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Current Electricity question

2008 · Shift 0 · Q67
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Current Electricity question

2008 · Shift 0 · Q67

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
Consider a block of conducting material of resistivity ′ρ′'\rho '′ρ′ shown in the figure. Current ′I′'I'′I′ enters at ′A′'A'′A′ and leaves from ′D′'D'′D′. We apply superposition principle to find voltage ′ΔV′'\Delta V'′ΔV′ developed between ′B′'B'′B′ and ′C′'C'′C′. The calculation is done in the following steps: (i) Take current ′I′'I'′I′ entering from ′A′'A'′A′ and assume it to spread over a hemispherical surface in the block. (ii) Calculate field E(r)E(r)E(r) at distance ′r′'r'′r′ from A by using Ohm's law E=ρj,E = \rho j,E=ρj, where jjj is the current per unit area at ′r′'r'′r′. (iii) From the ′r′'r'′r′ dependence of E(r)E(r)E(r), obtain the potential V(r)V(r)V(r) at rrr. (iv) Repeat (i), (ii) and (iii) for current ′I′'I'′I′ leaving ′D′'D'′D′ and superpose results for ′A′'A'′A′ and ′D′.'D'.′D′.AIEEE 2008 Physics - Current Electricity Question 310 EnglishΔV\Delta VΔV measured between BBB and CCC is
  1. A
    ρIπa−ρIπ(a+b){{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}πaρI​−π(a+b)ρI​
  2. B
    ρIa−ρI(a+b){{\rho I} \over a} - {{\rho I} \over {\left( {a + b} \right)}}aρI​−(a+b)ρI​
  3. C
    ρI2πa−ρI2π(a+b){{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}2πaρI​−2π(a+b)ρI​
  4. D
    ρI2π(a−b){{\rho I} \over {2\pi \left( {a - b} \right)}}2π(a−b)ρI​
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