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JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Afternoon)

  • question_answer
    During compression of a spring the work done is 10kJ and 2kJ escaped to the surroundings as heat. The change in internal energy, \[\Delta U\](in kJ) is: [JEE Main 9-4-2019 Afternoon]

    A) 8         

    B) 12

    C) - 12                 

    D) -8

    Correct Answer: A

    Solution :

    \[\Delta U=q+\text{w}\] \[q=-2kJ,\,W=10kJ\] \[\Delta U=8kJ\]                  


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