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Chapter 13: Problem 162

Statement I The temperature dependence of resistance is usually given as \(R=R_{0}(1+\Delta t)\). The resistance of a wire changes from \(100 \Omega\) to \(150 \Omega\) when its temperature is increased from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C}\). This implies that \(\alpha=2.5 \times 10^{-3} \mathrm{C}^{-1}\). Statement II \(R=R_{0}(1+\alpha \Delta t)\) is valid only when the change in the temperature is small and \(\Delta R=\left(R-R_{0}\right) \ll

Short Answer

Expert verified
(d) Statement I is true, Statement II is false.

Step by step solution

01

Understanding the Formula

The formula for the temperature dependence of resistance is given by \[ R = R_0 (1 + \alpha \Delta t) \]where \( R \) is the resistance at temperature \( t \), \( R_0 \) is the initial resistance, \( \alpha \) is the temperature coefficient of resistance, and \( \Delta t \) is the change in temperature.
02

Calculate \( \Delta t \)

The change in temperature \( \Delta t \) is given by \[ \Delta t = 227^{\circ} \text{C} - 27^{\circ} \text{C} = 200^{\circ} \text{C} \]
03

Substitute Values into Formula

Substitute \( R = 150 \Omega \), \( R_0 = 100 \Omega \), and \( \Delta t = 200^{\circ}\text{C} \) into the formula to find \( \alpha \). \[ 150 = 100 (1 + \alpha \times 200) \]
04

Solve for \( \alpha \)

Rearrange the equation to solve for \( \alpha \). \[ \frac{150}{100} = 1 + \alpha \times 200 \] \[ 1.5 = 1 + 200\alpha \] \[ 0.5 = 200\alpha \] \[ \alpha = \frac{0.5}{200} = 2.5 \times 10^{-3} \text{C}^{-1} \]
05

Evaluate Statement I

Statement I is confirmed as true. The calculation shows that \( \alpha = 2.5 \times 10^{-3} \text{C}^{-1} \), which matches the given value.
06

Evaluate Statement II

Statement II asserts that the formula is only valid when the change in temperature is small and \( \Delta R \ll R_0 \). In this problem, \( \Delta R = 50 \Omega \), which is not small compared to \( R_0 = 100 \Omega \), and \( \Delta t = 200^{\circ}\text{C} \) is not small.
07

Conclusion

Since the criteria of Statement II are not met (large \( \Delta t \) and \( \Delta R \)), Statement II is false.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Coefficient of Resistance
The temperature coefficient of resistance, denoted by \( \alpha \), indicates how much a material's electrical resistance changes with temperature. A positive \( \alpha \) implies that resistance increases with temperature, typical in conductors like metals.
For example, when a wire's temperature rises from \( 27^{\circ} \mathrm{C} \) to \( 227^{\circ} \mathrm{C} \), it shows a resistance change from \( 100 \Omega \) to \( 150 \Omega \).
The relationship that describes this is:
  • \( R = R_{0}(1 + \alpha \Delta t) \), where:
  • \( R \) is resistance at new temperature
  • \( R_{0} \) is original resistance
  • \( \Delta t \) is temperature change
  • \( \alpha \) is the coefficient
This highlights how even minor variations in \( \alpha \) can significantly impact the measurements, emphasizing the need for precision when evaluating how resistance shifts with temperature changes.
Ohm's Law
Ohm's Law is fundamental in understanding the relationship between voltage, current, and resistance in an electrical circuit.
It states that the current \( I \) flowing through a conductor between two points is directly proportional to the voltage \( V \) across the two points and inversely proportional to the resistance \( R \) of the conductor. The formula is:
  • \( V = IR \)
This means:
  • If the resistance increases (due to temperature, for instance), less current will flow for the same applied voltage unless adjustments are made.
  • In the context where temperature changes affect resistance, Ohm's Law helps predict how an electrical system's behavior will adjust or need correction.
  • It emphasizes that careful measurement and understanding of resistance changes are crucial, especially under varying temperature conditions.
AIEEE Physics Problems
The AIEEE (All India Engineering Entrance Examination), now part of JEE, featured challenging physics problems to assess students' understanding and application of fundamental principles.
In this instance, the emphasis was on understanding how resistance changes with temperature, requiring knowledge of both the temperature coefficient and Ohm's Law.
Such problems are designed to test both theoretical knowledge and practical application.
  • They involve calculations and interpretations that require familiarity with how real-world systems respond to changes.
    • For example, knowing when formulas are valid is crucial, as seen in evaluating Statement II about the small \( \Delta t \).
  • In preparing for such exams, problems like this ensure students understand how physics principles apply in realistic scenarios.
Engaging deeply with concepts like these helps hone problem-solving skills crucial for engineering studies and beyond.

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Most popular questions from this chapter

A metal ball immersed in alcohol weighs \(w_{1}\) at \(0^{\circ} \mathrm{C} \quad \overline{5}\) and \(w_{2}\) at \(59^{\circ} \mathrm{C}\). The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol, it can be shown that (a) \(w_{1}>w_{2}\) (b) \(w_{1}=w_{2}\) (c) \(w_{1}

A wheel is \(80.3 \mathrm{~cm}\) in circumference. An iron tyre measures \(80.0 \mathrm{~cm}\) around its inner face. If the coefficient of linear expansion for iron is \(1.2 \times 0^{-5}{ }^{\circ} \mathrm{C}^{-1}\), the temperature of the tyre must be raised by (a) \(105^{\circ} \mathrm{C}\) (b) \(417^{\circ} \mathrm{C}\) (c) \(312^{+} \mathrm{C}\) (d) \(223^{\circ} \mathrm{C}\)

A flask of volume \(10^{3} \mathrm{cc}\) is completely filled with mercury at \(0^{\circ} \mathrm{C}\). The coefficient of cubical expansion of mercury is \(1.80 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\) and that of glass is \(1.4 \times 10^{-6} \mathrm{C}^{-1}\). If the flask is now placed in boiling water at \(100^{\circ} \mathrm{C}\), how much mercury will overflow? (a) \(7 \mathrm{cc}\) (b) \(1.4 \mathrm{cc}\) (c) \(21 \mathrm{cc}\) (d) \(28 \mathrm{cc}\)

An oxygen cylinder of volume \(30 \mathrm{~L}\) has an initial gauge pressure of \(15 \mathrm{~atm}\) and a temperature of \(27^{\circ} \mathrm{C}\). After some oxygen is withdrawn from the cylinder the gauge pressure drops to 11 atm and its temperature drops to \(17^{\circ} \mathrm{C} .\) The mass of oxygen taken out of the cylinder \(\left(R=8.31 . \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\). molecular mas of \(\mathrm{O}_{2}=32 \mathrm{u}\) ) is [NCERT] (a) \(0.14 \mathrm{~g}\) (b) \(0.02 \mathrm{~g}\) (c) \(0.14 \mathrm{~kg}\) (d) \(0.014 \mathrm{~kg}\)

At a certain temperature, the ratio of the rms velocity of \(\mathrm{H}_{2}\) molecules to \(\mathrm{O}_{2}\) molecule is (a) \(1: 1\) (b) \(1: 4\) (c) \(4: 1\) (d) \(16: 1\)
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