Answer :
Let ρ represent the density of a block and let height of the block be L
Then, we can write the expression for the mass of the block as follows:
m = Vρ = L3ρ
Then, the weight of the block is
mg = L3ρg
Let x represent the height of the cube submerged.
Case II
When the vessel is placed on an elevator that accelerates upwards at a constant rate, the acceleration is equal to (g + a).
The weight of the block is given in the follwoing manner
m(g + a) = L3 ρ(g + a)
Therefore, the expression for effective weight becomes
m (g + a)
Let x1 be the new fraction of block submerged in water. Then, we have –
${{x}_{1}}=\frac{m}{{{L}^{3}}{{\rho }_{w}}}=\frac{{{L}^{3}}\rho }{{{L}^{3}}{{\rho }_{w}}}=\frac{\rho }{{{\rho }_{w}}}=x$