JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Self Evaluation Test - Work, Energy and Power

  • question_answer
    If stretch in a spring of force constant k is tripled then the ratio of elastic potential energy in the two cases will be

    A) 9:1

    B) 1:6  

    C) 3:1

    D) 1:3

    Correct Answer: A

    Solution :

    [a] For a given spring, \[\operatorname{u} =\frac{1}{2} k{{x}^{2}}\] \[\therefore \frac{{{\operatorname{u}}_{2}}}{{{u}_{1}}} =\frac{\frac{1}{2} k{{x}_{2}}^{2}}{\frac{1}{2} k{{x}_{1}}^{2}}=\frac{{{\left( 3x \right)}^{2}}}{{{x}^{2}}}=9:1\]


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