Ask Questions, Get Answers

Want to ask us a question? Click here
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

Brass at $\;30^{0}C\;$ is observed to be 1 m long when measured by a steel scale which is connect at $\;0^{0}C\;$ . Correct length of the brass of at $\;0^{0}C\;$ is $\;(\alpha_{steel}=12 \times 10^{-6}/^{0}C\;and \; \alpha_{brass}=18 \times 10^{-6}/^{0}C\;)$


Can you answer this question?

1 Answer

0 votes
Answer : $\;99.98\;cm$
Explanation :
Correct length of brass at $\;30^{0}C\;$ = length at $\;0^{0}C\;(1 + \alpha_{b} \bigtriangleup t)$
$=l_{0} (1 + \alpha_{b} \bigtriangleup t)$
Correct length at $\;30^{0}C\;$ = observed length $\;(1 + \alpha_{s} \bigtriangleup t)$
Therefore , $\;l_{0}(1+ \alpha_{s} \bigtriangleup t)\;$
$l_{0}= 1 (1+(\alpha_{s} - \alpha_{b}) \bigtriangleup t )$
$=900 [1 + (12-18) \times 10^{-6}]\;$ cm
answered Mar 24, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App