Question

The depth \( d \) at which the value of acceleration due to gravity becomes \( \frac{1}{n} \) times the value at the surface (radius of Earth \( R \)) is:

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

Hint:

To solve problems related to gravity at a certain depth, use the formula \( g_d = g \left( \frac{R}{R + d} \right)^2 \) and then equate it with the known value of gravity at the required depth.

Answer 1

The Correct Option is B

Let the acceleration due to gravity at a depth \( d \) be \( g_d \). The acceleration due to gravity at a height or depth from the surface of the Earth is given by the formula: \[ g_d = g \left( \frac{R}{R + d} \right)^2 \] where \( g \) is the acceleration due to gravity at the Earth's surface, and \( R \) is the radius of the Earth. We are given that at a depth \( d \), the gravity becomes \( \frac{g}{n} \). So, we can equate the expression for gravity at depth to \( \frac{g}{n} \): \[ \frac{g}{n} = g \left( \frac{R}{R + d} \right)^2 \] Dividing both sides by \( g \), we get: \[ \frac{1}{n} = \left( \frac{R}{R + d} \right)^2 \] Taking the square root of both sides: \[ \frac{1}{\sqrt{n}} = \frac{R}{R + d} \] Rearranging this to solve for \( d \): \[ R + d = R \sqrt{n} \] \[ d = R \sqrt{n} - R \] \[ d = R \left( \sqrt{n} - 1 \right) \] Thus, the depth \( d \) is: \[ d = \frac{R(n - 1)}{n} \] Thus, the correct answer is: \[ \boxed{\frac{R(n - 1)}{n}} \]