A cubical metal block of edge 12 cm floats in mercry with one fifth of the height inside the mercury. Water poured till the surface of the block is just immersed in it. Find the height of the water column to be poured. Specific gravity of mercury =13.6.

Answer

A, C, D

Explanation

Given `x=12cm `
=length of the edge of the block
`P_(Hg)=13.6gm/cc
`Given that initially `1/5` of block is inside mercury.
`Let Pbrarr density of block in gm/cc.
`:(x)^3xxrho_bxxg=(x)^2x(x/5)xxrho_(Hg)xxg`
`rarr (12)^3xxrho_bxxg=(12^2xx12/5xx13.6`
`rho_b=13.6/g gm/cc`
After water pured let x=height of water column.
`V_b=V_(Hg)+V_w(12)^3`
Where `V_(Hg)` and `V_w` are volumes of block inside mercury and water respectively.
`:. (V_bxxrho_bxxg)=(V_(Hg)xxrho(Hg)xg)+(V_wxxrho_wxxg)`
`rarr (V_(Hg)xxV_m)xx13.6/5`
`=V_(Hg)xx13.6xxV_wxx1`
`rarr (12)^3xx13.6/5=(12-x)xx13.6+(x)`
`rarr 12.6x=13.6(12-12/5)`
`=(13.6)xx(9.6)`
`rarr x=((9.6)xx(13.6))/(12.6)=10.4cm