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

Answer: (D)

$a(a^2\rho g+\frac{K}{2})$

In first case, the block just touches the spring i.e. spring has its natural length (no compression)
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the block of density $\rho$ floats in liquid of density $2\rho$, so we have according to law of floatation.
weight of the block= weight of water displaced
$a^3\rho g=x\times a^2\times 2\rho g$
$\Rightarrow x=\frac{a}{2}$ i.e. block is half submerged in first case.
in second case, when the block is completely submerged spring is compressed by $\frac{a}{2}$
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block and extra weight W are in rest. i.e Net force is zero
$\Rightarrow$ net downward force = net upward force
$\Rightarrow$ weight of block + extra weight = buoyant force+ spring force
$\Rightarrow a^3\rho g+W=a^3\times 2\rho g+K\left(\frac{a}{2}\right)$
$W=a^3\rho g+\frac{Ka}{2}=a(a^2\rho g+\frac{K}{2})$