Gravitation — Page 10 | JEE Physics

Q91

If the gravitational field in the space is given as

\left(-\dfrac{K}{r^{2}}\right)

. Taking the reference point to be at

\mathrm{r}=2 \mathrm{~cm}

with gravitational potential

\mathrm{V}=10 \mathrm{~J} / \mathrm{kg}

. Find the gravitational potential at

\mathrm{r}=3 \mathrm{~cm}

in SI unit (Given, that

\mathrm{K}=6 \mathrm{~Jcm} / \mathrm{kg}

)

A 9

B 11

C 10

D 12

Correct Answer

Option B

Solution

E = - {K \over {{r^2}}}

\Delta V = - \int\limits_{r = 2\,cm}^{3\,cm} {E.\,dr}

= \int\limits_2^3 {{k \over {{r^2}}}dr}

= \left[ { - {K \over r}} \right]_2^3 = \left( {{K \over 6}} \right) = {6 \over 6} = 1

J/kg

{V_f} - {V_i} = 1

\Rightarrow {V_f} - 10 = 1

{V_f} = 11

J/kg

Q92

The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.

A 12 hours

B 3 hours

C 6 hours

D 4 hours

Correct Answer

Option B

Solution

\because {T^2} \propto {R^3}

$\therefore$

{{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}

{{{{24}^2}} \over {T_2^2}} = {{R_1^3} \over {{{\left( {{{{R_1}} \over 4}} \right)}^3}}}

{{{{24}^2}} \over {T_2^2}} = {4^3}

{T_2} = {{24} \over {{2^3}}} = 3

hours

Q93

Two particles of equal mass '

m

' move in a circle of radius '

r

' under the action of their mutual gravitational attraction. The speed of each particle will be :

A

\sqrt{\dfrac{G m}{4 r}}

B

\sqrt{\dfrac{G m}{2 r}}

C

\sqrt{\dfrac{G m}{r}}

D

\sqrt{\dfrac{4 G m}{r}}

Correct Answer

Option A

Solution

$\dfrac{\mathrm{Gm}^{2}}{4 \mathrm{r}^{2}}=\dfrac{\mathrm{mv}^{2}}{\mathrm{r}}$

v=\sqrt{\frac{G m}{4 r}}

Q94

Every planet revolves around the sun in an elliptical orbit :- A. The force acting on a planet is inversely proportional to square of distance from sun. B. Force acting on planet is inversely proportional to product of the masses of the planet and the sun. C. The Centripetal force acting on the planet is directed away from the sun. D. The square of time period of revolution of planet around sun is directly proportional to cube of semi-major axis of elliptical orbit. Choose the correct answer from the options given below :

A C and D only

B B and C only

C A and D only

D A and C only

Correct Answer

Option C

Solution

A.

The force acting on a planet is inversely proportional to the square of the distance from the sun.

This is known as the inverse square law and is described by Newton's law of gravitation.

B.

Force acting on a planet is inversely proportional to the product of the masses of the planet and the sun.

This is also described by Newton's law of gravitation, but it is not directly related to the planet's elliptical orbit.

C.

The centripetal force acting on the planet is directed towards the sun, not away from it.

D.

The square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of the elliptical orbit.

This is known as Kepler's third law.

Therefore, only A and D are correct.

Q95

A body of mass is taken from earth surface to the height h equal to twice the radius of earth (R

_e

), the increase in potential energy will be : (g = acceleration due to gravity on the surface of Earth)

A

\dfrac{1}{2}mgR_e

B

3~mgR_e

C

\dfrac{1}{3}mgR_e

D

\dfrac{2}{3}mgR_e

Correct Answer

Option D

Solution

\begin{aligned} & V_{\text{surface }}=-\left(\frac{G M m}{R_{e}}\right) \\\\ & V_{p}=-\frac{G M m}{3 R_{e}} \\\\ & \Delta V=\frac{G M m}{R_{e}}\left(1-\frac{1}{3}\right) \\\\ & =\frac{2}{3} \frac{G M m}{\left(R_{e}^{2}\right)} \times R_{e} \\\\ & \Delta V=\frac{2}{3} m g R_{e} \end{aligned}

Q96

Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately) (Take g = 10 m s

^{-2}

, radius of earth = 6400 km)

A 12 hours

B 1 hour 24 minutes

C 24 hours

D 1 hour 40 minutes

Correct Answer

Option B

Solution

Gravitational acceleration at a distance of $r$ from centre of earth is given by

g^{\prime}=\frac{g}{R} r

Where $R$ is the radius of earth So, $\dfrac{d^{2} r}{d t^{2}}=-\dfrac{g}{R} r$ $\Rightarrow \quad T=2 \pi \sqrt{\dfrac{R}{g}}=2 \pi \sqrt{\dfrac{6400000}{10}}$ $=2 \pi \times 800 \mathrm{sec}$ $=5024 ~ \mathrm{sec}$ = 1 hour 24 minutes (approx.)

