Gravitation — Page 6 | JEE Physics

Q51

The acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about the axis passing through the pole is

\omega

. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is (h << R, where R is the radius of the earth)

A

{{{R^2}{\omega ^2}} \over {2g}}

B

{{{R^2}{\omega ^2}} \over g}

C

{{{R^2}{\omega ^2}} \over {8g}}

D

{{{R^2}{\omega ^2}} \over {4g}}

Correct Answer

Option A

Solution

At equator, g1 = g - R

{\omega ^2}

At height h, g2 =

g\left( {1 - {{2h} \over R}} \right)

[as given h << R] $\because$ Weight same at poles and at h (so g1 = g2) $\therefore$ g - R

{\omega ^2}

=

g\left( {1 - {{2h} \over R}} \right)

$\Rightarrow$

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

$\Rightarrow$

{h = {{{R^2}{\omega ^2}} \over {2g}}}

Q52

The mass density of a spherical galaxy varies as

{K \over r}

over a large distance ‘r’ from its centre. In that region, a small star is in a circular orbit of radius R. Then the period of revolution, T depends on R as :

A T2

\propto

R

B T2

\propto

R3

C T

\propto

R

D T2

\propto

{1 \over {{R^3}}}

Correct Answer

Option A

Solution

dm = $\rho$ dv $\Rightarrow$ dm =

\left( \frac{k}{r} \right)

(4 $\pi$ r2dr) $\Rightarrow$ dm = 4 $\pi$ krdr M =

\int\limits^{R}_{0} dm

=

\int\limits^{R}_{0} 4\pi krdr

$\Rightarrow$ M =

4\pi k\left[ \frac{r^{2}}{2} \right]^{R}_{0}

$\Rightarrow$ M = 2 $\pi$ kR2 For circular motion gravitational force will provide required centripetal force.

\frac{GMm}{R^{2}}

=

\frac{mv^{2}}{R}

$\Rightarrow$

\frac{G\left( 2\pi kR^{2}\right) m}{R^{2}}

=

\frac{mv^{2}}{R}

$\Rightarrow$ v =

\sqrt{2\pi GkR}

Time period, T =

\frac{2\pi R}{v}

=

\frac{2\pi R}{\sqrt{2\pi GkR} }

$\propto$

\sqrt{R}

$\Rightarrow$ T2 $\propto$ R

Q53

The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

A

{R \over 2}

B

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

C

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

D

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

Correct Answer

Option B

Solution

M = mass of earth M1 = mass of shaded portion Re = Radius of earth M1 =

{M \over {{4 \over 3}\pi {R^3}}}.{4 \over 3}\pi {\left( {R - h} \right)^3}

=

{{M{{\left( {R - h} \right)}^3}} \over R^3}

Given, Weight of body is same at P and Q $\therefore$ mgP = mgQ $\Rightarrow$ gP = gQ $\Rightarrow$

{{G{M_1}} \over {{{\left( {R - h} \right)}^2}}} = {{GM} \over {{{\left( {R + h} \right)}^2}}}

$\Rightarrow$

{{GM{{\left( {R - h} \right)}^3}} \over {{R^3}{{\left( {R - h} \right)}^2}}} = {{GM} \over {{{\left( {R + h} \right)}^2}}}

$\Rightarrow$ (R – h) (R + h)2 = R3 $\Rightarrow$ R3 – hR2 + h2R – h3 + 2R2h – 2Rh2 = R3 $\Rightarrow$ R2h – Rh2 – h3 = 0 $\Rightarrow$ R2 – Rh – h2 = 0 $\Rightarrow$ h2 + Rh – R2 = 0 $\Rightarrow$ h =

{{ - R \pm \sqrt {{R^2} + 4{R^2}} } \over 2}

$\Rightarrow$ h =

{{ - R + \sqrt 5 R} \over 2}

Q54

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass

{m \over {10}}

so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth) :

