Question

A wire of resistance 3Ω is stretched to twice its original length. The resistance of the new wire will be

• A. 1.5 Ω
• B. 3 Ω
• C. 6 Ω
• D. 12 Ω

Answer 1

The Correct Option is D

1. Relationship between resistance, resistivity, length, and area:

The resistance ($R$) of a wire is given by:

$R = \frac{\rho l}{A}$

where:

• $\rho$ is the resistivity of the material,
• $l$ is the length of the wire, and
• $A$ is the cross-sectional area of the wire.

2. Conservation of volume:

When the wire is stretched, its volume remains constant. The volume of a cylindrical wire is given by $V = Al$. Let $l_1$ and $A_1$ be the original length and area, and $l_2$ and $A_2$ be the new length and area. Then:

$A_1 l_1 = A_2 l_2$

We are given that $l_2 = 2l_1$. Therefore:

$A_1 l_1 = A_2 (2l_1)$

$A_2 = \frac{A_1}{2}$

3. New Resistance:

The original resistance is $R_1 = \frac{\rho l_1}{A_1} = 3\,\Omega$. The new resistance is:

$R_2 = \frac{\rho l_2}{A_2} = \frac{\rho (2l_1)}{\frac{A_1}{2}} = \frac{4\rho l_1}{A_1} = 4R_1 = 4(3\,\Omega) = 12\,\Omega$

The correct answer is (D) 12 Ω.