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A cylinder of height 20 m is completely filled with water. The velocity of efflux of water (in m/s) through a small hole on the side wall of the cylinder near its bottom is





(b) v=2gh=2×10×20=20 m/s

There is a hole in the bottom of tank having water. If total pressure at bottom is 3 atm (1 atm=105 N/m2) then the velocity of water flowing from hole is 





(a) Pressure at the bottom of tank P=hρg=3×105Nm2 

Pressure due to lipid column PI=3×105-1×105=2×105

and velocity of water v=2gh

v=2PIρ=2×2×105103=400 m/s

A cylindrical tank has a hole of 1 cm2 in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of 70 cm3/sec. then the maximum height up to which water can rise in the tank is





(a) The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second.
Volume of water flowing out per second

=Av=A2gh                             ....(i) 

Volume of water flowing in per second

=70 cm3/sec                             .....(ii)

From (i) and (ii) we get

A2gh=701×2gh=701×2×980×h=70h=49001960=2.5 cm

A square plate of 0.1 m side moves parallel to a second plate with a velocity of 0.1 m/s, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise, distance between the plates in m is





(d) 

A=(0.1)2=0.01 m2η=0.01      Poise=0.001 decapoise   (M.K.S. unit)dv=0.1 m/s and F=0.002 NF=ηAdvdxdx=ηAdvF=0.001×0.01×0.10.002=0.0005 m                                    

Spherical balls of radius 'r' are falling in a viscous fluid of viscosity 'η' with a velocity 'v'. The retarding viscous force acting on the spherical ball is 





(b) F=6πηrv

A small sphere of mass m is dropped from a great height. After it has fallen 100 m, it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100 m of fall is





(b) In the first 100 m body starts from rest and its velocity goes on increasing and after 100 m it acquire maximum velocity (terminal velocity). Further, air friction i.e. viscous force which is proportional to velocity is low in the beginning and maximum at v=vT.
Hence work done against air friction in the first 100 m is less than the work done in next 100 m.

Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be 






(c) If two drops of same radius r coalesce then radius of new drop is given by R

43πR3=43πr3+43πr3R3=2r3R=21/3r

If drop of radius r is falling in viscous medium then it acquire a critical velocity v and vr2

v2v1=Rr2=21/3r2v2=22/3×v1=22/3×5=5×41/3 cm/s

The rate of steady volume flow of water through a capillary tube of length 'l' and radius 'r' under a pressure difference of P is V. This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P)





(b) Rate of flow of liquid V=PR

where liquid resistance R=8ηlπr4

For another liquid resistance

R'=8ηlπr24=8ηlπr4.16=16R

For the series combination

VNew=PR+R'=PR+16R=P17R=V17

 

A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is





 (d) From

V=Pπr48ηlP=V8ηlπr4P2P1=V2V1×l2l1×r1r24=2×2×124=14P2=P14=P4

We have two (narrow) capillary tubes T1 and T2. Their lengths are l1 and l2 and radii of cross-section are r1 and r2 respectively. The rate of flow of water under a pressure difference P through tube T1 is 8cm3/sec. If l1 = 2l2 and r1 =r2, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)





(b) V=Pπr48ηlP=V8ηlπr4

For composite tube 

V1=Pπr48ηl+l2=23πPr48ηl=23×8=163cm3sec l1=l=2l2 or l2=l2

The Reynolds number of a flow is the ratio of





(c) Re=evdηor Re=ev2d×AA×v×ηRe=ev2AηAv/d

Here, ηAv/d=viscous force and ev2A=Inertial force

Hence, Re=Inertial forceviscous force

Water is flowing through a tube of non-uniform cross-section ratio of the radius at entry and exit end of the pipe is 3 : 2. Then the ratio of velocities at entry and exit of liquid is -





(a) If velocities of water at entry and exit points are v1 and v2, then according to equation of continuity,

A1V1=A2V2 V1V2 =A2A1 =r2r12=232=49

 

An application of Bernoulli's equation for fluid flow is found in 





Bernoulli's equation relates the pressure, velocity, and height of a flowing fluid. It is used to explain the dynamic lift of an aeroplane, where the faster air flow over the curved upper surface creates a lower pressure, resulting in an upward force on the wing.

The Working of an atomizer depends upon





(a)

Atomizer works on the principle of liquid flow, It follows bernaulis principle because here, horizontal air passing causes the vertices liquid to come out.

The pans of a physical balance are in equilibrium. Air is blown under the right hand pan; then the right hand pan will





(b) According to Bernoulli's theorem.