Two blocks ( 1 and 2) of equal mass $m$ are connected by an ideal string (see figure shown) over a frictionless pulley. The blocks are attached to the ground by springs having spring constants $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$ such that $\mathrm{k}_{1}>\mathrm{k}_{2}$
Initially, both springs are unstretched. The block 1 is slowly pulled down a distance $\mathrm{x}$ and released. Just after the release the possible values of the magnitude of the acceleration of the blocks $a_{1}$ and $a_{2}$ can be-

• A

  Either $\left(a_{1}=a_{2}=\frac{\left(k_{1}+k_{2}\right) x}{2 m}\right)$ or $\left(a_{1} \frac{k_{1} x}{m}-g\right.$ and $\left.a_{2}=\frac{k_{2} x}{m}+g\right)$
• Correct answer
  B

  $\left(a_{1}=a_{2} \frac{\left(k_{1}+k_{2}\right) x}{2 m}\right)$ only
• C

  $\left(a_{1}=a_{2}=\frac{\left(k_{1}-k_{2}\right) x}{2 m}\right)$ only
• D

  Either $\left(\mathrm{a}_{1}=\mathrm{a}_{2}=\frac{\left(\mathrm{k}_{1}-\mathrm{k}_{2}\right) \mathrm{x}}{2 \mathrm{~m}}\right)$ or $\left(\mathrm{a}_{1}=\mathrm{a}_{2}=\frac{\left(\mathrm{k}_{1} \mathrm{k}_{2}\right) \mathrm{x}}{\left(\mathrm{k}_{1}+\mathrm{k}_{2}\right) \mathrm{m}}-\mathrm{g}\right)$

Solution

$\begin{array}{l} \mathrm{T}+\mathrm{k}_{1} \mathrm{x}-\mathrm{mg}=\mathrm{ma}_{1} \\ \mathrm{k}_{2} \mathrm{x}+\mathrm{mg}-\mathrm{T}=\mathrm{ma}_{2} \end{array}$ By constraint relation $\mathrm{a}_{1}=\mathrm{a}_{2}$

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