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A potentiometer wire has length 4 m and resistance 8Ω. The resistance that must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient 1 mV per cm on the wire is
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- 40 Ω
- 44 Ω
- 48 Ω
- 32 Ω
Correct Option: D
Total potential difference across potentiometer wire
= 10–3 × 400 volt = 0.4 volt
potential gradient = 1mv/cm
= 10–3 v/cm = 10–1(v/3)
Let resistance of RΩ connected in series.
⇒R + 8 = 40 or,R = 32 Ω