Gravitation — Page 13 | JEE Physics

Q121

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The angular speed of the moon in its orbit about the earth is more than the angular speed of the earth in its orbit about the sun. Reason (R) : The moon takes less time to move around the earth than the time taken by the earth to move around the sun. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both (A) and (R) are correct but (R) is not the correct explanation of (A)

B (A) is correct but (R) is not correct

C Both (A) and (R) are correct and (R) is the correct explanation of (A)

D (A) is not correct but (R) is correct

Correct Answer

Option C

Solution

The angular speed $\omega$ of an object in circular motion is given by the equation $\omega = \dfrac{2\pi}{T}$ where

T

is the period of the motion - the time it takes to make one complete revolution.

Assertion (A) states that the angular speed of the Moon in its orbit around the Earth is more than the angular speed of the Earth in its orbit around the Sun.

We can compare these angular speeds by comparing their periods.

The Moon takes approximately 27.3 days to orbit the Earth (this is its sidereal period, not its synodic period, which accounts for the Earth's motion around the Sun as well).

The Earth takes approximately 365.25 days to orbit the Sun.

Since the period of the Moon’s orbit is much less than the period of the Earth’s orbit, the angular speed of the Moon is much larger than that of the Earth.

This makes Assertion (A) correct.

Reason (R) states that the Moon takes less time to move around the Earth than the time taken by the Earth to move around the Sun.

This is also correct, as the previously mentioned periods demonstrate: 27.3 days for the Moon to orbit Earth vs.

365.25 days for Earth to orbit the Sun.

Furthermore, the reason (R) directly explains why assertion (A) is true.

Since angular speed is inversely proportional to the period of orbit, the fact that the Moon takes less time to complete an orbit implies it has a higher angular speed relative to the Earth’s angular speed in its orbit around the Sun.

Therefore, the correct answer is: Option C Both (A) and (R) are correct and (R) is the correct explanation of (A)

Q122

Four identical particles of mass

m

are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is

\left(\dfrac{2 \sqrt{2}+1}{32}\right) \dfrac{\mathrm{Gm}^2}{L^2}

, the length of the sides of the square is

A 4L

B 3L

C 2L

D

\dfrac{L}{2}

Correct Answer

Option A

Solution

\begin{aligned} & F_{\text{net }}=\sqrt{2} F+F^{\prime} \\ & F=\frac{G m^2}{a^2} \text{ and } F^{\prime}=\frac{G^2}{(\sqrt{2} \mathrm{a})^2} \\ & F_{\text{net }}=\sqrt{2} \frac{\mathrm{Gm}^2}{\mathrm{a}^2}+\frac{\mathrm{Gm}^2}{2 \mathrm{a}^2} \\ & \left(\frac{2 \sqrt{2}+1}{32}\right) \frac{\mathrm{Gm}^2}{\mathrm{~L}^2}=\frac{\mathrm{Gm}^2}{\mathrm{a}^2}\left(\frac{2 \sqrt{2}+1}{2}\right) \\ & \mathrm{a}=4 \mathrm{~L} \end{aligned}

Q123

At what distance above and below the surface of the earth a body will have same weight. (take radius of earth as

R

.)

A

\dfrac{\sqrt{3} R-R}{2}

B

\dfrac{R}{2}

C

\dfrac{\sqrt{5} R-R}{2}

D

\sqrt{5} R-R

Correct Answer

Option C

Solution

\begin{aligned} & g_p=\frac{g R^2}{(R+h)^2} \\ & g_q=g\left(1-\frac{h}{R}\right) \\ & g_p=g_q \\ & \frac{g}{\left(1+\frac{h}{R}\right)^2}=g\left(1-\frac{h}{R}\right) \\ & \left(1-\frac{h^2}{R^2}\right)\left(1+\frac{h}{R}\right)=1 \end{aligned}

Take

\frac{\mathrm{h}}{\mathrm{R}}=\mathrm{x}

So

\begin{aligned} & \mathrm{x}^3-\mathrm{x}+\mathrm{x}^2=0 \\ & \mathrm{x}=\frac{\sqrt{5}-1}{2} \\ & \mathrm{~h}=\frac{\mathrm{R}}{2}(\sqrt{5}-1) \end{aligned}

Q124

A planet takes 200 days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution :

A 20

B 50

C 100

D 25

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{T}^2 \propto \mathrm{r}^3 \\ & \frac{\mathrm{T}_1^2}{\mathrm{r}_1^3}=\frac{\mathrm{T}_2^2}{\mathrm{r}_2^3} \\ & \frac{(200)^2}{\mathrm{r}^3}=\frac{\mathrm{T}_2^2}{\left(\frac{\mathrm{r}}{4}\right)^3} \\ & \frac{200 \times 200}{4 \times 4 \times 4}=\mathrm{T}_2^2 \\ & \mathrm{~T}_2=\frac{200}{4 \times 2} \\ & \mathrm{~T}_2=25 \text{ days } \end{aligned}

Q125

Escape velocity of a body from earth is

11.2 \mathrm{~km} / \mathrm{s}

. If the radius of a planet be onethird the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is :

A 7.9 km/s

B 8.4 km/s

C 4.2 km/s

D 11.2 km/s

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{\mathrm{E}}}{3}, \mathrm{M}_{\mathrm{P}}=\frac{\mathrm{M}_{\mathrm{E}}}{6} \\ & \mathrm{~V}_{\mathrm{c}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{e}}}{\mathrm{R}_{\mathrm{e}}}} \quad \text{.... (i)}\\ & \mathrm{V}_{\mathrm{P}}=\sqrt{\frac{2 \mathrm{GM}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{P}}}} \quad \text{.... (ii)}\\ & \frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{p}}}=\sqrt{2} \\ & \mathrm{~V}_{\mathrm{P}}=\frac{\mathrm{V}_{\mathrm{e}}}{\sqrt{2}}=\frac{11.2}{\sqrt{2}}=7.9 \mathrm{~km} / \mathrm{sec} \end{aligned}

Q126

The gravitational potential at a point above the surface of earth is

-5.12 \times 10^7 \mathrm{~J} / \mathrm{kg}

and the acceleration due to gravity at that point is

6.4 \mathrm{~m} / \mathrm{s}^2

. Assume that the mean radius of earth to be

6400 \mathrm{~km}

. The height of this point above the earth's surface is :

A 1600 km

B 1200 km

C 540 km

D 1000 km

Correct Answer

Option A

Solution

-\frac{G M_E}{R_E+h}=-5.12 \times 10^{-7}

.... (i)

\frac{G M_E}{\left(R_E+h\right)^2}=6.4

..... (ii) By (i) and (ii)

\Rightarrow h=16 \times 10^5 \mathrm{~m}=1600 \mathrm{~km}

Q127

Assuming the earth to be a sphere of uniform mass density, a body weighed

300 \mathrm{~N}

on the surface of earth. How much it would weigh at R/4 depth under surface of earth ?

A 75 N

B 375 N

C 300 N

D 225 N

Correct Answer

Option D

Solution

To solve this question, we first need to understand how gravitational force (and hence weight) changes with depth under the surface of the Earth.

The gravitational force at a depth $d$ is given by the formula:

F = F_0 \left(1 - \frac{d}{R}\right)

where $F_0$ is the gravitational force (or the weight) at the surface, $R$ is the radius of the Earth, and $d$ is the depth below the Earth's surface.

In your question, the body weighs 300 N on the surface, so $F_0 = 300$ N.

It is taken to a depth of $\dfrac{R}{4}$ under the surface.

Therefore, $d = \dfrac{R}{4}$ .

Substituting these values into our formula, we get:

F = 300 \left(1 - \frac{\frac{R}{4}}{R}\right)

F = 300 \left(1 - \frac{1}{4}\right)

F = 300 \left(\frac{3}{4}\right)

F = 225 \, \text{N}

Therefore, at a depth of $\dfrac{R}{4}$ under the surface of the Earth, the body would weigh 225 N.

The correct option is Option D.

Q128

A satellite of

10^3 \mathrm{~kg}

mass is revolving in circular orbit of radius

2 R

. If

\dfrac{10^4 R}{6} \mathrm{~J}

energy is supplied to the satellite, it would revolve in a new circular orbit of radius (use

g=10 \mathrm{~m} / \mathrm{s}^2, R=

radius of earth)

A 4 R

B 6 R

C 2.5 R

D 3 R

Correct Answer

Option B

Solution

To determine the new radius of the orbit after the energy is supplied to the satellite, we need to compare the initial and final energies of the satellite in its orbit around the Earth.

We will use the formula for total energy (the sum of kinetic and potential energy) of a satellite in a circular orbit: The total energy,

E

, of a satellite of mass

m

orbiting at a distance

r

from the center of the Earth is given by:

E = -\frac{GMm}{2r}

where:

G

is the gravitational constant

M

is the mass of the Earth

r

is the radius of the orbit Given, the initial radius of orbit is

2R

. Therefore, the initial total energy

E_i

is:

E_i = -\frac{GMm}{2 \times 2R} = -\frac{GMm}{4R}

Now, the energy supplied to the satellite is given as

\frac{10^4 R}{6} \mathrm{~J}

. The new total energy

E_f

will be the sum of the initial energy and the supplied energy:

E_f = E_i + \text{Energy Supplied}

Substituting the values:

E_f = -\frac{GMm}{4R} + \frac{10^4 R}{6}

The final energy formula for a new orbit radius

r_f

is similar to the initial energy formula:

E_f = -\frac{GMm}{2r_f}

By equating the two expressions for

E_f

, we get:

-\frac{GMm}{2r_f} = -\frac{GMm}{4R} + \frac{10^4 R}{6}

Rearrange the equation to solve for

r_f

:

\frac{GMm}{2r_f} = \frac{GMm}{4R} - \frac{10^4 R}{6}

Since

GMm = gR^2 m

(using gravitational acceleration

g

and radius of Earth

R

), we can substitute this to simplify the expression:

\frac{gR^2 m}{2r_f} = \frac{gR^2 m}{4R} - \frac{10^4 R}{6}

By cancelling the common terms and rearranging:

\frac{1}{2r_f} = \frac{1}{4R} - \frac{10^4 R}{6gR^2 m}

Recall that

g = 10 \mathrm{~m/s^2}

and the mass of the satellite

m = 10^3 \mathrm{~kg}

:

\frac{1}{2r_f} = \frac{1}{4R} - \frac{10^4 \times R}{6 \times 10 \times R^2 \times 10^3}

Simplifying the second term further:

\frac{10^4 \times R}{60 \times R^2 \times 10^3} = \frac{10 \times R}{60 \times R^2} = \frac{1}{6R}

So, the final equation becomes:

\frac{1}{2r_f} = \frac{1}{4R} - \frac{1}{6R}

Finding a common denominator for the right-hand side:

\frac{1}{2r_f} = \frac{3 - 2}{12R} = \frac{1}{12R}

Therefore:

2r_f = 12R

Thus:

r_f = 6R

Hence, the new radius of the circular orbit is 6R. The correct answer is: Option B: 6R

Q129

An astronaut takes a ball of mass

m

from earth to space. He throws the ball into a circular orbit about earth at an altitude of

318.5 \mathrm{~km}

. From earth's surface to the orbit, the change in total mechanical energy of the ball is

x \dfrac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}

. The value of

x

is (take

\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km})

:

A 12

B 11

C 9

D 10

Correct Answer

Option B

Solution

At earth surface,

E_1=-\frac{G M_m}{R_e}

in the orbit,

E_2=-\frac{G M_m}{2 r}

\begin{aligned} \Delta E & =G M_m\left[\frac{1}{R_e}-\frac{1}{2 r}\right] \\\\ & =G M_m\left[\frac{1}{R_e}-\frac{1}{2.1 R_e}\right] \\\\ & =\frac{11}{21} \frac{G M_m}{R_e} \\\\ \Rightarrow & x=11 \end{aligned}

Q130

A metal wire of uniform mass density having length

L

and mass

M

is bent to form a semicircular arc and a particle of mass

\mathrm{m}

is placed at the centre of the arc. The gravitational force on the particle by the wire is :

A

\dfrac{\mathrm{GmM} \pi^2}{\mathrm{~L}^2}

B

\dfrac{\mathrm{GMm} \pi}{2 \mathrm{~L}^2}

C 0

D

\dfrac{2 \mathrm{GmM} \pi}{\mathrm{L}^2}

Correct Answer

Option D

Solution

Field at centre due to arc,

I=\frac{2 G M \pi}{L^2}

\therefore \quad

Net force on mass,

F=\frac{2 G M m \pi}{L^2}