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Current Electricity question

2015 · Q119
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Current Electricity question

2015 · Q119

NEETPhysicsCurrent ElectricityMCQ+4 / −1
A potentiometer wire has length 4 m and resistance 8 Ω\OmegaΩ. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2 V, so as to get a potential gradient 1 mV per cm on the wire is
  1. A
    44 Ω\OmegaΩ
  2. B
    48 Ω\OmegaΩ
  3. C
    32 Ω\OmegaΩ
  4. D
    40 Ω\OmegaΩ
View written solutionFree

Correct answer: C

Total potential difference across potentiometer wire

= 10–3 × 400 volt = 0.4 volt

potential gradient = 1mvcm{{1mv} \over {cm}}cm1mv​

=10−3v/cm=10−1vm = {10^{ - 3}}v/cm = {10^{ - 1}}{v \over m}=10−3v/cm=10−1mv​

Let resistance of RΩ\Omega Ω connected in series.

AIPMT 2015 Cancelled Paper Physics - Current Electricity Question 99 English Explanation


∴\therefore∴ 2R+8=10−1×48=120{2 \over {R + 8}} = {{{{10}^{ - 1}} \times 4} \over 8} = {1 \over {20}}R+82​=810−1×4​=201​

⇒\Rightarrow⇒ R + 8 = 40 or, R = 32 Ω\Omega Ω

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