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Chapter 12: Problem 1778

The temperature co-efficient of resistance of a wire is $0.00125^{\circ} \mathrm{k}^{-1}\(. Its resistance is \)1 \Omega\( at \)300 \mathrm{~K}$. Its resistance will be \(2 \Omega\) at. (A) \(1400 \mathrm{~K}\) (B) \(1200 \mathrm{~K}\) (C) \(1000 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)

Short Answer

Expert verified
The correct answer is (B) 1200 K, which is the closest value to the calculated temperature of 1100 K, using the formula \( R = R_0 (1 + \alpha (T - T_0)) \) with the given values of \(\alpha\), \(R_0\), and \(T_0\).

Step by step solution

01

Understand the Formula

The formula for calculating the resistance (R) of a wire at a specific temperature (T) based on the initial resistance (R0) at a reference temperature (T0) is: \( R = R_0 (1 + \alpha (T - T_0)) \) Where: - \(R\) is the resistance at temperature T, - \(R_0\) is the initial resistance at the reference temperature \(T_0\), - \(\alpha\) is the temperature coefficient of resistance, - \(T\) is the temperature we want to find, and - \(T_0\) is the reference temperature. In this exercise, we are given \(\alpha = 0.00125 \, k^{−1}\), \( R_0 = 1 \, \Omega\), and \(T_0 = 300 \, K\). We want to find the temperature \(T\) at which the resistance becomes \(2 \, \Omega\).
02

Plug in the Given Values

First, plug in the given values into the formula: \(2 = 1 (1 + 0.00125 (T - 300)) \)
03

Simplify the Equation and Solve for T

Now let's simplify the equation and solve for the temperature, T: \(2 = 1 + 0.00125 (T - 300) \) Subtract 1 from both sides: \(1 = 0.00125 (T - 300)\) Divide both sides by 0.00125: \(\frac{1}{0.00125} = T - 300\) \(800 = T - 300\) Add 300 to both sides: \(800 + 300 = T\) \(T = 1100 \, K\) Now let's check which answer option is closest to 1100 K: (A) 1400 K (B) 1200 K (C) 1000 K (D) 800 K
04

Choose the Correct Answer

The correct answer is (B) 1200 K, as it is the closest value to the calculated temperature of 1100 K.

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Most popular questions from this chapter

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