Question

A stone is lying at rest in a river. The minimum mass of stone, $m = {kρv}^{x} g^{- 3}$ is needed for remaining at rest. Here, k = constant having no unit, g = acceleration due to gravity, v = river flow velocity, $ρ$ = density of water. The value of x is

Moderate

Solution

The unit of LHS is kg. The unit of RHS should be kg.
Here, ${kρv}^{x} g^{- 3} = (\frac{kg}{m^{3}}) ({\frac{m}{s}}^{x}) {(\frac{m}{s^{2}})}^{- 3}$
= (kg) ${(m)}^{- 3 + x - 3} {(s)}^{+ 6 - x}$
$∴$ -6+x = 0
$\Rightarrow$ x = 6

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