Pulley Constraint PYQ | JEE Main Physics 2025 | Acceleration Trick

 

❓ Question

In the given pulley system,

• Acceleration of block A is

$$a_A=2\ \text{m/s}^2 \quad (\text{upward})$$

• Acceleration of block B is

$$a_B=3\ \text{m/s}^2 \quad (\text{upward})$$

Find the acceleration of block C.

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✍️ Solution

Take

• Upward as positive
• Downward as negative

For an ideal pulley system, the principle of constraint relation is

$$\boxed{\sum T_i\,a_i=0}$$

where $T_i$ is the tension in each string segment attached to the moving body.

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Step 1: Count the tension segments

From the figure,

• Block C is attached to 1 tension segment.
• Block B contributes through 3 tension segments.
• Block A contributes through 2 pulleys, giving a total upward pull of

$$2\times 2T=4T.$$

Hence,

$$(-T)a_C+(3T)a_B+(4T)a_A=0.$$

(Subtracting for $C$ because it moves downward.)

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Step 2: Substitute the given accelerations

$$-Ta_C+3T(3)+4T(2)=0.$$
$$-Ta_C+9T+8T=0.$$
$$-Ta_C+17T=0.$$

Dividing by $T$,

$$a_C=17\ \text{m/s}^2.$$

Since the positive direction for $C$ was taken downward in the equation,

$$\boxed{a_C=17\ \text{m/s}^2\ \text{downward}.}$$

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✅ Final Answer

$$\boxed{17\ \text{m/s}^2\ \text{downward}}$$

Note: The handwritten solution in the image substitutes $3$ and $2$ directly instead of the given accelerations $a_B=3\,\text{m/s}^2$ and $a_A=2\,\text{m/s}^2$, leading to $3\times3+4\times2=17$. If the intended values were different, the numerical answer would change accordingly.