Question

A uniform rod of length $2l$ and mass ${}^{\prime }{m}^{\prime }$ is suspended from one end by inextence string an other end lies on smooth ground. The angle made by rod with vertical is $\theta ={\sin }^{-1}\left(1/\sqrt{3}\right)$. If $N_1$ and $N_2$ represents the contact force from ground on rod just before and just after cutting string then find the ratio of $N_1/N_2$

• A. $1$
• B. $1/2$
• C. $3/2$
• D. $4$

Answer 1

The correct option is B $1/2$

Equation of force T+N=Mg

Equation of torque Tsin$\theta \times$ l=Nsin$\theta \times l\to T=N$

So N${}_1$=$\frac{Mg}{2}$

After cutting string N${}_2$=Mg

Thus $\frac{N_1}{N_2}=\frac{1}{2}$