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**Hint:**Two blocks are tied with the same spring implies their tension force is equal. Firstly, we shall analyze the motion of the blocks before and after being released. Further, we will look at the forces acting on each block and write their separate translational equation of motion. Then, on simultaneously solving both the obtained equations, we will calculate the acceleration on each block.

**Complete answer:**

Since both blocks are connected by the same string and have the same tension force, therefore, their acceleration will also be the same. One of the blocks will be going up with that acceleration and the other would be coming down.

Initially, the spring is at its natural length and it is neither extended nor compressed. Let block A be connected to the spring and let the tension force in the string be $T$.

Thus, we shall now write the translational equations of motion for both blocks A and B.

For block A;

$T-mg=ma$ ……………………. Equation (1)

For block B;

$2mg-T=2ma$ …………………….. Equation (2)

Adding equation (1) and equation (2), we get

$ \Rightarrow 2mg-T+\left( T-mg \right)=2ma+ma $

$ \Rightarrow 2mg-T+T-mg=3ma $

$ \Rightarrow mg=3ma $

Cancelling m from both sides and dividing both sides by 3, we get

$ \Rightarrow g=3a $

$ \Rightarrow \dfrac{g}{3}=\dfrac{3a}{3} $

$\Rightarrow a=\dfrac{g}{3}$

**Therefore, the acceleration of block A is $\dfrac{g}{3}$ in downward direction and the acceleration of block B is $\dfrac{g}{3}$ in upward direction. So, option (B) is correct.**

**Note:**

Whenever two or more blocks are tied with the same string, then the tension of each block due to the spring is taken to be the same. Also, when such a problem consisting of two or more bodies performing motion related to each other is given, in that case we must draw a free-body diagram of each body individually and write the equation of their translational motion.

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