
A vertical column $50\;cm$ long at $50^{\circ}C$ balances another column of same liquid $60\;cm$ long at $100^{\circ}C$. The coefficient of absolute expansion of the liquid is:
A. $0.005/^{\circ}C$
B. $0.0005/^{\circ}C$
C. $0.002/^{\circ}C$
D.$0.0002/^{\circ}C$
Answer
520.5k+ views
Hint: 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:
$\rho=\dfrac{m}{v_0(1+\gamma T)}$
$P=\rho gh$
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 $\alpha=\dfrac{\Delta V}{V\Delta T}$, where $\alpha$ is the coefficient of thermal expansion of $V$ initial volume due to $\Delta T$ change in temperature and $\Delta V$ change in volume.
Let us consider the mass of the liquid be $m$, then the density $\rho=\dfrac{m}{v}$ where the volume is expressed in terms of temperature $T$ $v=v_0(1+\gamma T)$, where $\gamma$ is the coefficient of absolute expansion .
Then, we have
$\implies \rho=\dfrac{m}{v_0(1+\gamma T)}$
Given that the columns balance each other, then let us consider the pressure of the liquid column, given as $P=\rho gh$
Then the pressure are equal, then we have:
$P_1=P_2$, where $P_1$ is the pressure at $50^{\circ}C$ and $P_2$ is the pressure at $100^{\circ}C$
Substituting, we have
$\implies \rho_1 gh_1=\rho_2 gh_2$
$\implies \dfrac{50mg}{v_0(1+50\gamma )}=\dfrac{60mg}{v_0(1+100\gamma )}$
$\implies \dfrac{1+50\gamma}{1+100\gamma}=\dfrac{5}{6}$
$\implies 6+300\gamma=5+500\gamma$
$\implies 1=200\gamma$
$\implies \gamma=\dfrac{1}{200}=0.005/^{\circ}C$
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.
Formula used:
$\rho=\dfrac{m}{v_0(1+\gamma T)}$
$P=\rho gh$
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 $\alpha=\dfrac{\Delta V}{V\Delta T}$, where $\alpha$ is the coefficient of thermal expansion of $V$ initial volume due to $\Delta T$ change in temperature and $\Delta V$ change in volume.
Let us consider the mass of the liquid be $m$, then the density $\rho=\dfrac{m}{v}$ where the volume is expressed in terms of temperature $T$ $v=v_0(1+\gamma T)$, where $\gamma$ is the coefficient of absolute expansion .
Then, we have
$\implies \rho=\dfrac{m}{v_0(1+\gamma T)}$
Given that the columns balance each other, then let us consider the pressure of the liquid column, given as $P=\rho gh$
Then the pressure are equal, then we have:
$P_1=P_2$, where $P_1$ is the pressure at $50^{\circ}C$ and $P_2$ is the pressure at $100^{\circ}C$
Substituting, we have
$\implies \rho_1 gh_1=\rho_2 gh_2$
$\implies \dfrac{50mg}{v_0(1+50\gamma )}=\dfrac{60mg}{v_0(1+100\gamma )}$
$\implies \dfrac{1+50\gamma}{1+100\gamma}=\dfrac{5}{6}$
$\implies 6+300\gamma=5+500\gamma$
$\implies 1=200\gamma$
$\implies \gamma=\dfrac{1}{200}=0.005/^{\circ}C$
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.
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