Question

A portion of length $\pi \text{m}$ is cut out of a conical solid wire. The two ends of this portion have circular cross-sections of radii $r_1=5\text{cm}$ and $r_2=10\text{cm}$. It is connected lengthwise to a circuit and a current $I=2\text{A}$ is flowing through it. The resistivity of the material of the wire is $\rho =1.5\times {10}^{-3}\Omega \text{m}$. Calculate the resistance $R$, of the considered portion and the voltage, $V$ developed across it.

• A. $R=0.3\Omega ,V=0.6\text{volts}$
• B. $R=0.6\Omega ,V=0.6\text{volts}$
• C. $R=0.6\Omega ,V=0.3\text{volts}$
• D. $R=3\Omega ,V=6\text{volts}$

Answer 1

The correct option is A $R=0.3\Omega ,V=0.6\text{volts}$

It follows from the figure, that

$\tan \theta =\frac{r_2-r_1}{L}$

Therefore, radius at distance $x$ from the left end, $r=r_1+x\tan \theta =r_1+x(\frac{r_2-r_1}{L})=\frac{r_{1}L+x(r_2-r_1)}{L}$


Cross-sectional area at this point,

$\text{A}=\pi (\frac{r_{1}L+x(r_2-r_1)}{L}{)}^2$

Resistance for the element of length $dx$

$dR=\frac{\rho L^{2}dx}{\pi [r_{1}L+(r_2-r_1)x{]}^2}$

So, the total resistance offered by the wire,

$\Rightarrow R={\int }_0^{L}dR=\frac{\rho L^2}{\pi }{\int }_0^L\frac{dx}{[r_{1}L+(r_2-r_1)x{]}^2}$

$\Rightarrow R=\frac{\rho L^2}{\pi }[{\{r_{1}L+(r_2-r_1\left)x\right\}}^{-1}{]}_0^L(-\frac{1}{(r_2-r_1)})$

$\Rightarrow R=\frac{-\rho L}{\pi (r_2-r_1)}(\frac{1}{r_2}-\frac{1}{r_1})=\frac{\rho L}{\pi \left(r_1{r}_2\right)}$

Substituting the given values we get,

$R=\frac{1.5\times {10}^{-3}\times \pi }{\pi \times 5\times 10\times {10}^{-4}}=0.3\Omega$

And the voltage developed across it,

$V=IR=\frac{I\rho L}{\pi r_1{r}_2}$

By susbtituting the given values we get,

$V=2\times 0.3=0.6\text{volts}$

Hence, option $\left(\text{a}\right)$ is the correct answer.

Why this question? But this formula can be applied directly if conical shape is given. If the area is nonuniform we have to take an element and then solve using integration.