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  4. A body of uniform cross-sectional area floats in a liquid of density thrice its value. The portion of exposed height will be:

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Q. A body of uniform cross-sectional area floats in a liquid of density thrice its value. The portion of exposed height will be:

NTA AbhyasNTA Abhyas 2020

Solution:

For floating body $W=Th=V_{\text{in }}\rho _{L}g$
$AH\left(\rho \right)_{B}g=\left(\right.Ah\left.\right)\times 3\left(\rho \right)_{B}g$
$\frac{h}{H}=\frac{1}{3}$
Exposed height $=\frac{H - h}{H}=\frac{3 - 1}{3}=\frac{2}{3}$