Question

A wire of resistance 160 Ω is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be

• A. 10 Ω
• B. 16 Ω
• C. 40 Ω
• D. 640 Ω

Answer 1

The Correct Option is A

To find the new resistance of the wire after it is drawn into one-fourth of its original length, we apply the concept of resistivity and factors affecting resistance.

Step-by-Step Solution:

1. We start with the formula for resistance: R = \rho \frac{L}{A}, where \rho is the resistivity, L is the length, and A is the cross-sectional area of the wire.
2. Initially, the wire has a resistance of R_1 = 160 \, \Omega and length L_1=L.
3. When the wire is melted and redrawn into one-fourth of its initial length, the new length is: L_2 = \frac{L_1}{4} = \frac{L}{4}.
4. The volume of the wire remains constant, so A_1L_1 = A_2L_2, where A_1 and A_2 are the original and new cross-sectional areas. Substituting for L_2, we get: A_1L = A_2 \cdot \frac{L}{4} \implies A_2 = 4A_1.
5. Let’s compute the new resistance: R_2 = \rho \frac{L_2}{A_2} = \rho \frac{\frac{L}{4}}{4A_1} = \rho \frac{L}{16A_1} = \frac{R_1}{16}.
6. Plug in the known values: R_2 = \frac{160}{16} = 10 \, \Omega.

Therefore, the new resistance of the wire is 10 Ω.

Conclusion:

The correct answer is 10 Ω. As the length is reduced to one-fourth, the cross-sectional area becomes four times, while resistance changes inversely proportional to the square of these changes, leading to a much reduced resistance.