A solid uniform ball having volume V and density `rho` floats at the interface of two unmixible liquids as shown in Fig. 7(CF).7. The densities of the upper and lower liquids are `rho_(1)` and `rho_(2)` respectively, such that `rho_(1) lt rho lt rho_(2)`. What fractio9n of the volume of the ball will be in the lower liquid.

Options

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

Answer

C

Explanation

Let `V_(1)` and `V_(2)` be the volumes of the ball in the upper and lower liquids respectively. So
`V_(1)+V_(2)=V`
A ball = upthrust on the ball due to two liquids
`V rho g=V_(1)rho_(1)g+V_(2)rho_(2)g`
or `V rho= V_(1)rho_(1)+V_(2)rho_(2)`
or `V rho= V_(1)rho_(1)+(V-V_(1))rho_(2)`
or `V_(1)=((rho-rho_(2))/(rho_(1)-rho_(2)))V`
`:.` Fraction in the upper liquid `(V_(1))/(V)=(rho-rho_(2))/(rho_(1)-rho_(2))`
Fraction in the lower liquid `=1-(V_(1))/(V)`
`=1-(rho-rho_(2))/(rho_(1)-rho_(2))=(rho_(1)-rho)/(rho_(1)-rho_(2))`