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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+43πr3⇒R3=2r3⇒R=21/3r

If drop of radius r is falling in viscous medium then it acquire a critical velocity v and v∝r2

v2v1=Rr2=21/3r2⇒v2=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=Pπr48ηl⇒P=V8ηlπr4⇒P2P1=V2V1×l2l1×r1r24=2×2×124=14⇒P2=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=Pπr48ηl⇒P=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×η⇒Re=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.

According to Bernoulli's equation

Pρg+h+12v2g=constant

The terms A, B and C are generally called respectively:





(c) 

A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasoline. The tank was sealed at 3.10 atm. The stored gasoline has a density of 660 kgm-3. The velocity with which gasoline begins to shoot out of the hole is





The velocity of the gasoline shooting out of the hole can be calculated using Bernoulli's equation. The pressure difference between the inside of the tank and the atmospheric pressure provides the driving force for the flow, and the velocity depends on the height of the gasoline column.

A rectangular vessel when full of water takes 10 minutes to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water





(b) Time taken to be emptied for h height, t=2hg

and for h2 height , t=2h/2g=hg∴t't=12⇒t'=t2=102=7 minute 

A streamlined body falls through air from a height h on the surface of a liquid. If d and D(D > d) represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is





(d) Upthrust – weight of body = apparent weight

VDg-Vdg = Vda,

Where a = retardation of body ∴a=D-ddg

The velocity gained after fall from h height in air,

v=2gh

Hence, time to come in rest,

t=va=2gh×d(D-d)g=2hg×d(D-d)

A large tank of cross-section area A is filled with water to a height H. A small hole of area 'a' is made at the base of the tank. It takes time T1 to decrease the height of water to Hη(η>1) ; and it takes T2 time to take out the rest of water. If T1=T2, then the value of η is





(c) t=Aa2gH1-H2Now, T1=Aa2gH-Hηand T2=Aa2gHη-0

According to problem 

T1=T2∴H-Hη=Hη-0⇒H=2Hη⇒η=4