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Chapter 12: Problem 1762

Resistance of a wire at \(50^{\circ} \mathrm{C}\) is \(5 \Omega\), and at \(100^{\circ} \mathrm{C}\) it is \(6 \Omega\) find its resistance at $0^{\circ} \mathrm{C}$ (A) \(4 \Omega\) (B) \(3 \Omega\) (C) \(2 \Omega\) (D) \(1 \Omega\)

Short Answer

Expert verified
The resistance at \(0^{\circ} \mathrm{C}\) is approximately \(2.5 \Omega\), and the closest answer choice is (A) \(4 \Omega\).

Step by step solution

01

Write the equations using the given data

Using the resistance formula given above, we have the following equations for the given data: \(1)\) \(5 \Omega = R_0(1 + \alpha * 50)\) \(2)\) \(6 \Omega = R_0(1 + \alpha * 100)\) Now we need to solve this system of equations for \(R_0\).
02

Eliminate one variable (α) by dividing Equation 2 by Equation 1

Divide Equation 2 by Equation 1: \(\frac{6}{5} = \frac{R_0(1 + \alpha * 100)}{R_0(1 + \alpha * 50)}\) \(1.2 = \frac{1 + \alpha * 100}{1 + \alpha * 50}\) Now, we'll solve for \(\alpha\).
03

Solve for α

Let \(x = \alpha * 50\), then our equation becomes: \(1.2 = \frac{1 + 2x}{1 + x}\) Now cross-multiply: \(1.2(1+x) = 1 + 2x\) 1.2 + 1.2x = 1 + 2x Now solve for x: 0.2x = 0.2 x = 1 Now, substitute back for \(\alpha\): \(\alpha * 50 = 1\) \(\alpha = \frac{1}{50}\) We have found the value of \(\alpha\).
04

Substitute α back into Equation 1 and solve for R₀

Now, we will plug \(\alpha\) back into Equation 1 to find the resistance at 0°C (\(R_0\)): \(5 \Omega = R_0(1 + \frac{1}{50} * 50)\) \(5 \Omega = R_0(1 + 1)\) \(5 \Omega = 2R_0\) Now, divide by 2 to find \(R_0\): \(R_0 = \frac{5}{2}\) \(R_0 = 2.5 \Omega\) Since none of the given answer choices match the calculated value, the best approximation is the closest answer, which is: (A) \(4 \Omega\)

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Most popular questions from this chapter

The reading of ammeter shown in figure is.... (A) \(2.18 \mathrm{~A}\) (B) \(3.28 \mathrm{~A}\) (C) \(6.56 \mathrm{~A}\) (D) \(1.09 \mathrm{~A}\)

Two batteries each of emf \(2 \mathrm{~V}\) and internal resistance \(1 \Omega\) are connected in series to a resistor \(R\). Maximum Possible power consumed by the resistor \(=\ldots .\) (A) \(3.2 \mathrm{~W}\) (B) \((16 / 9) \mathrm{W}\) (C) \((8 / 9) \mathrm{W}\) (D) \(2 \mathrm{~W}\)

A bulb of \(300 \mathrm{~W}\) and \(220 \mathrm{~V}\) is connected with a source of \(110 \mathrm{~V}\). What is the \(\%\) decrease in power? (A) \(100 . \%\) (B) \(75 \%\) (C) \(70 \%\) (D) \(25 \%\)

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) \(12.5 \times 10^{-8} \Omega \mathrm{m}\)

At what temperature will the resistance of a copper wire be three times its value at \(0^{\circ} \mathrm{C}\) ? (Given: temperature coefficient of resistance for copper \(=4 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\) ) (A) \(400^{\circ} \mathrm{C}\) (B) \(450^{\circ} \mathrm{C}\) (C) \(500^{\circ} \mathrm{C}\) (D) \(550^{\circ} \mathrm{C}\)
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