1. Tardigrade
  2. Question
  3. Physics
  4. A stone is lying at rest in a river. The minimum mass of stone, m=kρ vxg-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

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

Physical World, Units and Measurements

Solution:

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