Resistance of $100\, cm$ long potentiometer wire is $10 \,\Omega$, it is connected to a battery ($2\, volt$) and a resistance $R$ in series. A source of $10\, mV$ gives null point at $40\, cm$ length, then external resistance $R$ is ........... $\Omega $
Medium
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(b) $E = \frac{e}{{(R + {R_h} + r)}}.\frac{R}{L} \times l$
$ \Rightarrow 10 \times {10^{ - 3}} = \frac{2}{{(10 + R + 0)}} \times \frac{{10}}{1} \times 0.4$ $ \Rightarrow $ $R = 790 $ $\Omega$
$ \Rightarrow 10 \times {10^{ - 3}} = \frac{2}{{(10 + R + 0)}} \times \frac{{10}}{1} \times 0.4$ $ \Rightarrow $ $R = 790 $ $\Omega$
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