A potentiometer having the potential gradient of $2\, mV/cm$ is used to measure the difference of potential across a resistance of $10 \,\Omega$. If a length of $50\, cm$ of the potentiometer wire is required to get the null point, the current passing through the $10 \,\Omega$ resistor is (in $mA$)
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(d) $V = xl \Rightarrow iR = xl$
$ \Rightarrow \,\,\,i \times 10 = \left( {\frac{{2 \times {{10}^{ - 3}}}}{{{{10}^{ - 2}}}}} \right) \times 50 \times {10^{ - 2}} = 0.1$
$ \Rightarrow \,\,\,\,i = 10 \times {10^{ - 3}}A = 10\,mA$.
$ \Rightarrow \,\,\,i \times 10 = \left( {\frac{{2 \times {{10}^{ - 3}}}}{{{{10}^{ - 2}}}}} \right) \times 50 \times {10^{ - 2}} = 0.1$
$ \Rightarrow \,\,\,\,i = 10 \times {10^{ - 3}}A = 10\,mA$.
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