JEE Physics 2026 Question with Solution

Given: The process $ABC$ is cyclic, and the work done in a cyclic process equals the area enclosed in the $P\!-\!V$ diagram.

Find: The total work done by the system during the cycle.

For a cyclic process:

$$W=\text{Area enclosed in the }P\!-\!V\text{ diagram}$$

From the diagram, the cycle forms a triangle.

Step 1: Identify base and height

$$\text{Base}=(5-2)=3\,\text{m}^3$$ $$\text{Height}=(300-100)=200\,\text{Pa}$$

Step 2: Calculate work done

$$W=\frac{1}{2}\times \text{base} \times \text{height}$$ $$W=\frac{1}{2}\times 3 \times 200$$ $$W=300\,\text{J}$$

Therefore, the total work done by the system during the cycle is $300\,\text{J}$.

Given: The loop $ABC$ on the $P\!-\!V$ diagram encloses a triangular region.

Find: The numerical value of the work done in one complete cycle.

The net work done in any closed thermodynamic cycle is equal to the enclosed area on the $P\!-\!V$ graph.

Here the triangle has vertices corresponding to pressure values $100\,\text{Pa}$ and $300\,\text{Pa}$, and volume values $2\,\text{m}^3$ and $5\,\text{m}^3$.

So,

$$\Delta V = 5-2 = 3\,\text{m}^3$$ $$\Delta P = 300-100 = 200\,\text{Pa}$$

Now the enclosed area is

$$\text{Area} = \frac{1}{2}(\Delta V)(\Delta P)$$ $$\text{Area} = \frac{1}{2}\times 3 \times 200 = 300$$

Since $1\,\text{Pa}\cdot\text{m}^3 = 1\,\text{J}$,

$$W=300\,\text{J}$$

Therefore, the required numerical answer is 300.