Question

A wooden stick of length L, radius R and density $\rho$ has a small metal piece of mass m (of negligible volume) attached to its one end. Find the minimum value for the mass m (in terms of given parameters) that would make the stick float vertically in stable equilibrium in a liquid of density $\sigma$ .

• A. $2\pi R^{2}L\rho \left(\sqrt{\frac{\sigma }{\rho }}-1\right)$
• B. $\pi R^{2}L\rho \left(\sqrt{\frac{2\sigma }{\rho }}-1\right)$
• C. $\pi R^{2}L\rho \left(\sqrt{\frac{\sigma }{\rho }}-1\right)$
• D. $\pi R^{2}L\rho \left(\sqrt{\frac{\sigma }{2\rho }}-1\right)$

Answer 1

Answer: (C)

$\left(M+m\right)g=\pi R^{2}h\sigma g$ .... (i)
$M=\pi R^{2}L\rho$
The height of center of mass from bottom
$\frac{\left(M\right)L/2+m\times 0}{m+M}=\frac{ML}{2\left(m+M\right)}$

For rotatory equilibrium and for minimum m, thus should be equal to $h/2$
$\therefore \frac{h}{2}=\frac{ML}{2\left(m+M\right)}$ .... (ii)
$\therefore h=\frac{ML}{2\left(m+M\right)};m=\pi R^{2}L\sqrt{\sigma \rho }-\pi R^{2}L\rho =\pi R^{2}L\rho \left(\sqrt{\frac{\sigma }{\rho }}-1\right)$