Question

A cubical block of wood of side $a$ and density $\rho$ floats in water of density $2\rho$. The lower surface of the cube just touches the free end of a massless spring of force constant $K$ fixed at the bottom of the vessel. The weight $W$ put over the block so that it is completely immersed in water without wetting the weight is

• A. $a(a^2\rho g+K)$
• B. $a(a^{}\rho g+2K)$
• C. $a(\frac{a\rho g}{2}+2K)$
• D. $a(a^2\rho g+\frac{K}{2})$

Answer 1

The correct option is D $a(a^2\rho g+\frac{K}{2})$

Since density of block = $\frac{1}{2}$(density of water), 50% of its volume is immersed in water.

In the second case, half of the volume of the block is further immersed in water.
Therefore,
$W$ = Extra upthrust + Spring force
$=a\times a\times \frac{a}{2}\times 2\rho \times g+K\left(\frac{a}{2}\right)$
$=a(a^2\rho g+\frac{K}{2})$