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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A mercury thermometer is transferred from melting ice to hot liquid . The mercury rises $\;\large\frac{9}{10}\;$ of the distance between the lower fixed point and the upper fixed point . The temperature of the liquid is $\;^{0}C\;$ is

$(a)\;90^{0}C\qquad(b)\;94^{0}C\qquad(c)\;45^{0}C\qquad(d)\;491^{0}F$

Can you answer this question?
 
 

1 Answer

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Answer : $\;90^{0}C$
Explanation :
$t=\large\frac{9}{10} \;[100-0]=90^{0}C$
answered Mar 24, 2014 by yamini.v
 

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