Answers

2014-11-16T21:12:13+05:30
 \alpha = \frac{\Delta\l}{l_1\Delta\ T}
Where, α=expansion on heating=2.5*(10)^{-5} °C
l_1 = initial length (100cm)
\Delta\l = Change in length, l_1=100cm l_2=?
\Delta\T =  change in temperature = 100°C-20°C=80°C
Using the above formula,
 \alpha l_1\Delta\T=\Delta\l
\alpha l_1\Delta\T=l_2-l_1
Substituting values:-
2.5*10^{-5}*100*80=l_2-100
Solve, you get
l_2=100.2cm
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2014-11-19T03:47:25+05:30

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\Delta L= \alpha *L*\Delta T\\\\.\ \ \ \ =2.5*10^{-5}*100*(100-20)\\\\.\ \ \ \ =0.2\ cm\\

So the aluminium bar of length 100 cm will expand by 0.2 cm and the bar will expand to 100.2 cm total length.

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