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Physics MCQs for NEET — Practice Questions with Answers

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Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e. , $F \propto 1 / r^{3}$ but still remaining a central force. Then -

Keplers law of areas still holds

Keplers law of period still holds

Keplers law of areas and period still hold

Neither the law of areas, nor the law of period still hold

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Explanation

(d) According to Keplers second law,

$\frac{∆ A}{∆ t} = \frac{L}{2 M}$

Since force is central, $τ = 0$

Hence L=constant

Now, $T α r^{3} / 2$

$F = \frac{G M m}{r^{3}} = \frac{m v^{2}}{r} \Rightarrow v = \frac{\sqrt{G M}}{r} T = \frac{2 {πr}^{2}}{r} \Rightarrow T α r^{2}$

So, Keplers law doesnot hold

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The distance of a planet from the sun is 5 times the distance between the earth and the sun. The time period of the planet is -

$5^{3 / 2}$ years

$5^{2 / 3}$ years

$5^{1 / 3}$ years

$5^{1 / 2}$ years

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Explanation

(a)

$\frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{{R_{1}}^{3}}{{R_{2}}^{3}} T_{2} = \sqrt{\frac{{R_{2}}^{3}}{{R_{1}}^{3}} {T_{1}}^{2}} = 5^{\frac{2}{3}} y e a r s$

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In an elliptical orbit under gravitational force, in general

Tangential velocity is constant

Angular velocity is constant

Radial velocity is constant

Areal velocity is constant

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Explanation

(d) Areal velocity = $\frac{d A}{d t} = \frac{L}{2 M}$

Here, L and M is cosntant

Areal velocity is constant

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Earth is revolving around the sun, if the distance of the Earth from the Sun is reduced to 1/4^th of the present distance then the present year length reduced to -

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{1}{6}$

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Explanation

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Two satellite are revolving around the earth with velocities $v_{1}$ and $v_{2}$ and in radii $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. Then -

$v_{1} = v_{2}$

$v_{1} > v_{2}$

$v_{1} < v_{2}$

$\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}$

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Explanation

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If orbital velocity of planet is given by $v = G^{a} M^{b} R^{c}$ , then -

$a = 1 / 3, b = 1 / 3, c = ‐ 1 / 3$

$a = 1 / 2, b = 1 / 2, c = ‐ 1 / 2$

$a = 1 / 2, b = ‐ 1 / 2, c = 1 / 2$

$a = 1 / 2, b = ‐ 1 / 2, c = ‐ 1 / 2$

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Explanation

(b) $v = \sqrt{\frac{G M}{R}} = G^{1 / 2} M^{1 / 2} R^{‐ 1 / 2}$

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The mass of a planet that has a moon whose time period and orbital radius are T and R respectively can be written as -

$4 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$8 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$12 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$16 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

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Explanation

(a) $m ω^{2} R = \frac{G M m}{R^{2}} \Rightarrow {(\frac{2 π}{T})}^{2} R = \frac{G M}{R^{2}} \Rightarrow M = \frac{4 π^{2} R^{3}}{G T^{2}}$

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The eccentricity of earth's orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is

2.507

1.033

8.324

1.000

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Explanation

(b) $\frac{v_{m a x}}{v_{m i n}} = \frac{1 + e}{1 - e} = \frac{1 + 0.0167}{1 - 0.0167} = 1.033$

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The radius of orbit of a planet is two times that of the earth. The time period of planet is -

4.2 years

2.8 years

5.6 years

8.4 years

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Explanation

(b) $T_{2} = T_{1} {[\frac{R_{2}}{R_{1}}]}^{3 / 2} = 1 \times {(2)}^{3 / 2} = 2.8 y e a r s$

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The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r, then the new angular momentum will be -

16 L

64 L

$\frac{L}{4}$

4 L

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Explanation

(d) $L = m v r = m \sqrt{\frac{G M}{r} r} = m \sqrt{G M r} ∴ L \propto \sqrt{r}$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.