Gravitation — Page 9 | JEE Physics

Q81

An object is taken to a height above the surface of earth at a distance

{5 \over 4}

R from the centre of the earth. Where radius of earth, R = 6400 km. The percentage decrease in the weight of the object will be :

A 36%

B 50%

C 64%

D 25%

Correct Answer

Option A

Solution

The weight of an object at a distance d from the center of the Earth is given by:

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

where: W' is the weight at distance d, W is the weight at the Earth's surface, R is the radius of the Earth, d is the distance from the center of the Earth.

In this problem, we are asked to find the percentage decrease in weight, which can be calculated by:

\Delta W = \frac{W - W'}{W} \times 100\%

where ΔW is the percentage change in weight. Substituting the weight formula into the percentage change formula gives:

\Delta W = \left(1 - \left(\frac{R}{d}\right)^2\right) \times 100\%

The radius of the Earth R is 6400 km and the distance d from the center of the Earth is given as $\dfrac{5}{4}R$ .

Substituting these values into the equation gives:

\Delta W = \left(1 - \left(\frac{6400~km}{\frac{5}{4} \cdot 6400~km}\right)^2\right) \times 100\% = \left(1 - \left(\frac{4}{5}\right)^2\right) \times 100\% = 36\%

Therefore, the percentage decrease in the weight of the object when taken to a height of $\dfrac{1}{4}R$ above the surface of the Earth is 36%.

Q82

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

Q83

Two satellites

\mathrm{A}

and

\mathrm{B}

, having masses in the ratio

4: 3

, are revolving in circular orbits of radii

3 \mathrm{r}

and

4 \mathrm{r}

respectively around the earth. The ratio of total mechanical energy of

\mathrm{A}

to

\mathrm{B}

is :

A 9 : 16

B 16 : 9

C 1 : 1

D 4 : 3

Correct Answer

Option B

Solution

U = - {{G{M_e}m} \over {2r}}

So,

{{{U_A}} \over {{U_B}}} = {{{m_A}} \over {{m_B}}} \times {{{r_B}} \over {{r_A}}}

= {4 \over 3} \times {4 \over 3} = {{16} \over 9}

Q84

A body of mass

\mathrm{m}

is projected with velocity

\lambda \,v_{\mathrm{e}}

in vertically upward direction from the surface of the earth into space. It is given that

v_{\mathrm{e}}

is escape velocity and $$\lambda (R : radius of earth)

A

\dfrac{\mathrm{R}}{1+\lambda^{2}}

B

\dfrac{R}{1-\lambda^{2}}

C

\dfrac{R}{1-\lambda}

D

\dfrac{\lambda^{2} \mathrm{R}}{1-\lambda^{2}}

Correct Answer

Option B

Solution

Using energy conservation

- {{G{M_e}m} \over {{R_e}}} + {1 \over 2}m{\left( {\lambda \sqrt {{{2G{M_e}} \over {{R_e}}}} } \right)^2} = - {{G{M_e}m} \over r}

{{G{M_e}m} \over r} = {{G{M_e}m} \over {{R_e}}} - {{G{M_e}m} \over {{R_e}}}{\lambda ^2}

r = {{{R_e}} \over {1 - {\lambda ^2}}}

Q85

If the radius of earth shrinks by

2 \%

while its mass remains same. The acceleration due to gravity on the earth's surface will approximately :

A decrease by

2 \%

B decrease by

4 \%

C increase by

2 \%

D increase by

4 \%

Correct Answer

Option D

Solution

The acceleration due to gravity (g) on the surface of a planet is given by the formula:

g = \frac{G M}{R^2}

where: G is the gravitational constant, M is the mass of the planet, R is the radius of the planet.

If the radius of the Earth shrinks by 2% but its mass remains the same, the new acceleration due to gravity (g') will be:

g' = \frac{G M}{(0.98R)^2} = \frac{G M}{0.9604 R^2} = \frac{g}{0.9604}

This implies that g' is approximately 1.0412 times g, or an increase of approximately 4.12%.

Therefore, the acceleration due to gravity on the Earth's surface will approximately increase by 4%.

Q86

The escape velocities of two planets

\mathrm{A}

and

\mathrm{B}

are in the ratio

1: 2

. If the ratio of their radii respectively is

1: 3

, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be :

A

\dfrac{4}{3}

B

\dfrac{2}{3}

C

\dfrac{3}{4}

D

\dfrac{3}{2}

Correct Answer

Option C

Solution

The escape velocity of a planet is given by the formula:

\mathrm{v}_{\text{escape}} = \sqrt{\frac{2GM}{R}}

where G is the gravitational constant, M is the mass of the planet, and R is its radius.

