JEE Main Physics Class 11 Physical world and Measurements Part 1 Questions

Given, radius of small drop =r
Radius of big drop =R
Surface tension =T
and mechanical equivalent of heat =J
As, small drops combine to form big drop.
∴ Volume of big drop (VB)=n× Volume of small drop (VS)
⇒‌‌‌

4

3

πR3=n⋅‌

4

3

πr3
⇒‌‌nr3=R3
⇒‌‌r=‌

R

n1∕3

. . . (i)
Surface energy of small drop (ES)= Surface tension (T)× Area (A)
⇒‌‌ES=n×4πr2T and EB=4πR2T
Now, change in energy will be
∆E=EB−ES=4πT(nr2−R2)
∴ Heat energy per unit volume =‌

∆E

V

=‌

4πT(nr2−R2)

J×‌ 4 3 πR3

=‌

3T

J

(‌

nr2

R3

−‌

1

R

)=‌

3T

J

(n‌

R2

n2∕3R3

−‌

1

R

)
=‌

3T

J

[‌

n1∕3

R

−‌

1

R

]‌‌[from Eq. (i)]
=‌

3T

J

[‌

1

r

−‌

1

R

]