[" 18.) Energy needed in breaking a drop of radius "R" into "],[" drops of radii r is given by "],[qquad [" U "4 pi T(nr^(2)-R^(2)),[2](4)/(3)pi r^(3)n-R^(2)],[[3]4 pi T(R^(2)-nr^(2)),[4]4 pi T(nr^(2)+R^(2))]],[hline]

Energy needed in breaking a drop of radius R into n drops of radii r is given by

Energy needed in breaking a drop of radius R into n drops of radii r is given by

Energy needed in breaking a drop of radius R into n drops of radii r is given by

If A=pi(R^(2)-r^(2)) , then R is equal to

[" A small circular coil of radius " "r" and having "n' turns carries a current "7" .The coefficient of self-induction of the coil is "],[[" (A) "mu_(0)pi nr/2],[" (B) ",mu_(0)pi nr],[" (C) ",mu_(o)pi n^(2)r/2],[" (D) ",mu_(o)pi nr^(2)/2]]

A soap bubble of radius r and surface tension T is given a potential V. Show that the new radius R of the bubble is related with the initial radius by the equation P(R^(3) -r^(3)) +4T (R^(2)-r^(2)) = (in_(0)V^(2)R)/(2) where P is the atmospheric pressure.

A soap bubble of radius r and surface tension T is given a potential V. Show that the new radius R of the bubble is related with the initial radius by the equation P(R^(3) -r^(3)) +4T (R^(2)-r^(2)) = (in_(0)V^(2)R)/(2) where P is the atmospheric pressure.

The excess of pressure in a liquid drop is 2T/R while in soap bubble it is 4T/R, why?