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(C) Let the total mass on the left side be $M_L = 3\,kg + 1\,kg = 4\,kg$ and the mass on the right side be $M_R = 3\,kg$.The acceleration $a$ of the s

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$\frac{g}{7}$ downwards,$\frac{6g}{7}$

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(C) Let the total mass on the left side be $M_L = 3\,kg + 1\,kg = 4\,kg$ and the mass on the right side be $M_R = 3\,kg$.The acceleration $a$ of the system is given by $a = \frac{M_L - M_R}{M_L + M_R} g = \frac{4 - 3}{4 + 3} g = \frac{g}{7}$.Since $M_L > M_R$,the $1\,kg$ mass (block $A$) moves downwards with acceleration $a = \frac{g}{7}$.Now,consider the free body diagram of block $A$ ($1\,kg$ mass). The forces acting on it are its weight $mg$ downwards and tension $T$ upwards.The equation of motion is $mg - T = ma$.Substituting $m = 1\,kg$ and $a = \frac{g}{7}$,we get $1 \cdot g - T = 1 \cdot \frac{g}{7}$.$T = g - \frac{g}{7} = \frac{6g}{7}$.

In the system shown in the figure,the acceleration of the $1\,kg$ mass and the tension in the string connecting between $A$ and $B$ is

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