Calculate the work done in breaking a water drop of radius 1mm to 1000 droplets
Answers
Answer:
Here,
T=525×10
−3
Nm
−1
R=1×10
−3
Now, the surface area of the bigger drop.
=4πR
2
=4×3.14×(1×10
−3
)
2
then,
S
1
=0.1256×10
−4
m
2
Now, the total initial volume =
3
4
πR
3
and the volume smaller drops =
1000
3
4
πR
3
consider r be the radius of smaller drops,
Hence, volume of droplets =
3
4
πr
3
and
3
4
πr
3
=
1000
3
4
πR
3
therefore, r=
10
R
Now, the surface area of small drop =4πr
2
=4π(R/10)
2
=4×3.14×(1×10
−4
)
2
=12.56×10
−8
m
2
Total surface area vof 1000 drops S
2
=1000×12.56×10
−8
=1.256×10
−4
m
2
So, change of surface area =S
2
−S
1
ΔS=1.256×10
−4
−0.1256×10
−4
=1.13×10
−4
and, we know that
workdone=TΔS
=525×10
−3
×1.13×10
−4
=593.25×10
−7
=5.9×10
−5
Hence, the work done in breaking a mercury drop of radius 1mm into 1000 droplets of same size will be 5.9×10
−5
J