Question

A wire has a resistance of $10$ ohm. Its resistance if it is stretched by one tenth of its original length is:
(A) $12.1\Omega$
(B) $7.9\Omega$
(C) $11\Omega$
(D) $9\Omega$

Answer 1

Hint : To solve this question, we have to know about resistance. We know that resistance is a measure of the tendency of a material to resist the flow of an electrical current. It is dependent on the nature of the material, its thickness and length, and on temperature. We know that, Resistance is low in substances, such as metals, that are good conductors, and high in materials, such as plastic and rubber, that are insulators. When an electrical current encounters resistance, part of its energy is converted into heat, and sometimes light, reducing the current. This phenomenon can be a problem, but also has many uses.

Complete Step By Step Answer:
let us consider that the resistance is R of a wire of length l is given by,
$R = \dfrac{{\rho l}}{A}$
Now if the wire is stretched keeping the volume intact then,
$R = \dfrac{{\rho {l^2}}}{V}$
Now, we can say, $R\alpha {l^2}$
Now, according to the question, we will get,
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{l_2}^2}}{{{l_1}^2}}$
But, ${R_1} = 10\Omega$ ,
$\dfrac{{{l_2}}}{{{l_1}}} = \dfrac{{11}}{{10}}$
\Rightarrow \dfrac{{{R_2}}}{{10}} = \dfrac{{121}}{{100}} \\\ \Rightarrow {R_2} = 12.1\Omega \\\
So, the right option will be option number A.

Note :
we have to keep this in our mind that the resistance of a wire depends both on the cross-sectional area and length of the wire and on the nature of the material of the wire. Thick wires have less resistance than thin wires. Longer wires have more resistance than short wires. Copper wire has less resistance thin steel wire of the same size.