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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. 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» Correct Answer - C 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))` |
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