Question

A homogeneous solid cylinder of length $L(L< H / 2)$ and cross sectional area $A / 5$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length $L / 4$ in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure $P_{0}$. Then density $D$ of solid is given by:

• A. $\frac{5}{4} d$
• B. $\frac{4}{5} d$
• C. $\frac{d}{5}$
• D. None of these

Answer 1

Answer: (A)

Weight of cylinder = upthrust due to both liquids
$V \times D \times g =\left(\frac{ A }{5} \cdot \frac{3}{4} L \right) \times d \times g +\left(\frac{ A }{5} \cdot \frac{ L }{4}\right) \times 2 d \times g$
$\Rightarrow \left(\frac{ A }{5} \cdot L \right) D \cdot g =\frac{ ALdg }{4} $
$\Rightarrow \frac{ D }{5}=\frac{ d }{4}$
$ \therefore D =\frac{5}{4} d$