Question

A ring is cut from platinum tube $8.5cm$ internal and $8.7cm$ external diameter. It is supported horizontally from the pan of a balance so that it comes in contact with the water in a glass vessel. If an extra $3.103gf$ is required to pull it away from water, then what is the surface tension of water?
A) $56.27dyne/cm$
B) $70.80dyne/cm$
C) $63.35dyne/cm$
D) $60dyne/cm$

Answer 1

Hint: Here we have to find the total force due to the surface tension using the formula. In the formula, internal and external radius has to be used but the internal diameter and external diameter are given in the data. We will have to change them into internal radius and external radius. Then using it we can find the surface tension.

Formula Used:
$F = T(2\pi {r_1} + 2\pi {r_2})$
$T$ is the surface tension, ${r_1}$ and ${r_2}$ is the internal and external radius.

Complete step by step solution:
The tangential force acting on a liquid per unit length, acting at right angles on either side of a line, imagined to be drawn on the free liquid surface in equilibrium is called surface Tension. Here the free surface behaves like a stretched elastic membrane. It also supports the small objects placed over it. At the surface of a liquid, the molecule experiences only net inward cohesive forces. Force of attraction among the molecules of the same substance is called cohesive force. Here, a ring is cut from the platinum tube having an internal diameter ${d_1} = 8.5cm$ and external diameter ${d_2} = 8.7cm$.
Now the internal and external radius,
${r_1} = \dfrac{{{d_1}}}{2} = \dfrac{{8.5}}{2} = 4.25cm$
${r_2} = \dfrac{{{d_2}}}{2} = \dfrac{{8.7}}{2} = 4.35cm$
The force required to pull away from water
$Rightarrow$ $F = 3.103 \times 980 = 3040.94dyne$
Here the value of $g = 980cm/{s^2}$
Now substituting the values to find surface tension:
$\Rightarrow$ $F = T(2\pi {r_1} + 2\pi {r_2})$
$\Rightarrow$ $3040.94 = T(2 \times 3.14 \times 4.25 + 2 \times 3.14 \times 4.35)$
$\Rightarrow$ $T = \dfrac{{3040.94}}{{54.01}} = 56.2dyne/cm$

Hence the correct answer is in option, $\left( A \right) \Rightarrow 56.27dyne/cm$.

Note: Due to the cohesive nature of its molecules, the surface of a liquid will allow resisting an external force, this property is called surface tension. The high surface tension helps the paper with much higher density to float on the water. The free surface is like a stretched elastic membrane. At the surface of a liquid, the molecule experiences only net inward cohesive forces.