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Physics NEET Questions — Free MCQ Practice

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Potential energy of a satellite having mass ‘m’ and rotating at a height of $6.4 \times 10^{6} m$ from the earth surface is -

A.

$- 0.5 m g R_{e}$

B.

$- m g R_{e}$

C.

$- 2 m g R_{e}$

D.

$4 m g R_{e}$

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Explanation

(a) Potential energy = $= \frac{- G M m}{r} = \frac{G M m}{R_{e} + h} = \frac{- G M m}{2 R_{e}}$

$= - \frac{g R^{2} m}{2 R_{e}} = - \frac{1}{2} m g R_{e} = - 0.5 m g R_{e}$

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When a satellite going round the earth in a circular orbit of radius r and speed v loses some of its energy, then r and v change as

A.

r and v both with increase

B.

r and v both will decrease

C.

r will decrease and v will increase

D.

r will decrease and v will decrease

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Explanation

(c) $E_{T o t a l} = - \frac{G M m}{2 r} w h e r e r i s t h e r a d i u s o f o r b i t S o i f e n e r g y d e c r e a s e s, r a l s o d e c r e a s e s . A l s o, K . E . = \frac{G M m}{2 r} S o, i f r a d i u s d e c r e a s e s, K . E . i n c r e a s e s$ If B.E. decreases then r also decreases and v increases as

$v \propto \frac{1}{\sqrt{r}}$

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Which of the following quantities does not depend upon the orbital radius of the satellite ?

A.

$\frac{T}{R}$

B.

$\frac{T^{2}}{R}$

C.

$\frac{T^{2}}{R^{2}}$

D.

$\frac{T^{2}}{R^{3}}$

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Explanation

(d) $T^{2} \propto R^{3} ∴ \frac{T^{2}}{R^{3}}$ = constant

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A satellite moves round the earth in a circular orbit of radius R making one revolution per day. A second satellite moving in a circular orbit, moves round the earth once in 8 days. The radius of the orbit of the second satellite is -

A.

8 R

B.

4R

C.

2R

D.

R

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Explanation

(b) Given that, $T_{1} = 1$ day and $T_{2} = 8$ days

$∴ \frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{\frac{3}{2}} \Rightarrow \frac{r_{2}}{r_{1}} = {(\frac{T_{2}}{T_{1}})}^{\frac{2}{3}} = {(\frac{8}{1})}^{\frac{2}{3}} = 4 \Rightarrow r_{2} = 4 r_{1} = 4 R$

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A satellite moves in a circle around the earth. The radius of this circle is equal to one half of the radius of the moon’s orbit. The satellite completes one revolution in

A.

$\frac{1}{2}$ lunar month

B.

$\frac{2}{3}$ lunar month

C.

$2^{\frac{- 3}{2}}$ lunar month

D.

$2^{\frac{3}{2}}$ lunar month

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Explanation

(c) Time period of revolution of moon around the earth = 1 lunar month.

$\frac{T_{s}}{T_{m}} = {(\frac{r_{s}}{r_{m}})}^{\frac{3}{2}} = {(\frac{1}{2})}^{\frac{3}{2}} \Rightarrow T_{s} = 2^{\frac{- 3}{2}}$ lunar month.

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A satellite of mass m is placed at a distance r from the centre of earth (mass M). The mechanical energy of the satellite is

A.

$- \frac{G M m}{r}$

B.

$\frac{G M m}{r}$

C.

$\frac{G M m}{2 r}$

D.

$- \frac{G M m}{2 r}$

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Explanation

(d) Mechanical energy = Kinetic energy + potential energy

$\Rightarrow \frac{1}{2} m v^{2} + (\frac{- G M m}{r}) = \frac{1}{2} m (\frac{G M}{r}) - \frac{G M m}{r} [a s v = \sqrt{\frac{G M}{r}}]$

Hence, mechanical energy = $\frac{- G M m}{2 r}$

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The acceleration due to gravity at a height 1km above the earth is the same as at a depth d below the surface of earth.Then

A.

$d = \frac{1}{2} k m$

B.

d=1km

C.

$d = \frac{3}{2} k m$

D.

d=2km

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Explanation

(d) Thinking process $g_{h}$ = Acceleration due to gravity at height above earth's surface

= $g (\frac{R}{R + h})$

= $g (1 - \frac{2 h}{R})$

$g_{d}$ =Acceleration at depth d below earth's surface

= $g (1 - \frac{d}{R})$

Given, when h=1km, $g_{d}$ = $g_{h}$

or $g (1 - \frac{d}{R})$ =g $(1 - \frac{2 h}{R})$

$\Rightarrow$ d=2h

0r d=2km

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Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will

A.

keep floating at the same distance between them

B.

move towards each other

C.

move away from each other

D.

will become stationary

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Explanation

(b) In the space, there is no external gravity. Due to masses of the astronauts , there will be small gravitational attractive force between them. Thus, these astronauts will move towards each other.

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A satellite of mass m is orbiting the earth [of radius R] at a height h from its surface. The total energy of the satellite in terms of $g_{o}$ , the value of acceleration due to gravity at the earth's surface is -

A.

$\frac{m g_{o} R^{2}}{2 (R + h)}$

B.

$- \frac{m g_{o} R^{2}}{2 (R + h)}$

C.

$\frac{2 m g_{o} R^{2}}{R + h}$

D.

$- \frac{2 m g_{o} R^{2}}{2 (R + h)}$

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Explanation

$(b) T o t a l e n e r g y o f a s a t e l l i t e a t h e i g h t h = K E + P E = \frac{G M m}{2 (R + h)} - \frac{G M m}{(R + h)}$

$= \frac{- G M m}{2 (R + h)} = \frac{- G M m R^{2}}{2 R^{2} (R + h)}$

= $\frac{- m g_{o} R^{2}}{2 (R + h)} (∵ g_{o} = \frac{G M}{R^{2}})$

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At what height from the surface of earth the gravitation potential and the value of g are $- 5.4 \times 10^{7} J k g^{- 2}$ and $6.0 m s^{- 2}$ respectively? (Take, the radius of earth as 6400 km.)

A.

1600 km

B.

1400 km

C.

2000 km

D.

2600 km

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Explanation

(d) Gravitational potential at some height h from the surface of the earth is given by

$V = - \frac{G M}{R + h}$ ...(i)

And accleration due to gravity at some height h from the earth surface can be given as

$g' = \frac{G M}{{(R + h)}^{2}}$ ...(ii)

From eq.(i) and (ii), we get

$\frac{|V|}{g'} = \frac{G M}{(R + h)} \times \frac{{(R + h)}^{2}}{G M}$

$\Rightarrow \frac{|V|}{g'} = R + h$ ...(iii)

$∵ V = 5.4 \times 10^{7} J k g^{- 2} a n d g' = 6.0 m s^{- 2}$

Radius of earth, R=6400 km.

Substitute these values in eq. (iii), we get

$\frac{5.4 \times 10^{7}}{6.0} = R + h \Rightarrow 9 \times 10^{6} = R + h \Rightarrow h = (9 - 6.4) \times 10^{6} = 2.6 \times 10^{6} m \Rightarrow h = 2600 k m$

Physics MCQs for NEET — Practice Questions with Answers

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