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Question

A vertical glass tube, closed at the bottom, contains a mercury column of length L0 at 0 C. If γ is the coefficient of cubical expansion of mercury α the coefficient of linear expansion of glass, the length of the mercury column when the temperature rises to t C is (assuming that t not more than 100 C)

A
Lt=L0[1+(γ3α)t]
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B
Lt=L0[1+(γ+3α)t]
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C
Lt=L0[1+(γ+2α)t]
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D
Lt=L0[1+(γ2α)t]
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Solution

The correct option is D Lt=L0[1+(γ2α)t]
If V0 is the volume of mercury at 0 C and A0 the cross-sectional area of the tube at 0 C, then the length of the mercury column at 0 C is
L0=V0A0 (1)
The cross-sectional are at t C is given by
At=A0(1+βt)
where β is the coefficient of superficial expansion of glass. Since β=2α, we have
At=A0(1+2αt)
If Vt is the volume of mercury at t C, the length of the mercury column at t C is
Lt=VtAt=V0(1+γt)A0(1+2αt)
Using (1) we have
Lt=L0(1+γt)(1+2αt)1
Since t is not too high (100 C) and γ are of the order of 105.γt and αt will be very small compared to unity.
Hence we can expand (1+2αt)1 binomially and retain terms upto order αt. Thus
(1+2αt)1=(12αt)
Lt=L0(1+γt)(12αt)
or Lt=L0[1+(γ2α)t]

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