Question

The values of the acceleration due to gravity at depths of \(\frac{R}{2}\) and \(\frac{2R}{3}\) from the surface of the earth are in the ratio

• A. 9 : 10
• B. 3 : 2
• C. 3 : 1
• D. 1 : 3
• E. 2 : 1

Hint:

Gravity inside the earth is directly proportional to the distance from the center (\(R-d\)). Ratio = \((R - R/2) : (R - 2R/3) = R/2 : R/3 = 3:2\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Acceleration due to gravity (\(g\)) decreases as we go below the surface of the earth. At the center of the earth, gravity is zero.

Step 2: Key Formula or Approach:
Gravity at depth \(d\): \(g_d = g \left(1 - \frac{d}{R}\right)\), where \(R\) is the radius of the earth.

Step 3: Detailed Explanation:

Step 1: Find \(g_1\) at depth \(d_1 = R/2\).
\[ g_1 = g \left(1 - \frac{R/2}{R}\right) = g \left(1 - \frac{1}{2}\right) = \frac{g}{2} \]

Step 2: Find \(g_2\) at depth \(d_2 = 2R/3\).
\[ g_2 = g \left(1 - \frac{2R/3}{R}\right) = g \left(1 - \frac{2}{3}\right) = \frac{g}{3} \]

Step 3: Calculate the ratio.
\[ \frac{g_1}{g_2} = \frac{g/2}{g/3} = \frac{3}{2} \]

Step 4: Final Answer:
The values are in the ratio 3 : 2.