A resistance thermometer has resistance 95\ at the ice point and 140\ at the steam point. The resistance obtained if the actual temperature is 30^ ^ C is

Correct answer: B. 108.5\ Using the formula for temperature on the resistance scale: t = R_t - R_0 R_ 100 - R_0 × 100 Given R_0 = 95\ , R_ 100 = 140\ , and t = 30^ ^ C . Substituting the values: 30 = (R_t - 95)/(140 - 95) × 100 30 = (R_t - 95)/(45) × 100 R_t - 95 = (30 × 45)/(100) R_t - 95 = 13.5 R_t = 95 + 13.5 = 108.5\ Answer: 108.5\

MHT CET 202625 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 25 April
Shift: Morning Shift
Subject: Physics
Chapter: Thermal Properties of Matter

MHT CET 2026 · 25 April · Morning Shift · Physics · Thermal Properties of Matter

A resistance thermometer has resistance 95\ at the ice point and...

A resistance thermometer has resistance

95\ \Omega

at the ice point and

140\ \Omega

at the steam point. The resistance obtained if the actual temperature is

30^{}^{\circ}\text{C}

Options

A $117.5\ \Omega$
B $108.5\ \Omega$
C $113.5\ \Omega$
D $13.5\ \Omega$

Correct answer

B. $108.5\ \Omega$

Step-by-step solution

Using the formula for temperature on the resistance scale:

t = \dfrac{R_t - R_0}{R_{100} - R_0} \times 100

Given

R_0 = 95\ \Omega

R_{100} = 140\ \Omega

, and

t = 30^{}^{\circ}\text{C}

.

Substituting the values:

30 = \dfrac{R_t - 95}{140 - 95} \times 100

30 = \dfrac{R_t - 95}{45} \times 100

R_t - 95 = \dfrac{30 \times 45}{100}

R_t - 95 = 13.5

R_t = 95 + 13.5 = 108.5\ \Omega

Answer:

108.5\ \Omega

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