Question

In the experiment of Ohm's law, a potential difference of $5.0\text{V}$ is applied across the end of a conductor of length $10.0\text{cm}$ and diameter of $5.00\text{mm}$. The measured current in the conductor is $2.00\text{A}$. The maximum permissible percentage error in the resistivity of the conductor is:

• A. $7.5$
• B. $3.9$
• C. $8.4$
• D. $3.0$

Answer 1

The correct option is B $3.9$

We know that$V=IR$ and $R=\rho \frac{l}{A}$
So, $V=I\left(\frac{\rho l}{A}\right)$
$\Rightarrow \rho =\frac{VA}{Il}$
Also, $A=\frac{\pi d^2}{4}$
So $\Rightarrow \rho =\frac{V\pi d^2}{4Il}$

Now, $\frac{\Delta \rho }{\rho }=\frac{\Delta V}{V}+\frac{\Delta I}{I}+\frac{\Delta l}{l}+2\frac{\Delta d}{d}$

Since $V=5.0\text{V}$ There is one zero after decimal, so $\Delta V=0.1\text{V}$

$\frac{\Delta \rho }{\rho }=\frac{0.1}{5}+\frac{0.01}{2}+\frac{0.1}{10}+2\times \frac{0.01}{5}$

$\frac{\Delta \rho }{\rho }=0.02+0.005+0.01+0.004$

$\frac{\Delta \rho }{\rho }=0.039$

$\frac{\Delta \rho }{\rho }\times 100=0.039\times 100=3.9$%