Equal potentials are applied on an iron and c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given that the potential difference $V$ and length $L$ are the same for both wires, and the current $I$ is also the same.From Ohm's law, $V = IR$,

Vedclass · Accepted Answer

About $2.4$

Vedclass · Answer

(B) Given that the potential difference $V$ and length $L$ are the same for both wires, and the current $I$ is also the same.From Ohm's law, $V = IR$, so the resistance $R$ must be the same for both wires.The resistance is given by $R = ho \frac{L}{A} = ho \frac{L}{\pi r^2}$.Since $R$, $L$, and $\pi$ are constant, we have $\frac{ho_{	ext{iron}}}{r_{	ext{iron}}^2} = \frac{ho_{	ext{copper}}}{r_{	ext{copper}}^2}$.Therefore, $\frac{r_{	ext{iron}}^2}{r_{	ext{copper}}^2} = \frac{ho_{	ext{iron}}}{ho_{	ext{copper}}}$.Taking the square root, $\frac{r_{	ext{iron}}}{r_{	ext{copper}}} = \sqrt{\frac{ho_{	ext{iron}}}{ho_{	ext{copper}}}} = \sqrt{\frac{1.0 	imes 10^{-7}}{1.7 	imes 10^{-8}}} = \sqrt{\frac{10}{1.7}} \approx \sqrt{5.88} \approx 2.42$.Thus, the ratio is about $2.4$.

Similar Questions

What length of the wire of specific resistance $48 \times 10^{-8} \, \Omega \, m$ is needed to make a resistance of $4.2 \, \Omega$ (in $.1$)? (Diameter of wire = $0.4 \, mm$)

$A$ uniform wire of resistance $R$ is uniformly compressed along its length,until its radius becomes $n$ times the original radius. Now,the resistance of the wire becomes:

The positive temperature coefficient of resistance is for

$A$ conducting wire has length $L_1$ and diameter $d_1$. After stretching,the same wire's length becomes $L_2$ and diameter $d_2$. The ratio of resistance before and after stretching is:

The resistance ${R_t}$ of a conductor varies with temperature $t$ as shown in the figure. If the variation is represented by ${R_t} = {R_0}[1 + \alpha t + \beta {t^2}]$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module