The depth d at which the value of acceleration due to gravity becomes 1 n times the value at the earth's surface is ( R = radius of the earth )

At depth d we have g ′ = g 1 - d R g ′ = g n g n = g 1 - d R ∴ 1 n = 1 - d R ∴ d R = 1 - 1 n = n - 1 n d = R n - 1 n

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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationField & ForceMedium2 minPYQ_2017

PhysicsMediumsingle choice

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the earth's surface is $(R$ = radius of the earth $)$

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Answer:

Solution:

At depth d we have
$g^{'} = g (1 - \frac{d}{R})$
$g^{'} = \frac{g}{n}$
$\frac{g}{n} = g (1 - \frac{d}{R})$
$∴ \frac{1}{n} = 1 - \frac{d}{R}$
$∴ \frac{d}{R} = 1 - \frac{1}{n} = \frac{n - 1}{n}$
$d = R (\frac{n - 1}{n})$

Stream:MHT_CETSubject:PhysicsTopic:GravitationSubtopic:Field & Force

⏱ 2mℹ️ Source: PYQ_2017

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