A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075N/m. In this process the gain in surface energy will be

Option 1 -

2.8 * 10^-4J

Option 2 -

1.5 * 10^-3 J

Option 3 -

1.9 * 10^-4 J

Option 4 -

9.4 * 10^-5 J

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1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

Gain in surface energy, $Δ U = T Δ A$

from volume centenary, $\frac{4}{3} π R^{3} = 64 \times \frac{4}{3} π r^{3}$

$\Rightarrow r = \frac{R}{4}$

Initial surface area, Ai = 4pR²

final surface area, $A_{f} = 64 \times 4 π r^{2}$

$∴ Δ U = 12 π R^{2} . T = 0.075 \times 12 \times 3.14 \times 1 0^{- 4} = 2.8 \times 1 0^{- 4} J$

Gain in surface energy, <math> <mrow> <mi>Δ</mi> <mi>U</mi> <mo>=</mo> <mi>T</mi> <mi>Δ</mi> <mi>A</mi> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1813751775"> </o:OLEObject></xml><! [endif]-> from volume centenary, <math> <mrow> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>π</mi> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>=</mo> <mn>6</mn> <mn>4</mn> <mo>×</mo> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mi>π</mi> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1813751776"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mo>⇒</mo> <mi>r</mi> <mo>=</mo> <mfrac> <mrow> <mi>R</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math>                                           <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1813751777"> </o:OLEObject></xml><! [endif]-> Initial surface area, Ai = 4pR2final surface area, <math> <mrow> <msub> <mrow> <mi>A</mi> </mrow> <mrow> <mi>f</mi> </mrow> </msub> <mo>=</mo> <mn>6</mn> <mn>4</mn> <mo>×</mo> <mn>4</mn> <mi>π</mi> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1813751778"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mo>∴</mo> <mi>Δ</mi> <mi>U</mi> <mo>=</mo> <mn>1</mn> <mn>2</mn> <mi>π</mi> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>.</mo> <mi>T</mi> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>0</mn> <mn>7</mn> <mn>5</mn> <mo>×</mo> <mn>1</mn> <mn>2</mn> <mo>×</mo> <mn>3</mn> <mo>.</mo> <mn>1</mn> <mn>4</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> <mo>=</mo> <mn>2</mn> <mo>.</mo> <mn>8</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> <mi>J</mi> </mrow> </math>

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