Get Answers to all your Questions

header-bg qa

Q. 11.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :

 R=R_{0}\left [ 1+\alpha (T-T_{0}) \right ]

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 ?

                

Answers (1)

best_answer

R=R_{0}\left [ 1+\alpha (T-T_{0}) \right ]

R0 = 101.6\Omega    T0 = 273.16 K

R = 165.5\; \Omega    R = 600.5 K

Putting the above values in the given equation we have

\\\alpha =\frac{165.5-101.6}{600.5-27316}\\ \alpha =1.92\times 10^{-2}\ K^{-1}

For R = 123.4\; \Omega

\\T=T_{0}+\frac{1}{\alpha }\left ( \frac{R}{R_{0}}-1 \right )\\ T=273.16+\frac{1}{1.92\times 10^{-3}}\left ( \frac{123.4}{101.6}-1 \right )\\ T=384.75K

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads