If earth has a mass nine times and radius twice to that of a planet P. Then $$\frac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1}$$ will be the minimum velocity required by a rocket to pull out of gravitationa

$$ \begin{aligned} & M_E=9 M_P \\\\ & R_E=2 R_P \end{aligned} $$ Escape velocity $=\sqrt{\frac{2GM}{R}}$ For earth $v_e=\sqrt{\frac{2 G M_E}{R_E}}$ $$ \begin{aligned} & \text { For } P, v_e=\sqrt{\frac{\frac{2 G M_E}{9}}{\frac{R_E}{2}}}=\sqrt{\frac{2 G M_E}{R_E} \times \frac{2}{9}} \\\\ & =\frac{v_e \sqrt{2}}{3} \end{aligned} $$

Gravitation — Page 8 | NEET Physics

Q71

The radii of two planets A and B are in the ratio 2 : 3. Their densities are 3

\rho

and 5

\rho

respectively. The ratio of their acceleration due to gravity is :

A 9 : 4

B 9 : 8

C 9 : 10

D 2 : 5

Correct Answer

Option D

Solution

Given,

{{{r_A}} \over {{r_B}}} = {2 \over 3}

{{{\rho _A}} \over {{\rho _B}}} = {3 \over 5}

We know, Acceleration due to gravity

g = {{GM} \over {{r^2}}} = {G \over {{r^2}}} \times {4 \over 3}\pi {r^3} \times \rho

= {{4\pi Gr\rho } \over 3}

$\therefore$

g \propto \rho r

$\therefore$

{{{g_A}} \over {{g_B}}} = {{{\rho _A}} \over {{\rho _B}}} \times {{{r_A}} \over {{r_B}}}

= {3 \over 5} \times {2 \over 3}

= {2 \over 5}

Q72

A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be : (Take radius of earth

=6400 \mathrm{~km}

and

\mathrm{g}=10 \mathrm{~ms}^{-2}

)

A 800 km

B 1600 km

C 2133 km

D 4800 km

Correct Answer

Option A

Solution

Applying conservation of energy

- {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {{1 \over 3}\sqrt {{{2G{m_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over {{R_e} + h}}

- {{G{M_e}m} \over {{R_e}}} + {{G{M_e}m} \over {9{R_e}}} = - {{G{M_e}m} \over {{R_e} + h}}

{8 \over {9{R_e}}} = {1 \over {{R_e} + h}}

\Rightarrow h = {{{R_e}} \over 8} = {{6400} \over 8} = 800

km

Q73

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.

Q74

Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be :

A 2W

B W

C

{2^{{1 \over 3}}}

W

D

\sqrt 2

W

Correct Answer

Option C

Solution

Density is same

M = {4 \over 3}\pi {R^3}\rho ,2m = {4 \over 3}\pi R{'^3}\rho

R' = {2^{1/3}}R

\omega = {{GMm} \over {{R^2}}}

{\omega _2} = {{G2Mm} \over {R{'^2}}}

{\omega _2} = {2^{1/3}}\omega

Q75

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then :

A T

\propto

Rn/2

B T

\propto

R3/2 for any n

C T

\propto

Rn/2 +1

D T

\propto

R(n+1)/2

Correct Answer

Option D

Solution

We know, Central force in circular motion, F =

m{\omega ^2}R

According to the question,

F \propto {1 \over {{R^n}}}

$\therefore$

m{\omega ^2}R

\propto {1 \over {{R^n}}}

\Rightarrow m{\omega ^2}R = {k \over {{R^n}}}

\Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}

$\therefore$

\omega \propto {1 \over {{R^{{{n + 1} \over 2}}}}}

.......(1) And we know,

T = {{2\pi } \over \omega }

$\therefore$

T \propto {1 \over \omega }

...... (2) From (1) and (2) we can conclude that,

T \propto {R^{{{n + 1} \over 2}}}

Q76

Given below are two statements: Statement I: If

\mathrm{E}

be the total energy of a satellite moving around the earth, then its potential energy will be

\dfrac{E}{2}

. Statement II: The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy

\mathrm{E}

. In the light of the above statements, choose the most appropriate answer from the options given below

A Both Statement I and Statement II are incorrect

B Statement I is incorrect but Statement II is correct

C Statement I is correct but Statement II is incorrect

D Both Statement I and Statement II are correct

Correct Answer

Option A

Solution

A satellite in orbit around a planet is subject to two main forces: gravitational force, which is trying to pull it towards the planet, and its own kinetic energy or inertia, which is trying to keep it moving in a straight line.

The balance of these two forces results in the satellite moving in a circular or elliptical orbit.

The gravitational potential energy ( $U$ ) of the satellite is given by the formula: $U = -\dfrac{GMm}{R}$ where $G$ is the gravitational constant, $M$ is the mass of the Earth, $m$ is the mass of the satellite, and $R$ is the radius of the orbit.

The negative sign indicates that work would have to be done to remove the satellite from the Earth's gravitational influence.