If the escape velocity of planet A is vA and the escape velocity of planet B is vB, then we can write the following relationship:

{{{v_B}} \over {{v_A}}} = {{\sqrt {{{2G{M_B}} \over {{R_B}}}} } \over {\sqrt {{{2G{M_A}} \over {{R_A}}}} }} = \sqrt {{{{R_A}{M_B}} \over {{M_A}{R_B}}}}

$\Rightarrow$

\sqrt {{{{R_A}{M_B}} \over {{M_A}{R_B}}}} = 2

$\Rightarrow$

{{{M_B}} \over {{M_A}}} \times {1 \over 3} = 4

$\Rightarrow$

{{{M_B}} \over {{M_B}}} = 12

We know, The acceleration due to gravity on a planet can be calculated using the formula:

\mathrm{g} = \frac{\mathrm{G} \mathrm{M}}{\mathrm{R}^2}

where G is the gravitational constant, M is the mass of the planet, and R is its radius.

{{{g_A}} \over {{g_B}}} = {{{M_A}R_B^2} \over {{M_B}R_A^2}} = {1 \over {12}} \times {\left( {{3 \over 1}} \right)^2} = {9 \over {12}} = {3 \over 4}

Q87

A body weight

\mathrm{W}

, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be :

A

\dfrac{W}{91}

B

\dfrac{\mathrm{W}}{3}

C

\dfrac{\mathrm{W}}{100}

D

\dfrac{\mathrm{W}}{9}

Correct Answer

Option C

Solution

The weight of an object varies with altitude due to the change in gravitational force.

The force of gravity decreases with the square of the distance from the center of the Earth.

The gravitational force (weight) at a height h from the surface of the Earth 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 height h is given as nine times the radius of the Earth (h = 9R).

Substituting h = 9R into the equation gives:

W' = W \left(\frac{R}{{R + 9R}}\right)^2 = W \left(\frac{1}{10}\right)^2

Simplifying this gives:

W' = \frac{W}{100}

Therefore, the weight of the body at that height will be 1/100th of its weight on the surface of the Earth.

The correct answer is Option C: $\dfrac{W}{100}$ .

Q88

Given below are two statements: Statement I: Acceleration due to gravity is different at different places on the surface of earth. Statement II: Acceleration due to gravity increases as we go down below the earth's surface. 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 false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option D

Solution

Statement I is true.

The acceleration due to gravity varies slightly over the surface of the Earth, being affected by factors such as latitude (because of the Earth's oblate shape or its equatorial bulge), altitude (it decreases with height above the Earth's surface), and local geological variations in the Earth's density.

Statement II is false.

The acceleration due to gravity actually decreases as we go down below the Earth's surface.

This is because the mass that is "above" or outside the location begins to pull the object away from the Earth's center, and the net gravitational acceleration decreases.

Therefore, Option D: Statement I is true but Statement II is false, is the correct answer.

Q89

At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height

\mathrm{3 R}

above earth surface. Where

\mathrm{R}

is Radius of earth (Take

\mathrm{R}=6400 \mathrm{~km}

). The depth

\mathrm{d}

is equal to

A 5260 km

B 2560 km

C 640 km

D 4800 km

Correct Answer

Option D

Solution

The acceleration due to gravity

g

at a distance

d

below the surface of the earth is given by : $g_{d}=\dfrac{G M}{R^{3}}(R-d)$ (depth variation) where

G

is the gravitational constant and

M

is the mass of the Earth. At a height

3R

above the surface of the Earth, the acceleration due to gravity $g_{h}$ is given by:

g_{h}=\frac{G M}{(R+3R)^{2}}

Given, $g_{d}=4 g_{h} \\\\$ , so we can write :

\begin{aligned} & \frac{G M}{R^{3}}(R-d)=4 \frac{G M}{(R+3 R)^{2}} \\\\ & \Rightarrow R-d=\frac{R}{4} \\\\ & \Rightarrow d=\frac{3 R}{4} \\\\ & \Rightarrow d=4800 \mathrm{~km} \end{aligned}

Q90

An object is allowed to fall from a height

R

above the earth, where

R

is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be

A

\sqrt{\dfrac{g R}{2}}

B

\sqrt{g R}

C

\sqrt{2 g R}

D

2 \sqrt{g R}

Correct Answer

Option B

Solution

{U_P} = - {{GMm} \over {2R}}

{U_S} = - {{GMm} \over R}

$\Rightarrow$ Energy conservation

{1 \over 2}m{v^2} - {{GMm} \over R} = - {{GMm} \over {2R}}

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

v = \sqrt {{{GM} \over R}} = \sqrt {gR}