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(D) The potential difference across the potentiometer wire $(\Delta V_P)$ is given by the voltage divider rule: $\Delta V_P = \frac{R_P}{R_P + R_{int}

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(D) The potential difference across the potentiometer wire $(\Delta V_P)$ is given by the voltage divider rule: $\Delta V_P = \frac{R_P}{R_P + R_{int} + R_{box}} 	imes E$.Here, $E = 20\,V$, $R_{int} = 5\,\Omega$, and $R_P = 75\,\Omega$.For the minimum resistance of the box $(R_{box} = 120\,\Omega)$:$\Delta V_{P,max} = \frac{75}{75 + 5 + 120} 	imes 20 = \frac{75}{200} 	imes 20 = 7.5\,V$.For the maximum resistance of the box $(R_{box} = 170\,\Omega)$:$\Delta V_{P,min} = \frac{75}{75 + 5 + 170} 	imes 20 = \frac{75}{250} 	imes 20 = 6\,V$.The potentiometer can measure any potential difference $V$ such that $V < \Delta V_P$. Since the maximum potential drop across the wire is $7.5\,V$, it can measure any voltage less than $7.5\,V$. Thus, $5\,V$, $6\,V$, and $7\,V$ can all be measured.

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