What equal length of an iron wire and a coppe | vedclass

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

Question

(B) The resistance $R$ of a wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is resistivity, $L$ is length, and $A$ is the cross-sectional area.G

Vedclass · Accepted Answer

$97$

Vedclass · Answer

(B) The resistance $R$ of a wire is given by $R = ho \frac{L}{A}$, where $ho$ is resistivity, $L$ is length, and $A$ is the cross-sectional area.Given diameter $d = 2 \; mm = 2 	imes 10^{-3} \; m$, so radius $r = 1 	imes 10^{-3} \; m$.Area $A = \pi r^2 = \pi 	imes (10^{-3})^2 = \pi 	imes 10^{-6} \; m^2$.Resistivity of iron $ho_1 = 12 \; \mu\Omega \cdot cm = 12 	imes 10^{-6} 	imes 10^{-2} \; \Omega \cdot m = 1.2 	imes 10^{-7} \; \Omega \cdot m$.Resistivity of alloy $ho_2 = 51 \; \mu\Omega \cdot cm = 51 	imes 10^{-6} 	imes 10^{-2} \; \Omega \cdot m = 5.1 	imes 10^{-7} \; \Omega \cdot m$.For parallel connection, $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \Rightarrow R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = 3 \; \Omega$.$R_1 = \frac{ho_1 L}{A}$ and $R_2 = \frac{ho_2 L}{A}$.$R_{eq} = \frac{(ho_1 L / A) \cdot (ho_2 L / A)}{(ho_1 L / A) + (ho_2 L / A)} = \frac{ho_1 ho_2 L}{A(ho_1 + ho_2)} = 3$.$L = \frac{3 A (ho_1 + ho_2)}{ho_1 ho_2} = \frac{3 	imes \pi 	imes 10^{-6} 	imes (1.2 + 5.1) 	imes 10^{-7}}{1.2 	imes 10^{-7} 	imes 5.1 	imes 10^{-7}} = \frac{3 	imes \pi 	imes 6.3 	imes 10^{-13}}{6.12 	imes 10^{-14}} \approx 97 \; m$.

Similar Questions

The equivalent resistance of two resistors in series is $S$. When they are connected in parallel,the equivalent resistance is $P$. If $S = nP$,then the minimum possible value of $n$ is:

$A$ student has $10$ resistors of resistance $r$. The minimum resistance made by him from given resistors is

$10$ resistors,each of resistance $10\,\Omega$,can be connected to obtain maximum and minimum equivalent resistance. The ratio of maximum to minimum equivalent resistance will be $..........$.

In the network shown in the adjoining figure,each resistance is $1\,\Omega$. The effective resistance between $A$ and $B$ is

$A$ regular hexagon is formed by six wires each of resistance $r \Omega$ and the corners are joined to the centre by wires of the same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module