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(D) According to the law of floatation,a body floats if the weight of the body is equal to the weight of the liquid displaced.Let $A$ be the cross-sec

Vedclass · Accepted Answer

$d_1 > \frac{3}{4} d$

Vedclass · Answer

(D) According to the law of floatation,a body floats if the weight of the body is equal to the weight of the liquid displaced.Let $A$ be the cross-sectional area of the cylinders.The total weight of the combined cylinder is $W = (d_1 L A + d_2 L A)g = (d_1 + d_2) L A g$.The length of the cylinder inside the liquid is $2L - L/2 = 3L/2$.The weight of the liquid displaced is $W_{disp} = (3L/2) A d g$.Equating the two,we get $(d_1 + d_2) L A g = (3L/2) A d g$,which simplifies to $d_1 + d_2 = \frac{3}{2} d$.Thus,$d_2 = \frac{3}{2} d - d_1$.Given $d_1 > d_2$,we substitute the expression for $d_2$: $d_1 > \frac{3}{2} d - d_1$.Adding $d_1$ to both sides: $2 d_1 > \frac{3}{2} d$.Dividing by $2$: $d_1 > \frac{3}{4} d$.

Two cylinders of same cross-section and length $L$ but made of two materials 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:

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