A cubical block of density rho is floating on the surface of water .Out of its height L, fraction x is submerged in water . The vessel is in an elevator accelerating upward with acceleration a . What is the fraction immersed ?

A cubical block of density rho is floating on the surface of water .Ou

1.	A cubical block of density rho is floating on the surface of water .Out of its height L, fraction x is submerged in water . The vessel is in an elevator accelerating upward with acceleration a . What is the fraction immersed ?
Answer» Solution :Let density of water is `rho_(w)`. A BLOCK of height L float on it .x be the height of block submerged in water . Volume of block `V=L^(3)` Mass of block `m=Vrho=L^(3)rhog` WEIGHT of the block `=mg=L^(3)rhog` First Case : Volume ofpart of cube submerged in water `=xL^(2)` `therefore` Weight of water displaced by block `=xL^(2)rho_(w)G` Weight of block =weight of water displaced by block . `L^(3)rhog=xL^(2)rho_(w)g` `therefore(x)/(L)=(rho)/(rho_(w))` `thereforex=(rho)/(rho_(w))L`....(1) Second Case : When vessel is PLACED in an elevator moving upward with acceleration a , then effective acceleration =g=(g+a) More acceleration a is due to Pseudo force `therefore` Weight of block =mg `=m(g+a)` Suppose , and elevator is moving upward .Let new fraction of block submerged in water is `x_(1)` For floating of block , Weight of block =Weight of displaced water , `L^(3)rho(g+a)=x_(1)L^(2)rho_(w)(g+a)` `therefore(x_(1))/(L)=(rho)/(rho_(w))`....(2) `x_(1)=(rho)/(rho_(w))*L` From equation (1)and (2) , `x=x_(1)` Hence , the fraction of the block submerged is independent of acceleration.