A block A of mass 7 kg is placed on a frictionless table. A thread tied to it passes over a frictionless pulley and carries a body B of mass 3 kg at the other end. The acceleration of the system is (given g = 10 ms⁻²)

The correct option is 3 ms⁻².

Step-by-step Explanation:

Consider the system consisting of two blocks connected by a light, inextensible thread passing over a frictionless pulley.

Let:
Mass of block A (resting on the frictionless table),
$m_A=7\text{kg}$
Mass of body B (hanging vertically),
$m_B=3\text{kg}$
Acceleration due to gravity,
$g=10{\text{ms}}^{-2}$
Let $T$ be the tension in the thread and $a$ be the common acceleration of the system.

1. Equation of motion for block A:
Since the table is frictionless, the only horizontal force acting on block A is the tension $T$ from the thread. Applying Newton's second law:
$T=m_{A}a$
Substituting the mass of block A:
$T=7a$
(Equation 1)

2. Equation of motion for body B:
Body B moves vertically downwards. The forces acting on it are its weight ($m_{B}g$) acting downwards and the tension ($T$) acting upwards. The equation of motion is:
$m_{B}g-T=m_{B}a$
Substituting the values of $m_B$ and $g$:
$3\left(10\right)-T=3a$
$30-T=3a$
(Equation 2)

3. Solving for acceleration:
Substituting the value of $T$ from Equation 1 into Equation 2:
$30-7a=3a$
Rearranging the equation:
$30=10a$
$a=3{\text{ms}}^{-2}$