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(A) The resistance per unit length of the potentiometer wire is $\lambda = \frac{R}{L} = \frac{5 \ \Omega}{10 \ m} = 0.5 \ \Omega/m$.The resistance of

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$1.6$

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(A) The resistance per unit length of the potentiometer wire is $\lambda = \frac{R}{L} = \frac{5 \ \Omega}{10 \ m} = 0.5 \ \Omega/m$.The resistance of the wire segment of length $6.6 \ m$ $(660 \ cm)$ is $R' = 0.5 \ \Omega/m 	imes 6.6 \ m = 3.3 \ \Omega$.The potential difference across this segment is $V' = I 	imes R'$,where $I$ is the current in the potentiometer wire.Given $V' = 1.1 \ V$,we have $1.1 = I 	imes 3.3$,which gives $I = \frac{1.1}{3.3} = \frac{1}{3} \ A$.The current in the circuit is given by $I = \frac{E}{R + r}$,where $E = 2.2 \ V$,$R = 5 \ \Omega$,and $r$ is the internal resistance.Substituting the values: $\frac{1}{3} = \frac{2.2}{5 + r}$.$5 + r = 3 	imes 2.2 = 6.6$.$r = 6.6 - 5 = 1.6 \ \Omega$.

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