Question

In the experiment of Ohm's law, a potential difference of $5.0V$ is applied across the end of a conductor of length $10.0cm$ and diameter of $5.00mm$. The measured current in the conductor is $2.00A$. 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

$3.9$

Step 1: Given Information

Potential difference = $5.0V$

Length = $10.0cm$

Diameter = $5.00mm$

Step 2: Calculate the maximum permissible percentage error in the resistivity of the conductor

Knows, formulas $V=IR$

$R=\rho \left(\frac{\mathcal{l}}{A}\right)$

$V=I\left(\frac{\rho \mathcal{l}}{A}\right)$

$\rho =\frac{VA}{l\mathcal{l}}$

$A=\frac{\pi d^2}{4}$

Where, $R$= Resistance, $\rho$= Resistivity, $\mathcal{l}$=Length, $A$= Area,$I$= Current and $V$= Potential difference

Therefore,

$$\begin{gathered} \frac{∆p}{p}=\frac{∆V}{v}+\frac{∆I}{I}+\frac{∆I}{I}+2\frac{∆d}{d} \\ \frac{∆\rho }{\rho }=\frac{0.01}{5}+\frac{0.001}{2}+\frac{0.01}{10}+2\frac{0.001}{5} \\ \frac{∆\rho }{\rho }=0.02+0.005+0.01+0.004 \\ \frac{∆\rho }{\rho }=0.039 \end{gathered}$$

Hence, the correct option is (B).