Question

The depth d at which the value of acceleration due to gravity becomes $\frac { 1 } { n }$ times the value at the surface, is [R = radius of the earth]

• A. $\frac { R } { n }$
• B. $R \left( \frac { n - 1 } { n } \right)$
• C. $\frac { R } { n ^ { 2 } }$
• D. $R \left( \frac { n } { n + 1 } \right)$

Answer 1

Answer: (B)

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

Answer 2

$g ^ { \prime } = g \left( 1 - \frac { d } { R } \right) \Rightarrow \frac { g } { n } = g \left( 1 - \frac { d } { R } \right)$ $\Rightarrow d = \left( \frac { n - 1 } { n } \right) R$