The spring block system lies on a smooth horizontal surface. The free end of the spring is being pulled towards right with constant speed $v_0=2m/s$. At $t=0\text{sec}$, , the spring of constant $k=100N/cm$ is unstretched and the block has a speed $1m/s$ to left. The maximum extension of the spring is
Diagram

The correct option is C $6cm$
In the frame (inertial w.r.t. earth) of free end of spring, the initial velocity of block is $3m/s$ to left and the spring unstretched.
Hence, by applying conservation of energy between initial and maximum extension state.
Kinetic Energy of block $=$ Potential Energy of the spring
$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$
$A=\sqrt{\frac{m}{k}}v$
Therefore, by putting the values we get,
$A=\sqrt{\frac{4\times 100}{100}}\times 3=6cm$

Final Answer: (c)