# 7 - Fluid in motion Questions Answers

**Asked By: KANDUKURI ASHISH**

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**Joshi sir comment**

By Bernoullis theorem, velocity of water jet = $\sqrt{2gh}$Water jet comes out perpendicular to surface.

make its horizontal and vertical components

By vertical component, calculate time, then use it with

horizontal velocity component to get distance along x axis

Finally calculate distance from centre, along x axis

Inform for any confusion

can the linear speed of two objects be different if they have the same angular speed?

**Asked By: VIBHU GOEL**

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**Joshi sir comment**

angular speed depends on the distance from pivot also so v may be same or different

formula affilated is v = r ω

THE TERMINAL SPEED ATTAINED BY AN Al SPERE OF RADIUS 1mm FALLING THROUGH H2O AT 20^{0}C WILL BE CLOSE TO ??

ANS 4.6 m/s

(assume laminar flow , specific gravity of Al=2.7 & η H2O =8X 10^{-4 }PL.)

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

v = 2r^{2}g(D-d)/9η

so v = 2(.001)^{2}*9.8(2700-1000)/9*8*10^{-5}

solve it

Q IF IN THE SAME QUESTn OF BAROMETER THE ELEVATOR IS ACCELERATING UPWARD READS 76cm. THEN AIR PRESSURE IN ELEVATOR IS >76cm PLS EXPLAIN THE CONCEPT BEHIND SUCH CASES.

**Asked By: SARIKA**

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**Joshi sir comment**

when elevator will move in upward direction the extra Hg level will feel a down force and given that in movable condition it is 76 cm. Hg so really it will be more than 76 cm.

#### There is some amount of water in the beaker, and a vertical rod is passing through the centre of the circular base of the cylindrical beaker which is rotated with angular velocity ω , then water takes a particular shape(miniscus is formed). Find the equation of the miniscus of water when the mid point of the miniscus just touches the bottom surface of the beaker?

**Asked By: AMIT DAS**

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**Joshi sir comment**

after making the diagram, complete the miniscus as a sphere, then following conditions will be obtained

1) h = R(1-cosθ) here h, R and θ are height of miniscus, radius of the sphere and contact angle.

2) R = r/cosθ, r is the radius of beaker

3) tanθ = g/rw^{2}

4) πr^{2}hρ - πR^{3}ρcos^{2}θ = 2πrTcosθ

There are 4 equations, remove R, θ and T and get a relation between h and r in terms of g, w, ρ and π