- #1
Puneeth423
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Consider a rod of length 'l'. When heated, its temperature got increased by 't'.
Increase in its length = l[itex]\alpha[/itex]t.
total length be (l1) = l + l[itex]\alpha[/itex]t.
Again the rod is heated and the temperature further got increased by 't'.
Total length be l2.
l2 can be calculated in two ways,
l2 = l1 + l1[itex]\alpha[/itex] t...(a)
0r l2 = l + l[itex]\alpha[/itex](2t)...(b)
Equating (a) and (b).
l1 + l1[itex]\alpha[/itex] t = l+l[itex]\alpha[/itex](2t)
{l + l[itex]\alpha[/itex]t} + {l + l[itex]\alpha[/itex]t}{[itex]\alpha[/itex] t}
= l+l[itex]\alpha[/itex](2t).
Solving above equation we get l([itex]\alpha[/itex]t)2 = 0.
Where am i going wrong?
Increase in its length = l[itex]\alpha[/itex]t.
total length be (l1) = l + l[itex]\alpha[/itex]t.
Again the rod is heated and the temperature further got increased by 't'.
Total length be l2.
l2 can be calculated in two ways,
l2 = l1 + l1[itex]\alpha[/itex] t...(a)
0r l2 = l + l[itex]\alpha[/itex](2t)...(b)
Equating (a) and (b).
l1 + l1[itex]\alpha[/itex] t = l+l[itex]\alpha[/itex](2t)
{l + l[itex]\alpha[/itex]t} + {l + l[itex]\alpha[/itex]t}{[itex]\alpha[/itex] t}
= l+l[itex]\alpha[/itex](2t).
Solving above equation we get l([itex]\alpha[/itex]t)2 = 0.
Where am i going wrong?
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