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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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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 $\;(\alpha_{steel}=12 \times 10^{-6}/^{0}C\;and \; \alpha_{brass}=18 \times 10^{-6}/^{0}C\;)$

$(a)\;99.89\;cm\qquad(b)\;98.99\;cm\qquad(c)\;99.98\;cm\qquad(d)\;99.67\;cm$

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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
$=99.98\;cm$
answered Mar 24, 2014 by yamini.v
 

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