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A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then:

 

  • Option 1)

    energy =4VT \left( \frac{1}{r}- \frac{1}{R} \right )is\:\:released

  • Option 2)

    energy =3VT \left( \frac{1}{r}+ \frac{1}{R} \right )is\:\:absorbed

  • Option 3)

    energy =3VT \left( \frac{1}{r}- \frac{1}{R} \right )is\:\:released

  • Option 4)

    Energy is neither released nor absorbed.

 

Answers (1)

best_answer

As discussed

Surface Energy -

It is defined as the amount of work done in increasing the area of the liquid against surface tension.

-

 

 As surface area decreases so energy is released energy released = 4 \pi R^{2}T \left [ n^{\frac{1}{3}}-1 \right ]

R= n^{\frac{1}{3}}r

= 4 \pi R^{2}T\left [ \frac{1}{r} - \frac{1}{R}\right ] = 3 v \left [ \frac{1}{r}-\frac{1}{R} \right ]


Option 1)

energy =4VT \left( \frac{1}{r}- \frac{1}{R} \right )is\:\:released

This solution is incorrect

Option 2)

energy =3VT \left( \frac{1}{r}+ \frac{1}{R} \right )is\:\:absorbed

This solution is incorrect

Option 3)

energy =3VT \left( \frac{1}{r}- \frac{1}{R} \right )is\:\:released

This solution is correct

Option 4)

Energy is neither released nor absorbed.

This solution is incorrect

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perimeter

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