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Physics MCQs for NEET — Practice Questions with Answers

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A geostationary satellite

Revolves about the polar axis

Has a time period less than that of the near-earth satellite

Moves faster than a near-earth satellite

Is stationary in the space

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Explanation

(a) A geosstationary satellite revolved in the same direction the earth rotates about the polar axis i.e. west to east.

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A small satellite is revolving near earth's surface. Its orbital velocity will be nearly

8 km/sec

11.2 km/sec

4 km/sec

6 km/sec

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Explanation

(a) Orbital velocity $v_{0} = \sqrt{g R_{e}}$

Here, g=9.8 m/ $s^{2}$ , $R_{e}$ = $6.4 \times 10^{6} m$

$H e n c e, v_{0} = \sqrt{9.8 \times 6.4 \times 106} = 7.91 \times 10^{3} m / s = 7.91 K m / s = 8 K m / s (a p p r o x)$

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The distance of neptune and saturn from sun are nearly $10^{13}$ and $10^{12}$ meters respectively. Assuming that they move in circular orbits, their periodic times will be in the ratio

$\sqrt{10}$

100

$10 \sqrt{10}$

$1 / \sqrt{10}$

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Explanation

(c) $\frac{T_{1}}{T 2} = {(\frac{R_{1}}{R_{2}})}^{3 / 2} = {(\frac{10^{13}}{10^{12}})}^{3 / 2} = {(1000)}^{1 / 2} = 10 \sqrt{10}$

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The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is v. For a satellite orbiting at an altitude of half of the earth's radius, the orbital velocity is

$\frac{3}{2} v$

$\sqrt{\frac{3}{2}} v$

$\sqrt{\frac{2}{3}} v$

$\frac{2}{3} v$

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Explanation

For first satellite $h = 0$ , $v_{1} = \sqrt{\frac{G M}{R}}$

For second satellite , $h = \frac{R}{2}, v_{2} = \sqrt{\frac{2 G M}{3 R}}$

$v_{2} = \sqrt{\frac{2}{3}} v_{1} = \sqrt{\frac{2}{3}} v$

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In a satellite if the time of revolution is T, then K.E. is proportional to

$\frac{1}{T}$

$\frac{1}{T^{2}}$

$\frac{1}{T^{3}}$

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

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Explanation

(d) $v = \sqrt{\frac{G M}{r}} ∴ K . E . \propto v^{2} \propto \frac{1}{r}$ and $T^{2} \propto r^{3} ∴ K . E . \propto T^{\frac{- 2}{3}}$

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The period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4R is

$\frac{T}{4}$

$\frac{T}{8}$

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Explanation

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If the height of a satellite from the earth is negligible in comparison to the radius of the earth R, the orbital velocity of the satellite is

gR/2

$\sqrt{g / R}$

$\sqrt{g R}$

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Explanation

(d) Orbital velocity near to earth = $\frac{v_{e}}{\sqrt{2}}$

$\Rightarrow v_{0} = \frac{\sqrt{2 g r}}{\sqrt{2}} = \sqrt{g r}$

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Choose the correct statement from the following : The radius of the orbit of a geostationary satellite depends upon -

Mass of the satellite, its time period and the gravitational constant

Mass of the satellite, mass of the earth and the gravitational constant

Mass of the earth, mass of the satellite, time period of the satellite and thegravitational constant

Mass of the earth, time period of the satellite and the gravitational constant

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Explanation

(d)

$T = 2 π \sqrt{\frac{r^{3}}{GM}} \Rightarrow r^{3} = \frac{G M T^{2}}{4 π^{2}} \Rightarrow r = {[\frac{G M T^{2}}{4 π^{2}}]}^{\frac{1}{3}}$

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If a body describes a circular motion under inverse square field, the time taken to complete one revolution T is related to the radius of the circular orbit as

$T \propto r$

$T \propto r^{2}$

$T^{2} \propto r^{3}$

$T \propto r^{4}$

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Explanation

(c)

$In a circular motion : F = \frac{{mv}^{2}}{r} = {mrω}^{2} In a inverse square field; F \propto \frac{1}{r^{2}} \Rightarrow {mrω}^{2} \propto \frac{1}{r^{2}} \Rightarrow ω^{2} \propto \frac{1}{r^{3}} \Rightarrow \frac{1}{T^{2}} \propto \frac{1}{r^{3}} \Rightarrow T^{2} \propto r^{3}$

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A planet moves around the sun. At a given point P, it is closest from the sun at a distance $d_{1}$ and has a speed $v_{1}$ . At another point Q, when it is farthest from the sun at a distance $d_{2}$ , its speed will be

$\frac{{d_{1}}^{2} v_{1}}{{d_{2}}^{2}}$

$\frac{d_{2} v_{1}}{d_{1}}$

$\frac{d_{1} v_{1}}{d_{2}}$

$\frac{{d_{2}}^{2} v_{1}}{{d_{1}}^{2}}$

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Explanation

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