JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    A body of mass starts moving from rest along x-axis so that its velocity varies as\[v=a\sqrt{s}\] where is a constant and is the distance covered by the body. The total work done by all the forces acting on the body in the first seconds after the start of the motion is [JEE Main 16-4-2018]

    A)  \[\frac{1}{8}m{{a}^{4}}{{t}^{2}}\]                 

    B) \[4m{{a}^{4}}{{t}^{2}}\]

    C) \[8m{{a}^{4}}{{t}^{2}}\]                       

    D) \[\frac{1}{4}m{{a}^{4}}{{t}^{2}}\] 

    Correct Answer: A

    Solution :

     Velocity of body                 \[v=a\sqrt{s}\] acceleration\[a'=\frac{dv}{dt}=\frac{a}{2\sqrt{s}}\frac{ds}{dt}=\frac{{{a}^{2}}}{2}\] displacement\[s'=\frac{1}{2}a'{{t}^{2}}=\frac{1}{2}\frac{{{a}^{2}}}{2}{{t}^{2}}\] Work done\[W=force\times displacement\] \[W=ma'\times s'=\frac{1}{8}m{{a}^{4}}{{t}^{2}}\]


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