Mechanical Properties of Fluids - Result Question 17

17. A certain number of spherical drops of a liquid of radius ’ r ’ coalesce to form a single drop of radius ’ R ’ and volume ’ V ‘. If ’ T ’ is the surface tension of the liquid, then :

[2014]

(a) energy =4VT(1r1R) is released

(b) energy =3VT(1r+1R) is absorbed

(c) energy =3VT(1r1R) is released

(d) energy is neither released nor absorbed

Show Answer

Answer:

Correct Answer: 17. (c)

Solution:

  1. (c) Volume same n(43πr3)=43πR3

n=R3r3

Change in energy =TΔA=T[4πR2n4πr2]

=4πT[R2R3r3r2]

=3(43πR3)T[1R1r]

$=3 VT\frac{1}{R}-\frac{1}{r}$

=3VT[1r1R] is released

Energy released when n drops of radius r coalesec to form a body drop of radius R,

Energy released =4πR3T[1r1R]

If this energy get converted into kinetic energy of big drop, then

12mv2=4πR3T[1r1R]12[43πR3d]V2=4πR3T[1r1R]v2=6Td[1r1R]

Velocity of big drop V=6Td(1r1R)

(Where, d= density of big liquid drop)



NCERT Chapter Video Solution

Dual Pane