Physics - Current Electricity Question with Solution | TestHub

Given here, $I=2\text{A},\mathrm{V}=3.4\mathrm{V},m=8.92\times 10^{-3}kg,$

$d=8.92\times 10^{3}kg{m}^{-3}and\rho =1.7\times 10^{-8}\Omega -m$

From Ohm's law, resistance is $R=\frac{V}{I}=\frac{3.4}{2}=1.7\Omega$

As resistance of wire is given by $R=\frac{\rho l}{A}$, where, $A$ is area and $l$ is length of wire.

$\Rightarrow 1.7=\frac{\rho l}{A}$ $\Rightarrow l=\frac{1.7\text{A}}{\rho }$

Now, volume$=\frac{m}{d}=\frac{8.92\times 10^{-3}}{8.92\times 10^3}=10^{-6}{m}^3$

Thus,

$$\begin{gathered} l^2=\frac{1.7\text{A}}{\rho }\times l \\ =\frac{1.7\left(\frac{\text{volume}}{l}\right)}{\rho }\times l \\ =\frac{1.7}{\rho }\left(10^{-6}\right) \\ =\frac{1.7}{1.7\times 10^{-8}}\times 10^{-6} \end{gathered}$$

Hence, $l=10m$