JEE Main & Advanced Physics Work, Energy, Power & Collision Question Bank Work Done by Variable Force

  • question_answer
    If a long spring is stretched by 0.02 m, its potential energy is U. If the spring is stretched by 0.1 m, then its potential energy will be                    [MP PMT 2002; CBSE PMT 2003; UPSEAT 2004]

    A)             \[\frac{U}{5}\]

    B)               \[U\]    

    C)             5U      

    D)             25U

    Correct Answer: D

    Solution :

                    \[U\propto {{x}^{2}}\] Þ \[\frac{{{U}_{2}}}{{{U}_{1}}}={{\left( \frac{{{x}_{2}}}{{{x}_{1}}} \right)}^{2}}={{\left( \frac{0.1}{0.02} \right)}^{2}}=25\]\\[{{U}_{2}}=25U\]

You need to login to perform this action.
You will be redirected in 3 sec spinner