Question

In the experiment of Ohm's law, when potential difference of 10.0 V is applied, current measured is 1.00 A. If length of wire is found to be 10.0cm and diameter of wire 2.50 mm, then find maximum permissible percentage error in resistivity.

​

Answer 1

V = 10 V

I = 1 A

R = V/I = 10/1 = 10 $\Omega$

Length = 10 cm = 0.1 m

Diameter = 2.5 mm = 25 × ${10}^{ - 4}$m

Area = πd²/4 = 22×25×25×${10}^{ - 4}$×${10}^{ - 4}$/7×4

Area = 13750×${10}^{ - 8}$/28

Area = 491.07 × ${10}^{ - 8}$ m²

$\boxed{Resistivity( \rho) = \frac{ R\times A }{l} }$

$\rho = \frac{ 10 \times 491.07 × {10}^{ - 8} }{0.1}$

$\rho = \frac{ 491.07 × {10}^{ - 7} }{0.1}$

$\rho = \frac{ 491.07 × {10}^{ - 6} }{1}$

$\red{ \rho = 4.91 × {10}^{ - 4}}$

Answer 2

$\underbrace{\boxed{ \rm{ \dfrac{ \dfrac{ V = 10 V}{</p><p>I = 1 A}} {\dfrac{ R = \frac{ V}{I} = 10 \Omega}{Length = 10 cm = 0.1 m}}}}}$

$Diameter = 2.5 mm = 25 × {10}^{ - 4}m$

$Area = \dfrac{πd²}{4} = 22×25×25×{10}^{ - 4}$

$Area = 13750×{10}^{ - 8}$

$Area = 491.07 × {10}^{ - 8}m²$

$\boxed{Resistivity( \rho) = \frac{ R\times A }{l} }$

$\begin{gathered}\rho = \frac{ 10 \times 491.07 × {10}^{ - 8} }{0.1}\\\end{gathered}$

$\begin{gathered}\rho = \frac{ 491.07 × {10}^{ - 7} }{0.1}\\\end{gathered}$

$\begin{gathered}\rho = \frac{ 491.07 × {10}^{ - 6} }{1}\\\end{gathered}$

$\begin{gathered}\red{ \rho = 4.91 × {10}^{ - 4}} \\\end{gathered}$