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NEET Sample Paper NEET Sample Test Paper-73

  • question_answer
    When a 1.0 kg mass hangs attached to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition, if \[\operatorname{g}= 10 m/{{s}^{2}}\]

    A) 1.5 Joule                      

    B) 2.0 Joule

    C) 2.5 Joule                      

    D) 3.0 Joule

    Correct Answer: C

    Solution :

    Force constant of a spring \[k=\frac{F}{x}=\frac{mg}{x}=\frac{1\times 10}{2\times {{10}^{-2}}}\,\,\,\Rightarrow \,\,k=500\,N/m\] Increment in the length \[= 60 - 50 = 10 cm\] \[\operatorname{U}=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}\,\,500{{(10\times 1{{0}^{-2}})}^{2}}=2.5\,J\]


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