The depth which the value of acceleration due to gravity becomes -dfrac-1-n- times the value the surface is -R be the radius of the earth-dfrac-R-n-dfrac-R-n-2-dfrac-Rn-n-1-dfrac-R-n-1-n-

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Toppr Admin · Accepted Answer

Option Cg1-g-1-x2212-dR-x27F9-gn-g-1-x2212-dR-x27F9-d-n-x2212-1n-R

The depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is (R be the radius of the earth):
$\frac{R}{n}$
$\frac{R}{n^{2}}$
$\frac{R n}{(n - 1)}$
$\frac{R (n - 1)}{n}$

option C
$g^{1} = g [1 - \frac{d}{R}]$

$⟹ \frac{g}{n} = g [1 - \frac{d}{R}] ⟹ d = [\frac{n - 1}{n}] R$

The depth at which the value of acceleration due to gravity becomes 1n times the value at the surface is (R be the radius of the earth):RnRn2Rn(n−1)R(n−1)n

option Cg1=g[1−dR]⟹gn=g[1−dR]⟹d=[n−1n]R

The depth at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface is (R be the radius of the earth):
$\frac{R}{n}$
$\frac{R}{n^{2}}$
$\frac{R n}{(n - 1)}$
$\frac{R (n - 1)}{n}$

option C
$g^{1} = g [1 - \frac{d}{R}]$

$⟹ \frac{g}{n} = g [1 - \frac{d}{R}] ⟹ d = [\frac{n - 1}{n}] R$