A vessel contains two immisible liquids of density `rho_(1) = 1000 kg m^(-3)` and `rho_(2) =1500 kg m(-3)`. A solid block of volume V=`10m^(3)` and density d = `800 kg m^(-3)` is tied to on end of a string and other end is tied to the bottom of the vessel. The block is immersed with `2//5^(th)` of its volume in the liquid of higher density and `3//5^(th)` in the liquid of lower density. The entire system is kept in an elevator which is moving upwards with an acceleration ao a = g/2. The tension in the string is (take g = `10 m s^(-2)`).
Text Solution
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We will analyse this problem from the reference frame of elavator. Total buoyant force on the block is `F_(B)= (2/5Vrho_(2)+3/5Vrho_(1))(g+a)` From the condition of equilibrium `F_(B)=T+Vd(g+a)` `impliesT=F_(B)-Vd(g+a)=(g+a)V[2/5rho_(2)+3/5rho_(1)-d]` `=15xx10^(-3)[2/5xx1500+3/5xx1000-800]=6N`
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