27000 small liquid droplets each of radius 1 mmcoalesce to form a big drop find the amount of work done in suface tention of liquid is 70 dyne /cm
Answers
Answer:
r=0.1mm=0.1×10
−3
m, T=0.072N/m
Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury. Volume of 27 droplets = volume of the single drop as the volume of the liquid remains constant.
∴27×
3
4
πr
3
=
3
4
πR
3
∴27r
3
=R
3
∴3r=R
Surface area of 27 droplets = 27×4πr
2
Surface area of single drop = 4πR
2
∴ Decrease in surface area = 27×4πr
2
−4πR
2
=4π(27r
2
−R
2
)=4π[27r
2
−(3r)
2
]=4π×18r
2
∴ The energy released = surface tension × decrease in surface area = T×4π×18r
2
=0.072×4×3.142×18×(1×10
−4
)
2
=1.628×10
−7
J.
Explanation:
r=0.1mm=0.1×10
−3
m, T=0.072N/m
Let R be the radius of the single drop formed due to the coalescence of 27 droplets of mercury. Volume of 27 droplets = volume of the single drop as the volume of the liquid remains constant.
∴27×
3
4
πr
3
=
3
4
πR
3
∴27r
3
=R
3
∴3r=R
Surface area of 27 droplets = 27×4πr
2
Surface area of single drop = 4πR
2
∴ Decrease in surface area = 27×4πr
2
−4πR
2
=4π(27r
2
−R
2
)=4π[27r
2
−(3r)
2
]=4π×18r
2
∴ The energy released = surface tension × decrease in surface area = T×4π×18r
2
=0.072×4×3.142×18×(1×10
−4
)
2
=1.628×10
−7
J.