Question

The tension in a string holding a solid block below the surface of a liquid (Where ${\rho }_{liquid}>{\rho }_{block})$ as in shown in the figure is $T$ when the system is at rest. Then what will be the tension in the string if the system has upward acceleration $a$?

• A. $T(1-\frac{a}{g})$
• B. $T(1+\frac{a}{g})$
• C. $T(\frac{a}{g}-1)$
• D. $\frac{a}{g}T$

Answer 1

The correct option is B $T(1+\frac{a}{g})$

When the system is at rest, block is in equilibrium i.e. Net force on block is zero.
$\Rightarrow \text{Weight of the block+Tension in the string=Buoyant force}$
$\Rightarrow V{\rho }_{block}g+T=V{\rho }_{liquid}g$ here $V$ is the volume of the block as well as volume of liquid displaced.
$\Rightarrow T=V{\rho }_{liquid}g-V{\rho }_{block}g....\left(I\right)\Rightarrow \frac{T}{g}=V{\rho }_{liquid}-V{\rho }_{block}....\left(II\right)$
When the system is accelerated upwards, buoyant force increases due to increase in weight of water displaced i.e. Now weight of displaced water becomes $m_w(g+a)$ and net force on the block becomes equal to the $m_{b}a$
$\Rightarrow V{\rho }_{block}a=V{\rho }_{liquid}(g+a)-V{\rho }_{block}g-T^{\prime }\Rightarrow V{\rho }_{block}a=V{\rho }_{liquid}g+V{\rho }_{liquid}a-V{\rho }_{block}g-T^{\prime }\Rightarrow 0=(V{\rho }_{liquid}g-V{\rho }_{block}g)+(V{\rho }_{liquid}-V{\rho }_{block})a-T^{\prime }$
$\Rightarrow 0=T+\left(\frac{T}{g}\right)a-T^{\prime }$ Using the values of $T$ and $\frac{T}{g}$, from $\left(I\right)$ and $\left(II\right)$
$\Rightarrow T^{\prime }=T+\left(\frac{T}{g}\right)a$
$\Rightarrow T^{\prime }=T(1+\frac{a}{g})$