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A satellite is moving very close to a planet of density $ρ$ . The time period of the satellite is:

$\sqrt{\frac{3 π}{ρG}}$

${(\frac{3 π}{ρG})}^{3 / 2}$

$\sqrt{\frac{3 π}{2 ρG}}$

${(\frac{3 π}{2 ρG})}^{3 / 2}$

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Explanation

$T = \frac{2 πR}{v} = \frac{2 πR}{\sqrt{\frac{G M}{R}}} = \frac{2 {πR}^{\frac{3}{2}}}{\sqrt{G M}} T = \frac{2 π R^{\frac{3}{2}}}{\sqrt{G \frac{4}{3} {πR}^{3} ρ}} = \frac{2 {πR}^{\frac{3}{2}}}{2 \sqrt{\frac{π}{3} ρ G} R^{\frac{3}{2}}} = \sqrt{\frac{3 π}{ρ G}}$

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A projectile is fired upwards from the surface of the earth with a velocity ${kv}_{e}$ where $v_{e}$ is the escape velocity and k < 1. If r is the maximum distance from the center of the earth to which it rises and R is the radius of the earth, then r equals

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

$\frac{R}{1 - k^{2}}$

$\frac{2 R}{1 - k^{2}}$

$\frac{2 R}{1 + k^{2}}$

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Explanation

$(2) K_{i} + U_{i} = K_{f} + U_{f} \frac{1}{2} m {(k v_{e})}^{2} - \frac{G M m}{R} = 0 - \frac{G M m}{r} \frac{1}{2} m {(k \sqrt{\frac{2 G M}{R}})}^{2} - \frac{G M m}{R} = 0 - \frac{G M m}{r} \frac{k^{2}}{R} - \frac{1}{R} = - \frac{1}{r} r = \frac{R}{1 - k^{2}}$

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The gravitational potential difference between the surface of a planet and 10 m above is 5 J/kg. If the gravitational field is supposed to be uniform, the work done in moving a 2 kg mass from the surface of the planet to a height of 8 m is

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Explanation

$P o t e n t i a l d i f f e r e n c e a t 10 m = 5 J / k g P o t e n t i a l d i f f e r e n c e a t 8 m = \frac{5}{10} \times 8 = 4 J / k g W o r k d o n e = m ∆ V = 2 \times 4 = 8 J$

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A planet is moving in an elliptical orbit. If T, V, E, and L stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the center of orbit, then

T is conserved

V is always positive

E is always negative

the magnitude of L is conserved but its direction changes continuously

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Explanation

The direction of angular momentum does not change. It remains same at each and every point of the path. We can find the direction of angular momentum using right hand thumb rule.

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In planetary motion the areal velocity of the position vector of a planet depends on the angular velocity ( $ω$ ) and the distance of the planet from the sun (r). The correct relation for areal velocity is:

$\frac{d A}{d t} α ω r$

$\frac{d A}{d t} α ω^{2} r$

$\frac{d A}{d t} α ω r^{2}$

$\frac{d A}{d t} α \sqrt{ω r}$

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Explanation

$(3) A = {πr}^{2} \frac{dA}{dt} = 2 πr \frac{dr}{dt} = 2 πrv \frac{dA}{dt} = 2 {πr}^{2} ω$

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If A is the areal velocity of a planet of mass M, its angular momentum is

$\frac{M}{A}$

2MA

$A^{2} M$

${AM}^{2}$

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Explanation

Angular momentum

= 2 x mass x Areal speed = 2 MA

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Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

$- \frac{5 Gm}{r}$

$- \frac{6 Gm}{r}$

$- \frac{9 Gm}{r}$

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Explanation

$(3) L e t g r a v i t a t i o n a l f i e l d i s z e r o a t x d i s \tan c e f r o m m a s s m . \Rightarrow \frac{G m}{x^{2}} = \frac{4 G m}{{(r - x)}^{2}} \Rightarrow {(r - x)}^{2} = 4 x^{2} \Rightarrow r - x = 2 x \Rightarrow x = \frac{r}{3} P o t e n t i a l a t t h a t p o i n t : V = \frac{- G m}{x} - \frac{4 G m}{r - x} V = \frac{- G m}{\frac{r}{3}} - \frac{4 G m}{\frac{2 r}{3}} V = \frac{- 9 G m}{r}$

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The satellite of mass m orbiting around the earth in a circular orbit with a velocity v. The total energy will be:

$\frac{3}{4} {mv}^{2}$

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

$- \frac{1}{2} {mv}^{2}$

${mv}^{2}$

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Explanation

$E = - K = - \frac{m v^{2}}{2}$

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Magnitude of potential energy (U) and time period (T) of a satellite are related to each other as:

$T^{2} α \frac{1}{U^{3}}$

$T α \frac{1}{U^{3}}$

$T^{2} α U^{3}$

$T^{2} α \frac{1}{U^{2}}$

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Explanation

$|U| = \frac{G M m}{r} o r r α \frac{1}{U} T = \frac{2 πr}{V} = \frac{2 {πr}^{3 / 2}}{\sqrt{G m}} T^{2} α r^{3} α \frac{1}{U^{3}}$

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A projectile fired vertically upwards with a speed v escapes from the earth. If it is to be fired at 45 $°$ to the horizontal, what should be its speed so that it escapes from the earth?

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

$\sqrt{2} v$

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Explanation

Escape velocity is independent of the angle of projection.

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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.