Physics - NLM Question with Solution | TestHub

Since $M_2>M_1$, therefore $M_2$ moves downwards and $M_1$ moves upwards with an acceleration $a$ as shown in the figure.
$$\text{ Free body diagram of }{M}_1$$
The equation of motion for $M_1$ is $$T-M_{1}g=M_{1}a$$ Free body diagram of $M_2$
The equation of motion for $M_2$ is $$M_{2}g-T=M_{2}a$$ Adding (i) and (ii), we get $$a=\frac{\left(M_2-M_1\right)g}{M_2+M_1}=\frac{(10-5)g}{(10+5)}=\frac{g}{3}$$