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JEE Main & Advanced AIEEE Solved Paper-2008

  • question_answer
    Directions: Questions No. 75 are based on the following paragraph Consider a block of conducting material of resistivity '\[\rho \]' shown in the figure. Current 'I' enters at 'A' and leaves from 'D' . We apply superposition principal to find voltage '\[\Delta V\]' developed between 'B' and 'C' .  The calculation is done in the following steps: (i) Take current 'I' entering from 'A' and assume it to spread over a hemispherical surface in the block. (ii) Calculate field E(r) at distance 'r' from A by using Ohm's law \[E=\rho j\], where 'j' is the current per unit area at 'r'. (iii) From the 'r' dependence of E(r), obtain the potential V(r) at 'r'. (iv) Repeat (i), (ii) and (iii) for current 'I' leaving 'D' and superpose results for 'A' and 'D'.        \[\Delta V\] measured between B and C is     AIEEE  Solved  Paper-2007

    A) \[\frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi \left( a+b \right)}\]

    B) \[\frac{\rho I}{2\pi \left( a-b \right)}\]   

    C) \[\frac{\rho I}{\pi a}-\frac{\rho I}{\pi \left( a+b \right)}\]

    D) \[\frac{\rho I}{a}-\frac{\rho I}{\left( a+b \right)}\]

    Correct Answer: C

    Solution :

    \[E=\rho \,j=\rho \frac{I}{2\pi {{r}^{2}}}\] Potential difference due to current at A \[{{V}_{B}}-{{V}_{C}}=-\int\limits_{C}^{B}{\vec{E}.\,\,d\vec{l}=-\int\limits_{a+b}^{a}{\rho \frac{I}{2\pi {{r}^{2}}}.\,dr\,;}}\]\[\Delta V'=-\frac{\rho I}{2\pi }\left[ -\frac{1}{r} \right]_{a+b}^{a}=\frac{\rho I}{2\pi a}-\frac{\rho I}{2\pi \left( a+b \right)}\] By principle of superposition, \[\Delta V=2\Delta V'=\frac{\rho I}{\pi a}-\frac{\rho I}{\pi \left( a+b \right)}\]


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