Question

A wire of resistance \(160\Omega\) is melted and drawn into a wire of one-fourth of its length. The new resistance of the wire will be:

• A. \(10\Omega\)
• B. \(640\Omega\)
• C. \(40\Omega\)
• D. \(16\Omega\)

Hint:

When a wire is redrawn with a reduced length, its cross-sectional area increases, reducing resistance.

Answer 1

The Correct Option is A

Step 1: {Understanding volume conservation}
Since the wire is melted and redrawn, its volume remains the same: \[ A_1 l_1 = A_2 l_2 \] Given that the new length is \(\frac{1}{4}\)th of the original: \[ A_2 = 4 A_1 \] Step 2: {Finding new resistance}
Resistance is given by: \[ R = \rho \frac{l}{A} \] Using the transformation, \[ R_2 = \frac{l_2}{A_2} R_1 \] \[ R_2 = \frac{\frac{1}{4} l_1}{4 A_1} R_1 = \frac{1}{16} R_1 \] Substituting \(R_1 = 160\Omega\): \[ R_2 = \frac{160}{16} = 10\Omega \] Thus, the new resistance is \(10\Omega\).

Answer 2

Step 1: Understand the relation of resistance with dimensions
Resistance of a wire is given by:
\( R = \rho \frac{L}{A} \)
where:
- \( R \) = resistance
- \( \rho \) = resistivity (remains constant for the same material)
- \( L \) = length of the wire
- \( A \) = cross-sectional area

Step 2: Volume remains constant when wire is redrawn
When a wire is melted and redrawn, its volume stays the same:
Original volume = New volume ⇒ \( L_1 A_1 = L_2 A_2 \)

Let the original length be \( L \), and new length be \( L' = \frac{L}{4} \)
Then using volume conservation:
\( LA = \frac{L}{4} A' \Rightarrow A' = 4A \)

Step 3: Find new resistance
Original resistance: \( R = \rho \frac{L}{A} = 160\Omega \)
New resistance:
\[ R' = \rho \frac{L'}{A'} = \rho \frac{L/4}{4A} = \rho \frac{L}{16A} \]
\[ R' = \frac{1}{16} \cdot \rho \frac{L}{A} = \frac{1}{16} \cdot 160 = 10\Omega \]

Final Answer:
\( 10\Omega \)