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

  • question_answer
    A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:                [JEE Online 15-04-2018 (II)]

    A) Increased 8 times           

    B) Doubled              

    C)          Halved 

    D)          Unchanged

    Correct Answer: A

    Solution :

    Rate of heat developed in the wire= \[P=\frac{{{V}^{2}}}{R}\] \[{{R}_{1}}=\frac{\rho L}{A}=\frac{\rho L}{\pi {{r}^{2}}}\] \[{{P}_{1}}=\frac{{{V}^{2}}}{{{R}_{1}}}\] \[{{R}_{2}}=\frac{\rho \frac{L}{2}}{\pi {{(2r)}^{2}}}=\frac{\rho L}{\pi 8{{r}^{2}}}=\frac{{{R}_{1}}}{8}\] \[{{P}_{2}}=\frac{V}{{{R}_{1}}}=\frac{8V}{{{R}_{1}}}\] \[{{P}_{2}}=8{{P}_{1}}\]


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