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Chapter 12: Problem 1752

Area of cross-section of two wires of same length carrying same current is in the ratio of \(1: 2\). Then the ratio of heat generated per second in the wires \(=\ldots\) (A) \(1: \sqrt{2}\) (B) \(1: 1\) (C) \(1: 4\) (D) \(2: 1\)

Short Answer

Expert verified
The ratio of heat generated per second in the wires is 2:1, as found by analyzing the power generated in each wire using the formula \(P = I^2R\) and considering their resistances with the given ratio of cross-sectional areas. The correct answer is (D).

Step by step solution

01

Write the formula for resistance of a wire

The formula for the resistance of a wire is given by: \(R = \rho(L/A)\), where \(R\) is resistance, \(\rho\) is the resistivity of the wire material, \(L\) is the length of the wire, and \(A\) is the area of the wire's cross-section.
02

Express resistance of both wires in terms of given information

The cross-sectional area ratio is given as \(1: 2\). Let's denote the cross-sectional areas of the wires as \(A_1\) and \(A_2\) respectively. Since the length and resistivity are the same for both wires, we can write their resistances as: \(R_1 = \rho(L / A_1)\) and \(R_2 = \rho(L / A_2)\).
03

Use the formula for power generated as heat

Recall the formula for power generated as heat: \(P = I^2R\). Write the power generated for both wires: \(P_1 = I^2R_1\) and \(P_2 = I^2R_2\).
04

Substitute the resistance equations into the power equations

Substitute the resistance formulas from step 2: \(P_1 = I^2(\rho(L / A_1))\) and \(P_2 = I^2(\rho(L / A_2))\).
05

Find the ratio of power generated in the two wires

Divide the power equations to find the ratio of power generated in the two wires: \(\frac{P_1}{P_2} = \frac{I^2(\rho(L / A_1))}{I^2(\rho(L / A_2))}\). Notice that \(\rho\), \(L\), and \(I^2\) are the same for both wires, and the ratio of cross-sectional areas is the given as 1:2 which is \(\frac{A_1}{A_2} = \frac{1}{2}\). Now rewrite the ratio in terms of the cross-section areas: \(\frac{P_1}{P_2} = \frac{A_2}{A_1}\). Plug in the values for the ratio \(\frac{A_1}{A_2}\) into the equation: \(\frac{P_1}{P_2} = \frac{2}{1}\). Therefore, the ratio of heat generated per second in the wires is 2:1. The correct option is (D).

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