Question

A rod is formed by joining two cylinders each having a length $\ell$ and cross sectional area $S$. The densities of cylinders are $\rho$ and $2\rho$ respectively. The rod is now horizontally suspended in a liquid of density $4\rho$ with help of two string as shown in the figure. The entire setup is kept inside a lift. For the quantities given in List I, select the correct value from those mentioned in List II.
$\begin{array}{ccc}\text{List I}& \text{List II}& \\ \text{(P) Tension in string 1 if the lift is moving upwards}& \text{(1)}\frac{11}{8}\rho S\ell g& \\ \text{with constant velocity.}& & \\ \text{(Q) Tension in string 2 if the lift is moving upwards}& \text{(2)}\frac{9}{8}\rho S\ell g& \\ \text{with constant velocity.}& & \\ \text{(R) Tension in string 1 if the lift is moving downwards}& \text{(3)}\frac{11}{4}\rho S\ell g& \\ \text{with an acceleration of}\frac{g}{2}& & \\ \text{(S) Tension in string 2 if the lift is moving downwards}& \text{(3)}\frac{9}{4}\rho S\ell g& \\ \text{with an acceleration of}\frac{g}{2}& & \end{array}$

• A. P -3, Q - 4, R - 1, S - 2
• B. P -1, Q - 2, R - 3, S - 4
• C. P -4, Q - 3, R - 2, S - 1
• D. P -2, Q - 1, R - 4, S - 3

Answer 1

The correct option is A P -3, Q - 4, R - 1, S - 2

Tension in string 1 is $T_1$ and in string 2 is $T_2$.
Consider the FBD shown in the figure.
Force balance
$\Rightarrow T_1+T_2=5\rho S\ell g$
Torque balance about point P
$\Rightarrow T_1\times 2\ell +\rho S\ell g\times \frac{3\ell }{2}-8\rho S\ell g\times \ell +2\rho S\ell g\times \frac{\ell }{2}=0$
$T_1=\frac{11}{4}\rho S\ell g$
$\Rightarrow T_2=\frac{9\rho S\ell g}{4}$
Now if lift is moving with an acceleration of $\frac{g}{2}$, then
$g_{eff}=g-\frac{g}{2}=\frac{g}{2}$
$\therefore T_1=\frac{11\rho S\ell g}{8}{T}_2=\frac{9\rho S\ell g}{8}$