Question

The length of a given cylindrical wire is increased by $100\% $. Due to the consequent decrease in diameter, the change in the resistance of the wire will be
(a) $300\% $ (b) $200\% $ (c) $100\% $ (d) $50\% $

Answer 1

Hint:We have to use the formula of specific resistance to find out the answer. If the length increases and thus diameter decreases the volume of the wire still remains the same. By using this relation we will find out the change in resistance of the wire.

Complete step-by-step solution:
Resistance is defined as the property of a material to obstruct or resist the passage of electrons or electricity through themselves.
Here, we have to use the expression $R = \rho \dfrac{l}{A} - - - - - \left( 1 \right)$
The variables are identified as,
$R = $ Resistance of the wire used.
$l = $ length of the wire used.
$A = $area of cross-section of the wire.
$\rho = $ specific resistance.
Now multiplying and dividing $l$ in R.H.S. of the equation $\left( 1 \right)$ we get,
$R = \rho \times \dfrac{{{l^2}}}{V}$ as $A \times l = V$ where $V$ is volume.
Since specific resistance $\rho $ and volume $V$ are constant then,
$R \propto {l^2} - - - - \left( 2 \right)$
Let the initial resistance of the wire be ${R_1}$ and final resistance of the wire be ${R_2}$.
Let the original length be $l$.
After $100\% $ increment in length, new length$ = l + l = 2l$
Using equation $\left( 2 \right)$, and comparing it we get,
$\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{l^2}}}{{{{\left( {2l} \right)}^2}}} = \dfrac{{{l^2}}}{{4{l^2}}} = \dfrac{1}{4}$
So, now we get,
${R_2} = 4{R_1}$
Change in resistance$ = {R_2} - {R_1} = 4{R_1} - {R_1} = 3{R_1}$
Percentage of change in resistance$ = \dfrac{{3{R_1}}}{{{R_1}}} \times 100\% = 300\% $
So, the percentage of change in resistance is $300\% $ and the correct option is (a).


Note:The specific resistance of a wire never changes as it is the specific resistance per unit length. We must consider the change in resistance rather than the new resistance and then find the percentage change. Consequent increment or decrement refers to the same amount of increment or decrement.