Question

A uniform chain of length 2m is held on a smooth horizontal table so that half of it hangs over the edge. If it is released from rest, the velocity with which it leaves the table will be nearest to

• A. 2 m/s
• B. 4 m/s
• C. 6 m/s
• D. 8 m/s

Answer 1

The correct option is B

4 m/s

Taking surface of table as a reference level (zero potential energy)
Potential energy of chain when 1/n^th length hanging from the edge
$=\frac{-MgL}{2n^2}$
Potential energy of chain when it leaves the table $=-\frac{MgL}{2}$
Kinetic energy of chain = loss in potential energy
$\Rightarrow \frac{1}{2}M{v}^2=\frac{MgL}{2}-\frac{MgL}{2n^2}\Rightarrow \frac{1}{2}M{v}^2=\frac{MgL}{2}[1-\frac{1}{n^2}]$

$\therefore Velocityofchainv=\sqrt{gL(1-\frac{1}{n^2})}=\sqrt{10\times 2(1-\frac{1}{(2{)}^2})}=\sqrt{15}=3.87\simeq 4m/s(approx.)$