Question

At a place, the value of 'g' is less by $1$% than its value on the surface of the Earth (Radius of Earth, $R=6400km$). The place is:

• A. $64km$ below the surface of the earth
• B. $64km$ above the surface of the earth
• C. $30km$ above the surface of the earth
• D. $32km$ below the surface of the earth

Answer 1

The correct option is A $64km$ below the surface of the earth

$g_d=g(1-\frac{d}{R})$

Now, $g_d=g-\frac{g}{100}=\frac{99}{100}g\therefore \frac{99}{100}g=g(1-\frac{d}{6400\times {10}^3})or,\frac{99}{100}=1-\frac{d}{64\times {10}^5}\Rightarrow \frac{1}{100}=\frac{d}{64\times {10}^5}\Rightarrow d=\frac{64\times {10}^5}{100}=64\times {10}^{3}m=64km$.