Question

A solid cylinder of length H, diameter D and of relative density S floats in neutral equilibrium in water with its axis vertical. What is the ratio of H to D if S = 0.6?

• A. 0.86
• B. 0.72
• C. 0.52
• D. 0.46

Answer 1

The correct option is B 0.72

$F_B=M_b.g$

$\left({10}^3\right)\left(\frac{\pi D^2}{4}\right).l.g=\left(600\right).g.H\left(\frac{\pi D^2}{4}\right)$
$l=0.6H$ .... (i)
For Neutral equilibrium,
$GM=0$
$\frac{l}{\forall }-BG=0$

$\frac{\frac{\pi D^4}{64}}{\frac{\pi D^2.l}{4}}-\frac{1}{2}[H-l]=0$

$\frac{D^2}{16}-\frac{1}{2}[H-l]=0$

$\frac{D^2}{8}=H-l$

$\frac{D^2}{8l}=H[1-0.6]$

$\frac{H^2}{D^2}=\frac{1}{8\times 0.6\times 0.4}$

$\frac{H}{D}=0.721$