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If a wire of resistance R is stretched to double of its length, then new resistance will be:
A) R/2
B) 2R
C) 4R
D) 16R

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Last updated date: 23rd Apr 2024
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Answer
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Hint: Resistance is defined as the property of a material to show hindrance against the current flowing in the conductor. It always moves opposite in the direction of the current.
The expression for the resistance of a wire is given as,
$R = \rho \dfrac{l}{A}$ where,
ρ is the resistivity of a material.
l is the length of wire.
A is the cross-sectional area.
In these types of questions, we will first find the resistance of wire having new length and divide it with the resistance of wire having original length so that we can find out the value of new resistance.

Complete step-by-step answer:
The resistance of wire having length (l) and area of cross-section (A) is given by-
$R = \rho \dfrac{l}{A}$
Now, on stretching a wire,
(a) volume of wire remains constant
(b) length of wire increases
(c) Area of cross section decreases
(d) Resistivity of wire remains constant.
If the length of the wire gets doubled, the new length of wire will be,
$l' = 2l$
As the length of wire gets doubled, the cross-sectional area will become half of its previous value because volume of wire remains constant.
So, the new cross-sectional area becomes:
$A' = \dfrac{A}{2}$
Now, the new resistance of the wire is given as:
$R' = \rho \dfrac{{l'}}{{A'}}$
Put all the values in formula, we get
⇒$R' = \rho \dfrac{{2l}}{{\dfrac{A}{2}}}$
⇒$R' = 4\rho \dfrac{l}{A}$
Now we will divide new resistance with previous resistance,
⇒$\dfrac{{R'}}{R} = \dfrac{{4\rho \dfrac{l}{A}}}{{\rho \dfrac{l}{A}}}$
⇒$\dfrac{{R'}}{R} = 4$
$R' = 4R$
Hence, we can see that the new resistance is four times the previous resistance.

Option C is correct.

Note: Due to increase in the length of wire it will become thinner and longer. If the wire is thinner, it will be difficult for the charge to move through, and so the resistance will increase. If the wire is longer, then the charge has to move further, so its resistance has increased.