A wire of resistance x Omega is drawn out,so | 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 $R = ho \frac{l}{A}$.When a wire is stretched,its volume $V = A 	imes l$ remains constant.Let the original

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.When a wire is stretched,its volume $V = A 	imes l$ remains constant.Let the original length be $l$ and area be $A$. The new length is $l' = 2l$.Since volume is constant,$A 	imes l = A' 	imes l' \implies A 	imes l = A' 	imes (2l) \implies A' = \frac{A}{2}$.The new resistance $R'$ is given by $R' = ho \frac{l'}{A'} = ho \frac{2l}{A/2} = 4 \left( ho \frac{l}{A} ight) = 4R$.Given $R' = 20 \ \Omega$ and $R = x$,we have $20 = 4x$.Therefore,$x = \frac{20}{4} = 5 \ \Omega$.

$A$ wire of resistance $x \ \Omega$ is drawn out,so that its length is increased to twice its original length,and its new resistance becomes $20 \ \Omega$. Then $x$ will be ........ $\Omega$.

Similar Questions

The temperature coefficient of resistance for a wire is $0.00125\,^{\circ}C^{-1}$. At $300\,K$ its resistance is $1\,\Omega$. The temperature at which the resistance becomes $2\,\Omega$ is .......... $K$.

The resistance of a conductor at $20\,^{\circ}C$ and $500\,^{\circ}C$ is $20\,\Omega$ and $60\,\Omega$ respectively. At what temperature will the resistance be $25\,\Omega$?

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

$A$ wire is stretched to increase its length by $2\%$,then the percentage change in its resistance is . . . . . . . (in $\%$)

Find the resistance of a cube of edge $60 \text{ cm}$,made of a material of specific resistance $60 \times 10^{-8} \Omega \text{ m}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module