Potentiometer wire of length $1 \,m$ is connected in series with $490\,\Omega $ resistance and $2\,V$ battery. If $0.2\, mV/cm $ is the potential gradient, then resistance of the potentiometer wire is ................ $\Omega$
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(a) Potential gradient $x = \frac{e}{{(R + {R_h} + r)}}.\frac{R}{L}$
$ \Rightarrow $ $\frac{{0.2 \times {{10}^{ - 3}}}}{{{{10}^{ - 2}}}} = \frac{2}{{(R + 490 + 0)}} \times \frac{R}{1}$ $ \Rightarrow $ $R = 4.9 \,\Omega$.
$ \Rightarrow $ $\frac{{0.2 \times {{10}^{ - 3}}}}{{{{10}^{ - 2}}}} = \frac{2}{{(R + 490 + 0)}} \times \frac{R}{1}$ $ \Rightarrow $ $R = 4.9 \,\Omega$.
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