Question

What is the radius of the biggest aluminum coin of thickness t and density \[\rho \] which will still be able to float on the water surface of surface tension S?
A.\[\dfrac{{4S}}{{3\rho gt}}\]
B.\[\dfrac{{3S}}{{4\rho gt}}\]
C.\[\dfrac{{2S}}{{\rho gt}}\]
D.\[\dfrac{S}{{\rho gt}}\]

Answer 1

Hint: To answer this question we need to use the two forms of force and substitute the given formula. Surface tension can be defined as the tendency of the surface film of the liquid that tends to shrink into the minimum surface area possible. It is the reason that allows objects with a higher density than water to float on the water. We need to use the formula of the force with respect to surface tension and the other form of the force in terms of mass and acceleration and equate both to find the radius of the aluminum coin.
Formula used:
The force due to the surface tension is given by,
\[S = \dfrac{F}{L}\]
The circumference of the circle=\[2\pi r\]
Also from newton’s second law,
\[F = mg\]

Complete answer:
Given that the surface tension of the water is S. And the biggest aluminum coin has a thickness t and density \[\rho \]. And also given that the coin is floating on the surface of the water. So the force on the coin due to the surface tension is given as,
\[S = \dfrac{F}{L}\]
Here S is said to be the surface tension of the liquid
          F is said to be the force per unit length
          L is said to be the length in which the force acts.
Now rearranging the above formula we get,
\[S \times L = F\]
When we that the shape of the coin is a circle. If we open a circle and make a straight line out of it then the total length of the line is equal to the circumference of the circle. Therefore substituting the circumference of the circle in the above formula in the place of length we get,
\[S \times (2\pi r) = F\] …….. (1)
Also, we know another form of force according to Newton’s second law,
\[F = mg\]
We know that mass can be calculated by multiplying the volume by the density. That is \[m = \rho V\]. Therefore substituting this in the above equation we get,
\[F = \rho V \times g\]
We have given that the thickness of the coin is t.
Therefore we replace the formula for the volume in terms of thickness we get,
\[F = \rho (\pi {r^2} \times t) \times g\] ………… (2) (Since volume is equal to area multiplied by thickness)
Equating equations (1) and (2)
\[S \times (2\pi r) = \rho (\pi {r^2} \times t) \times g\]
Solving the above equation we get,
\[r = \dfrac{{2S}}{{\rho gt}}\]:

Therefore the radius of the aluminum coin that is floating in the water is \[\dfrac{{2S}}{{\rho gt}}\]. Therefore the correct option is C.

Note:
We have used the formula \[m = \rho V\]. Rearranging the above formula we get, \[\dfrac{m}{V} = \rho \]. Here \[\rho \]is called mass density. Mass density can be defined as the mass divide by volume. It is a representation of the amount of mass of an object in relation to the space it occupies. This measurement will not be constant in all cases. The value of the mass density mainly depends on temperature and pressure.