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Posted by Rohan Jonko 5 years, 7 months ago
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Harender Singh 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
Coefficient of viscosity is defined as tangential force required to maintain a unit velocity gradient between two parallel layers of liquid of unit area.
>Mathematically,
Coefficient of viscosity (η)= Fr/Av where F = tangential force, r = distance between the layers , v = velocity.
Dimensional Formula of Force = M1L1T-2
Dimensional Formula of Area= M0L2T0
Dimensional Formula of distance= M0L1T0
Dimensional Formula of velocity= M0L1T-1
Putting these values in above equation we get,
[η]= [M1L1T-2][M0L1T0] / [M0L2T0] [M0L1T-1] = [M1L-1T-1]
Dimensional Formula of Coefficient of viscosity (η)=[M1L-1T-1]
SI unit of Coefficient of viscosity (η) is Pascal-second
Posted by Binod Kumar 5 years, 7 months ago
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Medical applications
- Ultrasonic waves are used to capture the images of the heart and its features and problems associated with the heart, if any (for treatment purpose). This is called echocardiography.
- Ultrasound scans/Ultrasonography are very commonly used to get images of internal body organs such as liver, kidney, uterus. It helps the doctor to diagnose and treat problems in the body of the patient. In this method, ultrasound waves are made to travel through the body to the organ under consideration. They get reflected and these are converted to electrical signals which can be monitored on a screen.
- Ultrasonography is used to observe the growth of the fetus inside the uterus. It can also used to monitor the abnormalities.
- It is used to break kidney stones into fine grains which later get flushed out through urine.
Industrial Applications
- Ultrasound is used to clean machine parts located in places which are not easily accesible - electronic components, internal parts, spiral parts etc. The objects are placed in a cleaning solution and waves are passed through it. Because of the high frequency of the waves, the dust particles just fall out.
- It is used to detect defects, flaws, cracks in machine parts, bridges, building etc. Ultrasound waves are passed at one end and monitored using detectors. If there are laws or cracks, then the ultrasound waves are reflected back indicating the presence of a defect.
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Ayushi Ayushi 5 years, 7 months ago
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Posted by Mr Tom Pawara 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
The dimensional formula of density is given by,
[M1 L-3 T0]
Where,
- M = Mass
- L = Length
- T = Time
Posted by Jishan Mansuri 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
as,1 Atm = 101325 pascals,
while 1 dyn/cm2 = 0.1 pascals.
so,
1 Atm = 1013250 dyne/cm2
Posted by Motiram Choudhary 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
Work energy theorem states that the change in kinetic energy of an object is equal to the net work done on it by the net force.Let us suppose that a body is initially at rest and a force F is applied on the body to displace it through dS along the direction of the force. Then, small amount of work done is given by
dW = FdS
Also, according to Newton's second law of motion, we have
F ma
where a is acceleration produced (in the direction of force) on applying the force. Therefore,
dW = MadS = M dv/st dS
Or dW = M Ds/dt dv = Mvdv
Now, work done by the force in order to increase its velocity from u (initial velocity) to v (final velocity) is given by
W = 1/2 mV2 - 1/2 Mu2
Hence, work done on a body by a force is equal to the change in its kinetic energy.
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Gaurav Seth 5 years, 7 months ago
A rigid body performs a pure rotational motion, if each particle of the body moves in a circle, and the centre of all the circles lie on a straight line called the axes of rotation.
Posted by Syed Arsalan 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
The porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through. Thus they play an important role in facilitated diffusion.
Posted by Bhaven Kumar 5 years, 7 months ago
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Yogita Ingle 5 years, 7 months ago
- Consider the fluid initially lying between B and D. In an infinitesimal timeinterval Δt, this fluid would have moved.
- Suppose v1= speed at B and v2= speedat D, initial distance moved by fluid from to C=v1Δt.
- In the same interval Δtfluid distance moved by D to E = v2Δt.
- P1= Pressureat A1, P2=Pressure at A2.
- Work done on the fluid atleft end (BC) W1 = P1A1(v1Δt).
- Work done by the fluid at the other end (DE)W2 = P2A2(v2Δt)
- Net work done on the fluid is W1 – W2 = (P1A1v1Δt− P2A2v2Δt)
- By the Equation of continuity Av=constant.
- P1A1 v1Δt - P2A2v2Δt where A1v1Δt =P1ΔV and A2v2Δt = P2ΔV.
- Therefore Work done = (P1− P2) ΔVequation (a)
- Part of this work goes in changing Kinetic energy, ΔK = (½)m (v22 – v12) and part in gravitational potential energy,ΔU =mg (h2 − h1).
- The total change in energy ΔE= ΔK +ΔU = (½) m (v22 – v12) + mg (h2 − h1). (i)
- Density of the fluid ρ =m/V or m=ρV
- Therefore in small interval of time Δt, small change in mass Δm
- Δm=ρΔV (ii)
- Putting the value from equation (ii) to (i)
- ΔE = 1/2 ρΔV (v22 – v12) + ρgΔV (h2 − h1) equation(b)
- By using work-energy theorem: W = ΔE
- From (a) and (b)
- (P1-P2) ΔV =(1/2) ρΔV (v22 – v12) + ρgΔV (h2 − h1)
- P1-P2 = 1/2ρv22 - 1/2ρv12+ρgh2 -ρgh1(By cancelling ΔV from both the sides).
- After rearranging we get,P1 + (1/2) ρ v12 + ρg h1 = (1/2) ρ v22 + ρg h2
- P+(1/2) ρv2+ρg h = constant.
- This is the Bernoulli’s equation.
Posted by Mia Shekar 5 years, 7 months ago
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Mr. Singh 5 years, 7 months ago
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