Question

A vessel contains two immiscible liquids of density $\rho_{1}=1000 \, kg \, m^{-3}$ and $\rho_{2}=1500 \, kg \, m^{-3}$. A solid block of volume $V=10^{-3} \, m^{3}$ and density $d=800 \, kg \, m^{-3}$ is tied to one end of a string and other end is tied to the bottom of the vessel. The block is immersed with $2/5^{th}$ of its volume in the liquid of higher density and $3/5^{th}$ in the liquid of lower density. The entire system is kept in an elevator which is moving upwards with an acceleration of $a=g/2$. The tension in the string is (Take $g=10\, m\, s^{-2}$).

• A. $8\, N$
• B. $6\, N$
• C. $10\, N$
• D. $12\, N$

Answer 1

Answer: (B)

$6\, N$

Answer 2

We analysed this problem from the reference frame of elevator Total buoyant force on the block, $F_{B}=\left(\frac{2}{5}V\rho_{2}+\frac{3}{5}V \rho_{1}\right)\left(g+a\right)$ For equilibrium, $F_{B}=T+Vd \left(g+a\right)$ or $T=F_{B}-Vd\left(g+a\right)$ $=\left(g+a\right)V \left[\frac{2}{5}\rho_{2}+\frac{3}{5}\rho_{1}-d\right]$ $T=\left(10+\frac{10}{2}\right)\times10^{-3}\left[\frac{2}{5}\times1500+\frac{3}{5}\times1000-800\right]$ $=6N$