Q97

If the distance of the earth from Sun is 1.5

\times

10

^6

km. Then the distance of an imaginary planet from Sun, if its period of revolution is 2.83 years is :

A

6\times10^6

km

B

3\times10^7

km

C

6\times10^7

km

D

3\times10^6

km

Correct Answer

Option D

Solution

We can use Kepler's third law to solve this problem.

Kepler's third law states that the square of the period of revolution of a planet around the Sun is proportional to the cube of its average distance from the Sun.

Let $T$ be the period of revolution of the imaginary planet in years, and let $d$ be its average distance from the Sun in kilometers.

We can use the following equation to solve for $d$ :

\frac{T_1^2}{T_2^2} = \frac{d_1^3}{d_2^3}

where $T_1$ and $d_1$ are the period of revolution and an average distance of the Earth from the Sun, respectively, and $T_2$ is the period of revolution of the imaginary planet.

Substituting the given values, we get:

\begin{aligned} & \frac{T_{1}}{T_{2}}=\left(\frac{d_{1}}{d_{2}}\right)^{\frac{3}{2}} \\\\ & \Rightarrow\left(\frac{1 \text{ year }}{2.83 \text{ year }}\right)^{\frac{2}{3}}=\left(\frac{1.5 \times 10^{6} \mathrm{~km}}{d_{2}}\right) \\\\ & \Rightarrow \frac{1}{2}=\frac{1.5 \times 10^{6} \mathrm{~km}}{d_{2}} \\\\ & d_{2}=3 \times 10^{6} \mathrm{~km} \end{aligned}

Q98

Given below are two statements: Statement I : Acceleration due to earth's gravity decreases as you go 'up' or 'down' from earth's surface. Statement II : Acceleration due to earth's gravity is same at a height 'h' and depth 'd' from earth's surface, if h = d. In the light of above statements, choose the most appropriate answer from the options given below

A Statement I is correct but statement II is incorect

B Both Statement I and II are correct

C Statement I is incorrect but statement II is correct

D Both Statement I and Statement II are incorrect

Correct Answer

Option A

Solution

The most appropriate answer is Statement I is correct but statement II is incorrect.

Statement I is correct as acceleration due to Earth's gravity decreases as you move away from its surface either upward or downward.

This is because the gravity of the Earth follows an inverse square law, which means that the gravitational force decreases as the square of the distance between two objects increases.

However, statement II is incorrect because acceleration due to Earth's gravity is not the same at a height and depth from Earth's surface, even if they are equal in magnitude.

This is because the Earth is not a perfect sphere and has a non-uniform distribution of mass, which causes variations in the strength of gravity at different locations.

Therefore, the acceleration due to Earth's gravity will be different at a certain height and depth, even if they are equal.

Q99

The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth's surface is (given, radius of earth

\mathrm{R_e=6400~km}

) :

A 9.8 N

B 4.9 N

C 19.6 N

D 8 N

Correct Answer

Option D

Solution

The weight of an object at a height h from the Earth's surface is given by:

W' = W \left(\frac{R}{{R + h}}\right)^2

where: W' is the weight at height h, W is the weight at the Earth's surface, R is the radius of the Earth, h is the height above the Earth's surface.

In this problem, the weight at the Earth's surface W is given as 18 N, the radius of the Earth R is given as 6400 km, and the height h is given as 3200 km.

Substituting these values into the equation gives:

W' = 18 N \left(\frac{6400~km}{{6400~km + 3200~km}}\right)^2 = 18 N \left(\frac{2}{3}\right)^2 = 18 N \cdot \frac{4}{9} = 8 N

Therefore, the weight of the body at an altitude of 3200 km above the Earth's surface is 8 N.

Q100

Two identical particles each of mass '

m

' go round a circle of radius

a

under the action of their mutual gravitational attraction. The angular speed of each particle will be :

A

\sqrt{\dfrac{G m}{2 a^{3}}}

B

\sqrt{\dfrac{G m}{a^{3}}}

C

\sqrt{\dfrac{G m}{8 a^{3}}}

D

\sqrt{\dfrac{G m}{4 a^{3}}}

Correct Answer

Option D

Solution

The gravitational force between two particles of mass $m$ separated by a distance $r$ is given by:

F = \frac{Gm^2}{r^2}

where $G$ is the gravitational constant.

In this problem, the two particles are moving in a circular orbit of radius $a$ under the influence of their mutual gravitational attraction.

Therefore, the gravitational force between the two particles provides the necessary centripetal force to keep them in circular motion.

The centripetal force required for a particle of mass $m$ moving in a circle of radius $a$ with angular speed $\omega$ is given by:

F_{\text{centripetal}} = m\omega^2a

Setting the gravitational force equal to the centripetal force, we get:

\frac{Gm^2}{r^2} = m\omega^2a

Substituting $r = 2a$ (since the two particles are separated by a distance equal to twice the radius of the circle), we get:

\frac{Gm^2}{(2a)^2} = m\omega^2a

Simplifying, we get:

\omega^2 = \frac{Gm}{4a^3}

Taking the square root of both sides, we get:

\omega = \sqrt{\frac{Gm}{4a^3}}

Therefore, the angular speed of each particle is $\sqrt{\dfrac{Gm}{4a^3}}$ .