Question

When a current of $\left(2.5\pm 0.5\right)$ A flows through a wire, it develops a c of $\left(20\pm 1\right)V,$ the resistance of wire is

• A. $\left(8\pm 2\right)\Omega$
• B. $\left(8\pm 1.6\right)\Omega$
• C. $\left(8\pm 1.5\right)\Omega$
• D. $\left(8\pm 3\right)\Omega$

Answer 1

Answer: (A)

Here,
Current, $I=(2.5\pm 0.5)A$
Potential difference, $V=(20\pm 1)V$
Resistance, $R=\frac{V}{I}=\frac{20V}{2.5A}=8\Omega$
$\frac{\Delta R}{R}=\pm \left(\frac{\Delta V}{V}+\frac{\Delta I}{I}\right)$
$=\pm \left(\frac{1}{20}+\frac{0.5}{2.5}\right)$
$\pm \left(0.05+0.2\right)=\pm \left(0.25\right)$
$\Delta R=\pm \left(0.25\right)\left(8\right)=\pm 2\Omega$
$\therefore$ Resistance of the wire is $\left(8\pm 2\right)\Omega$.