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Current Electricity question

2004 · Q121
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Current Electricity question

2004 · Q121

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A 6 volt battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of 100 ohm. The difference of potential between two points on the wire separated by a distance of 50 cm will be
  1. A
    2 volt
  2. B
    3 volt
  3. C
    1 volt
  4. D
    1.5 volt
View written solutionFree

Correct answer: C

R α ℓR\,\alpha \,\ell Rαℓ

For 300 cm, R = 100 Ω\Omega Ω

For 50 cm, R′=100300×50=503ΩR' = {{100} \over {300}} \times 50 = {{50} \over 3}\Omega R′=300100​×50=350​Ω

∴\therefore∴ IR = 6

⇒IR′=6R×R′=6100×503=1 volt. \Rightarrow IR' = {6 \over R} \times R' = {6 \over {100}} \times {{50} \over 3} = 1\,volt.⇒IR′=R6​×R′=1006​×350​=1volt.

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