Eight copper wire of length $l$ and diameter $d$ are joined in parallel to form a single composite conductor of resistance $R$ If a single copper wire of length $2 l$ have the same resistance $(R)$ then its diameter will be ______$d$
To solve the problem, let's break it down step-by-step.
Step 1: Understand the problem.
We have eight copper wires each of length \( l \) and diameter \( d \), joined in parallel to form a composite conductor with resistance \( R \). We need to find the diameter of a single copper wire of length \( 2l \) that has the same resistance \( R \).
Step 2: Determine the resistance of the parallel wires.
The resistance of a single wire is given by \( R_1 = \frac{\rho l}{A} \), where \( A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \). So, \( R_1 = \frac{4\rho l}{\pi d^2} \).
Since the wires are in parallel, the equivalent resistance \( R \) is given by \( \frac{1}{R} = 8 \times \frac{1}{R_1} \). Hence, \( R = \frac{R_1}{8} = \frac{4\rho l}{8\pi d^2} = \frac{\rho l}{2\pi d^2} \).
Step 3: Determine the resistance of the single wire of length \( 2l \).
The resistance of this single wire is given by \( R = \frac{\rho \times 2l}{A'} \), where \( A' = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \). Thus, \( R = \frac{8\rho l}{\pi D^2} \).
Step 4: Equate the resistances.
Set the resistance of the combined wires equal to the resistance of the single wire: \(\frac{\rho l}{2\pi d^2} = \frac{8\rho l}{\pi D^2}\).
Cancel out \(\rho \) and \( l \) from both sides: \(\frac{1}{2\pi d^2} = \frac{8}{\pi D^2}\).
Cross-multiply to solve for \( D \): \( D^2 = 16d^2 \).
Take the square root of both sides: \( D = 4d \).
Conclusion: The diameter \( D \) of the single wire will be \( 4d \).
This value (4) is within the expected range (4,4) provided.
