Physics - Gravitation - Quiz-4

A spring balance is graduated on sea level. If a body is weighed with this balance at

consecutively increasing heights from earth's surface, the weight indicated by the

balance

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Will go on increasing continuously
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Will go on decreasing continuously
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Will remain same
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Will first increase and then decrease

The gravitational field due to a mass distribution is $E = K / x^{3}$ in the x-direction. (K is a

constant). Taking the gravitational potential to be zero at infinity, its value at a distance x

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K/x
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K/2x
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$K / x^{2}$
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K/2 $x^{2}$

The change in the potential energy, when a body of mass m is raised to a height nR from the

Earth's surface is: (R = Radius of the Earth)

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$m g R (\frac{n}{n - 1})$
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nmgR
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mgR $(\frac{n^{2}}{n^{2} + 1})$
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$m g R (\frac{n}{n + 1})$

The orbital speed of an artificial satellite very close to the surface of the earth is $V_{o}$ . Then the orbital speed of another artificial satellite at a height equal to three times the radius of the earth is

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() $4 V_{0}$
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() $2 V_{0}$
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() $0.5 V_{0}$
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() $4 V_{0}$

The masses and radii of the earth and moon are $M_{1,} R_{1}$ and $M_{2}, R_{2}$ respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escapes to infinity is:

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$2 \sqrt{\frac{G}{D} (M_{1} + M_{2})}$
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$2 \sqrt{\frac{2 G}{D} (M_{1} + M_{2})}$
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$2 \sqrt{\frac{G M}{D} (M_{1} + M_{2})}$
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$2 \sqrt{\frac{G M (M_{1} + M_{2})}{D (R_{1} + R_{2})}}$

If the mass of the earth is M, the radius is R and the gravitational constant is G, then work done to take

1 kg mass from earth surface to infinity will be:

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$\sqrt{\frac{G M}{2 R}}$
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$\frac{G M}{R}$
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$\sqrt{\frac{2 G M}{R}}$
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$\frac{G M}{2 R}$

If $v_{e}$ and $v_{o}$ represent the escape velocity and orbital velocity of a satellite corresponding

to a circular orbit of radius R, then

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$v_{e} = v_{o}$
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$\sqrt{2} v_{o} = v_{e}$
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$v_{e} = v_{o} / \sqrt{2}$
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$v_{e}$ and $v_{o}$ are not related

If r represents the radius of the orbit of a satellite of mass m moving around a planet of

mass M, the velocity of the satellite is given by:

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$v^{2} = g \frac{M}{r}$
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$v^{2} = \frac{G M m}{r}$
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$v = \frac{G M}{r}$
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$v^{2} = \frac{G M}{r}$

An earth satellite of mass m revolves in a circular orbit at a height h from the surface of

the earth. R is the radius of the earth and g is acceleration due to gravity at the surface

of the earth. The velocity of the satellite in the orbit is given by:

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$\frac{g R^{2}}{R + h}$
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gR
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$\frac{g R}{R + h}$
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$\sqrt{\frac{g R^{2}}{R + h}}$

A satellite which is geostationary in a particular orbit is taken to another orbit. Its

distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time

period in the second orbit is:

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4.8 hours
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$48 \sqrt{2}$ hours
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24 hours
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$24 \sqrt{2}$ hours

The ratio of the K.E. required to be given to the satellite to escape earth's gravitational

field to the K.E. required to be given so that the satellite moves in a circular orbit just

above earth atmosphere is:

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One
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Two
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Half
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Infinity

An astronaut orbiting the earth in a circular orbit 120 km above the surface of earth, gently drops a spoon out of space-ship. The spoon will

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Fall vertically down to the earth
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3. Move towards the moon
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4. Will move along with space-ship
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Will move in an irregular way then fall down

The period of a satellite in a circular orbit around a planet is independent of

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The mass of the planet
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The orbital radius of satellite around the planet
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The mass of the satellite
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All the three parameters 1, 2 and 3

A geostationary satellite:

