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A potentiometer wire, $10 \mathrm{~m}$ long, has a resistance of $40 \Omega$. It is connected in series with a resistance box and a $2 \mathrm{~V}$ storage cell. If the potential gradient along the wire $0.1 \mathrm{~m}$ is $\mathrm{V} / \mathrm{cm}$, the resistance unplugged in the box is
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The correct answer is:
$760 \Omega$
Potential gradient along wire
$$
=\frac{\text { potential difference along wire }}{\text { length of wire }}
$$
or, $0.1 \times 10^{-3}=\frac{\mathrm{I} \times 40}{1000} \mathrm{~V} / \mathrm{cm}$
or, Current in wire, $\mathrm{I}=\frac{1}{400} \mathrm{~A}$
or,$\frac{2}{40+\mathrm{R}}=\frac{1}{400}$ or $\mathrm{R}=800-40=760 \Omega$
$$
=\frac{\text { potential difference along wire }}{\text { length of wire }}
$$
or, $0.1 \times 10^{-3}=\frac{\mathrm{I} \times 40}{1000} \mathrm{~V} / \mathrm{cm}$
or, Current in wire, $\mathrm{I}=\frac{1}{400} \mathrm{~A}$
or,$\frac{2}{40+\mathrm{R}}=\frac{1}{400}$ or $\mathrm{R}=800-40=760 \Omega$
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