A liquid drop having surface energy ‘E’ is spread into 216 droplets of the same size. The final surface energy of the droplets is __.
Surface area of drop, A1 = 4πR2
Surface area of droplets, A2 = 216(4πr2)
Volume of drops = n x (Volume of droplets)
\(\frac {4}{3}\) πR3 = 216 (\(\frac {4}{3}\) πr3)
R = 6r
Now A2 = 216(4π(\(\frac {R}{6}\))2)
A2 = 6(4πR2)
E ∝ A
\(\frac {E_1}{E_2 }\)= \(\frac {A_1}{A_2}\)
\(\frac {E_1}{E_2 }\) = \(\frac {4πR^2}{6(4πR^2)}\)
\(\frac {E_1}{E_2 }\) = \(\frac {1}{6}\)
6E1 = E2
E2 = 6E1
E2 = 6E {E1 = E}
Therefore, the correct option is (D) 6E
Find the surface tension at critical velocity?
Which one of the following statements is ‘NOT’ true about the angle of contact of a liquid?
The amount of energy required to increase the liquid's surface area by one unit area is known as surface tension. In other words, it is a property of the liquid surface to resist force.
Surface tension is defined as,
The ratio of the surface force F to the length L along which the force acts.
Mathematically, the surface tension formula can be expressed as follows:
T=F/L
Where,
Read More: Detergents and Surface Tension
The SI unit of Surface Tension is Newton per Meter or N/m.