Question

A sphere of mass m is held between two smooth inclined walls. For $sin{37}^o=\frac{3}{5}$, the normal reaction of the wall (2) is equal to

• A. $\frac{16mg}{25}$
• B. $\frac{5Mg}{6}$
• C. $\frac{39mg}{21}$
• D. None of these

Answer 1

The correct option is B $\frac{5Mg}{6}$

$Mg=2R\sin {37}^0$

$R=\frac{Mg}{2\sin {37}^0}$

$R=\frac{Mg}{2\times \frac{3}{5}}\left(\sin {37}^0=\frac{3}{5}\right)$------------$given$

therefore$,$ $R=\frac{5Mg}{6}$

this sphere is lying between $2$ inclined planes so there are $2$ normal reactions on the sphere by the plane$.$

if we resolve these reactions the vertical forces will be $R\sin {37}^0\times 2$ this system is under equilibrium$.$

So$,$ vertical forces $R\sin {37}^0$ is equal to weight of he sphere $Mg$ $($ here assume normal reaction $=R)$

Hence,

option $\left(B\right)$ is correct answer.