NEET 2026 (Phase 1) · PhysicsPrevious Year Question
At a certain temperature, during a process, $500$ J is absorbed by the system and $200$ J of work is done by the system. The change in internal energy of the system is
- A.$400$ J
- B.$300$ J✓
- C.$700$ J
- D.$500$ J
Correct Answer
(B) $300$ J
Solution & Explanation
\textbf{Answer:} (b) $300$ J \textbf{Solution:} By the first law of thermodynamics, $$\Delta U=Q-W,$$ where $Q$ is the heat absorbed by the system and $W$ is the work done by the system. Given $Q=+500$ J (absorbed) and $W=+200$ J (done by the system), $$\Delta U=500-200=300\text{ J}.$$ This matches option (b).
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