Question

Find the percentage error in specific resistance given by p=(pi*r^2*R)/l where r is the radius having value between (0.2+0.02)cm and (0.2-0.02)cm, R is the resistance of values between (60+2) and (60-2)ohm and l is the length of values between (150+0.1) and (150-0.1)cm.

Answer 1

Answer:

The percentage error in the specific resistance = 23.37 %.

Explanation:

The specific resistance is given by

$\rm p=\dfrac{\pi r^2R}{l}$

where,

• r = radius.
• R = resistance.
• l = length.

Given:

$\rm r\pm \Delta r = (0.2+0.02)\ cm\ and\ (0.2-0.02)\ cm = (0.2\pm0.02)\ cm.\\R\pm \Delta R = (60+2)\ \Omega\ and\ (60-2)\ \Omega = (60\pm2)\ \Omega.\\l\pm \Delta l = (150+0.1)\ cm\ and\ (150-0.1)\ cm = (150\pm0.1)\ cm.$

where,

$\rm \Delta r,\ \Delta R,\ \Delta l$ are the uncertainties in the radius, resistance and the length respectively, such that,

$\rm r = 0.2\ cm,\ \Delta r = 0.02\ cm.\\R = 60\ \Omega,\ \Delta R = 2\ \Omega.\\l=150\ cm,\ \Delta l = 0.1\ cm.$

The uncertainty in specific resistnace is related with the uncertainties of all the other values as:

$\rm \dfrac{\Delta p}{p}=\dfrac{\Delta(r^2) }{r}+\dfrac{\Delta R}{R}+\dfrac{\Delta l}{l}=\dfrac{2\Delta r }{r}+\dfrac{\Delta R}{R}+\dfrac{\Delta l}{l}\\=\dfrac{2\times 0.02}{0.2}+\dfrac{2}{60}+\dfrac{0.1}{150}\\=0.2+0.033+0.00067\\=0.23367\\\approx0.2337.$

The percentage error in the specific resistance is given by

$\rm Percentage\ error = \dfrac{\Delta p}{p}\times 100\%=0.2337\times 100\%=23.37\%.$

Answer 2

Answer:

applying logarithm on both sides of the given expression and differentiating,

we get

ρ

δρ

=±(2 r/δr + l/δl + r/δr )

given : δr=0.02cm, δr=2 ohm, δl=0.1cm

substituting the values in above expression,

ρ

δρ

=±(2 *0.2/0.02 + 150/0.1 + 60/2 )=0.234=23.4