How much deep inside the earth radius should a an go ,so that his wight become on fourth of that one the earth's surface/

\frac{R}{4}

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\frac{R}{2}

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\frac{3 R}{4}

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\frac{2 R}{3}

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Solution

The correct option is C $\frac{3 R}{4}$
If the weight has to be $\frac{1}{4}$ of the original, the new weight is $\frac{m g}{4}$

Therefore, the gravitational force $= \frac{g}{4} = \frac{m g}{\frac{4}{m}}$

Now, $\frac{g}{4} = g (1 - \frac{h}{R}) \Rightarrow 1 - \frac{h}{R} = \frac{1}{4} \Rightarrow R - \frac{h}{R} = \frac{1}{4} \Rightarrow 4 (R - h) = R \Rightarrow h = \frac{3}{4} R$

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How much deep inside the earth radius should a an go ,so that his wight become on fourth of that one the earth's surface/

The correct option is C 3R4If the weight has to be 14 of the original, the new weight ismg4Therefore, the gravitational force =g4=mg4mNow, g4=g(1−hR)⇒1−hR=14⇒R−hR=14⇒4(R−h)=R⇒h=34R