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A liquid column of height 80cm at 0°C balances the same liquid of height 80.4cm at 100°C. γ_R is

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3 x 10^–5/°C

5 x 10^–5/°C

4 x 10^–5/°C

6 x 10^–5/°C

answer is C.

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Detailed Solution

Given that h₁ = 80cm, h₂ = 80.4cm; t₁ = 0⁰C, t₂ = 100⁰C
We have $\large \gamma_R=\frac {h_2-h_1}{h_1t_2-h_2t_1}$
$\large =\frac {80.4-80}{(80)(100)-0}=5\times 10^{-5}/^0C$

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