The resistance of a wire is 10,Omega. Its len | 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 volume $V = A 	imes l$ remains constant during stretching,we can write $A =

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) The resistance of a wire is given by $R = ho \frac{l}{A}$.Since the volume $V = A 	imes l$ remains constant during stretching,we can write $A = \frac{V}{l}$.Substituting this into the resistance formula,we get $R = ho \frac{l^2}{V}$.Since $ho$ and $V$ are constant,$R \propto l^2$.If the length is increased by $10\%$,the new length $l' = l + 0.1l = 1.1l$.The new resistance $R'$ is given by $R' = R \left( \frac{l'}{l} ight)^2$.$R' = 10 	imes (1.1)^2 = 10 	imes 1.21 = 12.1\,\Omega$.However,for small changes,the approximation $\Delta R/R \approx 2(\Delta l/l)$ gives $2 	imes 10\% = 20\%$,leading to $10 + 2 = 12\,\Omega$. Given the options,$12\,\Omega$ is the intended answer.

The resistance of a wire is $10\,\Omega$. Its length is increased by $10\%$ by stretching. The new resistance will now be .......... $\Omega$.

Similar Questions

The temperature coefficient of resistance of a conductor is:

When the length and area of cross-section both are doubled,then its resistance

The resistivity of a wire depends on its

The variation of the resistance of a conductor with temperature is as shown in the graph. The temperature coefficient $(\alpha)$ of the conductor is:

The dimension of $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is equal to that of (where $\mu_0 = \text{vacuum permeability}$ and $\epsilon_0 = \text{vacuum permittivity}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module