Asked: I.A.S. (Pre) 2001medium Level
QUESTION STATEMENT:द्विभाषी / Bilingual
दो तारों की लंबाइयां, व्यास और प्रतिरोधकताएं सभी 1:2 के अनुपात में हैं। अगर पतले तार का प्रतिरोध 10 ओम है, तो मोटे तार का प्रतिरोध होगा-
The lengths, diameters, and resistivities of two wires are all in the ratio 1:2. If the resistance of the thinner wire is 10 ohms, the resistance of the thicker wire will be:
SELECT CORRECT OPTION:
A
5 ओम
5 ohms
B
Correct10 ओम
10 ohms
C
20 ओम
20 ohms
D
40 ओम
40 ohms
Official Solution & Explanation:
Correct Answer: Option B(10 ओम)
प्रतिरोध का सूत्र $R = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \frac{4\rho l}{\pi d^2}$ होता है।
दिया है: $\frac{l_1}{l_2} = \frac{1}{2}$, $\frac{d_1}{d_2} = \frac{1}{2}$, $\frac{\rho_1}{\rho_2} = \frac{1}{2}$।
$$\frac{R_1}{R_2} = \left(\frac{\rho_1}{\rho_2}\right) \left(\frac{l_1}{l_2}\right) \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times (2)^2 = \frac{1}{4} \times 4 = 1$$
अतः $R_2 = R_1 = 10\text{ }\Omega$।
दिया है: $\frac{l_1}{l_2} = \frac{1}{2}$, $\frac{d_1}{d_2} = \frac{1}{2}$, $\frac{\rho_1}{\rho_2} = \frac{1}{2}$।
$$\frac{R_1}{R_2} = \left(\frac{\rho_1}{\rho_2}\right) \left(\frac{l_1}{l_2}\right) \left(\frac{d_2}{d_1}\right)^2 = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times (2)^2 = \frac{1}{4} \times 4 = 1$$
अतः $R_2 = R_1 = 10\text{ }\Omega$।
Resistance $R = \rho \frac{l}{A} \propto \frac{\rho l}{d^2}$.
Given $\frac{\rho_1}{\rho_2} = \frac{1}{2}$, $\frac{l_1}{l_2} = \frac{1}{2}$, and $\frac{d_1}{d_2} = \frac{1}{2}$:
$$\frac{R_1}{R_2} = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{2}{1}\right)^2 = 1$$
Therefore, $R_2 = R_1 = 10\text{ }\Omega$.
Given $\frac{\rho_1}{\rho_2} = \frac{1}{2}$, $\frac{l_1}{l_2} = \frac{1}{2}$, and $\frac{d_1}{d_2} = \frac{1}{2}$:
$$\frac{R_1}{R_2} = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times \left(\frac{2}{1}\right)^2 = 1$$
Therefore, $R_2 = R_1 = 10\text{ }\Omega$.