Question

A hollow metal cylinder of length 10 cm floats in alcohol of density 0.80 g cm^-3, with 1 cm of its length above it. What length of cylinder will be above copper sulphate solution of density 1.25 g cm^-3?

Answer 1

Length of hollow metal cylinder = x= 10 cm
1 cm length of cylinder is above the alcohol
∴Length of the cylinder below alcohol = (10-1) = 9cm
Density of alcohol = `ρ_"alcohol"` = 0.80 g cm-3
Density of copper sulphate solution = ρCuS04 = 1.25 g cm^-3  

When block floats in alcohol By the law of floatation:

`"h"_"block" xx rho_"block" = "h"_"alcohol" xx rho_"alcohol"`

`10 xx rho_"block" = 9 xx 0.80`

`rho_"block" = (9 xx 0.80)/10 = 0.72` g cm^-3 

When block is floats in copper sulphate solution:

By law floatation:

`"h"_"block" xx rho_"block" = "h"_"CuSO4" xx rho_"CuSO4"`

`10 xx 0.72 = "h"_"CuSO4" xx 1.25`

`"h"_"CuSO4" = 7.2/1.25 = 5.76` cm

Length of metal block below copper sulphate solution = 5.76 cm
So, length of metal block above copper sulphate solution
= 10 – 5.76 = 4.24 cm