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Two cyllinders of same cross-section and length $L$ but made of two material of densities $d_1$ and $d_2$ are cemented together to form a cylinder of length $2L$. The combination floats in a liquid of density $d$ with $a$ length $L/2$ above the surface of the liquid. If $d_1 > d_2$ then:
$d$ $>$ $\frac{3}{4}$$ d$
$\frac{d}{2}$ $>$ $d_1$
$\frac{d}{4}$ $>$ $d_1$
$d$ $<$ $d_1$
Solution
According to law of floatation, a body floats if weight of the body is equal to the weight of water displaced. So we have
$\left(d_{1}+d_{2}\right) L A g=\frac{3}{2} L A d g \quad$ since $\left(2 L-\frac{L}{2}\right)=\frac{3 L}{2}$ length of cylinder is inside liquid.
$\left(d_{1}+d_{2}\right)=\frac{3}{2} d$
$\Rightarrow d_{2}=\frac{3}{2} d-d_{1}$
since $d_{1}>d_{2},$ we have $d_{1}>\frac{3}{2} d-d_{1}$
$\Rightarrow 2 d_{1}>\frac{3}{2} d$
$\Rightarrow d_{1}>\frac{3}{4} d$
