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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 emf 2 V, so as to get a potential gradient 1 mV per cm on the wire is
A
32 $\Omega$
B
40 $\Omega$
C
44 $\Omega$
D
48 $\Omega$
Detailed Solution
Potential gradient $x = 1$ mV/cm $= \frac{10^{-3}}{10^{-2}}$ V/m $= 0.1$ V/m
$x = \frac{E}{R + R'}\times\frac{R}{L}$
$0.1 = \frac{2}{8 + R'}\times\frac{8}{4}$
$8 + R' = 40 \Rightarrow R' = 32\ \Omega$
$x = \frac{E}{R + R'}\times\frac{R}{L}$
$0.1 = \frac{2}{8 + R'}\times\frac{8}{4}$
$8 + R' = 40 \Rightarrow R' = 32\ \Omega$