Two plates each of the mass $\mathrm{m}$ are connected by a massless spring as shown.
A weight $W$ is put on the upper plate which compresses the spring further. When $W$ is removed, the entire assembly jumps up. The minimum weight $W$ needed for the assembly to jump up when the weight is removed is just more than -

• A

  $\mathrm{mg}$
• Correct answer
  B

  $2 \mathrm{mg}$
• C

  $3 \mathrm{mg}$
• D

  $4 \mathrm{mg}$

Solution

For lower block +ve lift, $\mathrm{kx} \geq \mathrm{mg}$ $\Rightarrow \mathrm{x} \geq \frac{\mathrm{mg}}{\mathrm{k}}$
W/E theorem $\begin{array}{l} -\mathrm{mg}(\mathrm{h}+\mathrm{x})+\left(\frac{1}{2} \mathrm{kh}^{2}-\frac{1}{2} \mathrm{kx}^{2}\right)=0-0 \\ \Rightarrow-\mathrm{mgh}-\frac{\mathrm{m}^{2} \mathrm{~g}^{2}}{\mathrm{k}}+\frac{1}{2} \mathrm{kh}^{2}-\frac{1}{2} \frac{\mathrm{m}^{2} \mathrm{~g}^{2}}{\mathrm{k}}=0 \\ \frac{\mathrm{kh}^{2}}{2}-\mathrm{mgh}-\frac{3 \mathrm{~m}^{2} \mathrm{~g}^{2}}{2 \mathrm{k}}=0 \\ \mathrm{~h}=\frac{\mathrm{mg} \pm \sqrt{\mathrm{m}^{2} \mathrm{~g}^{2}+3 \mathrm{~m}^{2} \mathrm{~g}^{2}}}{\mathrm{k}} \\ =\frac{\mathrm{mg} \pm 2 \mathrm{mg}}{\mathrm{k}}=\frac{3 \mathrm{mg}}{\mathrm{k}}, \frac{-\mathrm{mg}}{\mathrm{k}} \\ \therefore \mathrm{h}=\frac{3 \mathrm{mg}}{\mathrm{k}} \\ \mathrm{W}+\mathrm{mg}=\mathrm{kh} \\ \mathrm{W}+\mathrm{mg}=3 \mathrm{mg} \\ \mathrm{W}=2 \mathrm{mg} \end{array}$

Hence, the correct answer is option: 'B'.