A conductor has resistance values of 2.71 Om | vedclass

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

Question

(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_{t_0} [1 + \alpha(t - t_0)]$.Given $R_1 = 2.71 \ \Omega$ at $t_1 = 10^{\circ

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_{t_0} [1 + \alpha(t - t_0)]$.Given $R_1 = 2.71 \ \Omega$ at $t_1 = 10^{\circ}C$ and $R_2 = 3.70 \ \Omega$ at $t_2 = 100^{\circ}C$.Let $R_x = 3.26 \ \Omega$ at temperature $x$.Using the formula: $R_2 = R_1 [1 + \alpha(t_2 - t_1)]$$3.70 = 2.71 [1 + \alpha(100 - 10)]$$3.70 = 2.71 [1 + 90\alpha] \implies 1 + 90\alpha = \frac{3.70}{2.71} \implies 90\alpha = \frac{3.70}{2.71} - 1 = \frac{0.99}{2.71}$.Now,for temperature $x$: $R_x = R_1 [1 + \alpha(x - t_1)]$$3.26 = 2.71 [1 + \alpha(x - 10)]$$\frac{3.26}{2.71} = 1 + \alpha(x - 10) \implies \alpha(x - 10) = \frac{3.26}{2.71} - 1 = \frac{0.55}{2.71}$.Dividing the two expressions for $\alpha$:$\frac{\alpha(x - 10)}{90\alpha} = \frac{0.55 / 2.71}{0.99 / 2.71} \implies \frac{x - 10}{90} = \frac{0.55}{0.99} = \frac{55}{99} = \frac{5}{9}$.$x - 10 = 90 	imes \frac{5}{9} = 50$.$x = 50 + 10 = 60^{\circ}C$.

$A$ conductor has resistance values of $2.71 \ \Omega$ and $3.70 \ \Omega$ at $10^{\circ}C$ and $100^{\circ}C$ respectively. The temperature at which the resistance is $3.26 \ \Omega$ is .......... $^{\circ}C$.

Similar Questions

The resistance of a resistance wire is $5\,\Omega$ at $50\,^{\circ}\text{C}$ and $6\,\Omega$ at $100\,^{\circ}\text{C}$. The resistance at $0\,^{\circ}\text{C}$ will be ............. $\Omega$.

At what temperature will the resistance of a copper wire become three times its value at $0^{\circ}C$? (Temperature coefficient of resistance for copper = $4 \times 10^{-3} \,^{\circ}C^{-1}$)

Three copper wires of lengths and cross-sectional areas are $(l, A)$,$(2l, A/2)$,and $(l/2, 2A)$. Resistance is minimum in

$A$ wire of resistance $R_{1}$ is drawn out so that its length is increased by twice of its original length. The ratio of the new resistance to the original resistance is:

Masses of three wires of copper are in the ratio of $1: 3: 5$ and their lengths are in the ratio of $5: 3: 1$. The ratio of their electrical resistances is

Vedclass Test Series

Exam Paper Generator

Online Exam Module