Problem 162
Question
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
Step-by-Step Solution
Verified Answer
(d) Statement I is true, Statement II is false.
1Step 1: 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.
2Step 2: Calculate \( \Delta t \)
The change in temperature \( \Delta t \) is given by \[ \Delta t = 227^{\circ} \text{C} - 27^{\circ} \text{C} = 200^{\circ} \text{C} \]
3Step 3: Substitute Values into Formula
Substitute \( R = 150 \Omega \), \( R_0 = 100 \Omega \), and \( \Delta t = 200^{\circ}\text{C} \) into the formula to find \( \alpha \). \[ 150 = 100 (1 + \alpha \times 200) \]
4Step 4: Solve for \( \alpha \)
Rearrange the equation to solve for \( \alpha \). \[ \frac{150}{100} = 1 + \alpha \times 200 \] \[ 1.5 = 1 + 200\alpha \] \[ 0.5 = 200\alpha \] \[ \alpha = \frac{0.5}{200} = 2.5 \times 10^{-3} \text{C}^{-1} \]
5Step 5: 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.
6Step 6: 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.
7Step 7: Conclusion
Since the criteria of Statement II are not met (large \( \Delta t \) and \( \Delta R \)), Statement II is false.
Key Concepts
Temperature Coefficient of ResistanceOhm's LawAIEEE Physics Problems
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:
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
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:
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:
- \( V = IR \)
- 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.
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.
Other exercises in this chapter
Problem 159
Assertion When speed of sound in a gas is \(c\), then $$ c_{\mathrm{rms}}=\sqrt{\frac{3}{\gamma}} \times c $$ Reason \(c=\sqrt{\frac{\gamma p}{\rho}}\)
View solution Problem 160
Assertion The root mean speed (rms) of oxygen molecules at a certain absolute temperature \(T\) is \(c\).If the temperature is doubled and oxygen gas dissociate
View solution Problem 163
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
View solution Problem 164
\(1 \mathrm{~kg}\) of diatomic gas is at a pressure of \(8 \times 10^{4} \mathrm{Nm}^{-2}\). The density of the gas is \(4 \mathrm{~kg} \mathrm{~m}^{-3}\). What
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