Gujarati
9-1.Fluid Mechanics
normal

$\rho$ घनत्व के पानी में $R$ त्रिज्या का एक बुलवुला वेग $v$ से एकसमान रूप से प्रसारित हो रहा है। पानी को असंपीड्य मानते हुए बुलबुले द्वारा विस्थापित (pushed) पानी की गतिज ऊर्जा क्या होगी ?

A

zero

B

$2 \pi \rho R^{3} v^{2}$

C

$2 \pi \rho R^{3} v^{2} / 3$

D

$4 \pi \rho R^{3} v^{2} / 3$

(KVPY-2019)

Solution

$(b)$ Let bubble of radius $x$ expands and pushes water of density $\rho$ by a distance $d x$. Then, kinetic energy of water of strip $d x$ is

$d K=\frac{1}{2} d m \cdot v_{x}^{2} \Rightarrow d K=\frac{1}{2} 4 \pi x^{2} \cdot d x \cdot \rho \cdot v_{x}^{2}$

or $d K=2 \pi R^{2} \cdot \rho \cdot v_{x}^{2} \cdot d x \quad \ldots$ $(i)$

Considering laminar flow, for section $1$ and $2$ shown in figure, by equation of continuity

$A_{1} v_{1}=A_{2} v_{2} \Rightarrow 4 \pi x^{2} v_{x}=4 \pi R^{2} v$

$\Rightarrow \quad v_{x} =\frac{R^{2}}{x^{2}} \cdot v\quad \ldots$ $(ii)$

From Eqs. $(i)$ and $(ii)$, we have

$d K=2 \pi x^{2} \rho\left(\frac{R^{2}}{x^{2}} \cdot v \cdot\right)^{2} d x$

$\Rightarrow d K=2 \pi R^{4} \rho v^{2} \cdot\left(\frac{d x}{x^{2}}\right)$

So, kinetic energy of complete volume of water,

$K=\int \limits_{R}^{\infty} d K=2 \pi R^{4} \rho v^{2} \int \limits_{R}^{\infty} \frac{d x}{x^{2}}$

$\Rightarrow \quad K=2 \pi R^{4} \cdot \rho \cdot v\left[-\frac{1}{x}\right]_{R}^{\infty}$

$=2 \pi \rho R^{3} v^{2}$

Standard 11
Physics

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