Question

The tension in a string holding a solid block below the surface of a liquid (where $\rho_{\text {liquid }} > \rho_{\text {block }}$ ) as in shown in the figure is $T_{0}$ when the system is at rest. Then what will be the tension in the string if the system has upward acceleration $a$ ?

• A. $T_{0}\left(1-\frac{a}{g}\right)$
• B. $T_{0}\left(1+\frac{a}{g}\right)$
• C. $T_{0}\left(\frac{a}{g}-1\right)$
• D. $\frac{a}{g} T_{0}$

Answer 1

Answer: (B)

Let $v$ be the volume of the solid block of density $\rho$. Let $\rho_{1}$ be the density of water. Weight of body $=v \rho g$.
When the body is immersed in water,
Tension in the string = Upward thrust - Weight of the body
$\Rightarrow T=v \rho_{1} g-v \rho g=v g\left(\rho_{1}-\rho_{2}\right)$
When the lift is moving upwards with acceleration $a$, the tension in the string is $T=v\left(\rho_{1}-\rho\right)(g+a)$
From Eqs. (i) and (ii), $T=T_{0}(1+a / g)$.