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
$\frac{\left(M\right) L / 2 + m \times 0}{m + M}=\frac{M L}{2 \left(m + M\right)}$

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