The depth d at which the value of acceleration due to gravity becomes (1 / n) time the value at the surface is (R be the radius of the earth)

Question

The depth d at which the value of acceleration due to gravity becomes (1 / n) time the value at the surface is (R be the radius of the earth) (A) R/n (B) R/n2 (C) R((n-1)/n) (D) (Rn/(n-1))

Tardigrade Admin · Accepted Answer

At depth d from earth's surface, g'=g(1-(d/R) ) Given g'=(g/n) So, (g/n)=g(1-(d/R)) or (1/n)=1-(d/R) or (d/R)=1-(1/n) =((n-1)R/n) ∴ d=((n-1)R/n)

Q. The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ time the value at the surface is (R be the radius of the earth)

R/n

$R / n^{2}$

$R \frac{( n - 1 )}{n}$

$\frac{R n}{( n - 1 )}$

At depth d from earth's surface,
$g^{'} = g (1 - \frac{d}{R})$
Given $g^{'} = \frac{g}{n}$
So, $\frac{g}{n} = g (1 - \frac{d}{R})$
or $\frac{1}{n} = 1 - \frac{d}{R}$ or $\frac{d}{R} = 1 - \frac{1}{n}$
$= \frac{( n - 1 ) R}{n}$
$∴$ $d = \frac{( n - 1 ) R}{n}$

Q. The depth d at which the value of acceleration due to gravity becomes n1​ time the value at the surface is (R be the radius of the earth)

R/n

R/n2

Rn(n−1)​

(n−1)Rn​

At depth d from earth's surface, g′=g(1−Rd​) Given g′=ng​ So, ng​=g(1−Rd​) or n1​=1−Rd​ or Rd​=1−n1​ =n(n−1)R​ ∴ d=n(n−1)R​

Q. The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ time the value at the surface is (R be the radius of the earth)

$R / n^{2}$

$R \frac{( n - 1 )}{n}$

$\frac{R n}{( n - 1 )}$

At depth d from earth's surface,
$g^{'} = g (1 - \frac{d}{R})$
Given $g^{'} = \frac{g}{n}$
So, $\frac{g}{n} = g (1 - \frac{d}{R})$
or $\frac{1}{n} = 1 - \frac{d}{R}$ or $\frac{d}{R} = 1 - \frac{1}{n}$
$= \frac{( n - 1 ) R}{n}$
$∴$ $d = \frac{( n - 1 ) R}{n}$