NEET Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति NEET PYQ-Work Energy Power and Collision

  • question_answer
    When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy in it will be: [AIPMT 2003]  

    A) 10 U

    B) 25 U

    C) U/5

    D) 5 U

    Correct Answer: B

    Solution :

    Potential energy in a stretched spring is given by
                            \[U=\frac{1}{2}k\,{{x}^{2}}\]
                \[\therefore \]            \[\frac{{{U}_{1}}}{{{U}_{2}}}={{\left( \frac{{{x}_{1}}}{{{x}_{2}}} \right)}^{2}}\]
                Given, \[{{x}_{1}}=2\,cm\,=0.02\,m,\,\,{{x}_{2}}=10\,cm=0.1\,m\]
                Substituting the values, we have
                            \[\frac{{{U}_{1}}}{{{U}_{2}}}={{\left( \frac{0.02}{0.1} \right)}^{2}}={{\left( \frac{1}{5} \right)}^{2}}=\frac{1}{25}\]
                \[\Rightarrow \]   \[{{U}_{2}}=25\,{{U}_{1}}=25\,U\]


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