Question

If measured value of resistance$R=1.05$ $\Omega$, wire diameter$d=0.60$mm, and length $l$ = $75.3$ cm, then find max. Permissible error in resistivity = $\frac{R\left(\frac{\pi d^2}{4}\right)}{l}$

• A. $4$ %
• B. $6$ %
• C. $8$ %
• D. $1$ %

Answer 1

Answer: (A)

$4$ %

${\left(\frac{\Delta p}{p}\right)}_{max}=\frac{\Delta R}{R}+2\frac{\Delta d}{d}+\frac{\Delta l}{l}$
$R=1.05\Omega \to \Delta R=0.01\Omega$ (least count)
$d=0.60mm\to \to \Delta d=0.01mm$ (least count)
$l=75.3\to \Delta l=0.1cm$ (least count)
${\left(\frac{\Delta p}{p}\right)}_{max}=\left(\frac{0.01\Omega }{105\Omega }+2\frac{0.01mm}{0.60mm}+\frac{0.1cm}{75.3cm}\right)\times 100$ = 4 %