# 8 - Fluid in motion Questions Answers

A man travels one third part of his total distance with speed v1 and rest two third part with speed v2.find out average speed of man

**Asked By: GAURAV MANSINGHANI**4 year ago

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A cylinder of radius r full of liquid of density rho is rotated about its axis at w rad/s. What will be the increase in pressure at the centre of the cylinder ?

**Asked By: SAYANTANI DATTA**4 year ago

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

**Asked By: VIBHU GOEL**6 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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**7 year ago

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**Answer Strategies and trick by Manish sir**(it will help you to solve it by yourself)

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