CUET Preparation Today
The depth d at which the value of acceleration due to gravity becomes 1/n times the value at the surface, is [R = radius of the earth] : |
\(\frac{R}{n}\) \(\frac{n-1}{n}R\) \(\frac{R}{n^2}\) \(\frac{n}{n+1}\) |
\(\frac{n-1}{n}R\) |
\(g' = g(1-\frac{d}{R})\) \(\Rightarrow \frac{g}{n} = g(1-\frac{d}{R})\) \(\Rightarrow \frac{d}{R} = 1 - \frac{1}{n}\) \(\Rightarrow d = \frac{n-1}{n}R\) |
