a
(a) The voltage per unit light of the metre wire $PQ$ is $\left( {\frac{{6.00\,mV}}{{0.600\,m}}} \right)$ i.e. $10\,mV/m$.
Hence potential difference across the metre wire is $10\,mV{\rm{/}}m \times 1\,m = 10\,mV$.
The current drawn from the driver cell is $i = \frac{{10\,mV}}{{5\,\Omega }} = 2\,mA$.
The resistance $R = \frac{{(2\,V - 10\,mV)}}{{2\,mA}} = \frac{{1990\,mV}}{{2\,mA}} = 995\;\Omega $.