Answer
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Hint: In order to solve this numerical problem, we have to use the relationship between the acceleration due to gravity of the planet and the mass and radius of the planet. This will help us obtain the acceleration due to gravity on that planet and we can put this value of acceleration due to gravity to obtain the height of the liquid column in the capillary rise method of finding surface tension of the particular liquid.
Formula Used:
The acceleration due to gravity is related to mass and radius of earth is given by the following mathematical expression:
$g = \dfrac{{GM}}{{{R^2}}}$
Here, M denotes the mass of earth and R denotes the radius of earth.
Complete step by step answer:
We know that in an experiment to find the surface tension of a particular liquid by capillary rise method, the height of the liquid column is obtained by using the mathematical expression given below:
$h = \dfrac{{2T\cos \theta }}{{r\rho g}}$
In this expression, T denotes the surface tension of the liquid, r denotes the radius of the bore of the capillary tube, g denotes the acceleration due to gravity, $\theta $ denotes the angle of contact and $\rho $ denotes the density of the liquid.
Now, we have to obtain the acceleration due to gravity in the particular planet compared to the acceleration due to gravity on earth. It is given that the radius of the planet is twice that of earth and the mass of the planet is four times that of earth. Substituting these values in the mathematical formula to find acceleration due to gravity, we obtain:
$
g' = G\dfrac{{\left( {4M} \right)}}{{{{\left( {2R} \right)}^2}}} \\
\Rightarrow g' = G\dfrac{{4M}}{{4{R^2}}} = G\dfrac{M}{{{R^2}}} = g \\
$
Thus, we find that there is no change in the acceleration due to gravity on the planet. That is, the acceleration due to gravity on the planet is the same as that of earth.
Since, there is no change in the acceleration due to gravity on the planet; we can say that the height of the liquid column will be the same in that particular planet as well.
Thus, the height of the liquid column in the planet will also be equal to $4cm$.
Note: It is important to note that we cannot find the acceleration due to gravity on the particular planet unless the mass and radius of the planet is compared to that of earth. We can only use this relation when an adequate comparison is given.
Formula Used:
The acceleration due to gravity is related to mass and radius of earth is given by the following mathematical expression:
$g = \dfrac{{GM}}{{{R^2}}}$
Here, M denotes the mass of earth and R denotes the radius of earth.
Complete step by step answer:
We know that in an experiment to find the surface tension of a particular liquid by capillary rise method, the height of the liquid column is obtained by using the mathematical expression given below:
$h = \dfrac{{2T\cos \theta }}{{r\rho g}}$
In this expression, T denotes the surface tension of the liquid, r denotes the radius of the bore of the capillary tube, g denotes the acceleration due to gravity, $\theta $ denotes the angle of contact and $\rho $ denotes the density of the liquid.
Now, we have to obtain the acceleration due to gravity in the particular planet compared to the acceleration due to gravity on earth. It is given that the radius of the planet is twice that of earth and the mass of the planet is four times that of earth. Substituting these values in the mathematical formula to find acceleration due to gravity, we obtain:
$
g' = G\dfrac{{\left( {4M} \right)}}{{{{\left( {2R} \right)}^2}}} \\
\Rightarrow g' = G\dfrac{{4M}}{{4{R^2}}} = G\dfrac{M}{{{R^2}}} = g \\
$
Thus, we find that there is no change in the acceleration due to gravity on the planet. That is, the acceleration due to gravity on the planet is the same as that of earth.
Since, there is no change in the acceleration due to gravity on the planet; we can say that the height of the liquid column will be the same in that particular planet as well.
Thus, the height of the liquid column in the planet will also be equal to $4cm$.
Note: It is important to note that we cannot find the acceleration due to gravity on the particular planet unless the mass and radius of the planet is compared to that of earth. We can only use this relation when an adequate comparison is given.
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