If the length of a wire is increased by 10% b | vedclass

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

Question

(C) The resistance of a wire is given by $R = \rho \frac{\ell}{A}$.Since the volume $V = A \ell$ remains constant when the wire is stretched,$A = \fra

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(C) The resistance of a wire is given by $R = ho \frac{\ell}{A}$.Since the volume $V = A \ell$ remains constant when the wire is stretched,$A = \frac{V}{\ell}$.Substituting this into the resistance formula,we get $R = ho \frac{\ell^2}{V}$,which implies $R \propto \ell^2$.Given that the length increases by $10\%$,the new length $\ell_2 = 1.1 \ell_1$.Therefore,the ratio of the new resistance to the original resistance is $\frac{R_2}{R_1} = \left(\frac{\ell_2}{\ell_1}ight)^2 = (1.1)^2 = 1.21$.This means $R_2 = 1.21 R_1$.The percentage increase in resistance is $\Delta R \% = \frac{R_2 - R_1}{R_1} 	imes 100\% = (1.21 - 1) 	imes 100\% = 21\%$.

If the length of a wire is increased by $10\%$ by stretching it,the percentage increase in its resistance is ..............$\%$

Similar Questions

What is conductivity? On which factors does its value depend? Write its unit and dimension.

$A$ nichrome wire $50 \text{ cm}$ long and $1 \text{ mm}^2$ cross-section carries a current of $4 \text{ A}$ when connected to a $2 \text{ V}$ battery. The resistivity of the nichrome wire in $\Omega \cdot \text{m}$ is:

The resistance of a wire is $R \; \Omega$. If it is melted and stretched to $n$ times its original length,its new resistance will be

Specific resistance of a conductor increases with

$A$ cylindrical metallic wire is stretched to increase its length in such a way that the metallic wire changes its resistance by $6\%$. The percentage increase in its length is (in $\%$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module