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A spherical drop of oil of radius 1 cm is broken into 1000 droplets of equal radii. If the surface tension of oil is 50 dynes/cm, the work done is
  • a)
    18π ergs
  • b)
    180π ergs
  • c)
    1800π ergs
  • d)
    18000π ergs
Correct answer is option 'C'. Can you explain this answer?
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A spherical drop of oil of radius 1 cm is broken into 1000 droplets of...
To find the work done, we need to find the change in surface area as the drop is broken into 1000 droplets.

The original drop has a radius of 1 cm, so its surface area is given by the formula for the surface area of a sphere: A = 4πr^2.

The original surface area is therefore 4π(1^2) = 4π cm^2.

When the drop is broken into 1000 droplets, each droplet will have a radius of 1/10 cm.

The surface area of each droplet is given by the formula for the surface area of a sphere: A = 4πr^2.

The surface area of each droplet is therefore 4π((1/10)^2) = 4π/100 cm^2.

Since there are 1000 droplets, the total surface area of all the droplets is 1000 * (4π/100) = 40π cm^2.

The change in surface area is therefore 40π - 4π = 36π cm^2.

The work done is equal to the change in surface area multiplied by the surface tension:

Work done = 36π * 50 dynes/cm = 1800π dynes/cm.

Since π is a constant, the work done is equal to 1800 times π dynes/cm.

The value of π is approximately 3.14, so the work done is approximately 1800 * 3.14 dynes/cm.

Therefore, the work done is approximately 5652 dynes/cm or 56.52 ergs.

Since 1 erg = 1 dyne * cm, the work done is also approximately 56.52 ergs.

Therefore, the answer is approximately 56.52 or 18 when rounded to the nearest whole number.

So, the correct answer is a) 18.
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A spherical drop of oil of radius 1 cm is broken into 1000 droplets of equal radii. If the surface tension of oil is 50 dynes/cm, the work done isa)18π ergsb)180π ergsc)1800π ergsd)18000π ergsCorrect answer is option 'C'. Can you explain this answer?
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