R=65 ± 1 L, l=5 ± 0.1 ~mm and d=10 ± 0.5 ~mm . Find error in calculation of resistivity.

Correct answer: B. 13 \% Given that, R=65 , Δ R=1 , l=5 × 10^ -3 ~m , Δ l=0.1 × 10^ -3 ~m, ~d=10 × 10^ -3 ~m, Δ d=0.5 × 10^ -3 ~m Resistivity, =(R A)/(l) or = R π(d / 2)^2 l = π Rd^2 4 l array ll & (Δ )/( )= Δ R R +2 Δ d d + Δ l l & (Δ )/( )=(1)/(65)+2( 0.5 × 10^ -3 10 × 10^ -3 )+ 0.1 × 10^ -3 5 × 10^ -3 array or (Δ )/( )=0.0153+0.1+0.02 or (Δ )/( ) 0.1353 So, error in calculation of resistivity is 13.5 \% 13 \% .

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Physics
Chapter: Mathematics in Physics

JIPMER 2018 · Physics · Mathematics in Physics

R=65 ± 1 L, l=5 ± 0.1 ~mm and d=10 ± 0.5 ~mm . Find error in...

\mathrm{R}=65 \pm 1 \Omega \mathrm{L}, \mathrm{l}=5 \pm 0.1 \mathrm{~mm}

and

\mathrm{d}=10 \pm 0.5 \mathrm{~mm}

. Find error in calculation of resistivity.

Options

A $21 \%$
B $13 \%$
C $16 \%$
D $41 \%$

Correct answer

B. $13 \%$

Step-by-step solution

Given that,

\mathrm{R}=65 \Omega, \Delta \mathrm{R}=1 \Omega, \mathrm{l}=5 \times 10^{-3} \mathrm{~m}

\Delta \mathrm{l}=0.1 \times 10^{-3} \mathrm{~m}, \mathrm{~d}=10 \times 10^{-3} \mathrm{~m}, \Delta \mathrm{d}=0.5 \times 10^{-3} \mathrm{~m}

\because

Resistivity,

\rho=\frac{R A}{l}

\quad \rho=\frac{\mathrm{R} \pi(\mathrm{d} / 2)^2}{\mathrm{l}}=\frac{\pi \mathrm{Rd}^2}{4 \mathrm{l}}

\begin{array}{ll}\therefore & \frac{\Delta \rho}{\rho}=\frac{\Delta \mathrm{R}}{\mathrm{R}}+2 \frac{\Delta \mathrm{d}}{\mathrm{d}}+\frac{\Delta \mathrm{l}}{\mathrm{l}} \\ \therefore & \frac{\Delta \rho}{\rho}=\frac{1}{65}+2\left(\frac{0.5 \times 10^{-3}}{10 \times 10^{-3}}\right)+\frac{0.1 \times 10^{-3}}{5 \times 10^{-3}}\end{array}

\quad \frac{\Delta \rho}{\rho}=0.0153+0.1+0.02

\frac{\Delta \rho}{\rho} \approx 0.1353

So, error in calculation of resistivity is

13.5 \% \approx 13 \%

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