JCECE Engineering JCECE Engineering Solved Paper-2006

  • question_answer
    Two bodies of different masses \[{{m}_{1}}\] and \[{{m}_{2}}\] are dropped from different heights \[{{h}_{1}}\] and \[{{h}_{2}}\]. The ratio of the time taken by the two bodies to fall through these distances is:

    A) \[{{h}_{1}}:{{h}_{2}}\]                   

    B) \[\sqrt{{{h}_{1}}}:\sqrt{{{h}_{2}}}\]

    C) \[h_{1}^{2}:h_{2}^{2}\]                

    D) \[{{h}_{2}}:{{h}_{1}}\]

    Correct Answer: A

    Solution :

    Key Idea: Time taken by the body to fall through the height \[{{h}_{1}}\] is                 \[{{t}_{1}}=\sqrt{\frac{2{{h}_{1}}}{g}}\] Let \[{{t}_{1}}\] and \[{{t}_{2}}\] be the time taken by the two bodies to fall through the heights \[{{h}_{1}}\] and \[{{h}_{2}}\] respectively, then                 \[{{h}_{1}}=\frac{1}{2}gt_{1}^{2}\]and\[{{h}_{2}}=\frac{1}{2}gt_{2}^{2}\] \[\Rightarrow \]               \[{{t}_{1}}=\sqrt{\frac{2{{h}_{1}}}{g}},\,\,{{t}_{2}}=\sqrt{\frac{2{{h}_{2}}}{g}}\] \[\therefore \]    \[{{t}_{1}}:{{t}_{2}}=\sqrt{\frac{2{{h}_{1}}}{g}}:\sqrt{\frac{2{{h}_{2}}}{g}}\] \[\Rightarrow \]  \[{{t}_{1}}:{{t}_{2}}=\sqrt{{{h}_{1}}}:\sqrt{{{h}_{2}}}\]


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