A

{{3m} \over 8}{\left( {u + \sqrt {{{5GM} \over {6R}}} } \right)^2}

B

{m \over {20}}\left( {{u^2} + {{113} \over {100}}{{GM} \over R}} \right)

C

5m\left( {{u^2} - {{119} \over {100}}{{GM} \over R}} \right)

D

{m \over {20}}{\left( {u - \sqrt {{{2GM} \over {3R}}} } \right)^2}

Correct Answer

Option C

Solution

Using energy conservation Ki + Ui = Kf + Uf $\Rightarrow$

{1 \over 2}m{u^2} - {{GmM} \over R}

=

{1 \over 2}m{v^2} - {{GmM} \over {2R}}

$\Rightarrow$ v =

\sqrt {{u^2} - {{GM} \over R}}

After ejecting a rocket of mass

{m \over {10}}

the remaining part of mass

{{9m} \over {10}}

will rotate the earth with orbital velocity v0. $\therefore$ v0 =

\sqrt {{{GM} \over {2R}}}

Applying momentum conservation along radial direction, Before firing rocket momentum of satelite in radial direction = mv And after firing rocket momentum of satelite in radial direction = 0 and momentum of rocket in radial direction =

{m \over {10}}{v_2}

$\therefore$ mv =

{m \over {10}}{v_2}

$\Rightarrow$ v2 = 10v Now applying momentum conservation along tangential direction we get, 0 =

{m \over {10}}{v_1}

-

{{9m} \over {10}}{v_0}

$\Rightarrow$

{{9m} \over {10}}{v_0}

=

{m \over {10}}{v_1}

$\Rightarrow$ v1 = 9v0 $\therefore$ Total Kinetic Energy of rocket =

{1 \over 2}{m \over {10}}\left( {v_1^2 + v_2^2} \right)

=

{1 \over 2}{m \over {10}}\left( {81v_0^2 + 100{v^2}} \right)

=

{m \over {20}}\left( {81\left( {{{GM} \over {2R}}} \right) + 100\left( {{u^2} - {{GM} \over R}} \right)} \right)

=

{m \over {20}}\left( {100{u^2} + {{81GM} \over {2R}} - {{100GM} \over R}} \right)

=

{m \over {20}}\left( {100{u^2} - {{119GM} \over {2R}}} \right)

=

5m\left( {{u^2} - {{119GM} \over {200R}}} \right)

Q55

A box weight 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 ms–2 at the north pole and the radius of the earth = 6400 km) :

A 194.32 N

B 195.66 N

C 195.32 N

D 194.66 N

Correct Answer

Option C

Solution

At equator, weight W = Mg - M

{\omega ^2}

R = 196 -

\left( {19.6} \right){\left( {{{2\pi } \over {24 \times 3600}}} \right)^2} \times 6400 \times {10^3}

= 195.32 N

Q56

Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :

A

\sqrt {{G \over 2}(1 + 2\sqrt 2 )}

B

\sqrt {{G \over 2}(2\sqrt 2 - 1)}

C

\sqrt {G(1 + 2\sqrt 2 )}

D

{1\over2}\sqrt {G(1 + 2\sqrt 2 )}

Correct Answer

Option D

Solution

Given, m = 1 kg, R = 1 m We know that,

F = {{G{m_1}{m_2}} \over {{r^2}}}

$\because$

{F_1} = {{Gmm} \over {{{(2R)}^2}}} = {{G{m^2}} \over {4{R^2}}}

and

{F_2} = {{Gmm} \over {{{(\sqrt 2 R)}^2}}} = {{G{m^2}} \over {2{R^2}}}

Net force on one particle,

{F_{net}} = {F_1} + {F_2}\cos 45^\circ + {F_2}\cos 45^\circ

= {F_1} + 2{F_2}\cos 45^\circ

= {{G{m^2}} \over {4{R^2}}} + 2\left( {{{G{m^2}} \over {2{R^2}}}} \right).{1 \over {\sqrt 2 }}

= {{G{m^2}} \over {4{R^2}}} + {{G{m^2}} \over {\sqrt 2 {R^2}}}

= {{G{m^2}} \over {{R^2}}}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right]

As the gravitational force provides the necessary centripetal force, so

{F_{net}} = {F_C} = {{m{v^2}} \over R}

Here, FC = centripetal force.

