An object is placed one by one in three liqui | vedclass

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$(A)$ According to the law of flotation, the weight of the object is equal to the weight of the displaced liquid.Let $V$ be the total volume of the ob

Vedclass · Accepted Answer

$d_1 < d_2 < d_3$

Vedclass · Answer

$(A)$ According to the law of flotation, the weight of the object is equal to the weight of the displaced liquid.Let $V$ be the total volume of the object and $d$ be the density of the object.Weight of object = $V \cdot d \cdot g$.If a fraction $f$ of the volume is outside the liquid, then the volume inside the liquid is $V_{in} = V(1 - f)$.Weight of displaced liquid = $V(1 - f) \cdot d_{liquid} \cdot g$.Equating the two: $V \cdot d \cdot g = V(1 - f) \cdot d_{liquid} \cdot g$, which simplifies to $d = (1 - f) \cdot d_{liquid}$, or $d_{liquid} = \frac{d}{1 - f}$.For $d_1$, $f_1 = \frac{1}{9}$, so $d_1 = \frac{d}{1 - 1/9} = \frac{d}{8/9} = 1.125d$.For $d_2$, $f_2 = \frac{2}{11}$, so $d_2 = \frac{d}{1 - 2/11} = \frac{d}{9/11} = 1.222d$.For $d_3$, $f_3 = \frac{3}{7}$, so $d_3 = \frac{d}{1 - 3/7} = \frac{d}{4/7} = 1.75d$.Comparing the values, we get $1.125d < 1.222d < 1.75d$, which implies $d_1 < d_2 < d_3$.

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