Question:
A body weighs \(500\ \text{N}\) on the surface of the earth. How much would it weigh half way below the surface of the earth?
A body weighs \(500\ \text{N}\) on the surface of the earth. How much would it weigh half way below the surface of the earth?
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At the centre of earth (\(d = R\)), \(g' = 0\).
Updated On: Apr 23, 2026
- \(1000\ \text{N}\)
- \(500\ \text{N}\)
- \(250\ \text{N}\)
- \(125\ \text{N}\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Acceleration due to gravity at depth \(d\) below surface: \(g' = g\left(1 - \frac{d}{R}\right)\).
Step 2: Detailed Explanation:
Given weight on surface = \(mg = 500\ \text{N}\). Halfway below surface means \(d = R/2\).
\(g' = g\left(1 - \frac{R/2}{R}\right) = g\left(1 - \frac{1}{2}\right) = \frac{g}{2}\).
New weight = \(mg' = m \times \frac{g}{2} = \frac{500}{2} = 250\ \text{N}\).
Step 3: Final Answer:
Thus, weight = \(250\ \text{N}\).
Acceleration due to gravity at depth \(d\) below surface: \(g' = g\left(1 - \frac{d}{R}\right)\).
Step 2: Detailed Explanation:
Given weight on surface = \(mg = 500\ \text{N}\). Halfway below surface means \(d = R/2\).
\(g' = g\left(1 - \frac{R/2}{R}\right) = g\left(1 - \frac{1}{2}\right) = \frac{g}{2}\).
New weight = \(mg' = m \times \frac{g}{2} = \frac{500}{2} = 250\ \text{N}\).
Step 3: Final Answer:
Thus, weight = \(250\ \text{N}\).
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