JEE Main & Advanced JEE Main Paper (Held On 9 April 2017)

  • question_answer
    A uniform wire of length I and radius r has a resistance of \[100\,\Omega \]. It is recast into a wire of radius \[\frac{r}{2}\]. The resistance of new wire will be - [JEE Online 09-04-2017]

    A)  \[1600\,\Omega \]                        

    B)  \[100\,\Omega \]

    C)  \[200\,\Omega \]                           

    D)  \[400\,\Omega \]

    Correct Answer: A

    Solution :

                    \[R=\frac{\rho l}{A}\]                     \[Al=V\] \[R=\frac{\rho v}{{{A}^{2}}}\] \[\rho \to \] constant \[v\to \] constant \[R\propto \frac{1}{{{A}^{2}}}\propto \frac{1}{{{r}^{2}}}\] \[R\propto \frac{1}{{{r}^{4}}}\] \[{{R}_{2}}=16\,\,{{R}_{1}}=1600\Omega \]


You need to login to perform this action.
You will be redirected in 3 sec spinner