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CBSE Class 12-science Answered

The apparatus shown in figure consists of four glass columns connected by horizontal sections.The heights of two central columns B and C are 49 cm each. The two outer column A and D are open to atmosphere. A and C are maintained a a temperature of 95 degree C, while column B and D are maintained at 5 degree C. The heights of liquid in A and D measured from base line are 52.8 cm and 51 cm respectively. Determine the coefficient of thermal expansion of the liquid.
Asked by Sourav | 01 May, 2015, 05:29: PM
answered-by-expert Expert Answer
As the liquid columns in the glass tubes are in equilibrium:
Pressure diffference between the columns A and B = Pressure difference between the columns C and D

begin mathsize 14px style Let space straight rho space and space space straight rho apostrophe space be space the space densities space of space the space liquid space at space 5 degree straight C space and space 95 degree straight C If space straight h subscript straight A comma straight h subscript straight B comma straight h subscript straight C space and space straight h subscript straight D space are space the space heights space of space the space liquid space columns space in space the space tubes space straight A comma straight B comma straight C comma straight D colon therefore space colon straight h subscript straight A space end subscript straight rho apostrophe space straight g space minus straight h subscript straight B space end subscript straight rho space straight g space equals straight h subscript straight D space end subscript straight rho space straight g space minus straight h subscript straight C space end subscript straight rho apostrophe space straight g Given space that colon straight h subscript straight A space equals space 52.8 space cm space comma space straight h subscript straight B space end subscript equals space 49 space cm space comma space straight h subscript straight C space end subscript equals 49 space cm comma space straight h subscript straight D space equals space 51 space cm 52.8 cross times straight rho apostrophe cross times straight g space minus space 49 cross times straight rho cross times straight g space equals 51 cross times straight rho cross times straight g space minus space 49 cross times straight rho apostrophe cross times straight g 52.8 cross times straight rho apostrophe minus 49 cross times straight rho equals 51 cross times straight rho minus 49 cross times straight rho apostrophe open parentheses 52.8 plus 49 close parentheses space straight rho apostrophe space equals space open parentheses 51 plus 49 close parentheses space straight rho straight rho apostrophe space equals fraction numerator open parentheses 51 plus 49 close parentheses over denominator open parentheses 52.8 plus 49 close parentheses space end fraction straight rho space space minus negative negative negative negative negative negative negative negative negative negative negative negative negative negative left parenthesis 1 right parenthesis But space we space have space that colon straight rho apostrophe space space equals straight rho space open square brackets 1 minus straight gamma increment straight T close square brackets where space straight gamma space is space the space coefficient space of space cubical space expansion straight rho apostrophe space equals straight rho space open square brackets 1 minus straight gamma open parentheses 95 minus 5 close parentheses close square brackets straight rho apostrophe space equals straight rho space open square brackets 1 minus 90 straight gamma close square brackets space space minus negative negative negative negative negative negative negative negative negative negative negative negative negative negative negative left parenthesis 2 right parenthesis Thus space from space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis space we space get colon straight rho space open square brackets 1 minus 90 straight gamma close square brackets equals fraction numerator open parentheses 51 plus 49 close parentheses over denominator open parentheses 52.8 plus 49 close parentheses space end fraction straight rho 1 minus 90 straight gamma equals fraction numerator open parentheses 51 plus 49 close parentheses over denominator open parentheses 52.8 plus 49 close parentheses space end fraction 90 straight gamma equals 1 minus fraction numerator open parentheses 51 plus 49 close parentheses over denominator open parentheses 52.8 plus 49 close parentheses space end fraction 90 straight gamma space equals space 1 minus 0.982 space equals 0.018 straight gamma equals fraction numerator 0.018 over denominator 90 end fraction equals 2 cross times 10 to the power of negative 4 end exponent space degree straight C to the power of negative 1 end exponent We space know space that space coefficient space of space linear space expansionof space liquid comma space straight alpha space equals space straight gamma over 3 equals fraction numerator 2 cross times 10 to the power of negative 4 end exponent over denominator 3 end fraction equals 6.67 cross times 10 to the power of negative 5 end exponent degree straight C to the power of negative 1 end exponent Therefore space coefficient space of space linear space expansionof space liquid equals 6.67 cross times 10 to the power of negative 5 end exponent degree straight C to the power of negative 1 end exponent end style
Answered by Jyothi Nair | 04 May, 2015, 09:42: AM

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