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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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Most popular questions from this chapter

On applying an electric field of \(5 \times 10^{-8} \mathrm{Vm}^{-1}\) across a conductor, current density through it is \(2.5 \mathrm{Am}^{-2}\) The resistivity of the conductor is \(\ldots .\) (A) \(1 \times 10^{-8} \Omega \mathrm{m}\) (B) \(2 \times 10^{-8} \Omega \mathrm{m}\) (C) \(0.5 \times 10^{-8} \Omega \mathrm{m}\) (D) \(12.5 \times 10^{-8} \Omega \mathrm{m}\)

A wire of resistor \(R\) is bent into a circular ring a circular ring of radius \(\mathrm{r}\) Equivalent resistance between two points \(\mathrm{X}\) and \(\mathrm{Y}\) on its circumference, when angle xoy is \(\alpha\), can be given by (A) $\left\\{(\mathrm{R} \alpha) /\left(4 \pi^{2}\right)\right\\}(2 \pi-\alpha)$ (B) \((\mathrm{R} / 2 \pi)(2 \pi-\alpha)\) (C) \(\mathrm{R}(2 \pi-\alpha)\) (D) \((4 \pi / \mathrm{R} \alpha)(2 \pi-\alpha)\)

There are n resistors having equal value of resistance \(\mathrm{r}\). First they are connected in such a way that the possible minimum value of resistance is obtained. Then they are connected in such a way that possible maximum value of resistance is obtained the ratio of minimum and maximum values of resistances obtained in these way is.... (A) \((1 / n)\) (B) \(\mathrm{n}\) (C) \(\mathrm{n}^{2}\) (D) \(\left(1 / \mathrm{n}^{2}\right)\)

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