
Question 20
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Mass is converted into energy according to the relation

A
E = mgh

B
E = mc²

C
E = mg/h

D
E = (1/2) mv²

Solution

Mass is converted into energy according to the relation E = mc²