Physics, asked by sajalkhandelwal3710, 1 year ago

Calculate the work done in breaking a water drop of radius 1mm to 1000 droplets

Answers

Answered by reenadevi0703
0

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

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