"There is no seeing. Seeing is only being." Source Self is what you are, You are That Fathomlessness in which experience and concepts appear. Self is the Moment which has no coming or going. It is the Heart, Atman, Emptiness. It shines to Itself, by Itself, in Itself. Self is what gives breath to life, you need not search for It, It is Here. You are That through which you would search. You are what you are looking for! And That is all it is. Only Self is. You were never born, and though only desire takes birth, Nothing has ever happened, Nothing has ever existed! This Nothingness you are, and this is the Ultimate Truth. You are totally alone because Beauty alone is. Only Self is. You simply cannot deny that you are Consciousness. You dwell in the Lotus of the Heart as Joy in Bliss. Keep Quiet and you will reveal your Self to your Self. Self Knowledge is that which is worth sacrificing anything for, because everything else is just a mirage rising out of Cons...

Calculate the magnitude and direction of the terminal velocity of an air bubble of radius $1\;mm$ passing through an oil viscosity $0.2\;N\,s\, /m^2$ and specific gravity $0.9$ if the density of air is $1.29\;kg/m^3$ .

Given, Radius of the air bubble

$(r) = 1\;mm = 1 * 10^{-3}\;m$ Viscosity of the oil

$(\eta) = 0.2\; Ns/\;m^2$ Density of the air

$(\rho) = 1.92\;kg/m^3$ Density of the oil

$(\sigma) = 0.9 * 1000 = 900 \;kg/m^3$ Terminal velocity

$(v) = ?$ We have,

$v =$

$\frac{2\;r^2 (\rho \; - \; \sigma)\,g}{9\;\eta} = \frac{2 \; * \; (1 \; * \; 10^{-3})^2 \; * \;(1.26 \; - \; 900 ) \; * \; 10}{9 \; * \; 0.20}$

$= -0.0099 \; m/s$ The

$-\;ve$ sign indicates that the ball moves in the upward direction against the force of gravity. Return to Main Menu

Two drops of same liquid of same radius are falling through air with steady velocity of $2\;m/s$ . If the two drops coalesce what would be the terminal velocity?

Given, Terminal velocity of the each small raindrop

$(v) = 2 \; m/s$ Terminal velocity of the big raindrop

$(V) = ?$ Let

$r$ be the radius of the small rain droplets. Then the terminal velocity of each small raindrop is,

$v = \frac{2\,r^2 (\rho \; - \; \sigma)\,g}{9\; \eta}$ .......... (i) Let

$R$ be the radius of the combined drop. Then the terminal velocity of the large drop is,

$V = \frac{2\,R^2 (\rho \; - \; \sigma)\,g}{9\; \eta}$ .......... (ii) Since the volume of big raindrop = volume of eight small raindrops. Then

$\frac{4}{3}\; \pi\; R^3 = 2 *\;\frac{4}{3}\; \pi \; r^3$ ⇒

$R = 2^{1/3}\;r$ .......... (iii) Now, Dividing equation (ii) by (i), we get

$\frac{V}{v} = \frac{R^2}{r^2}$ .......... (iv) Using equation (iii) in equation (iv), then we get or,

$V = v * \frac{R^2}{r^2} = v * \frac{(2^{1/3} \; r)^2}{r^2} = 2 * 1.58 = 3.17 \;m/s$

$\therefore$ Terminal velocity of the big raindrop

$(V) = 3.17 \;m/s$ Return to Main ...

Water flows steadly through a horizontal pipe of non-uniform cross section. If the pressure of water is $4 * 10^6\;N/m^2$ at a point where the velocity of flow is $2\;m/s$ and cross section is $0.02\;m^2$ , what is the pressure at a point where cross-section reduces to $0.01\;m^2$ ?

Given, For one end of the pipe, For another end of the pipe, Pressure

$(P_1) = 4 * 10^4 \;Ns/m^2$ Pressure

$(P_2) = ?$ Velocity

$(v_1) = 2\;m/s$ Velocity

$(v_2) = ?$ Area

$(A_1) = 0.02\;m/s$ ...

Contact Form

Name

Email *

Message *