Question

A wire of resistance $10.0$ $ohm$ is stretched so as to increase its length by $20\%$. Its resistance then would be :

• A. $10.0$ $ohm$
• B. $12.0$ $ohm$
• C. $14.4$ $ohm$
• D. $10.2$ $ohm$

Answer 1

The correct option is C $14.4$ $ohm$

Answer is C.
The resistance of a wire can be expressed as $R=\frac{\rho L}{A}$where,
$\rho$ is the resistivity of the wire
L is the length of the wire
A is the area of cross section of the wire
In this case, the wire is stretched so as to increase its length by 20%.
Here, the wire's cross-sectional area(CSA) reduces by 20%, that is, 1.2 times of its original value and the length of the wire increases by 20%, that is 1.2 times. Since, resistance of the wire is inversely proportional to its CSA and is directly proportional to its length, the wire's resistance will increase from its original value after stretching.
That is, when the length of the wire is increased by n times then the resistance of the wire increases by $n^2$ times the old resistance.
So, the resistance of the new length will be $n^2={1.2}^2=1.44$ times more.
Hence, the new resistance is 1.44 * 10 = 14.4 ohms.