BITSAT 2018 Slot 2 Solved Paper

Since, volume remains unchanged, during this phenomenon, so

4

3

πR3=N×

4

3

πr3
N=

R3

r3

Now, change in surface area =4πR2−N4πr2
=4π(R2−Nr2)
Energy released (ΔU)=T× change in surface area
=T×4π[R2−Nr2]
Here, all this energy released is at the cost of lowering the temperature and mass of the big drop of liquid =

4

3

πR2ρ.
Now, change in temperature,
Δθ=

ΔU

ms

=

T×4π(R2−Nr2)

( 4 3 πR3ρ)S

=

3T

ρS

(

1

R

−

Nr2

R3

)
=

3T

ρS

(

1

R

−

R3×r2

r3×R3

)
=

3T

ρS

(

1

R

−

1

r

)