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Question: A vertical column \(50\;cm\) long at \(50^{\circ}C\) balances another column of same liquid \(60\;cm...

A vertical column 50  cm50\;cm long at 50C50^{\circ}C balances another column of same liquid 60  cm60\;cm long at 100C100^{\circ}C. The coefficient of absolute expansion of the liquid is:
A. 0.005/C0.005/^{\circ}C
B. 0.0005/C0.0005/^{\circ}C
C. 0.002/C0.002/^{\circ}C
D.0.0002/C0.0002/^{\circ}C

Explanation

Solution

We know that thermal expansion is the process of change in physical properties like shape, volume or density if the given matter due to change in temperature without any change in the phase of the given substance. Using this we can solve the above question.
Formula used:
ρ=mv0(1+γT)\rho=\dfrac{m}{v_0(1+\gamma T)}
P=ρghP=\rho gh

Complete step by step answer:
Since thermal expansion is the change of physical property due to the heat energy, the coefficient of thermal expansion, also describes the fractional change in shape or size of the given object with respect to the degree change in the temperature at a constant pressure.
It is mathematically expressed as α=ΔVVΔT\alpha=\dfrac{\Delta V}{V\Delta T}, where α\alpha is the coefficient of thermal expansion of VV initial volume due to ΔT\Delta T change in temperature and ΔV\Delta V change in volume.
Let us consider the mass of the liquid be mm, then the density ρ=mv\rho=\dfrac{m}{v} where the volume is expressed in terms of temperature TT v=v0(1+γT)v=v_0(1+\gamma T), where γ\gamma is the coefficient of absolute expansion .
Then, we have
    ρ=mv0(1+γT)\implies \rho=\dfrac{m}{v_0(1+\gamma T)}
Given that the columns balance each other, then let us consider the pressure of the liquid column, given as P=ρghP=\rho gh
Then the pressure are equal, then we have:
P1=P2P_1=P_2, where P1P_1 is the pressure at 50C50^{\circ}C and P2P_2 is the pressure at 100C100^{\circ}C
Substituting, we have
    ρ1gh1=ρ2gh2\implies \rho_1 gh_1=\rho_2 gh_2
    50mgv0(1+50γ)=60mgv0(1+100γ)\implies \dfrac{50mg}{v_0(1+50\gamma )}=\dfrac{60mg}{v_0(1+100\gamma )}
    1+50γ1+100γ=56\implies \dfrac{1+50\gamma}{1+100\gamma}=\dfrac{5}{6}
    6+300γ=5+500γ\implies 6+300\gamma=5+500\gamma
    1=200γ\implies 1=200\gamma
    γ=1200=0.005/C\implies \gamma=\dfrac{1}{200}=0.005/^{\circ}C

So, the correct answer is “Option A”.

Note: There are various types of coefficient of thermal expansion, based on the property which changes due to the temperature change, they are the volumetric change, area change or the linear change. This also depends on the nature of the material, if it is free to expand or is constrained due to various other reasons.