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Physics MCQs for NEET — Practice Questions with Answers

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A body of mass m is placed on the earth’s surface. It is taken from the earth’s surface to a height 3R . The change in gravitational potential energy of the body is -

$\frac{2}{3} m g R$

$\frac{3}{4} m g R$

$\frac{m g R}{2}$

$\frac{m g R}{4}$

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Explanation

(b) $∆ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g \times 3 R}{1 + \frac{3 R}{R}} = \frac{3}{4} m g R$

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Weight of 1 kg becomes 1/6 kg on moon. If radius of moon is $1.768 \times 10^{6} m$ , then the mass of moon will be -

$1.99 \times 10^{30} k g$

$7.56 \times 10^{22} k g$

$5.98 \times 10^{24} k g$

$7.65 \times 10^{22} k g$

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Explanation

(d) $g_{m} = \frac{G M_{m}}{{R_{m}}^{2}}$ and $g_{m} = \frac{g_{e}}{6} = \frac{9.8}{6} m / s^{2} = 1.63 m / s^{2}$

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A geo-stationary satellite is orbiting the earth at a height of 6 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is

10 hr

$(6 / \sqrt{2}) h r$

6 hr

$6 \sqrt{2} h r$

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Explanation

(d) Distances of the satellite from the centre are 7R and 3.5R respectively.

$\frac{T_{2}}{T_{1}} = {(\frac{R_{2}}{R_{1}})}^{\frac{3}{2}} \Rightarrow T_{2} = 24 {(\frac{3.5 R}{7 R})}^{\frac{3}{2}} = 6 \sqrt{2} h r$

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A body of mass m kg starts falling from a point 2R above the Earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the Earth’s surface [R - Radius of Earth, M - Mass of Earth, G - Gravitational Constant] -

$\frac{1}{2} \frac{G M m}{R}$

$\frac{1}{6} \frac{G M m}{R}$

$\frac{2}{3} \frac{G M m}{R}$

$\frac{1}{3} \frac{G M m}{R}$

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Explanation

(b) Potential energy, U is given by

$U = - \frac{G M m}{R} = - \frac{G M m}{R + h} U_{i n i t i a l} = - \frac{G M m}{3 R} a n d U_{f i n a l} = - \frac{- G M m}{2 R} L o s s i n P E = g a i n i n K E = \frac{G M m}{2 R} - \frac{G M m}{3 R} = \frac{G M m}{6 R}$

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Let g be the acceleration due to gravity at earth's surface and K be the rotational kinetic energy of the earth. Suppose the earth's radius decreases by 2% keeping all other quantities same, then

g decreases by 2% and K decreases by 4%

g decreases by 4% and K increases by 2%

g increases by 4% and K increases by 4%

g decreases by 4% and K increases by 4%

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Explanation

c) $g = \frac{G M}{R^{2}}$ and $K = \frac{L^{2}}{2 I}$

If mass of the earth and its angular momentum remains constant then $g \propto \frac{1}{R^{2}}$ and $K \propto \frac{1}{R^{2}}$

i.e. if radius of earth decreases by 2% then g and K both increases by 4%.

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The distance between centre of the earth and moon is 384000 km. If the mass of the earth is $6 \times 10^{24} k g$ and $G = 6.66 \times 10^{- 11} N m^{2} / k g^{2}$ . The speed of the moon is nearly -

1 km/sec

4 km/sec

8 km/sec

11.2 km/sec

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Explanation

(a) $v = \sqrt{\frac{G M}{r}} = \sqrt{\frac{6.67 \times 10^{- 11} \times 6 \times 10^{24}}{384000 \times 10^{3}}} = 1 k m / s$

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A body is projected vertically upwards from the surface of a planet of radius R with a velocity equal to half the escape velocity for that planet. The maximum height attained by the body is -

R/3

R/2

R/4

R/5

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Explanation

$Total energy on the surface = total enrgy on height h - \frac{GMm}{R} + \frac{1}{2} m \frac{2 GM}{4 R} = - \frac{GMm}{(R + h)} - \frac{1}{R} + \frac{1}{4 R} = - \frac{1}{(R + h)} R + h = \frac{4 R}{3} h = \frac{R}{3}$

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A satellite is launched into a circular orbit of radius ‘R’ around earth while a second satellite is launched into an orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is

0.7

1.0

1.5

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Explanation

$(d) T = \frac{2 {πr}^{\frac{3}{2}}}{\sqrt{GM}} \frac{∆ T}{T} \times 100 = \frac{3}{2} \times \frac{∆ r}{r} \times 100 = \frac{3}{2} \times \frac{0.02 R}{R} \times 100 = 3 %$

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If the radius of the earth shrinks by 1.5% (mass remaining same), then the value of acceleration due to gravity changes -

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Explanation

Percentage change in g = 2(percentage change in R)

= $2 \times 1.5 = 3 %$

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Distance of geostationary satellite from the surface of the earth $r a d i u s (R_{e} = 6400 k m)$ in terms of $R_{e}$ is

$13.76 R_{e}$

$10.76 R_{e}$

$6.56 R_{e}$

$2.56 R_{e}$

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Explanation

(c)The distance of a geostationary satellite from centre of earth, r=36000 + 6400 km=42400 km

Hence, $d = \frac{42400}{6400} R_{e} = 6.56 R_{e}$

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