Gravitation — Page 2 | NEET Physics

Q11

A satellite is orbiting just above the surface of the earth with period

T

. If

d

is the density of the earth and

G

is the universal constant of gravitation, the quantity

\dfrac{3 \pi}{G d}

represents :

A

T^{2}

B

T^{3}

C

\sqrt{T}

D

T

Correct Answer

Option A

Solution

For a satellite orbiting just above the surface of the Earth, we can use the formula for the period T of the satellite in terms of the Earth's density d and the gravitational constant G:

T = 2\pi\sqrt{\frac{a^3}{GM}}

where a is the semi-major axis of the orbit (which is approximately equal to the Earth's radius R for a satellite orbiting just above the surface) and M is the mass of the Earth.

We can express the mass of the Earth M in terms of its density d and volume:

M = dV = d\times\frac{4}{3}\pi R^3

Now, substitute this expression for M into the equation for T:

T = 2\pi\sqrt{\frac{a^3}{Gd\times\frac{4}{3}\pi R^3}}

Since the satellite is orbiting just above the Earth's surface, we can approximate a ≈ R:

T = 2\pi\sqrt{\frac{R^3}{Gd\times\frac{4}{3}\pi R^3}}

Simplify the equation:

T = 2\pi\sqrt{\frac{1}{Gd\times\frac{4}{3}\pi}}

Now, square both sides of the equation:

T^2 = 4\pi^2\frac{1}{Gd\times\frac{4}{3}\pi}

Simplify further:

T^2 = \frac{3\pi}{Gd}

Thus, the quantity

\frac{3\pi}{Gd}

represents

T^2

.

Q12

A gravitational field is present in a region and a mass is shifted from A to B through different paths as shown. If W 1 , W 2 and W 3 represent the work done by the gravitational force along the respective paths, then :

A W 1 2 3

B W 1 = W 2 = W 3

C W 1 > W 2 > W 3

D W 1 > W 3 > W 2

Correct Answer

Option B

Solution

Since the gravitational field is conservative in nature hence the work done would depend only on the initial and final positions and not on the path followed by the mass.

Hence, W 1 = W 2 = W 3

Q13

In a gravitational field, the gravitational potential is given by,

V = - {K \over x}

(J/Kg). The gravitational field intensity at point (2, 0, 3) m is

A

+ {K \over 4}

B

+ {K \over 2}

C

- {K \over 2}

D

- {K \over 4}

Correct Answer

Option D

Solution

Gravitational field intensity

= {{ - \partial V} \over {\partial x}}\widehat i + {{ - \partial V} \over {\partial y}}\widehat j + {{ - \partial V} \over {\partial x}}\widehat k

where V is gravitational potential $\Rightarrow$ Gravitational field intensity

= {{ - \partial } \over {\partial x}}\left[ {{{ - K} \over x}} \right] = {{ - K} \over {{x^2}}}\widehat i

at (2, 0, 3)

\overrightarrow {{E_G}} = {{ - K} \over {{2^2}}} = {{ - K} \over 4}\widehat i

Q14

A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is

A 0.05 N/kg

B 50 N/kg

C 20 N/kg

D 180 N/kg

Correct Answer

Option B

Solution

F = m{E_G}

3 = {{60} \over {1000}}{E_G}

{E_G} = 50

N/kg

Q15

The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :

A 4

\upsilon

B

\upsilon

C 2

\upsilon

D 3

\upsilon

Correct Answer

Option A

Solution

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

= \sqrt {{{8\pi G\rho } \over 3}{R^2}}

\Rightarrow {v_e} \propto R

\Rightarrow {{{v_e}} \over \upsilon } = {{4R} \over R} \Rightarrow {v_e} = 4v

Q16

A particle of mass 'm' is projected with a velocity

\upsilon

= kV e = (k < 1) from the surface of the earth. (V e = escape velocity) The maximum height above the surface reached by the particle is :

A

{{R{k^2}} \over {1 - {k^2}}}

B

R{\left( {{k \over {1 - {k^2}}}} \right)^2}

C

R{\left( {{k \over {1 + {k^2}}}} \right)^2}

D

{{{R^2}k} \over {1 + k}}

Correct Answer

Option A

Solution

- {{GMm} \over R} + {1 \over 2}m{k^2}{v_e}^2 = - {{GMm} \over r}

- {{GMm} \over R} + {1 \over 2}m{k^2}{{2GM} \over R} = - {{GMm} \over r}

- {1 \over R} + {{{k^2}} \over R} = - {1 \over r}

{1 \over r} = {1 \over R} - {{{k^2}} \over R}

{1 \over r} = {{1 - {k^2}} \over R}

r = {R \over {1 - {k^2}}}

R + h = {R \over {1 - {k^2}}}

h = {R \over {1 - {k^2}}} - R = {{{k^2}} \over {1 - {k^2}}}R

Q17

A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

A 32 N

B 30 N

C 24 N

D 48 N

Correct Answer

Option A

Solution

We know W = mg At a height h, above the surface of earth, weight is given as,

mgh = {{mg} \over {\left[ {{{\left( {1 + {h \over R}} \right)}^2}} \right]}}

= {{72} \over {{{\left( {1 + {{R/2} \over R}} \right)}^2}}}

= {{72} \over {{{\left( {{3 \over 2}} \right)}^2}}}

= {4 \over 9} \times 72 = 32N

Q18

The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is :

A mgR

B 2mgR

C

{1 \over 2}

mgR

D

{3 \over 2}

mgR

Correct Answer

Option C

Solution

Initial potential energy at Earths surface is U i =

- {{GMm} \over R}

Final potential energy at height h = R : U f =

- {{GMm} \over 2R}

work done = change in PE w = U f - U i =

{{GMm} \over 2R}

=

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

=

{1 \over 2}

mgR

Q19

A body weight 200 N on the surface of earth. How much will it weight half way down to the centre of the earth?

A 200 N

B 250 N

C 100 N

D 150 N

Correct Answer

Option C

Solution

Accelerattion due to gravity at a depth d from surface of Earth

g

' =

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

....(1) Multiplying by mass 'm' on both sides of equation (1) m

g

' = m

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

=

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

( $\because$

{d = {R \over 2}}

) =

{{200} \over 2} = 100

N

Q20

If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct?

A Raindrops will fall faster.

B Walking on the ground would become more difficult.

C Time period of a simple pendulum on the Earth would decrease.

D g on the Earth will not change.

Correct Answer

Option D

Solution

If universal gravitational constant becomes ten times, then G' = 10 G Acceleration due to gravity, g =

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

So, acceleration due to gravity increases, hence ‘g’ on the Earth will not change is not correct.