Physics - Fluid Question with Solution | TestHub

Let $L_{total}$ be the total length of the rod, so $L_{total}=2L$.

Let $L_{in}$ be the length of the rod immersed in water.

Let $L_{out}$ be the length of the rod out of water.

The specific gravity of the rod is $s=0.75$.

For a floating object, the buoyant force equals the weight of the object.

The ratio of the immersed volume to the total volume is equal to the specific gravity of the object.

For a homogeneous rod, this ratio is equivalent to the ratio of the immersed length to the total length.

So, $\frac{L_{in}}{L_{total}}=s$

$L_{in}=s\times L_{total}$

$L_{in}=0.75\times 2L=1.5L$

The length of the rod that extends out of water is $L_{out}=L_{total}-L_{in}$.

$L_{out}=2L-1.5L=0.5L=\frac{L}{2}$

The final answer is $\frac{L}{2}$.