A solid sphere of volume V and density `rho` floats at the interface of two immiscible liquids of densities `rho_(1) and rho_(2)` respectively. If `rho_(1) lt rho lt rho_(2)`, then the ration of volume of the parts of the sphere in upper and lower liquids is :

Options

A. `(rho-rho_(2))/(rho_(2)-rho)`
B. `(rho_(2)-rho)/(rho-rho_(1))`
C. `(rho+rho_(1))/(rho+rho_(2))`
D. `(rho+rho_(2))/(rho+rho_(1))`

Answer

A

Explanation

Let `V_(1) and V_(2)` be the volumes, then
`" "V_(1)+V_(2)=V`
As ball is floating.
Weight of ball = upthrust on ball due to two liquids
`" "Vrhog=V_(1)rho_(1)g+V_(2)rho_(2)g`
`rArr" "Vrho=V_(1)rho_(1)+(V-V_(1))rho_(2)`
`rArr" "V_(1)=((rho-rho_(2))/(rho_(1)-rho_(2)))`
Fraction in upper part `=V_(1)/V=(rho-rho_(2))/(rho-rho_(2))`
Fraction in lower part = `1-V_(1)/V=1-(rho-rho_(2))/(rho_(1)-rho_(2))`
`" "=(rho_(1)-rho)/(rho_(1)-rho_(2))`
`therefore` Ratio of lower and upper parts `=(rho-rho_(2))/(rho_(1)-rho)`