Two 30 kg blocks rest on a massless belt which passes over a fixed pulley and is attached to a 40 kg block. If coefficient of friction between the belt and the table as well as between the belt and the blocks B and C is $\mu$ and the system is released from rest from the position shown, the speed with which the block B falls off the belt is 

 

Maximum possible acceleration of block will be $= \mu g$.
(i) When $\mu = 0.2$ maximum acceleration possible so that block does not slip is $2 m/s^2$. Now acceleration of  system assuming blocks don’t slip is
 

$\begin{array}{l}a=\frac{40g-2\mu mg}{100}=3.8m/s^2\\ \mu =0.2\\ As3.8\text{\hspace{0.17em}}>2\end{array}$

Hence the assumption is wrong and blocks slip on the belt.
$\therefore$ Maximum force applicable on block B is µ mg
 $\therefore$  Acceleration of block $= 2 m/s^2$
Hence, velocity of B when it strikes pulley is

$\begin{array}{l}{V}^2=0^2+2\times 2\times 2\\ V^2=8\\ \therefore V=2\sqrt{2}m/s\end{array}$

(ii) When µ = 0.5 maximum acceleration possible so that block does not   
      slip is $5 m/s^2$
Now acceleration of system assuming blocks don’t slip is 
$\begin{array}{l}a=\frac{40g-2\mu mg}{100}=1m/s^2\\ \mu =0.5\\ as1<5\end{array}$

Hence, blocks do not slip and acceleration of block B is $1m/s^2$
$\therefore$ Hence velocity of B when it strikes pulley is
$\begin{array}{l}{V}^2=0^2+\mathrm{2.1.2}\\ V^2=4\\ \therefore V=2m/s\end{array}$