Verified Solution • 2025–26 Session moderate

Q10.3: NCERT Physics - Part 2 THERMAL PROPERTIES OF MATTER 10.3 Solution: The electrical resistance in ohms of a certain thermometer…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often compute the temperature coefficient as $\alpha=\frac{R-R_0}{T-T_0}$, omitting the required division by $R_0$. The correct formula is $\alpha=\frac{\frac{R}{R_0}-1}{T-T_0}=\frac{R-R_0}{R_0\,(T-T_0)}$; using the wrong expression gives a much larger $\alpha$ and an incorrect temperature estimate.

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:

R = R_0[1 + \alpha(T - T_0)]

. The resistance is 101.6

\Omega

at the triple-point of water 273.16 K, and 165.5

\Omega

at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4

\Omega

?

Step-by-Step Verified Solution

**Answer:**

T = 384.61 \text{ K}

**Step-by-Step Explanation:** Given:

R_0 = 101.6\ \Omega

at

T_0 = 273.15

K,

R = 165.5\ \Omega

at

T = 600.5

K. Using the resistance law:

165.5 = 101.6[1 + \alpha(600.5 - 273.15)]

1.629 = 1 + \alpha(327.35)

\alpha = \frac{0.629}{327.35} = 1.92 \times 10^{-3}\ \text{K}^{-1}

For

R = 123.4\ \Omega

:

123.4 = 101.6[1 + 1.92 \times 10^{-3}(T - 273.15)]

Solving for

T

:

T = 384.61\ \text{K}

[citation:7]

All Questions in this Chapter

Q10.1 Q10.2 Q10.3 Q10.4 Q10.5 Q10.6 Q10.7 Q10.8 Q10.9 Q10.10 Q10.11 Q10.12 Q10.13 Q10.14 Q10.15 Q10.16 Q10.17 Q10.18 Q10.19 Q10.20

Related Learning Resources & Exam Preparation

Chapter Practice Bank

Practice all topic-wise questions from THERMAL PROPERTIES OF MATTER

Reference Textbook Solutions

Explore complete chapter exercises for NCERT Physics - Part 2

Competitive PYQ Question Bank

Browse year-wise authentic previous years' questions with solutions

Authentic CBT Mock Tests

Simulate official exam environment with live timer and percentile scoring