Two cylindrical of equal cross-sectional are 16 cm² contain water upto height 100cm and 150 cm respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, it [Take, density of water = 10³ kg/m³ and g = 10 ms^-2]:

Correct Option - 2


Detailed Solution:

$\rho =10^{3}kg/m^3$

g = 10 m/s²

Final height in both vessels

$=\frac{100+150}{2}=125cm$

So, less in U = ${\displaystyle \int A\times 0.25\times g\times 0.25}$

$10^3\times 625\times 10^{-4}\times 10\times 16\times 10^{-4}$

= 625 × 16 × 104

= 1J.

<p><span contenteditable="false"> <math> <mrow> <mi>ρ</mi> <mo>=</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mi>k</mi> <mi>g</mi> <mo>/</mo> <msup> <mrow> <mi>m</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </math> </span></p><p>g = 10 m/s²</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753765715phpJISSAI.jpeg" alt="" width="338" height="110" loading="lazy"></picture></div></div><p>Final height in both vessels</p><p><span contenteditable="false"> <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>+</mo> <mn>1</mn> <mn>5</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mn>2</mn> <mn>5</mn> <mi>c</mi> <mi>m</mi> </mrow> </math> </span></p><p>So, less in U = <span contenteditable="false"> <math> <mrow> <mstyle displaystyle="true"> <mrow> <mo>∫</mo> <mrow> <mi>A</mi> <mo>×</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> <mo>×</mo> <mi>g</mi> <mo>×</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> </mrow> </mrow> </mstyle> </mrow> </math> </span></p><p><span contenteditable="false"> <math> <mrow> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>×</mo> <mn>6</mn> <mn>2</mn> <mn>5</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mo>×</mo> <mn>1</mn> <mn>6</mn> <mo>×</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>−</mo> <mn>4</mn> </mrow> </msup> </mrow> </math> </span></p><p>= 625 × 16 × 104</p><p>= 1J.</p>

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Correct Option - 2


Detailed Solution:

$?=10^{3}kg/m^3$

g = 10 m/s²

Final height in both vessels

$=\frac{100+150}{2}=125cm$

So, less in U = ${\displaystyle ?A*0.25*g*0.25}$

$10^3*625*10^{?4}*10*16*10^{?4}$

= 625 * 16 * 104

= 1J.

<p><span contenteditable="false"> <math> <mrow> <mi>? </mi> <mo>=</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mi>k</mi> <mi>g</mi> <mo>/</mo> <msup> <mrow> <mi>m</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </math> </span></p><p>g = 10 m/s²</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753765715phpJISSAI.jpeg" alt="" width="338" height="110" loading="lazy"></picture></div></div><p>Final height in both vessels</p><p><span contenteditable="false"> <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>+</mo> <mn>1</mn> <mn>5</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mn>2</mn> <mn>5</mn> <mi>c</mi> <mi>m</mi> </mrow> </math> </span></p><p>So, less in U = <span contenteditable="false"> <math> <mrow> <mstyle displaystyle="true"> <mrow> <mo>? </mo> <mrow> <mi>A</mi> <mo>*</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> <mo>*</mo> <mi>g</mi> <mo>*</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> </mrow> </mrow> </mstyle> </mrow> </math> </span></p><p><span contenteditable="false"> <math> <mrow> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>*</mo> <mn>6</mn> <mn>2</mn> <mn>5</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>4</mn> </mrow> </msup> <mo>*</mo> <mn>1</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <mn>6</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>4</mn> </mrow> </msup> </mrow> </math> </span></p><p>= 625 * 16 * 104</p><p>= 1J.</p>

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Correct Option - 2


Detailed Solution:

$?=10^{3}kg/m^3$

g = 10 m/s²

Final height in both vessels

$=\frac{100+150}{2}=125cm$

So, less in U = ${\displaystyle ?A*0.25*g*0.25}$

$10^3*625*10^{?4}*10*16*10^{?4}$

= 625 * 16 * 104

= 1J.

<p><span class="mathml" contenteditable="false"> <math> <mrow> <mi>? </mi> <mo>=</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mi>k</mi> <mi>g</mi> <mo>/</mo> <msup> <mrow> <mi>m</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </math> </span></p><p>g = 10 m/s²</p><div class="photo-widget-full"><div class="figure"><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753765715phpJISSAI.jpeg" alt="" width="338" height="110" loading="lazy"></picture></div></div><p>Final height in both vessels</p><p><span class="mathml" contenteditable="false"> <math> <mrow> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>+</mo> <mn>1</mn> <mn>5</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mn>2</mn> <mn>5</mn> <mi>c</mi> <mi>m</mi> </mrow> </math> </span></p><p>So, less in U = <span class="mathml" contenteditable="false"> <math> <mrow> <mstyle displaystyle="true"> <mrow> <mo>? </mo> <mrow> <mi>A</mi> <mo>*</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> <mo>*</mo> <mi>g</mi> <mo>*</mo> <mn>0</mn> <mo>.</mo> <mn>2</mn> <mn>5</mn> </mrow> </mrow> </mstyle> </mrow> </math> </span></p><p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>*</mo> <mn>6</mn> <mn>2</mn> <mn>5</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>4</mn> </mrow> </msup> <mo>*</mo> <mn>1</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <mn>6</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>4</mn> </mrow> </msup> </mrow> </math> </span></p><p>= 625 * 16 * 104</p><p>= 1J.</p>

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