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AIIMS AIIMS Solved Paper-2015

  • question_answer
    A uniform disc is acted by two equal forces of magnitude F. One of them, acts tangentially to the disc, while other one is acting at the central point of the disc. The friction between disc surface and ground surface is \[nF\]. If r be the radius of the disc, then the value of n would be (in N)

    A) 0 

    B) 1.2             

    C) 2.0                                         

    D) 3.2

    Correct Answer: A

    Solution :

                    Let \[\Rightarrow \] be the friction exerting between disc surface and ground surface, then for the motion of the disc we can write \[X=\sqrt{\frac{2mgh}{K}}=\sqrt{\frac{2\times 0.04\times 10\times 5}{400}}\]                 ...... (ii) and         \[=\frac{1}{10}m=10cm\approx 9.8cm\] \[V'\]     \[mv=2mv'\Rightarrow V'=\frac{V}{2}\]          ??...  (ii) Here, a = linear acceleration of the disc. Solving Eqs. (i) and (ii) we get, \[\text{e=}\frac{\text{Velocity of separation}}{\text{Velocity of approach}}\]                                


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