Two mercury droplets of radii 0.1 cm. and 0.2 cm. collapse into one single drop. What amount of energy is released? The surface tension of mercury T= 435.5 × 10^-3 Nm^-1 .

This is a short answer type question as classified in NCERT Exemplar

Consider the diagram

Radii of mercury droplets  r₁=0.1cm = 1 $\times {10}^{-3}m$

r₂=0.2cm=2 $\times {10}^{-3}m$

Surface tension T= 435.5 × 10^-3 Nm^-1

Let the radius of the big drop formed by collapsing be R

Therefore,

volume of big drop = volume of small droplets

4/3 $\pi$ R³=4/3 $\pi r_1^3+$ 4/3 $\pi r_2^3$

R³= $r_1^3+r_2^3$

= 0.1³+0.2³

= 0.001+0.008

=0.009

R= 0.21 cm = 2.1 $\times {10}^{-3}m$

Change in surface area $?A$ = 4 $\pi R^2-(4\pi r_1^2+4\pi r_2^2)$

=4 $\pi (R^2-(r_1^2+r_2^2\left)\right)$

Energy released =T $?A$

= T $\times 4\pi \left[\right(R^2-(r_1^2+r_2^2)\left)\right]$

= 435.5 $\times {10}^{-3}\times 4\times 3.14[2.1\times {10}^{-3}]$ 2-(1 $\times {10}^{-6}+4\times {10}^{-6}$ )

= 435.5 $\times 4\times 3.14\left[4.41-5\right]\times {10}^{-6}\times {10}^{-3}$

= -32.23 $\times {10}^{-7}$

Therefore -3.22 $\times {10}^{-6}J$ energy will be absorbed.

<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>Consider the diagram</p><p>Radii of mercury droplets&nbsp; r₁=0.1cm = 1<span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> </span></p><p>r₂=0.2cm=2<span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> </span></p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745474399phpWHzozo_480x360.jpeg" media="(max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745474399phpWHzozo.jpeg" alt="" width="202" height="171"></picture></div></div><p>Surface tension T= 435.5 × 10^-3 Nm^-1</p><p>Let the radius of the big drop formed by collapsing be R</p><p>Therefore,</p><p>volume of big drop = volume of small droplets</p><p>4/3<span contenteditable="false"> <math> <mi>π</mi> </math> </span>R³=4/3<span contenteditable="false"> <math> <mi>π</mi> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msubsup> <mo>+</mo> </math> </span>4/3<span contenteditable="false"> <math> <mi>π</mi> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msubsup> </math> </span></p><p>R³=<span contenteditable="false"> <math> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msubsup> </math> </span></p><p>= 0.1³+0.2³</p><p>= 0.001+0.008</p><p>=0.009</p><p>R= 0.21 cm = 2.1<span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>3</mn> </mrow> </mrow> </msup> <mi>m</mi> </math> </span></p><p>Change in surface area <span contenteditable="false"> <math> <mo>?</mo> <mi>A</mi> </math> </span> = 4<span contenteditable="false"> <math> <mi>π</mi> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mo>(</mo> <mn>4</mn> <mi>π</mi> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <mn>4</mn> <mi>π</mi> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>)</mo> </math> </span></p><p>=4<span contenteditable="false"> <math> <mi>π</mi> <mo>(</mo> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>-</mo> <mo>(</mo> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>)</mo> <mo>)</mo> </math> </span></p><p>Energy released =T<span contenteditable="false"> <math> <mo>?</mo> <mi>A</mi> </math> </span></p><p>= T&nbsp;<math><mo>×</mo><mn>4</mn><mi>π</mi><mo>[</mo><mo>(</mo><msup><mrow><mrow><mi>R</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup><mo>-</mo><mo>(</mo><msubsup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msubsup><mo>+</mo><msubsup><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msubsup><mo>)</mo><mo>)</mo><mo>]</mo></math></p><p>= 435.5&nbsp;<math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup><mo>×</mo><mn>4</mn><mo>×</mo><mn>3.14</mn><mo>[</mo><mn>2.1</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup><mo>]</mo></math>&nbsp;2-(1&nbsp;<math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mo>+</mo><mn>4</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup></math>&nbsp;)</p><p>= 435.5&nbsp;<math><mo>×</mo><mn>4</mn><mo>×</mo><mn>3.14</mn><mfenced open="[" close="]" separators="|"><mrow><mrow><mn>4.41</mn><mo>-</mo><mn>5</mn></mrow></mrow></mfenced><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>6</mn></mrow></mrow></msup><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math></p><p>= -32.23&nbsp;<math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>7</mn></mrow></mrow></msup></math></p><p>Therefore -3.22 <span contenteditable="false"> <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>6</mn> </mrow> </mrow> </msup> <mi>J</mi> </math> </span>energy will be absorbed.</p>

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