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A small ball of density `rho` is immersed in a liquid of density `sigma(gtrho)` to a depth `h` and released. The height above the surface of water up to which the ball will jump isA. `(sigma/rho-1)h`B. `(rho/sigma-1)h`C. `(rho/sigma+1)h`D. `(sigma/rho+1)h` |
Answer» Correct Answer - A Let `V` be the volume of the ball. Net upward acceleration. `a=(Vsigmag-Vrhog)/(Vrho)=((sigma-rho)gh)/rho` If `h_(a)` is the height in air to which the ball rises, then `0-(2(sigma-rho)gh)/rho=2(-g)h_(a)` `:. H_(a)=((sigma-r)gh)/(grho)=(sigma/rho-1)h` |
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