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NEET Sample Paper NEET Sample Test Paper-65

  • question_answer
    When a 4 kg mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by 2 cm. The work required to be done by an external agent in stretching this spring by 5 cm will be \[(g=9.8\,\,m/{{s}^{2}})\]

    A) 4.900 joule                    

    B) 2.450 joule

    C) 0.495 joule                    

    D) 0.245 joule

    Correct Answer: B

    Solution :

    \[K=\frac{F}{x}=\frac{40}{2\times {{10}^{-2}}}\,\,=\,\,0.2\,\,N/m\] Work done \[=\,\,\frac{1}{2}\,K{{x}^{2}}=\frac{1}{2}\,\times (0.2)\times {{(0.05)}^{2}}\,\,=\,\,2.5\,\,J\]


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