Physics - Gravitation - Quiz-1

If the gravitational potential on the surface of the earth is V₀, then the potential at a point at a height equal to half of the radius of the earth is:

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$\frac{V_{0}}{2}$
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$\frac{2}{3} V_{0}$
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$\frac{V_{0}}{3}$
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$\frac{3 V_{0}}{2}$

The total mechanical energy of an object of mass m projected from the surface of the earth with escape speed is:

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Zero
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Infinite
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$- \frac{GMm}{2 R}$
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$- \frac{GMm}{3 R}$

A body is thrown with a velocity equal to n times the escape velocity (v_e). The velocity of the body at a large distance away will be:

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$v_{e} \sqrt{n^{2} - 1}$
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$v_{e} \sqrt{n^{2} + 1}$
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$v_{e} \sqrt{1 - n^{2}}$
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None of these

If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of earth to a height equal to the radius of the earth R, is

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$\frac{1}{2} m g R$
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2 mgR
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mgR

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$\frac{1}{4} m g R$

Weightlessness experienced while orbiting the earth in space-ship is the result of

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   Inertia

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Acceleration

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Zero gravity

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Freefall towards the earth

A satellite is moving very close to a planet of density $ρ$ . The time period of the satellite is:

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   $\sqrt{\frac{3 π}{ρG}}$

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   ${(\frac{3 π}{ρG})}^{3 / 2}$

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   $\sqrt{\frac{3 π}{2 ρG}}$

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   ${(\frac{3 π}{2 ρG})}^{3 / 2}$

A projectile is fired upwards from the surface of the earth with a velocity ${kv}_{e}$ where $v_{e}$ is the escape velocity and k < 1. If r is the maximum distance from the center of the earth to which it rises and R is the radius of the earth, then r equals

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   $\frac{R}{k^{2}}$

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   $\frac{R}{1 - k^{2}}$

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   $\frac{2 R}{1 - k^{2}}$

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   $\frac{2 R}{1 + k^{2}}$

The gravitational potential difference between the surface of a planet and 10 m above is 5 J/kg. If the gravitational field is supposed to be uniform, the work done in moving a 2 kg mass from the surface of the planet to a height of 8 m is

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2J

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4J

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6J

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8J

A planet is moving in an elliptical orbit. If T, V, E, and L stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the center of the orbit, then

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T is conserved

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V is always positive

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E is always negative

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the magnitude of L is conserved but its direction changes continuously

In planetary motion, the areal velocity of the position vector of a planet depends on the angular velocity ( $ω$ ) and the distance of the planet from the sun (r). The correct relation for areal velocity is:

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$\frac{d A}{d t} α ω r$

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$\frac{d A}{d t} α ω^{2} r$

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$\frac{d A}{d t} α ω r^{2}$

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$\frac{d A}{d t} α \sqrt{ω r}$

If A is the areal velocity of a planet of mass M, its angular momentum is

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   $\frac{M}{A}$

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2MA

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   $A^{2} M$

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   ${AM}^{2}$

Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

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   $- \frac{5 Gm}{r}$

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$- \frac{6 Gm}{r}$

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   $- \frac{9 Gm}{r}$

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0

Magnitude of potential energy (U) and time period (T) of a satellite are related to each other as:

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$T^{2} α \frac{1}{U^{3}}$

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$T α \frac{1}{U^{3}}$

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$T^{2} α U^{3}$

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$T^{2} α \frac{1}{U^{2}}$

A projectile fired vertically upwards with a speed v escapes from the earth. If it is to be fired at 45 $°$ to the horizontal, what should be its speed so that it escapes from the earth?

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v

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   $\frac{v}{\sqrt{2}}$

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   $\sqrt{2} v$

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2v

Kepler's second law regarding constancy of the areal velocity of a planet is a consequence of the law of conservation of:

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Energy

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Linear momentum

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Angular momentum

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Mass

A point P lies on the axis of a ring of mass M and radius 'a' at a distance 'a' from its centre C. A small particle starts from P and reaches C under gravitational attraction. Its speed at C will be :

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$\sqrt{\frac{2 GM}{a}}$

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$\sqrt{\frac{2 GM}{a} (1 - \frac{1}{\sqrt{2}})}$

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$\sqrt{\frac{2 GM}{a} (\sqrt{2} - 1)}$

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zero

The escape velocity for a rocket from the earth is 11.2 km/sec. Its value on a planet where the acceleration due to gravity is double that on the earth and the diameter of the planet is twice that of the earth (in km/sec) will be:

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11.2

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5.6

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22.4

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53.6

The escape velocity from the earth is about 11 km/second. The escape velocity from a planet having twice the radius and the same mean density as the earth is

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22 km/sec

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11 km/sec

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5.5 km/sec

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15.5 km/sec

What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

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()   $7.4 \times 10^{- 4}$ rad/sac

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&nsp;(b) 6.4 $\times 10^{- 4}$ rad/sac

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()   $7.8 \times 10^{- 4}$ rad/sac

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() 8.7 $\times 10^{- 4}$ rad/sac

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 the body to escape out of the earth's gravitational field is:

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gr

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$\sqrt{2 g r}$

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g/r

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r/g

The escape velocity of a projectile from the earth is approximately

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(1)12 m/sec

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(2)112 km/sec

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(3)11.2 km/sec

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(4)11200 km/sec

The escape velocity of a particle of mass m varies as:

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$m^{2}$

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m

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$m^{0}$

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$m^{- 1}$

Acceleration due to gravity is ‘g’ on the surface of the earth. The value of acceleration due to gravity at a height of 32 km above earth’s surface is (Radius of the earth = 6400 km)

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0.9 g

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0.99g

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0.8 g

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1.01 g

The time period of a simple pendulum on a freely moving artificial satellite is

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Zero

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2 sec

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3 sec

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Infinite

For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of -

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2

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$\sqrt{2}$

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1/ $\sqrt{2}$

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4

The height of a point vertically above the earth’s surface, at which the acceleration due to gravity becomes 1% of its value at the surface is: (Radius of the earth = R)

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8R

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9R

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10R

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20R

The escape velocity of an object from the earth depends upon the mass of the earth (M), its mean density, its radius (R) and the gravitational constant (G). Thus the formula for escape velocity is:

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$v = R \sqrt{\frac{8 π}{3} G p}$

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$v = M \sqrt{\frac{8 π}{3} G R}$

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$v = \sqrt{2 G M R}$

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$v = \sqrt{\frac{2 G M}{R^{2}}}$

Escape velocity on a planet is $v_{e}$ . If radius of the planet remains same and mass becomes 4 times, the escape velocity becomes

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$4 v_{e}$

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$2 v_{e}$

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$v_{e}$

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$\frac{1}{2} v_{e}$

The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of earth to that on the surface of moon will be

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0.2

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2.57

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4.81

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0.39

If radius of earth is R then the height h’ at which value of ‘g’ becomes one-fourth is

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$\frac{R}{4}$

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$\frac{3 R}{4}$

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R

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$\frac{R}{8}$

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