Question

A coin is sliding down on a smooth hemi-spherical surface of radius R. The height from the bottom, where it looses contact with the surface is

• A. $R/3$
• B. 2$R/3$
• C. 3$R/4$
• D. 4$R/5$

Answer 1

Answer: (B)

2$R/3$

$\text{Here, initial velocity of projection is zero.}$

$\begin{array}{c}\text{Given, radius of hemisphere}=R\text{.}\\ \text{To find}\to \text{value of}h\text{.}\\ \left(h\text{is height of}Q\text{from ground.}\right)\end{array}$

$\begin{array}{c}\text{It is clear, that}h=R\cos \theta \\ \text{thus, the blocks covers}(R-h)=R(1-\cos \theta )\text{vertically}\\ \text{before leaving the contact of Q}\text{.}\\ \text{Let, the velocity attained by block at Q be}v\text{,}\end{array}$

$\text{then,}{v}^2=0^2+2g(R-h)\Rightarrow v^2=2gR(1-\cos \theta )$

$\text{Now,}\begin{array}{cc}\text{Now,}A+B& \to \mathrm{mgcos}\theta -N=\frac{m{v}^2}{R}\text{(But}N=0\text{at}B\text{, as contact}\\ & \Rightarrow g\cos \theta =\frac{v^2}{R}=2g\left(1\cos \theta \right)[\because v^2=2gR(1\cos \theta \left)\right]\\ & \Rightarrow \cos \theta =\frac{2}{3}\end{array}$

$\text{Hence,}h=R\cos \theta =\left[\frac{2R}{3}\right]\text{[So, option}B\text{is correct}]$