Question

A wire of resistance R and radius r is stretched till its radius became r/2. If new resistance of the stretched wire is x R, then value of x is _________.

Answer 1

Correct Answer: 16

We know that:

\[ R = \rho \frac{l}{A}, \quad R \propto \frac{l}{r^2} \]

As we stretch the wire, its length increases, and its radius decreases, keeping the volume constant:

\[ V_i = V_f \]

\[ \pi r^2 l = \pi r_f^2 l_f \implies l_f = 4l \]

Now:

\[ \frac{R_\text{new}}{R_\text{old}} = \frac{l_f}{l} \cdot \frac{r^2}{(r/2)^2} = \frac{4l}{l} \cdot \frac{r^2}{r^2/4} = 4 \cdot 4 = 16 \]

\[ R_\text{new} = 16R \]

Thus, $x = 16$.

Answer 2

This problem involves determining the new resistance of a wire after it has been stretched. We are given the initial resistance \(R\) and the initial radius \(r\). The wire is stretched until its radius becomes \(r/2\). The new resistance is given as \(xR\), and we need to find the value of \(x\).

Concept Used:

The solution is based on two main principles:

1. Resistance of a Wire: The resistance \(R\) of a wire is given by the formula: \[ R = \rho \frac{L}{A} \] where \(\rho\) is the resistivity of the material (a constant for a given material), \(L\) is the length of the wire, and \(A\) is its cross-sectional area. For a cylindrical wire of radius \(r\), the area is \(A = \pi r^2\).
2. Conservation of Volume: When a wire is stretched, its length increases and its cross-sectional area decreases, but its total volume remains constant (assuming the density of the material does not change). The volume \(V\) is given by: \[ V = A \times L = \text{constant} \]

Step-by-Step Solution:

Step 1: Express the initial resistance in terms of the initial dimensions.

Let the initial length of the wire be \(L_1\) and the initial radius be \(r_1 = r\). The initial cross-sectional area is \(A_1 = \pi r_1^2 = \pi r^2\).

The initial resistance \(R\) is:

\[ R = \rho \frac{L_1}{A_1} = \rho \frac{L_1}{\pi r^2} \]

Step 2: Determine the new dimensions of the wire after stretching.

The new radius is given as \(r_2 = r/2\).

The new cross-sectional area is \(A_2 = \pi r_2^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4}\).

Step 3: Apply the principle of conservation of volume to find the new length \(L_2\).

The initial volume \(V_1\) must equal the final volume \(V_2\).

\[ V_1 = A_1 L_1 \quad \text{and} \quad V_2 = A_2 L_2 \] \[ A_1 L_1 = A_2 L_2 \]

Substituting the expressions for the areas:

\[ (\pi r^2) L_1 = \left(\frac{\pi r^2}{4}\right) L_2 \]

Solving for the new length \(L_2\):

\[ L_2 = 4 L_1 \]

Step 4: Calculate the new resistance \(R_{\text{new}}\) using the new dimensions.

The new resistance is given by:

\[ R_{\text{new}} = \rho \frac{L_2}{A_2} \]

Substitute the expressions for \(L_2\) and \(A_2\) in terms of the initial dimensions:

\[ R_{\text{new}} = \rho \frac{4 L_1}{\frac{\pi r^2}{4}} = 16 \left(\rho \frac{L_1}{\pi r^2}\right) \]

Final Computation & Result:

From Step 1, we know that the initial resistance \(R = \rho \frac{L_1}{\pi r^2}\). Substituting this into the expression for the new resistance:

\[ R_{\text{new}} = 16 R \]

The problem states that the new resistance is \(xR\). By comparing our result with the given information:

\[ xR = 16R \]

This implies that \(x = 16\).

The value of x is 16.