work done in splitting a drop of water of 1mm radius into 10⁶ droplets is (surfsce tension of water =72×10-⁶J/m²)
Answers
Answer:
let R the radius of the big drop R= 10
−3
m
r is the radius of small drops
since volume remains constant ⟹
3
4
πR
3
= n(
3
4
πr
2
)
⟹ n = (
r
R
)
3
intial area of the big drop = 4πR
2
final drops (small) area = n×4πr
2
change in area = 4π(nr
2
−R
2
)
work done W =4πR
3
T(
r
1
−
R
1
)
=4πR
2
T[n
3
1
−1]
=4π×(10
−3
)
2
×72×10
−3
×[10
3
6
−1]
= 8.95×10
−5
J
Hari 19:36
let R the radius of the big drop R= 10
−3
m
r is the radius of small drops
since volume remains constant ⟹
3
4
πR
3
= n(
3
4
πr
2
)
⟹ n = (
r
R
)
3
intial area of the big drop = 4πR
2
final drops (small) area = n×4πr
2
change in area = 4π(nr
2
−R
2
)
work done W =4πR
3
T(
r
1
−
R
1
)
=4πR
2
T[n
3
1
−1]
=4π×(10
−3
)
2
×72×10
−3
×[10
3
6
−1]
= 8.95×10
−5
J