Question

A cubical block of wood of side and density $\rho$ float 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 $a(a^2\rho g+\frac{K}{2})$

Let y be length up to which cube is immersed initially. Then

$\rho a^{3}g=2\rho a^2\times y$

$y=\frac{a}{2}$

After putting weight on block due to compression in spring of ${}^{\prime }a/2^{\prime }$

$W+\rho a^{3}g=2\rho a^{3}g+\frac{Ra}{2}$

$W=\rho a^{3}g+\frac{Ra}{2}$

$W=a(\rho a^{2}g+\frac{R}{2})$