\Rightarrow {{G{m^2}} \over {{R^2}}}\left[ {{1 \over 4} + {1 \over {\sqrt 2 }}} \right] = {{m{v^2}} \over R}

\Rightarrow v = {1 \over 2}\sqrt {{{Gm} \over R}(1 + 2\sqrt 2 )}

\Rightarrow v = {1 \over 2}\sqrt {G(1 + 2\sqrt 2 )}

Q57

Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is :

A

{1 \over {2\pi }}\sqrt {{{{d^3}} \over {3Gm}}}

B

2\pi \sqrt {{{3Gm} \over {{d^3}}}}

C

{1 \over {2\pi }}\sqrt {{{3Gm} \over {{d^3}}}}

D

2\pi \sqrt {{{{d^3}} \over {3Gm}}}

Correct Answer

Option D

Solution

The given situation is shown below The gravitational force between these two stars provide the required centripetal force for rotation in a circle about their common centre.

Assuming 2 m at origin, the centre of mass of the system lies at

x = {{2m \times 0 + m \times d} \over {2m + m}} = {d \over 3}

Hence,

{F_G} = {F_C}

where, FG is gravitational force between them and FC is centripetal force.

\Rightarrow {{G{m_1}{m_2}} \over {{r^2}}} = 2m{\omega ^2}x

\Rightarrow {{G(2m)(m)} \over {{d^2}}} = 2m{\omega ^2} \times {d \over 3} \Rightarrow {\omega ^2} = {{3Gm} \over {{d^3}}}

\Rightarrow \omega = \sqrt {{{3Gm} \over {{d^3}}}}

We know that,

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

$\therefore$

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

\sqrt {{{3Gm} \over {{d^3}}}}

= 2\pi \sqrt {{{{d^3}} \over {3Gm}}}

[using Eq. (i)]

Q58

A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator? [Use

g = {{GM} \over {{R^2}}}

= 9.8 ms

-

2 and radius of earth, R = 6400 km.]

A 49 N

B 49.83 N

C 48.83 N

D 49.17 N

Correct Answer

Option C

Solution

Given, weight of body at North pole, wp = mg = 49 N Radius of Earth, R = 6400 km Let weight of body at equator be we.

At equator, ge = g $-$ R $\omega$ 2 $\therefore$ we = mge = m(g $-$ R $\omega$ 2) Since, wp > we $\Rightarrow$ we < 49 N Hence, above condition is satisfied by only option (b).

Q59

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R. Assertion A : The escape velocities of planet A and B are same. But A and B are of unequal mass. Reason R : The product of their mass and radius must be same. M1R1 = M2R2 In the light of the above statements, choose the most appropriate answer from the options given below :

A Both A and R are correct and R is the correct explanation of A

B Both A and R are correct but R is NOT the correct explanation of A

C A is correct but R is not correct

D A is not correct but R is correct

Correct Answer

Option C

Solution

{v_e}

= escape velocity

{v_e} = \sqrt {{{2GM} \over R}}

so for same

{v_e},{{{M_1}} \over {{R_1}}} = {{{M_2}} \over {{R_2}}}

A is true but R is false

Q60

A planet revolving in elliptical orbit has : A. a constant velocity of revolution. B. has the least velocity when it is nearest to the sun. C. its areal velocity is directly proportional to its velocity. D. areal velocity is inversely proportional to its velocity. E. to follow a trajectory such that the areal velocity is constant. Choose the correct answer from the options given below :

A D only

B E only

C C only

D A only

Correct Answer

Option B

Solution

According to Kepler’s second law of planetary motion, areal velocity of every planet moving around the sun should remain constant in elliptical orbit.