Gravitation NEET PYQ – Previous Year Questions with Solutions

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Topic Question Set

Q 1 :
The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury? [2025]

172 earth days
124 earth days
88 earth days
225 earth days

1

2

3

4

(3)

Given, $R_{martian} = 4 R_{mercury}$

$T_{martian} = 687$ Earth days

$T_{mercury} = ?$

By Kepler's Third Law

${(\frac{T_{Mercury}}{T_{Martian}})}^{2} = {(\frac{R_{Mercury}}{R_{Martian}})}^{3} = {(\frac{1}{4})}^{3} = \frac{1}{64}$

$T_{Mercury} = \frac{T_{Martian}}{8} = \frac{687}{8} = 85.8 days$

Nearest answer is 88 Earth days.

Q 2 :
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are $K_{A}$ , $K_{B}$ and $K_{C}$ , respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then [2018]

[IMAGE 73]

$K_{A} < K_{B} < K_{C}$
$K_{A} > K_{B} > K_{C}$
$K_{B} < K_{A} < K_{C}$
$K_{B} > K_{A} > K_{C}$

1

2

3

4

(2)

[IMAGE 74]

Point A is perihelion and C is aphelion.

So, $v_{A} > v_{B} > v_{C}$

As kinetic energy $K = (\frac{1}{2}) m v^{2}$

or $K \propto v^{2}$

So, $K_{A} > K_{B} > K_{C}$

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