Intensity of Gravitational Field | JEE Main previous year questions with Solutions

Q 1 :

A 90 kg body placed at 2R distance from surface of earth experiences gravitational pull of (R = Radius of earth, $g = 10 m s^{- 2}$ ) [2024]

100 N
300 N
120 N
225 N

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(1)

Acceleration due to gravity at height $h$ from earth's surface

$g_{h} = \frac{G M}{{(R + h)}^{2}}$

$g_{h} = \frac{G M}{9 R^{2}} = \frac{g_{s}}{9} (∵ h = 2 R)$

Here $g_{s} =$ gravitational acceleration at surface

Gravitational Pull $m g_{s} = 90 \times \frac{g_{s}}{9} = 100 N$

Q 2 :

Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the surface of earth. How much it would weigh at R/4 depth under surface of earth? [2024]

300 N
75 N
225 N
375 N

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(3)

At surface: $m g_{s} = 300 N \Rightarrow m = \frac{300}{g_{s}}$

Gravitational field at $\frac{R}{4}$ depth under surface, $g_{d} = g_{s} [1 - \frac{d}{R}]$

$g_{d} = g_{s} [1 - \frac{R}{4 R}] = \frac{3 g_{s}}{4}$

Weight at depth, $W_{d} = m \times g_{d} = m \times \frac{3 g_{s}}{4} = \frac{3}{4} \times 300 = 225 N$

Q 3 :

The acceleration due to gravity on the surface of earth is g. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be [2024]

g/4
2g
g/2
4g

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(4)

$g = \frac{G M}{R^{2}} \Rightarrow g \propto \frac{1}{R^{2}}$

$g_{2} = 4 g_{1} (R_{2} = \frac{R_{1}}{2})$

Q 4 :

If R is the radius of the earth and the acceleration due to gravity on the surface of earth is $g = π^{2} m / s^{2}$ , then the length of the second's pendulum at a height $h = 2 R$ from the surface of earth will be [2024]

$\frac{2}{9} m$
$\frac{1}{9} m$
$\frac{4}{9} m$
$\frac{8}{9} m$

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(2)

$g = \frac{G M}{R^{2}} = π^{2}$

$g' = \frac{G M}{9 R^{2}} = \frac{g}{9} = \frac{π^{2}}{9}$

We know, $T = 2 π \sqrt{\frac{l}{g}}$

Since the time period of second pendulum is 2 sec.

$2 = 2 π \sqrt{\frac{l \times 9}{π^{2}}}$

On solving, $l = \frac{1}{9} m$

Q 5 :

At what distance above and below the surface of the earth a body will have same weight? (take radius of earth as R). [2024]

$\sqrt{5} R - R$
$\frac{\sqrt{3} R - R}{2}$
$\frac{R}{2}$
$\frac{\sqrt{5} R - R}{2}$

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(4)

Below the surface of the earth, $W_{d e p t h} = \frac{G M m (R - h)}{R^{3}}$

Above the surface of the earth, $W_{a b o v e} = \frac{G M m}{{(R + h)}^{2}}$

If the body will have same weight, $g_{p} = g_{q}$

$\frac{G M m (R - h)}{R^{3}} = \frac{G M m}{{(R + h)}^{2}}$

$R^{3} = {(R + h)}^{2} (R - h) \Rightarrow h^{2} + R h - R^{2} = 0$

$\Rightarrow h = \frac{R}{2} (\sqrt{5} - 1)$

Q 6 :

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4m, then the time period of small oscillations will be ___________ s. $[t a k e g = π^{2}$ $m s^{- 2}]$ [2024]

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(8) $T = 2 π \sqrt{\frac{l}{g_{eff}}}$

$g = \frac{G M}{R^{2}}, at a height, h, g = \frac{G M}{{(R + h)}^{2}}$

$g_{eff} = g {(\frac{R}{R + h})}^{2}, g_{eff} = g {(\frac{R}{2 R})}^{2}$

$g_{eff} = \frac{g}{4}$

$∴ T = 2 π \sqrt{\frac{l}{g_{eff}}} = 2 π \sqrt{\frac{4 \times 4}{g}} = \frac{(2 π) (4)}{\sqrt{g}}$

