
A vertical column 50cm long at ${{50}^{o}}C$ balances ${{100}^{o}}C$ the coefficient of absolute expansion of the liquid is,
$\begin{align}
& A.\text{ }0.005{{/}^{o}}C \\
& B.\text{ }0.0005{{/}^{o}}C \\
& C.\text{ }0.002{{/}^{o}}C \\
& D.\text{ }0.0002{{/}^{o}}C \\
\end{align}$
Answer
552k+ views
Hint: Here in the question it is given that both vertical columns are balancing each other so we can say that their pressure will be equal by using this phenomena we can find the coefficient of absolute expansion of the liquid.
Formula used:
$P=\rho gh$
$\rho =\dfrac{{{\rho }_{o}}}{1+\lambda T}$
Complete step by step solution:
Now it is given that both vertical columns balance each other so we can say that pressure should be equal to both the vertical columns.
If ${{P}_{1}}\text{ and }{{P}_{2}}$ be the pressure of the columns of respectively then,
${{P}_{1}}={{P}_{2}}......\left( 1 \right)$
Now formula for the pressure is,
$P=\rho gh.....\left( 2 \right)$
Where P = pressure
$\rho $ = gravity
g = acceleration due to gravity
h = height
Now given that,
Height of vertical column is ${{h}_{1}}=50cm$
Other column is ${{h}_{2}}=60cm$
Now from the equation (2) we can write equation (1) as
${{\rho }_{1}}g{{h}_{1}}={{\rho }_{2}}g{{h}_{2}}....\left( 3 \right)$
Now if the $\lambda $ be the coefficient of expansion of liquid then formula for the density for respective column is
$\begin{align}
& {{\rho }_{1}}=\dfrac{{{\rho }_{o}}}{\left( 1+\lambda {{T}_{1}} \right)}......\left( 4 \right) \\
& {{\rho }_{2}}=\dfrac{{{\rho }_{o}}}{\left( 1+\lambda {{T}_{2}} \right)}......\left( 5 \right) \\
\end{align}$
Where, $\rho $ = density
${{\rho }_{o}}$ = density ${{0}^{o}}C$
$\lambda $ = the coefficient of absolute expansion of liquid.
T = temperature
Now put values of equation (4) and equation (5) into equation (3)
$\begin{align}
& \Rightarrow \dfrac{{{\rho }_{o}}}{\left( 1+\lambda {{T}_{1}} \right)}\times g\times {{h}_{1}}=\dfrac{{{\rho }_{o}}}{\left( 1+\lambda {{T}_{2}} \right)}\times g\times {{h}_{2}} \\
& \Rightarrow {{h}_{1}}\left( 1+\lambda {{T}_{2}} \right)={{h}_{2}}\left( 1+\lambda {{T}_{1}} \right) \\
\end{align}$
Here be careful while putting values,
$\begin{align}
& \Rightarrow 50\left( 1+\lambda \times 100 \right)=60\left( 1+\lambda \times 50 \right) \\
& \Rightarrow 50+5000\lambda =60+3000\lambda \\
& \Rightarrow 5000\lambda -3000\lambda =60-50 \\
& \Rightarrow 2000\lambda =10 \\
& \therefore \lambda =0.005{{/}^{o}}C \\
\end{align}$
Additional information:
Definition of The absolute coefficient of expansion,
The absolute coefficient of expansion of liquid is defined as the actual increase in the volume of liquid or we can say the sum of the apparent increase in the volume of liquid in a vessel and increase in the volume of the vessel per unit original volume of liquid per unit rise in temperature.
Note:
When we are putting values of the height and the temperature we have to be careful, because if any value from the given equation is replaced then it will lead us to an incorrect solution.
Formula used:
$P=\rho gh$
$\rho =\dfrac{{{\rho }_{o}}}{1+\lambda T}$
Complete step by step solution:
Now it is given that both vertical columns balance each other so we can say that pressure should be equal to both the vertical columns.
If ${{P}_{1}}\text{ and }{{P}_{2}}$ be the pressure of the columns of respectively then,
${{P}_{1}}={{P}_{2}}......\left( 1 \right)$
Now formula for the pressure is,
$P=\rho gh.....\left( 2 \right)$
Where P = pressure
$\rho $ = gravity
g = acceleration due to gravity
h = height
Now given that,
Height of vertical column is ${{h}_{1}}=50cm$
Other column is ${{h}_{2}}=60cm$
Now from the equation (2) we can write equation (1) as
${{\rho }_{1}}g{{h}_{1}}={{\rho }_{2}}g{{h}_{2}}....\left( 3 \right)$
Now if the $\lambda $ be the coefficient of expansion of liquid then formula for the density for respective column is
$\begin{align}
& {{\rho }_{1}}=\dfrac{{{\rho }_{o}}}{\left( 1+\lambda {{T}_{1}} \right)}......\left( 4 \right) \\
& {{\rho }_{2}}=\dfrac{{{\rho }_{o}}}{\left( 1+\lambda {{T}_{2}} \right)}......\left( 5 \right) \\
\end{align}$
Where, $\rho $ = density
${{\rho }_{o}}$ = density ${{0}^{o}}C$
$\lambda $ = the coefficient of absolute expansion of liquid.
T = temperature
Now put values of equation (4) and equation (5) into equation (3)
$\begin{align}
& \Rightarrow \dfrac{{{\rho }_{o}}}{\left( 1+\lambda {{T}_{1}} \right)}\times g\times {{h}_{1}}=\dfrac{{{\rho }_{o}}}{\left( 1+\lambda {{T}_{2}} \right)}\times g\times {{h}_{2}} \\
& \Rightarrow {{h}_{1}}\left( 1+\lambda {{T}_{2}} \right)={{h}_{2}}\left( 1+\lambda {{T}_{1}} \right) \\
\end{align}$
Here be careful while putting values,
$\begin{align}
& \Rightarrow 50\left( 1+\lambda \times 100 \right)=60\left( 1+\lambda \times 50 \right) \\
& \Rightarrow 50+5000\lambda =60+3000\lambda \\
& \Rightarrow 5000\lambda -3000\lambda =60-50 \\
& \Rightarrow 2000\lambda =10 \\
& \therefore \lambda =0.005{{/}^{o}}C \\
\end{align}$
Additional information:
Definition of The absolute coefficient of expansion,
The absolute coefficient of expansion of liquid is defined as the actual increase in the volume of liquid or we can say the sum of the apparent increase in the volume of liquid in a vessel and increase in the volume of the vessel per unit original volume of liquid per unit rise in temperature.
Note:
When we are putting values of the height and the temperature we have to be careful, because if any value from the given equation is replaced then it will lead us to an incorrect solution.
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