A vessel contains two immiscible liquids of densit `rho_(1)=1000 kg//m^(3)` and `rho_(2)=1500kg//m^(3)`. A solid block of volume `V=10^(3)m^(3)` and density `d=800kg//m^(3)` is tied to one end of a string and the outer end is tied to the bottom of the vessel as shown in figure. The block is immersed with two fifths of its volume in the liquid of lower density. The entire system is kept in an elavator which is moving upwards with an acceleration of `a=g/2`. Find the tension in the string. brgt
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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