Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density <math> <mi>d</mi> </math> . The area of the base of both vessels is <math> <mi>S</mi> </math> but the height of liquid in one vessel is <math> <mi>x</mi> <mn>1</mn> </math> and in the other, <math> <mi>x</mi> <mn>2</mn> </math> . When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is :

Sol. Uinitial =k (4q) (q) (d/2)+k (q) (-q) (d/2)⇒6kq2d⇒Ufinal =4 (4q) (q)3d2+k (q) (-q) (d/2)⇒23kq2d⇒ΔU=23-6kq2d⇒-163kq2d

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density <math> <mi>d</mi> </math> . The area of the base of both vessels is <math> <mi>S</mi> </math> but the height of liquid in one vessel is <math> <mi>x</mi> <mn>1</mn> </math> and in the other, <math> <mi>x</mi> <mn>2</mn> </math> . When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is : &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

<math> <mi mathvariant="normal">g</mi> <mi mathvariant="normal">d</mi> <mi mathvariant="normal">S</mi> <mo>⁡</mo> <mfenced separators="|"> <mrow> <mrow> <msubsup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <mo>+</mo> <msubsup> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> </mrow> </mrow> </mfenced> </math> (Fluids)

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Physics Class 12th Physics Mechanical Properties of Fluids Physics NCERT Exemplar Solutions Class 12th Chapter Ten

Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $d$ . The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x 1$ and in the other, $x 2$ . When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is :

Option 1 -

$\frac{3}{4} g d S {(x_{2} - x_{1})}^{2}$

Option 2 -

$g d S {(x_{2} + x_{1})}^{2}$

Option 3 -

$\frac{1}{4} g d S {(x_{2} - x_{1})}^{2}$

Option 4 -

$g d S (x_{2}^{2} + x_{1}^{2})$ (Fluids)

3 Views | Posted 5 months ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

5 months ago

Correct Option - 1

Detailed Solution:

Sol. $U_{initial} = \frac{k (4 q) (q)}{(d / 2)} + \frac{k (q) (- q)}{(d / 2)}$

$\begin{array}{r} \Rightarrow \frac{6 k q^{2}}{d} \\ \Rightarrow U_{final} = \frac{4 (4 q) (q)}{(\frac{3 d}{2})} + \frac{k (q) (- q)}{(d / 2)} \\ \Rightarrow \frac{2}{3} \frac{k q^{2}}{d} \\ \Rightarrow Δ U = (\frac{2}{3} - 6) \frac{k q^{2}}{d} \Rightarrow \frac{- 16}{3} \frac{{k q}^{2}}{d} \end{array}$

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