The kinetic energy ( $K$ ) of the satellite is given by the formula: $K = \dfrac{GMm}{2R}$ This is obtained from the fact that for a satellite in stable orbit, the gravitational force must be equal to the centripetal force required to keep the satellite moving in a circle.

From this, we can derive an expression for the velocity of the satellite, and hence its kinetic energy.

The total mechanical energy ( $E$ ) of the satellite, which is the sum of its kinetic and potential energy, is therefore: $E = K + U = \dfrac{GMm}{2R} - \dfrac{GMm}{R} = -\dfrac{GMm}{2R}$ So the potential energy $U$ is $-2E$ , and the kinetic energy $K$ is $-E$ .

Thus, the statement "If $E$ be the total energy of a satellite moving around the earth, then its potential energy will be $2E$ " is correct, and the statement "The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy $E$ " is incorrect.

Q77

If earth has a mass nine times and radius twice to that of a planet P. Then

\dfrac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1}

will be the minimum velocity required by a rocket to pull out of gravitational force of

\mathrm{P}

, where

v_{e}

is escape velocity on earth. The value of

x

is

A 1

B 3

C 2

D 18

Correct Answer

Option C

Solution

\begin{aligned} & M_E=9 M_P \\\\ & R_E=2 R_P \end{aligned}

Escape velocity $=\sqrt{\dfrac{2GM}{R}}$ For earth $v_e=\sqrt{\dfrac{2 G M_E}{R_E}}$

\begin{aligned} & \text{ For } P, v_e=\sqrt{\frac{\frac{2 G M_E}{9}}{\frac{R_E}{2}}}=\sqrt{\frac{2 G M_E}{R_E} \times \frac{2}{9}} \\\\ & =\frac{v_e \sqrt{2}}{3} \end{aligned}

Q78

The escape velocity of a body on a planet 'A' is 12 kms

-

1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is :

A 12 kms

-

1

B 24 kms

-

1

C 36 kms

-

1

D 6 kms

-

1

Correct Answer

Option A

Solution

{v_{esc}} - \sqrt {{{2GM} \over R}} = \sqrt {{{2G} \over R} \times \rho \times {4 \over 3}\pi {R^3}}

\Rightarrow {v_{esc}} \propto R\sqrt \rho

\Rightarrow {{{{({v_{esc}})}_B}} \over {{{({v_{esc}})}_A}}} = 1

\Rightarrow {({v_{esc}})_B} = 12

km/s

Q79

Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:

A 8.4

B 11.2

C 5.6

D 2.8

Correct Answer

Option C

Solution

The escape velocity ( $v_{\text{esc}}$ ) for a planet with mass $M$ and radius $R$ is given by the formula:

v_{\text{esc}} = \sqrt{\frac{2GM}{R}}

where $G$ is the gravitational constant.

For Earth, let's denote its mass as $M_E$ and its radius as $R_E$ .

The escape velocity is given as 11.2 km/s.

Therefore:

v_{\text{esc, Earth}} = \sqrt{\frac{2G M_E}{R_E}} = 11.2 \text{ km/s}

For the planet in question, its mass $M_P$ is $M_E/8$ and its radius $R_P$ is $R_E/2$ .

The escape velocity from the planet is:

v_{\text{esc, Planet}} = \sqrt{\frac{2G M_P}{R_P}} = \sqrt{\frac{2G (M_E/8)}{R_E/2}}

Simplifying the expression:

v_{\text{esc, Planet}} = \sqrt{\frac{2G (M_E/8)}{R_E/2}} = \sqrt{\frac{2G M_E \cdot 2}{8R_E}} = \sqrt{\frac{G M_E}{2R_E}}

Now, relate it to the escape velocity from Earth:

v_{\text{esc, Planet}} = \frac{1}{\sqrt{4}} \cdot v_{\text{esc, Earth}}

Calculating further:

v_{\text{esc, Planet}} = \frac{1}{2} \cdot 11.2 \text{ km/s} = 5.6 \text{ km/s}

Thus, the escape velocity from the planet is 5.6 km/s. Option C: 5.6 km/s is the correct answer.

Q80

Given below are two statements : Statement I : The law of gravitation holds good for any pair of bodies in the universe. Statement II : The weight of any person becomes zero when the person is at the centre of the earth. In the light of the above statements, choose the correct answer from the options given below.

A Both Statement I and Statement II are true

B Both Statement I and Statement II are false

C Statement I is true but Statement II is false

D Statement I is false but Statement II is true

Correct Answer

Option A

Solution

Statement I is true.

Newton's law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between their centers.

This law applies to any pair of bodies in the universe.

Statement II is also true.

The weight of an object is the force of gravity acting on it.

If a person were at the center of the Earth, the gravitational pull from all the surrounding mass of the Earth would cancel out, resulting in zero net gravitational force and therefore zero weight.

Therefore, Option A: Both Statement I and Statement II are true, is the correct answer.