© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the known information
Let the specific resistance (resistivity) of material A be $ \rho_A $ and that of material B be $ \rho_B $. We are given:
• $ \rho_B = 2 \rho_A $ (material B has twice the specific resistance of material A).
• A circular wire of material B has twice the diameter of a wire made of material A. So if the diameter of the wire made of A is $ d_A $, then the diameter of the wire made of B is $ d_B = 2 d_A $.
Step 2: Express the resistance of each wire
The resistance $ R $ of a wire is given by the formula
$$
R = \frac{\rho \, \ell}{A},
$$
where $ \rho $ is the resistivity, $ \ell $ is the length of the wire, and $ A $ is its cross-sectional area.
Step 3: Write the condition for equal resistance
For the two wires to have the same resistance, we set
$$
\frac{\rho_B \,\ell_B}{A_B} = \frac{\rho_A \,\ell_A}{A_A}.
$$
We want to find the ratio $ \frac{\ell_B}{\ell_A} $.
Step 4: Express the areas for each wire
Since both wires are circular in cross-section, the area is proportional to the square of the diameter. If $ d_A $ and $ d_B $ are the diameters for wires of A and B respectively, then
$$
A_A = \pi \left(\frac{d_A}{2}\right)^2 = \frac{\pi d_A^2}{4},
\quad
A_B = \pi \left(\frac{d_B}{2}\right)^2 = \frac{\pi d_B^2}{4}.
$$
Given $ d_B = 2 d_A $, we have
$$
A_B = \frac{\pi (2 d_A)^2}{4} = \frac{\pi \cdot 4 d_A^2}{4} = \pi d_A^2.
$$
Meanwhile,
$$
A_A = \frac{\pi d_A^2}{4}.
$$
Step 5: Substitute areas and resistivities into the equality condition
Substitute $ \rho_B = 2 \rho_A $, $ A_B = \pi d_A^2 $, and $ A_A = \frac{\pi d_A^2}{4} $ into
$$
\frac{\rho_B \,\ell_B}{A_B} = \frac{\rho_A \,\ell_A}{A_A}.
$$
We get
$$
\frac{(2 \rho_A)\,\ell_B}{\pi d_A^2}
= \frac{\rho_A \,\ell_A}{\frac{\pi d_A^2}{4}}.
$$
Step 6: Solve for the ratio $ \frac{\ell_B}{\ell_A} $
Canceling common terms and rearranging, we have
$$
2 \rho_A \,\ell_B \cdot \frac{4}{\rho_A \,\pi d_A^2}
= \ell_A \cdot \frac{1}{\pi d_A^2}.
$$
Simplifying step by step, we get:
$$
\frac{2 \rho_A \,\ell_B}{\pi d_A^2} = \frac{\rho_A \,\ell_A}{\frac{\pi d_A^2}{4}}
\;\;\Longrightarrow\;\;
2 \ell_B \left(\frac{1}{\pi d_A^2}\right) = \ell_A \left(\frac{4}{\pi d_A^2}\right).
$$
Cancel common factors $ \frac{1}{\pi d_A^2} $ on both sides:
$$
2 \ell_B = 4 \ell_A \;\;\Longrightarrow\;\; \ell_B = 2 \ell_A.
$$
Therefore,
$$
\frac{\ell_B}{\ell_A} = 2.
$$
Final Answer
The ratio of their lengths must be $ 2 $.
