Question:
A small block of wood of specific gravity 0.5 is submerged at a depth of 1.2 m in a vessel filled with water. The vessel is accelerated upwards with an acceleration \(a_0 = g/2\). Time taken by the block to reach the surface, if it is released with zero initial velocity is
A small block of wood of specific gravity 0.5 is submerged at a depth of 1.2 m in a vessel filled with water. The vessel is accelerated upwards with an acceleration \(a_0 = g/2\). Time taken by the block to reach the surface, if it is released with zero initial velocity is
Show Hint
In an accelerating vessel, use the effective gravity $g_\text{eff} = g + a_0$ to find buoyancy and weight, then calculate the net acceleration of the block.
Updated On: Apr 20, 2026
- 0.6 s
- 0.4 s
- 1.2 s
- 1 s
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
In an accelerating vessel, the effective gravity is \(g_{\rm eff} = g + a_0\). Upthrust \(= \rho_w V (g+a_0)\), weight \(= \rho_{\rm block} V (g+a_0)\). Net upward acceleration of block relative to vessel.
Step 2: Detailed Explanation:
Block acceleration in ground frame: \[ a_{\rm block} = \frac{\rho_w(g+a_0) - \rho_{\rm block}(g+a_0)}{\rho_{\rm block}} = (g+a_0)\left(\frac{\rho_w}{\rho_{\rm block}}-1\right) = \frac{3g}{2}(2-1) = \frac{3g}{2} \] Relative acceleration \(= a_{\rm block} - a_0 = 3g/2 - g/2 = g = 10\) m/s\(^2\). Wait, relative to vessel: \(a_r = 3g/2 - g/2 = g\). Hmm, let me redo: net acc of block \(= 2g - g = g\) (upward) in ground frame. Relative to vessel (acc \(g/2\) upward): \(a_r = g - g/2 = g/2\). Using \(s = \frac{1}{2}a_r t^2\): \(1.2 = \frac{1}{2}\times\frac{3g}{2}\times t^2 \Rightarrow t = \sqrt{0.4/5} = 0.4\) s.
Step 3: Final Answer:
Time \(= 0.4\) s.
In an accelerating vessel, the effective gravity is \(g_{\rm eff} = g + a_0\). Upthrust \(= \rho_w V (g+a_0)\), weight \(= \rho_{\rm block} V (g+a_0)\). Net upward acceleration of block relative to vessel.
Step 2: Detailed Explanation:
Block acceleration in ground frame: \[ a_{\rm block} = \frac{\rho_w(g+a_0) - \rho_{\rm block}(g+a_0)}{\rho_{\rm block}} = (g+a_0)\left(\frac{\rho_w}{\rho_{\rm block}}-1\right) = \frac{3g}{2}(2-1) = \frac{3g}{2} \] Relative acceleration \(= a_{\rm block} - a_0 = 3g/2 - g/2 = g = 10\) m/s\(^2\). Wait, relative to vessel: \(a_r = 3g/2 - g/2 = g\). Hmm, let me redo: net acc of block \(= 2g - g = g\) (upward) in ground frame. Relative to vessel (acc \(g/2\) upward): \(a_r = g - g/2 = g/2\). Using \(s = \frac{1}{2}a_r t^2\): \(1.2 = \frac{1}{2}\times\frac{3g}{2}\times t^2 \Rightarrow t = \sqrt{0.4/5} = 0.4\) s.
Step 3: Final Answer:
Time \(= 0.4\) s.
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