JEE Mains 2026 (28 Jan Shift 1) Physics Solution: Thermodynamics P-V Diagram Work Done скачать в хорошем качестве

JEE Mains 2026 (28 Jan Shift 1) Physics Solution: Thermodynamics P-V Diagram Work Done

Detailed step-by-step solution for the Thermodynamics (P-V Diagram and Work Done) problem from JEE Mains 2026 (28 January, Shift 1). 📝 THE QUESTION: In the following p - V diagram the equation of state along the curved path is given by (V - 2)^2 = 4ap where a is a constant. The total work done in the closed path is ________. Options: 1. 1/3a 2. -1/a 3. -1/3a 4. -1/2a ⏱️ TIMESTAMPS: 0:00 - Reading the Question 0:24 - Concept: Work Done in a Cyclic Process 0:55 - Finding the Pressure for the Straight Line AC 1:15 - Setting up the Work Done Equation (Integrals) 1:43 - Integrating the Pressure Equation along the Curve 2:20 - Calculating Limits and Final Subtraction 2:35 - Final Answer Calculation 💡 DETAILED STEP-BY-STEP SOLUTION: Step 1: Understand the Cyclic Process The process is a closed loop on a P-V diagram. The total work done in a cyclic process is the area enclosed by the loop. Since the direction of the cycle (A B C A) is anti-clockwise, the net work done will be negative. Total Work (W_net) = Work done along CA (straight line) + Work done along ABC (curve) Step 2: Find the Constant Pressure along path CA The straight horizontal line AC represents a constant pressure process (isobaric). We know the curve equation is: (V - 2)^2 = 4ap then p = (V - 2)^2 / 4a Points A and C lie on this curve at V = 1 and V = 3 respectively. Let's find the pressure at V = 1. p = (1 - 2)^2 / 4a = (-1)^2 / 4a = 1/4a So, the constant pressure along the line AC is P = 1/4a. Step 3: Calculate Work Done from C to A (W_CA) The path goes from C (V_initial = 3) to A (V_final = 1). Work Done = P * (V_final - V_initial) W_CA = (1/4a) * (1 - 3) W_CA = (1/4a) * (-2) = -2/4a = -1/2a Step 4: Calculate Work Done from A to C along the curve (W_ABC) The path goes from A (V_initial = 1) to C (V_final = 3). Work Done is the integral of P dV. W_ABC = Integral [from 1 to 3] of ( (V - 2)^2 / 4a ) dV Take the constant (1/4a) outside the integral: W_ABC = (1/4a) * Integral [from 1 to 3] of (V - 2)^2 dV Integrate using the power rule: W_ABC = (1/4a) * [ (V - 2)^3 / 3 ] evaluated from 1 to 3 Step 5: Apply the Limits of Integration Upper limit (V = 3): (3 - 2)^3 / 3 = (1)^3 / 3 = 1/3 Lower limit (V = 1): (1 - 2)^3 / 3 = (-1)^3 / 3 = -1/3 Subtract lower limit from upper limit: W_ABC = (1/4a) * [ 1/3 - (-1/3) ] W_ABC = (1/4a) * [ 1/3 + 1/3 ] W_ABC = (1/4a) * (2/3) W_ABC = 2 / 12a = 1/6a Step 6: Calculate Net Work Done W_net = W_CA + W_ABC W_net = (-1/2a) + (1/6a) To add these, find a common denominator (which is 6a): W_net = (-3/6a) + (1/6a) W_net = -2/6a W_net = -1/3a Correct Option: 3 (-1/3a)