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(A) The resistance of the rod is given by $R = \frac{\rho l}{A}$. Since the volume $V_{vol} = Al$ remains constant,we can write $R = \frac{\rho l^2}{V

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(A) The resistance of the rod is given by $R = \frac{\rho l}{A}$. Since the volume $V_{vol} = Al$ remains constant,we can write $R = \frac{\rho l^2}{V_{vol}}$.When the rod is stretched by a small amount $\Delta l$,the change in length is $l' = l + \Delta l$. The new resistance is $R' = \frac{\rho (l + \Delta l)^2}{V_{vol}} \approx \frac{\rho (l^2 + 2l \Delta l)}{V_{vol}} = R + \frac{2 \rho l \Delta l}{V_{vol}}$.The change in resistance is $\Delta R = R' - R = \frac{2 \rho l \Delta l}{V_{vol}} = \frac{2 \rho l^2}{V_{vol}} \cdot \frac{\Delta l}{l} = 2R \varepsilon$,where $\varepsilon = \frac{\Delta l}{l}$ is the strain.The voltage across the rod is $V = iR$. The change in voltage is $\Delta V = i \Delta R = i(2R \varepsilon) = (2iR) \varepsilon$.Since $i$ and $R$ are constants,$\Delta V$ is directly proportional to the strain $\varepsilon$ $(\Delta V \propto \varepsilon)$.Therefore,the total voltage $V_{total} = V_{initial} + \Delta V = V_{initial} + (2iR) \varepsilon$. This represents a linear relationship with a positive slope,starting from the initial voltage $V_{initial}$.

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