Question

A homogeneous solid cylinder of length L (L 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 fig. The lower
density liquid is open to atmosphere having pressure Po. Then density D of
solid is given by (a) 5d/4
(b) 4d/5 (c) Ad (d) d/5

Answer 1

Given:

• A homogeneous solid cylinder of length L and 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 (2d).
• The lower  density liquid (d) is open to atmosphere having pressure P₀.

To find:

• Density D of the solid.

Answer:

• By force balance,

           Weight of cylinder = $Upthrust_{upper liquid}+Upthrust_{lower liquid}$

       ⇒  m g = $V_1 d g +V_2 (2d) g$

       ⇒$D (\frac{A}{5} L) g = (\frac{3L}{4} *\frac{A}{5}) *d*g + (\frac{L}{4} * \frac{A}{5} ) *2d *g$

       ⇒ $D = \frac{3}4} d + \frac{1}{2} d$

       ⇒ D = $\frac{5}4} d$

• Density, D of the solid = 1.25 d