# NEET Questions Solved

NEET - 2014

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

(a) Energy = 4VT (1/r-1/R) is released

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

(c) Energy = 3VT (1/r-1/R)  is released

(d) Energy is neither released nor absorbed

Here, if the surface are a changes, it will change the surface energy as well. If the surface area decreases, it means that energy is released and vfce-versa.

Change in surface energy  ΔAxT                      ...(i)

Let we have 'n' number of drops initially.

So, ΔA=4πR2-n(4πr2)                         ...(ii)

Volume is constant

So, n4/3πR3-n(4πr2)                         ...(iii)

From Eqs. (ii) and (iii)

ΔA=3/R x 4π/3 x R3-3/r(n4π/r3)

=3/R x V-3/r x V

ΔA=3V(1/R-1/r)= -ve value

As K > r, so ΔA is negative It means surface area is decreased, so energy must be released.

Energy released=ΔA x T=-3VT(1/r-1/R)

Above expression shows the magnitude of energy released.

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