NEET Sample Paper NEET Sample Test Paper-34

  • question_answer
    A ring takes \[{{t}_{1}}\] second in rolling down an inclined plane without slipping. The same ring \[{{t}_{2}}\] second in sliding down without rolling on a similar but frictionless inclined plane. Then the ratio \[{{t}_{1}}:{{t}_{2}}\] is

    A)  \[1:\sqrt{2}\]            

    B)  \[\sqrt{2}:1\]

    C)  \[1:2\]             

    D)  \[2:1\]

    Correct Answer: B

    Solution :

    \[{{t}_{1}}=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}\left( 1+\frac{{{K}^{2}}}{{{r}^{2}}} \right)}\] For ring \[=\frac{{{K}^{2}}}{{{r}^{2}}}=1\] so \[{{t}_{1}}=\frac{1}{\sin \theta }\sqrt{\frac{4h}{g}}\] In sliding \[\frac{{{K}^{2}}}{{{r}^{2}}}=0\] \[{{t}_{2}}=\frac{I}{\sin \theta }\sqrt{\frac{2h}{g}}\]      so   \[{{t}_{1}}:{{t}_{2}}=\sqrt{2}:1\]

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