A cylindrical copper wire is stretched such that its diameter decreases by 0.01 %. The approximate percentage increase in its resistance is :-
A. 0.02%
B. 0.04%
C. 0.08%
D. 0.16%

Hint: The percentage change in length is not given so, in order to deal with constants only, we use a constant volume (area into length) as the volume of the wire remains unchanged on stretching it.
Formula used:
The formula for resistance to be used here:
$R = \rho \dfrac{l}{A}$.
The percentage error for the expression,
$Z = \dfrac{A^p}{B^q}$
can be written as:
$\dfrac{\Delta Z}{Z} = p \dfrac{\Delta A}{A} + q \dfrac{\Delta B}{B} $

Complete answer:
We first multiply and divide the resistance formula by A, so we may write:
$R = \rho \dfrac{l A}{A^2} = \rho \dfrac{V}{A^2}$.
By doing this, we just made sure the rest of the terms are constant so we may just evaluate the change in resistance with respect to change in area.
Now, the area for the case of cylindrical wire can be written as:
$A = \pi \dfrac{d^2}{4}$,
where we used d = 2r, for the purpose of substitution.
The resistance in terms of diameter can be written as:
$R = \rho \dfrac{16 V}{\pi^2 d^4} $.
Now, as the rest of the factors in our expression are constant we may write:
$\dfrac{\Delta R}{R} = 4 \dfrac{\Delta d}{d} $.
In finding the above expression, we just referred to the percentage error expression (in the formula used section) and substituted at the required places places like Z became R, B became d with power q =4 and A was not required.
We are given that $\Delta d$ = 0.01%d. Therefore, the percentage change in resistance is:
$\dfrac{\Delta R}{R} = 4 \times \dfrac{0.01 d}{d} $ = 0.04%

Therefore, the correct answer is option (B).

Note:
There is another simple way of doing the same thing is that we first write the formula for resistance and then differentiate both sides with respect to d. In using this trick, we don't even need to remember any particular formula. A very common mistake here is to directly differentiate 'A' without even considering the fact that length is also changing. Only to eliminate change in length, we substitute a constant volume in place of it.