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Q. If the buoyant force $F$ acting on an object depends on its volume $V$ immersed in a liquid, the density $\rho$ of the liquid and the acceleration due to gravity $g$. The correct expression for $F$ can be

Physical World, Units and Measurements

Solution:

$F \propto V^{a} \rho^{b} g^{c}$
$F=\left[ L ^{3}\right]^{a}\left[ ML ^{-3}\right]^{b}\left[ LT ^{-2}\right]^{c}$
$\left[ MLT ^{-2}\right]=F=\left[ M ^{b} L ^{3 s -3 b+c} T ^{-2 c}\right]$
On comparing,
$b=1,\, -2 c=-2$
$\Rightarrow c=1$
$3 a-3 b+ c=1$
$\Rightarrow 3 a-3+1=1$
$\Rightarrow 3 a-2=1$
$\Rightarrow 3 a=3$
$\Rightarrow 3 a=1$
So, on putting all these values,
$F=V \rho g$