Question

A wire of resistance 40 ohm is stretched to double it’s length. It’s new resistance is

Answer 1

Answer:

80 ohm because resistance is directly proportional to the length of the conductor

hence length is double then resistance is also double

please mark as brainlist please

Answer 2

Answer:

The new resistance is increased four times, when the length of a wire is doubled.

Explanation:

Given the resistance of wire, $R=40 \Omega$

The wire is stretched to double it’s length.

Resistance is the property of a conductor to resist the flow of current through it. And is given by

$R=\rho \frac{l}{A}$

where, ρ is the resistivity, $l$ is the length  and A is the cross-sectional area of the conductor

Since the material is same, therefore the  resistivity, ρ is constant.

Let the length of the wire be $l$.

When it is stretched to double it’s length, then the new length is $l^{'} =2l$.

As the volume of the wire is constant, if the length gets doubled, then the cross-sectional area will become half of its value, $A^{'}= \frac{A}{2}$

Therefore, the new resistance is,

$R^{'} =\rho \frac{l^{'} }{A^{'} }$

$=\rho \frac{2l}{\frac{A}{2} }=\rho \times {2l}\times {\frac{2}{A} }=\rho {\frac{4l}{A} }$

The ratio of resistances is  

$\frac{R^{'} }{R} =\frac{\rho \frac{4l}{A }}{\rho \frac{l}{A }} =4$

$\implies {R^{'} =4R$

Therefore, the new resistance is increased four times if the length of the wire is doubled.