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Uttarakhand PMT Uttarakhand PMT Solved Paper-2007

  • question_answer
    A body of mass 10 kg falls from a height of 5 m (g = 10 m/s) and is stopped within one-tenth of a second on the ground. The force of interaction is

    A)  100 N         

    B)  zero

    C)  1000 N        

    D)  1100N

    Correct Answer: D

    Solution :

     From the condition, \[(F-mg)t=mv\] \[\Rightarrow \] \[(F-100)\frac{1}{10}=10\times \sqrt{2\times 10\times 5}\] \[(\because v=\sqrt{2gh})\] \[\Rightarrow \] \[F-100=10\times 10\times 10\] \[\Rightarrow \] \[F-100=1000\] \[\therefore \] \[F=1000+100=1100N\]


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