Question

A spherical drop of water of radius 1mm and surface tension 72 x 10-3 N/m is split in to 1000 identical spherical droplets. Calculate the energy released in the process

Answer 1

Answer:

The energy released in the process is  8.95×10^−5 Joules

Explanation:

Correct statement:

What is the work done in splitting a drop of water 1 mm radius into 106 droplets (surface tension of water =72×10−3J/m^2)?

Solution:

• Radius of big drop  = R
• Radius of smaller drops  = r
• 4 / 3πR^3 = n 4 / 3πr^3

R^3 = n . r^3

n=(R^3 / r^3)

n = ( R/r)^3

• Initial area of drop =4πR^2
• Final area of drops =n 4πr^2

Changes in area =ΔA= nπr^2 − 4πR^2

                            = 4π( n r^2−R^2)

Work done = 4πT (n.r^2−R^2)

W = 4πR^3.T(1 / r − 1 /R)

or = 4πR^2.T[n^1/3 − 1]

    = 4π × (10^−3)^2×72×10^−3[10^6/3 − 1]

    = 8.95×10^−5 Joules

Hence the energy released in the process is  8.95×10^−5 Joules

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