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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2001

  • question_answer
    Two springs of force constants k1 = 500 N/m and \[{{k}_{2}}=3000\text{ }N/m\]stretched by same force. Therefore ratio of their potential energies will be

    A)  2 : 1            

    B)  1 : 2

    C)  4 : 1            

    D)  1 : 4

    Correct Answer: A

    Solution :

     Force, \[F=kx\] and potential energy, \[U=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}\frac{{{F}^{2}}}{k}\] \[\Rightarrow \] \[U\propto \frac{1}{k}\] \[\therefore \] \[\frac{{{U}_{1}}}{{{U}_{2}}}=\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{3000}{1500}=\frac{2}{1}\]


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