Question

A slender homogeneous rod of length $2L$ floats partly immersed in water, being supported by a string fastened to one of its end, as shown. The specific gravity of the rod is $0.75$. the length of rod that extends out of water is :

• A. $L$
• B. $\frac{1}{2}L$
• C. $\frac{1}{4}L$
• D. $3L$

Answer 1

The correct option is A $L$

Lets say $x$ length of the rod is dipped into the water.

Since the buoyant force acts through the center of gravity the displaced water , the condition for rotational equilibrium is, taking moments about a point $O$ along the line of action of $T,$

$0=\Sigma {\tau }_o$

$\Rightarrow 0=wl\cos \theta -F_B(2l-\frac{x}{2})\cos \theta$

$\Rightarrow 0={\rho }_{rod}gA\left(2l\right)\left(l\cos \theta \right)-{\rho }_{water}gAx(2l-\frac{x}{2})\cos \theta$

$\Rightarrow 0=\left(\frac{1}{2}{\rho }_{water}gA\cos \theta \right)$ $(x^2-4lx+4\frac{{\rho }_{rod}}{{\rho }_{water}}{l}^2)$ where $A=\text{cross section area}$

$\Rightarrow x^2-4lx+3l^2=0$

$\Rightarrow x=l,3l$.

$x=3l$ is not a possible solution, Hence $2l-x=2l-l=l$ length of the rod extends out of water.