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If the radius of a spherical liquid (of ...

If the radius of a spherical liquid (of surface tension S) drop increases from r to `r +Deltar` , the corresponding increase in the surface energy is

A

`8 pi r Delta rS`

B

`4pi r DeltarS`

C

`16 pi r DeltarS`

D

`8 pi r DeltarS`

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The correct Answer is:
To find the increase in surface energy when the radius of a spherical liquid drop increases from \( r \) to \( r + \Delta r \), we can follow these steps: ### Step 1: Understand the formula for surface energy The surface energy \( E \) of a spherical drop is given by the formula: \[ E = 4\pi r^2 S \] where \( S \) is the surface tension and \( r \) is the radius of the drop. ### Step 2: Calculate the initial surface energy For the initial radius \( r \), the surface energy \( E_1 \) is: \[ E_1 = 4\pi r^2 S \] ### Step 3: Calculate the final surface energy When the radius increases to \( r + \Delta r \), the new surface energy \( E_2 \) is: \[ E_2 = 4\pi (r + \Delta r)^2 S \] ### Step 4: Expand the expression for \( E_2 \) Using the binomial expansion, we can expand \( (r + \Delta r)^2 \): \[ (r + \Delta r)^2 = r^2 + 2r\Delta r + (\Delta r)^2 \] Thus, we can rewrite \( E_2 \) as: \[ E_2 = 4\pi (r^2 + 2r\Delta r + (\Delta r)^2) S \] \[ E_2 = 4\pi r^2 S + 8\pi r \Delta r S + 4\pi (\Delta r)^2 S \] ### Step 5: Calculate the increase in surface energy The increase in surface energy \( \Delta E \) is given by: \[ \Delta E = E_2 - E_1 \] Substituting the values we found: \[ \Delta E = (4\pi r^2 S + 8\pi r \Delta r S + 4\pi (\Delta r)^2 S) - 4\pi r^2 S \] This simplifies to: \[ \Delta E = 8\pi r \Delta r S + 4\pi (\Delta r)^2 S \] ### Step 6: Consider the case where \( \Delta r \) is very small If \( \Delta r \) is very small, the term \( 4\pi (\Delta r)^2 S \) becomes negligible compared to \( 8\pi r \Delta r S \). Therefore, we can approximate: \[ \Delta E \approx 8\pi r \Delta r S \] ### Conclusion The corresponding increase in the surface energy when the radius of the spherical liquid drop increases from \( r \) to \( r + \Delta r \) is: \[ \Delta E \approx 8\pi r \Delta r S \] ---

To find the increase in surface energy when the radius of a spherical liquid drop increases from \( r \) to \( r + \Delta r \), we can follow these steps: ### Step 1: Understand the formula for surface energy The surface energy \( E \) of a spherical drop is given by the formula: \[ E = 4\pi r^2 S \] where \( S \) is the surface tension and \( r \) is the radius of the drop. ...
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