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A solid uniform ball of volume V floats ...

A solid uniform ball of volume `V` floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid is `rho_(1)` and that of lower one is `rho_(2)` and the specific gravity of ball is `rho(rho_(1)gtrhogtrho_(2))` The fraction of the volume the ball in the `u` of upper liquid is

A

`(rho_(2))/(rho_(1))`

B

`(rho_(2)-rho)/(rho_(2)-rho_(1))`

C

`(rho-rho_(1))/(rho_(2)-rho_(1))`

D

`(rho_(1))/(rho_(2))`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `V` be the total volume of the ball and `v` be the volume of the ball in the upper liquid. Then `V-v` is the volume of the lower liquid displaced.
Using the law of floatation we have
`Vrhog=vrho_(1)+(V-v)rho_(2)g`
`Vrho=vrho_(1)+Vrho_(2)-vrho_(2)`
or `V(rho-rho_(2))=v(rho_(1)-rho_(2))`
`v/V=(rho-rho_(2))/(rho_(1)-rho_(2))=(rho_(2)-rho)/(rho_(2)-rho_(1))`
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