Question

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

• A. 3
• B. 5
• C. 6
• D. 8

Answer 1

Answer: (C)

The unit of $LHS$ is kg. The unit of $RHS$ should be kg.

Here, $k\rho v^x{g}^{-3}=\left(\frac{kg}{m^3}\right){\left(\frac{m}{s}\right)}^x{\left(\frac{m}{s^2}\right)}^{-3}$

$=\left(kg\right){\left(m\right)}^{-3+x-3}{\left(s\right)}^{+6-x}$

$\therefore -6+x=0$

$\Rightarrow x=6$