Two wires A and B of the same material and sa | vedclass

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

Question

(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the mass $m = 	ext{Volume} 	imes 	ext{Density} = (A \cdot l) \cdot d$,and bot

Vedclass · Accepted Answer

$544$

Vedclass · Answer

(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the mass $m = 	ext{Volume} 	imes 	ext{Density} = (A \cdot l) \cdot d$,and both wires have the same material and mass,the product $A \cdot l$ is constant.Thus,$l = \frac{m}{A \cdot d} \propto \frac{1}{A}$.Substituting this into the resistance formula: $R = ho \frac{l}{A} \propto \frac{1}{A^2}$.Since $A = \pi r^2$,we have $R \propto \frac{1}{(\pi r^2)^2} \propto \frac{1}{r^4}$.Therefore,$\frac{R_A}{R_B} = \left( \frac{r_B}{r_A} ight)^4$.Given $R_A = 34\,\Omega$,$r_A = 2r$,and $r_B = r$:$\frac{34}{R_B} = \left( \frac{r}{2r} ight)^4 = \left( \frac{1}{2} ight)^4 = \frac{1}{16}$.$R_B = 34 	imes 16 = 544\,\Omega$.

Two wires $A$ and $B$ of the same material and same mass have radii $2r$ and $r$,respectively. If the resistance of wire $A$ is $34\,\Omega$,then the resistance of wire $B$ will be ............ $\Omega$.

Similar Questions

Why are alloys used for making standard resistance coils?

The alloys constantan and manganin are used to make standard resistance due to they have

The coefficient of linear expansion of the material of a resistor is $\alpha$. Its temperature coefficient of resistivity and resistance are $\alpha_\rho$ and $\alpha_R$ respectively. Then the correct relation is:

$A$ wire of resistance $R$ is stretched until its radius is halved. What will be the new resistance (in $, R$)?

Explain the qualitative dependence of resistivity on temperature.

Vedclass Test Series

Exam Paper Generator

Online Exam Module