Question

A cubic block of side a is connected with two similar vertical springs as shown. Initially, bottom surface of the block of density $\sigma$ touches the surface of the fluid of density $2\sigma$ while floating. A weight is placed on the block so that it is immersed half in the fluid, find the weight:

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

Answer 1

The correct option is B $a(K+a^2\sigma g)$

In first case, the block just touches the liquid surface i.e. (no buoyant force) and lets say spring above the block has been streched by $x$ so the spring below the block will be compressed by $x$.
for equilibrium of block, weight of the block = upward spring force
$a^3\sigma g=Kx+Kx=2Kx\Rightarrow x=\frac{a^3\sigma g}{2K}$
in second case, when the block is half submerged due to extra weight $W$, spring above the block will now be stretched by $(x+\frac{a}{2})$ i.e. $(\frac{a^3\sigma g}{2K}+\frac{a}{2})$ and the spring below the block will be compressed by $(\frac{a^3\sigma g}{2K}+\frac{a}{2})$.
&
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
$\Rightarrow a^3\sigma g+W=\frac{a}{2}{a}^2\times 2\sigma g+2K(\frac{a^3\sigma g}{2K}+\frac{a}{2})$
$W=2K(\frac{a^3\sigma g}{2K}+\frac{a}{2})=a(K+a^2\sigma g)$