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.

Answer 2

Step 1: Recall the formula for resistance:

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

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

where $\rho$ is the resistivity of the material, $L$ is the length, and $A$ is the cross-sectional area.

Step 2: Consider the effect of stretching on length and area:

Let the initial length of the wire be $L_1$ and the initial area be $A_1$. The initial resistance $R_1$ is given as $3\Omega$.

$R_1 = \rho \frac{L_1}{A_1} = 3\Omega$

When the wire is stretched to twice its original length, the new length $L_2 = 2L_1$.

Since the volume of the wire remains constant during stretching, we have:

Initial Volume $V_1 = A_1 L_1$
Final Volume $V_2 = A_2 L_2$
$V_1 = V_2 \implies A_1 L_1 = A_2 L_2$

Step 3: Relate the new area to the initial area:

Substitute $L_2 = 2L_1$ into the volume equation:

$A_1 L_1 = A_2 (2L_1)$
Divide both sides by $L_1$:

$A_1 = 2A_2$
So, the new area $A_2 = \frac{A_1}{2}$.

Step 4: Calculate the new resistance $R_2$:

The new resistance $R_2$ will be:

$R_2 = \rho \frac{L_2}{A_2}$
Substitute $L_2 = 2L_1$ and $A_2 = \frac{A_1}{2}$ into the equation for $R_2$:

$R_2 = \rho \frac{2L_1}{\frac{A_1}{2}} = \rho \frac{2L_1 \times 2}{A_1} = 4 \rho \frac{L_1}{A_1}$

Step 5: Express $R_2$ in terms of $R_1$:

We know that $R_1 = \rho \frac{L_1}{A_1}$, so we can substitute $R_1$ into the equation for $R_2$:

$R_2 = 4 R_1$

Step 6: Calculate the value of $R_2$:

Given $R_1 = 3\Omega$:

$R_2 = 4 \times 3\Omega = 12\Omega$

Final Answer:

The new resistance is $12\Omega$, which corresponds to option (D).