JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    The relative error in the determination of the surface area of a sphere is\[\alpha \]. Then the relative error in the determination of its volume is:                                                             [JEE Online 15-04-2018]

    A) \[\frac{2}{3}\alpha \]             

    B) \[\frac{5}{2}\alpha \] 

    C) \[\frac{3}{2}\alpha \]             

    D) \[\alpha \]

    Correct Answer: C

    Solution :

    \[S=\] \[4\pi {{R}^{2}}\] In \[S=In(4\pi )+In({{R}^{2}})\] In \[S=2\] In \[R\] \[\frac{\Delta S}{S}=2\frac{\Delta R}{R}=\alpha \] \[\frac{\Delta R}{R}=\frac{\alpha }{2}..............(1)\] \[V=\frac{4}{3}\pi {{R}^{3}}\] In \[V=In(\frac{4}{3}\pi )+In\,\,{{R}^{3}}\] In \[V=3\,\,In\,\,R\] \[\frac{\Delta V}{V}=3\frac{\Delta R}{R}\] \[\frac{\Delta V}{V}=3(\frac{\alpha }{2})\]m


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