Question:
If \(10\) identical silver coins each of mass \(m\) are placed one over the other, then force on the \(6^{\text{th}}\) coin from the bottom is
If \(10\) identical silver coins each of mass \(m\) are placed one over the other, then force on the \(6^{\text{th}}\) coin from the bottom is
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In stacked objects:
- Force on a layer = weight of all layers above it
- Ignore layers below for force calculation on that layer
Updated On: Apr 30, 2026
- $10mg$
- $8mg$
- $6mg$
- $4mg$
- zero
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The Correct Option is D
Solution and Explanation
Concept:
Each coin supports the weight of all coins placed above it. Force on a coin is due to the weight of coins above it.
Step 1: Identify position of the coin.
The $6^{\text{th}}$ coin from the bottom has: \[ 10 - 6 = 4 \text{ coins above it} \]
Step 2: Calculate total force on the coin.
Each coin has weight $mg$, so total force due to coins above: \[ 4 \times mg = 4mg \]
Step 3: Conclusion.
Force on the $6^{\text{th}}$ coin is: \[ 4mg \]
Step 1: Identify position of the coin.
The $6^{\text{th}}$ coin from the bottom has: \[ 10 - 6 = 4 \text{ coins above it} \]
Step 2: Calculate total force on the coin.
Each coin has weight $mg$, so total force due to coins above: \[ 4 \times mg = 4mg \]
Step 3: Conclusion.
Force on the $6^{\text{th}}$ coin is: \[ 4mg \]
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