Question

The resistance of a wire is $10\,\Omega $ . Its length increased by 10% by stretching. The new resistance will now be?
A. $12.22\,\Omega $
B. $1.2\,\Omega $
C. $13\,\Omega $
D. $11\,\Omega $

Answer 1

Hint: Here the length of the given wire is increased by 10% so its radius will decrease as the wire gets stretched. So, also the area of the wire will decrease as it depends on the radius. Then find the resistance with this decrease value of the area of the given wire and also the increase value of length.

Formula used:
Resistance, $R = \rho \dfrac{l}{A}$
Where, $\rho $ is resistivity, A is the area of the given wire, and l is the length of the wire.

Complete step by step solution:
Original resistance of the wire, $R = 10\Omega $.
Length is increased by stretching = 10%
Let the length be $l$.
Then increase in length = ${l_1}$ = $l + 10\% \,of\,l$
${l_1} = 1.1\,l$.......(equation 1)
As the length of the wire is increased by stretching the wire then the radius of the given wire will decrease and hence the area will also decrease.

Let the original area be A.
New area will be = ${A_1} = 0.9A$
Now, we know that resistance is given as follows:
$R = \rho \dfrac{l}{A} \\ $
Putting the increased length and decreased area value in above formula, we get;
$R = \rho \dfrac{{1.1l}}{{0.9A}} \\ $
By solving, we get;
$R = \dfrac{{1.1 \times 10}}{{0.9}} \\$
$\therefore R = 12.22\,\Omega $

Hence, the correct answer is option A.

Note: Here the length was increased by stretching and we know that it will cause decrease in radius and hence the area of the wire will also decrease but we need to know that the volume of the wire will remain the same. The stretching does not cause any change in the volume of the given wire.