Question

A wooden block is floating in a liquid. About $50\%$ of its volume is inside the liquid when the vessel is stationary. Percentage volume immersed when the vessel moves upwards with acceleration $a=\frac{g}{2}$ is

• A. $75\%$
• B. $25\%$
• C. $50\%$
• D. $33.33\%$

Answer 1

The correct option is C $50\%$

When the vessel is stationary

$\text{weight}=\text{upthrust}\Rightarrow V{\rho }_{w}g=V^{\prime }{\rho }_{l}g\Rightarrow \frac{V^{\prime }}{V}=\frac{{\rho }_w}{{\rho }_l}..........\left(I\right){\rho }_w=\text{density of wood};{\rho }_l=\text{density of liquid}$

When the vessel moves upward with an acceleration $\frac{g}{2}$

$\text{upthrust}-\text{weight}=\text{mass}\times \text{acceleration}\Rightarrow V^{\prime \prime }(g+\frac{g}{2}){\rho }_l-V{\rho }_{w}g=\frac{V{\rho }_{w}g}{2}\Rightarrow \frac{3}{2}\left(V^{\prime \prime }{\rho }_{l}g\right)=\frac{3}{2}\left(V{\rho }_{w}g\right)\Rightarrow \frac{V^{\prime \prime }}{V}=\frac{{\rho }_w}{{\rho }_l}..........\left(II\right)$

From equation $\left(I\right)$ and $\left(II\right)$, We have $V^{\prime }=V^{\prime \prime }\Rightarrow$ Fraction of the volume immersed remains unchanged i.e. $50\%$.