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VMMC VMMC Medical Solved Paper-2003

  • question_answer
    A stone falls reel              such that the distance covered by it in the last second of              its motion is equal to the distance covered by it in the first 5 seconds. It remained in air for:

    A)  12 sec                  

    B)         13 sec                  

    C)         25 sec                  

    D)         26 sec

    Correct Answer: B

    Solution :

    Distance travelled in nth second \[{{s}_{n}}=u+\frac{1}{2}g(2n-1)\] For freely falling body \[u=0\] \[\therefore \]  \[{{s}_{n}}=\frac{1}{2}g(2n-1)\] Distance travelled in t sec. \[s=ut+\frac{1}{2}g{{t}^{2}},\]put \[t=5\] \[\therefore \]  \[s=\frac{1}{2}g{{t}^{2}}=\frac{g\times 25}{2}\] Now, according to the question \[{{s}_{n}}=s\]                 \[\therefore \]  \[\frac{1}{2}g(2n-1)=\frac{g}{2}\times 25\]                 So,          \[n=13\,\sec \]


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