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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceEasy2 minPYQ_2020

PhysicsEasysingle choice

The acceleration to gravity at a height 1/20th of the radius of the earth above the earth surface is $9 m s^{- 2}$ . Its value at a point at an equal distance below the surface of the earth in $m s^{- 2}$ is about

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Answer:

Solution:

At a height $h$ ,

$g_{h} = g (1 - \frac{2 h}{R})$

for $h = \frac{R}{20}$ we get

$9 = g (1 - \frac{1}{10}) = \frac{9}{10} g$

At a depth $d$ ,

$g_{d} = g (1 - \frac{d}{R})$

for $d = \frac{R}{20}$

$g_{d} = g (1 - \frac{1}{20}) = \frac{19}{20} g$

$\Rightarrow \frac{g_{d}}{9} = \frac{19}{18} \Rightarrow g_{d} = 9.5 m s^{- 2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2020

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