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Revolves about the polar axis
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Has a time period less than that of the near-earth satellite
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Moves faster than a near-earth satellite
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Is stationary in the space

A small satellite is revolving near earth's surface. Its orbital velocity will be nearly

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8 km/sec
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11.2 km/sec
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4 km/sec
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6 km/sec

The distance of neptune and saturn from sun are nearly $10^{13}$ and $10^{12}$ meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio

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$\sqrt{10}$
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100
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$10 \sqrt{10}$
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$1 / \sqrt{10}$

The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is v. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is

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$\frac{3}{2} v$
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$\sqrt{\frac{3}{2}} v$
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$\sqrt{\frac{2}{3}} v$
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$\frac{2}{3} v$

In a satellite if the time of revolution is T, then K.E. is proportional to

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$\frac{1}{T}$
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$\frac{1}{T^{2}}$
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$\frac{1}{T^{3}}$
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$T^{\frac{- 2}{3}}$

The period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4R is

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4T
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$\frac{T}{4}$
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8T
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$\frac{T}{8}$

If the height of a satellite from the earth is negligible in comparison to the radius of the earth R, the orbital velocity of the satellite is

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gR
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gR/2
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$\sqrt{g / R}$
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$\sqrt{g R}$

Choose the correct statement from the following.

The radius of the orbit of a geostationary satellite depends upon -

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Mass of the satellite, its time period and the gravitational constant
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Mass of the satellite, mass of the earth and the gravitational constant
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Mass of the earth, the mass of the satellite, time period of the satellite and the gravitational constant
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Mass of the earth, time period of the satellite and the gravitational constant

A planet moves around the sun. At a given point P, it is closest from the sun at a distance $d_{1}$ and has a speed $v_{1}$ . At another point Q, when it is farthest from the sun at a distance $d_{2}$ , its speed will be

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$\frac{{d_{1}}^{2} v_{1}}{{d_{2}}^{2}}$
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$\frac{d_{2} v_{1}}{d_{1}}$
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$\frac{d_{1} v_{1}}{d_{2}}$
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$\frac{{d_{2}}^{2} v_{1}}{{d_{1}}^{2}}$

A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1%, its speed will

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Increase by 1%
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Increase by 0.5%
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Decrease by 1%
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Decrease by 0.5%

Orbital velocity of an artificial satellite does not depend upon

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Mass of the earth
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Mass of the satellite
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Radius of the earth
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Acceleration due to gravity

The time period of a geostationary satellite is

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24 hours
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12 hours
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365 days
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One month

The orbital velocity of Earth's satellite near the surface is 7 km/s. When the radius of the orbit is 4 times more than that of Earth's radius, then orbital velocity in that orbit is:

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3.5 km/s
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7 km/s
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72 km/s
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14 km/s

A mass M is split into two parts, m and (M–m), which are then separated by a certain distance. What ratio of m/M maximizes the gravitational force between the two parts

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1/3
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1/2
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1/4
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1/5

Two identical satellites are at R and 7R away from earth surface, the wrong statement is (R = Radius of earth)

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Ratio of total energy will be 4
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Ratio of kinetic energies will be 4
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Ratio of potential energies will be 4
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Ratio of total energy will be 4 but ratio of potential and kinetic energies will be 2

For a satellite, the escape velocity is 11 km/s. If the satellite is launched at an angle of 60° with the vertical, then escape velocity will be:

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11 km/s
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$11 \sqrt{3} k m / s$
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$\frac{11}{\sqrt{3}} k m / s$
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33 km/s

The mean radius of the earth is R, its angular speed on its own axis is $ω$ and the acceleration due to gravity at the earth's surface is g. The cube of the radius of the orbit of a geostationary satellite will be -

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$R^{2} g / ω$
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$R^{2} ω^{2} / g$
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$R g / ω^{2}$
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$R^{2} g / ω^{2}$

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Practice Physics Quiz Questions and Answers

0:0:1

Practice Physics Quiz Questions and Answers

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