Question:
A cubical block of cork of side \(10\text{ cm}\) is floating in water. When the cork is depressed slightly and then released, if it oscillates with a time period of \(0.5\text{ s}\), then the mass of the block is nearly (Acceleration due to gravity \(=10\text{ ms}^{-2}\)):
A cubical block of cork of side \(10\text{ cm}\) is floating in water. When the cork is depressed slightly and then released, if it oscillates with a time period of \(0.5\text{ s}\), then the mass of the block is nearly (Acceleration due to gravity \(=10\text{ ms}^{-2}\)):
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A floating body oscillates in SHM because the excess buoyant force developed after displacement acts as a restoring force.
Updated On: Oct 6, 2026
- \(500\text{ g}\)
- \(625\text{ g}\)
- \(250\text{ g}\)
- \(175\text{ g}\)
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The Correct Option is B
Solution and Explanation
Concept:
When a floating body is displaced slightly and released, the change in buoyant force acts as a restoring force. The body executes Simple Harmonic Motion (SHM).
For a floating body,
\[
T=2\pi\sqrt{\frac{m}{\rho A g}}
\]
where
\[
m=\text{mass of floating body}
\]
\[
\rho=\text{density of liquid}
\]
\[
A=\text{cross-sectional area}
\]
\[
g=\text{acceleration due to gravity}
\]
Step 1: Calculate the cross-sectional area of the cube. Side of cube: \[ a=10\text{ cm}=0.1\text{ m} \] Therefore, \[ A=a^2=(0.1)^2 \] \[ A=0.01\text{ m}^2 \]
Step 2: Substitute the known quantities in the SHM formula. Given: \[ T=0.5\text{ s} \] \[ \rho=1000\text{ kg m}^{-3} \] \[ g=10\text{ ms}^{-2} \] Using \[ T=2\pi\sqrt{\frac{m}{\rho Ag}} \] \[ 0.5=2\pi\sqrt{\frac{m}{1000\times0.01\times10}} \] \[ 0.5=2\pi\sqrt{\frac{m}{100}} \] Squaring both sides, \[ 0.25=4\pi^2\frac{m}{100} \] \[ m=\frac{25}{4\pi^2} \] Using \[ \pi^2\approx 10 \] \[ m\approx\frac{25}{40} \] \[ m=0.625\text{ kg} \] \[ m=625\text{ g} \]
Step 3: Final answer. \[ \boxed{625\text{ g}} \]
Step 1: Calculate the cross-sectional area of the cube. Side of cube: \[ a=10\text{ cm}=0.1\text{ m} \] Therefore, \[ A=a^2=(0.1)^2 \] \[ A=0.01\text{ m}^2 \]
Step 2: Substitute the known quantities in the SHM formula. Given: \[ T=0.5\text{ s} \] \[ \rho=1000\text{ kg m}^{-3} \] \[ g=10\text{ ms}^{-2} \] Using \[ T=2\pi\sqrt{\frac{m}{\rho Ag}} \] \[ 0.5=2\pi\sqrt{\frac{m}{1000\times0.01\times10}} \] \[ 0.5=2\pi\sqrt{\frac{m}{100}} \] Squaring both sides, \[ 0.25=4\pi^2\frac{m}{100} \] \[ m=\frac{25}{4\pi^2} \] Using \[ \pi^2\approx 10 \] \[ m\approx\frac{25}{40} \] \[ m=0.625\text{ kg} \] \[ m=625\text{ g} \]
Step 3: Final answer. \[ \boxed{625\text{ g}} \]
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