Question

A beaker is filled with two non-mixing liquids. The lower liquid has density twice that of the upper one. A cylinder of height h and density 'D' floats with one-fourth of the height submerged in the lower liquid and half of its height submerged in the upper liquid. Another beaker is filled with the denser of the two liquids alone. If the same cylinder is kept in the second beaker, the height of the submerged position would be

• A. h
• B. $\frac{3h}{4}$
• C. $\frac{h}{2}$
• D. $\frac{h}{4}$

Answer 1

The correct option is C $\frac{h}{2}$



For the first beaker which is immersed in two liquids, we have from the arrchimedes' principle,
$VDg=V^{\prime }2\rho g+V^{\prime \prime }\rho g$
$\text{/}A\times \text{/}h\times D\times \text{/}g=\text{/}A\times \frac{\text{/}h}{4}\times 2\rho \times \text{/}g+\text{/}A\times \frac{\text{/}h}{2}\times \rho \times \text{/}g$

$⟹D=\frac{\rho }{2}+\frac{\rho }{2}=8$

$D=\rho$

For the Second beaker,
$VDg=V^{\prime }dg$
$A\times h\times \rho \times g=A\times h^{\prime }\times 2\rho \times g$
$h^{\prime }=\frac{h}{2}$