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InterviewSolution
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A stone of density `2.5 xx10^3 kg m^(-3)` completely immersed in sea water is allowed to sink from rest in 2 s. Neglect the effect of vicosity. Relative density of sea water is 1.025. |
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Answer» Correct Answer - 11.6 m Let m be the mass of stone. Volume of stone. `V = (m)/(2.5xx10^3)m^3` upward thrust on stone. ` = (m)/(2.5)xx10^3 xx(1.025 xx 10^3)xxg` Weight of stonw = mg , acting vertically downwards. The resultant downward force on stone is `F = mg - (1.025)/(2.5)mg = mg[1 - (1.025)/(2.5)]` The ac celeration of stone is `a = (F)/(m) = g[1 - (1.025)/(2.5)] = 9.8 xx (1.475)/(2.5)` ` =5.782 ms^(-2)` Starting from rest the distance travelled by stone is `S = (1)/(2)a t^2 = (1)/(2)xx(5.782)xx2^2 ~~11.6 m` |
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