$Given, g = π^{2}$

$T = 8 \sec$

Q 7 :

A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc, and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is _____. [2024]

0
$\frac{G M m π}{2 L^{2}}$
$\frac{2 G M m π}{L^{2}}$
$\frac{G M m π}{L^{2}}$

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(3)

$E_{g} = \frac{2 \cdot G \cdot λ}{R} = \frac{2 G M}{L \cdot R}$

$also, π R = L \Rightarrow R = \frac{L}{π}$

$∴ E_{g} = \frac{2 G M}{L \cdot \frac{L}{π}} = \frac{2 G M π}{L^{2}}$

$F = m E_{g} = \frac{2 G M m π}{L^{2}}$

Q 8 :

Acceleration due to gravity on the surface of earth is 'g'. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ________ g. [2025]

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(9)

$∵$ Acceleration due to gravity on surface, $g = \frac{G M}{R_{e}^{2}}$

Now since diameter is reduced to $1 / 3^{rd}$ , radius also reduces to $1 / 3^{rd}$ , keeping mass constant.

New value of acceleration due to gravity on Earth's surface,

$g' = \frac{G M}{{(\frac{R_{e}}{3})}^{2}} = 9 \frac{G M}{R_{e}^{2}} = 9 g$

Q 9 :

The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth’s surface is (given, radius of earth $R_{e} = 6400$ km) [2023]

19.6 N
9.8 N
4.9 N
8 N

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(4)

$Acceleration due to gravity at height h$

$g' = \frac{g}{{[1 + \frac{h}{R}]}^{2}}$

$So weight at given height$

$m g' = \frac{m g}{{[1 + \frac{h}{R}]}^{2}} = \frac{18}{{[1 + \frac{1}{2}]}^{2}} = 8 N$

Q 10 :

Given below are two statements: [2023]

Statement I: Acceleration due to earth’s gravity decreases as you go ‘up’ or ‘down’ from earth’s surface.

Statement II: Acceleration due to earth’s gravity is same at a height ‘ $h$ ’ and depth ‘ $d$ ’ from earth’s surface, if $h = d .$

In the light of above statements, choose the most appropriate answer from the options given below:

Statement I is correct but statement II is incorrect
Both Statement I and Statement II are incorrect
Both Statement I and II are correct
Statement I is incorrect but statement II is correct

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(1)

$g' = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$g' = g {1 - \frac{d}{R}}$

[IMAGE 69]-------------

Statement I is correct & Statement II is incorrect.

Topic Question Set

Q 1 :

A 90 kg body placed at 2R distance from surface of earth experiences gravitational pull of (R = Radius of earth, $g = 10 m s^{- 2}$ ) [2024]

Q 2 :

Assuming the earth to be a sphere of uniform mass density, a body weighed 300 N on the surface of earth. How much it would weigh at R/4 depth under surface of earth? [2024]

Q 3 :

The acceleration due to gravity on the surface of earth is g. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be [2024]

Q 4 :

If R is the radius of the earth and the acceleration due to gravity on the surface of earth is $g = π^{2} m / s^{2}$ , then the length of the second's pendulum at a height $h = 2 R$ from the surface of earth will be [2024]

Q 5 :

At what distance above and below the surface of the earth a body will have same weight? (take radius of earth as R). [2024]

Q 6 :

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4m, then the time period of small oscillations will be ___________ s. $[t a k e g = π^{2}$ $m s^{- 2}]$ [2024]

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Q 7 :

A metal wire of uniform mass density having length L and mass M is bent to form a semicircular arc, and a particle of mass m is placed at the centre of the arc. The gravitational force on the particle by the wire is _____. [2024]

Q 8 :

Acceleration due to gravity on the surface of earth is 'g'. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ________ g. [2025]

0%

Q 9 :

The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth’s surface is (given, radius of earth $R_{e} = 6400$ km) [2023]

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