The depth at which the effective value of acceleration due to gravity is g -4 is R - radius of the earthA- RB- 3 R-4C- R-2D- R-4

The correct option is B g'=g ( 1 d/R ) ⇒g/4=g ( 1 d/R ) ⇒ d= 3R/4 ...

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Standard XII

Physics

Newton's Law of Gravitation

The depth at ...

Question

The depth at which the effective value of acceleration due to gravity is $\frac{g}{4}$ is (R = radius of the earth)

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Solution

The correct option is B

$g^{'} = g (1 - \frac{d}{R}) \Rightarrow \frac{g}{4} = g (1 - \frac{d}{R}) \Rightarrow d = \frac{3 R}{4}$

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The depth at which the effective value of acceleration due to gravity is g4 is (R = radius of the earth)

The correct option is B g′=g(1−dR)⇒g4=g(1−dR)⇒d=3R4

The depth at which the effective value of acceleration due to gravity is $\frac{g}{4}$ is (R = radius of the earth)

The correct option is B

$g^{'} = g (1 - \frac{d}{R}) \Rightarrow \frac{g}{4} = g (1 - \frac{d}{R}) \Rightarrow d = \frac{3 R}{4}$