Question

The length of a conductor having resistance 160 Ω, is compressed to 25% of its initial value. The new resistance will be

• A. 10 Ω
• B. 20 Ω
• C. 15 Ω
• D. 17 Ω

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the resistance of a conductor and its dimensions.

Step 1: Understanding the relationship.

Resistance R of a conductor is given by the formula:

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

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

Step 2: Effect of compression on dimensions.

When the length L of the conductor is compressed to 25% of its original length, its new length L' is:

L' = 0.25L.

Step 3: Volume conservation.

Since the volume of the conductor remains constant, we have:

L \times A = L' \times A'

Substituting the value of L', the new area A' becomes:

A' = \frac{L \times A}{L'} = \frac{L \times A}{0.25L} = 4A

Step 4: Calculate the new resistance.

Using the resistance formula R' = \rho \frac{L'}{A'} for the new dimensions, we replace L' and A':

R' = \rho \frac{0.25L}{4A} = \frac{\rho L}{16A} = \frac{R}{16}

The original resistance R is 160 Ω. Thus, the new resistance R' is:

R' = \frac{160}{16} = 10 Ω

Conclusion:

The new resistance of the conductor when it is compressed to 25% of its original length is 10 Ω.