Current Electricity Ques 112
112. A potentiometer wire has length $4 m$ and resistance $8 \Omega$. 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
[2015]
(a) $40 \Omega$
(b) $44 \Omega$
(c) $48 \Omega$
(d) $32 \Omega$
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Answer:
Correct Answer: 112.(d)
Solution:
- (d) Total potential difference across potentiometer wire $=10^{-3} \times 400$ volt $=0.4$ volt potential gradient $=\frac{1 mv}{cm}$
$=10^{-3} v / cm=10^{-1} \frac{v}{m}$ Let a resistance of $R \Omega$ be connected in series.
So, $\frac{2}{R+8}=\frac{10^{-1} \times 4}{8}=\frac{1}{20}$
$\Rightarrow R+8=40$ or, $R=32 \Omega$
BETA
