Question

A uniform length of wire is $ 1.2m $ and has a resistance of $ 16\Omega $. It is stretched such that its resistance is $ 36\Omega $. The new length is
(A) $ 1.8m $
(B) $ 2.4m $
(C) $ 2.7m $
(D) $ 3m $

Answer 1

Hint: To solve this question we need to use the formula of the resistance of a wire in terms of its geometrical parameters. Then we have to conserve the volume of the wire in the two cases. Combining these two relations we will get the proportionality between the resistance and the length with which we will get the final answer.

Formula used: The formula used to solve this question s given by
 $\Rightarrow R = \dfrac{{\rho l}}{A} $, $ R $ is the resistance of a wire of length $ l $, cross sectional area $ A $, and having the resistivity $ \rho $.

Complete step by step answer
We know that the resistance of a wire is given by
 $\Rightarrow R = \dfrac{{\rho l}}{A} $ ………………….(1)
Now, when the wire is stretched, its volume will not change. We know that the volume of the wire is given by
 $\Rightarrow V = Al $
This means that the area is
 $\Rightarrow A = \dfrac{V}{l} $ ………………….(2)
Putting (2) in (1) we get
 $\Rightarrow R = \dfrac{{\rho {l^2}}}{V} $
Since the resistivity and the volume of the wire is constant, so we can write
 $\Rightarrow R \propto {l^2} $
 $ \Rightarrow R = k{l^2} $ ( $ k $ is a constant) ………………….(3)
According to the question, the original resistance of the wire is $ {R_1} = 16\Omega $ corresponding to the original length of $ {l_1} = 1.2m $. Substituting these in (3) we get
 $\Rightarrow 16 = k{\left( {1.2} \right)^2} $
 $ \Rightarrow 16 = 1.44k $ ………………….(4)
Now, the resistance of the wire after stretching it becomes $ {R_2} = 36\Omega $. Let $ {l_2} $ be the new length of the wire after being stretched. Putting these in (3) we get
 $\Rightarrow 36 = k{l_2}^2 $ ………………….(5)
Dividing (5) by (4) we have
 $\Rightarrow \dfrac{{36}}{{16}} = \dfrac{{k{l_2}^2}}{{1.44k}} $
 $ \Rightarrow {l_2}^2 = \dfrac{{36}}{{16}} \times 1.44 $
Taking square root on both the sides, we get
 $\Rightarrow {l_2} = \dfrac{6}{4} \times 1.2 $
 $ \Rightarrow {l_2} = 1.8m $
Thus, the new length of the wire is equal to $ 1.8m $.

Hence, the correct answer is option A.

Note
Substitute the cross sectional area in terms of the volume of the wire in the formula of the resistance. When the wire is stretched, its cross sectional area will decrease such that its volume remains conserved.