A body of mass $\mathrm{m}$ slides down along a frictionless inclined plane from height ' $h$ ' and just completes motion in a vertical circle of radius $2 \mathrm{~m}$ after reaching the bottom. What is the value of $h$ ? [Use $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right]$

• A

  2 m
• B

  $\frac{5}{2} \mathrm{~m}$
• Correct answer
  C

  5 m
• D

  10 m

Solution

To complete vertical circle, minimum velocity should be $\sqrt{5 \mathrm{Rg}}$, where $\mathrm{R}=$ Radius of circle
$\begin{aligned} & \text { Now, } \mathrm{P}_1+\mathrm{K}_1=\mathrm{P}_2+\mathrm{K}_2 \\ & \Rightarrow \quad \mathrm{mgh}+0=0+\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{Rg}})^2 \\ & \Rightarrow \quad \mathrm{mgh}=\frac{1}{2} \mathrm{~m} \times 5 \mathrm{Rg} \\ & \Rightarrow \quad \mathrm{h}=\frac{5 \mathrm{R}}{2}=\frac{5 \times 2}{2}=5 \mathrm{~m}\end{aligned}$

Hence, the correct answer is option: 'C'.