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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

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) \(550^{\circ} \mathrm{C}\)

Two electirc bulbs whose resistances are in the ratio of \(1: 2\) are connected in parallel to a constant voltage source the power dissipated in them have the ratio. (A) \(1: 2\) (B) \(1.1\) (C) \(2: 1\) (D) \(1: 4\)

A potentiometer wire has length \(10 \mathrm{~m}\) and resistance \(20 \Omega\). A \(2.5 \mathrm{~V}\) battery of negligible internal resistance is connected across the wire with an \(80 \Omega\) series resistance. The potential gradient on the wire will be: (A) \(2.5 \times 10^{-4} \mathrm{~V} / \mathrm{cm}\) (B) \(0.62 \times 10^{-4} \mathrm{~V} / \mathrm{mm}\) (C) \(1 \times 10^{-5} \mathrm{~V} / \mathrm{mm}\) (D) \(5 \times 10^{-5} \mathrm{~V} / \mathrm{mm}\)

The masses of three wires of copper are in the ratio of \(1: 3: 5\) and their lengths are in the ratio of \(5: 3: 1\). The ratio of their electrical resistance is: (A) \(1: 1: 1\) (B) \(1: 3: 5\) (C) \(5: 3: 1\) (D) \(125: 15: 1\)

A wire in a circular shape has \(10 \Omega\) resistance. The resistance per one meter is \(1 \Omega\) The resultant between \(A \& B\) is equal to \(2.4 \Omega\), then the length of the chord \(\mathrm{AB}\) will be equal to (A) \(2.4\) (B) 4 (C) \(4.8\) (D) 6
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