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A projectile is projected with velocity ${kv}_{e}$ in vertically upward direction from the ground into the space. ( $v_{e}$ is escape velocity and k<1). If air resistance is considered to be negligible then the maximum height from the centre of earth to which it can go, will be : (R = radius of earth)

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

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

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

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

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Explanation

(c)Loss in Kinetic energy = Gain in Potential energy

$\frac{1}{2} m ({kv}_{e})^{2} = \frac{mgh}{1 + \frac{h}{R}} \Rightarrow \frac{1}{2} {mk}^{2} 2 gR = \frac{mgh}{1 + \frac{h}{R}} \Rightarrow h = \frac{{Rk}^{2}}{1 - k^{2}}$

Height of projectile from the earth's surface = h

Height from the centre $r = R + h = R + \frac{{Rk}^{2}}{1 - k^{2}}$

By solving $r = \frac{R}{1 - k^{2}}$

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A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is $R_{0}$ and mass of the earth M, the angular momentum about the centre of the earth is

$m \sqrt{G M R_{0}}$

$M \sqrt{G m R_{0}}$

$m \sqrt{\frac{G M}{R_{0}}}$

$M \sqrt{\frac{G M}{R_{0}}}$

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Explanation

(a) Angular momentum = Mass ×Orbital velocity ×Radius

$= m \times (\sqrt{\frac{G M}{R_{0}}}) \times R_{0} = m \sqrt{G M R_{0}}$

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If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been

64.5

129

182.5

730

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Explanation

(b) According to Kepler's third law, the ratio of the squares of the periods of any two planets revolving about the sun is equal to the ratio of the cubes of their average distances from the sun i.e.

${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{r_{1}}{r_{2}})}^{3} = {[\frac{r_{1}}{\frac{1}{2} r_{1}}]}^{3} = 8 \Rightarrow \frac{T_{1}}{T_{2}} = 2 \sqrt{2} ∴ T_{2} = \frac{T_{1}}{2 \sqrt{2}} = \frac{365 days}{2 \sqrt{2}} = 129 days$

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According to Kepler, the period of revolution of a planet (T) and its mean distance from the sun (r) are related by the equation

$T^{3} r^{3} = c o n s \tan t$

$T^{2} r^{‐ 3} = c o n s \tan t$

$T r^{3} = c o n s \tan t$

$T^{2} r = c o n s \tan t$

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Explanation

(b) $\frac{T^{2}}{r^{3}} = c o n s \tan t \Rightarrow T^{2} r^{‐ 3} = c o n s \tan t$

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A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface ( $R_{earth}$ = 6400 km) will approximately be -

1/2 h

1 h

2 h

4 h

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Explanation

(c) $\frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{3 / 2} \Rightarrow T_{2} = 24 {(\frac{6400}{36000})}^{3 / 2} ≅ 2 hour$

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A planet revolves around sun whose mean distance is 1.588 times the mean distance between earth and sun. The revolution time of planet will be

1.25 years

1.59 years

0.89 years

2 years

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Explanation

$\frac{T_{p l a n e t}}{T_{e a r t h}} = {[\frac{r_{p l a n e t}}{r_{e a r t h}}]}^{3 / 2} = {(1.588)}^{3 / 2} = 2 ∴ T_{p l a n e t} = 2 y e a r s$

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The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun ?

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Explanation

(c) $\frac{T_{A}}{T_{B}} = {(\frac{r_{A}}{r_{B}})}^{3 / 2} \Rightarrow 8 = {(\frac{r_{A}}{r_{B}})}^{3 / 2} \Rightarrow r_{A} = {(8)}^{2 / 3} r_{B} = 4 r_{B}$

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If the radius of earth's orbit is made 1/4 th, the duration of an year will become -

8 times

4 times

1/8 times

1/4 times

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Explanation

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If mass of a satellite is doubled and time period remain constant, the ratio of orbit in the two cases will be

1 : 2

1 : 1

1 : 3

None of these

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Explanation

(b) Mass of satellite does not affects on orbital radius.

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The earth revolves round the sun in one year. If the distance between them becomes double, the new period of revolution will be

1/2 year

$2 \sqrt{2}$ years

4 years

8 years

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Explanation

(b) $\frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{3 / 2} = {(2)}^{3 / 2} = 2 \sqrt{2} \Rightarrow T_{2} = 2 \sqrt{2} y e a r s$

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Yes. You can attempt every NEET 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.