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Tuesday, 25 December 2012

P.No. 4.6:- 100meter cube of water is pumped from a reservoir into a tank, 10m higher than the reservoir, in 20 minutes. Find (a) The increase in P.E (b) The power delivered by the pump.


P.No.   4.6:- 100m³ of water is pumped from a reservoir into a tank, 10m higher than the reservoir, in 20 minutes. Find

       (a)  The increase in P.E

       (b)  The power delivered by the pump.

Solution:-

Given data:-

                 Volume of water   = V =     100m³

                  Height                  = h =      10m

                   Time taken          = t =      20min   =   20 × 60s   = 1200s

To determine:-

                          Potential energy =   P.E =?

                           Power                 =   P    =?

Calculation:-

 ( we know that work done is equal to change in energy it may be kinetic or potential but  in this case only P.E is involved. So,)

                 Work done = Increase in P.E     

                                    = mgh

                To determine mass ‘m’ we first take

                     Density = mass/volume

   Or,                 ρ = m/V



Sunday, 23 December 2012

P.No.15.11:- A square coil of side 16cm has 200 turns and rotates in a uniform magnetic field of magnitude 0.05T.If the peak emf is 12V, what is the angular velocity of the coil.


P.No.15.11:- A square coil of side 16cm has 200 turns and rotates in a uniform magnetic field of magnitude 0.05T.If the peak emf is 12V, what is the angular velocity of the coil.
Solution:-
Given Data:-
          Number of turns                 =     N     =     200
          Magnetic field                     =      B      =    0.05T
          Side of square coil              =      l       =   16cm

P.No.15.10:- Like any field, the earth’s magnetic field stores energy. Find the magnetic energy stored in space where strength of earth’s field is 0.00007T, if the space occupies an area of 100000000 meter square and has a height of 750m.

P.No.15.10:-   Like any field, the earth’s magnetic field stores energy. Find the magnetic energy stored in space where strength of earth’s field is 0.00007T, if the space occupies an area of 100000000 meter square and has a height of 750m.
Solution:-
Given Data:-


Saturday, 22 December 2012

P.No.15.9:- When current through a coil changes from 100mA to 200mA in 0.005s, an induced emf of 40mV is produced in the coil (a) what is the inductance of the coil? (b) Find the increase in energy stored in the coil.

P.No.15.9:- When current through a coil changes from 100mA to 200mA in 0.005s, an induced emf of 40mV is produced in the coil (a) what is the inductance of the coil? (b) Find the increase in energy stored in the coil.
Solution:-
Given Data:-







P.No.15.8:- A solenoid of length 8.0cm and cross section area 0.5 cm square has 520 turns. Find the self inductance of the solenoid when the core is air. If the current in the solenoid increases through 1.5A in 0.2s, find the magnitude of induced emf in it.


P.No.15.8:- A solenoid of length 8.0cm and cross section area 0.5  cm square has 520 turns. Find the self inductance of the solenoid when the core is air. If the current in the solenoid increases through 1.5A in 0.2s, find the magnitude of induced emf in it.
Solution:-







P.No.15.7:- A solenoid has 250 turns and its self inductance is 2.4mH. What is the flux through each turn when the current is 2A? What is the induced emf when the current changes at 20A/s? Solution:-



P.No.15.7:- A solenoid has 250 turns and its self inductance is 2.4mH. What is the flux through each turn when the current is 2A? What is the induced emf when the current changes at 20A/s?
Solution:-
Given Data:-
           No. of turns                          =     N       =      250
           Current                                 =      I       =     2A








P.No.15.6:- A pair of adjacent coils has a mutual inductance of 0.75H. If the current in the primary changes from 0 to 10A in 0.025s, what is the average induced emf in the secondary? What is the change in flux in it if the secondary has 500 turns?


P.No.15.6:- A pair of adjacent coils has a mutual inductance of 0.75H. If the current in the primary changes from 0 to 10A in 0.025s, what is the average induced emf in the secondary? What is the change in flux in it if the secondary has 500 turns?
Solution:-
Given Data:-
            Mutual inductance                             =     M      =   0.75H

P.No.15.5:- Two coils are placed side by side. An emf of 0.8V is observed in one coil when the current is changing at the rate of 200 A/s in the other coil. What is the mutual inductance of the coils?





P.No.15.5:- Two coils are placed side by side. An emf of 0.8V is observed in one coil when the current is changing at the rate of 200 A/s in the other coil. What is the mutual inductance of the coils?
Solution:-
Given data:-











P.No.15.4:- A circular coil has 15 turns of radius 2cm each. The plane of the coil lies at 40 degree to a uniform magnetic field of 0.2 T. If the field is increased to 0.5 T in 0.2s, find the magnitude of the induced emf.


P.No.15.4:- A circular coil has 15 turns of radius 2cm each. The plane of the coil lies at 40 degree  to a uniform magnetic field of 0.2 T. If the field is increased to 0.5 T in 0.2s, find the magnitude of the induced emf.



P.No.15.3:- A coil of wire has 10 loops. Each loop has an area of 0.0015 meter spuare. A magnetic field is perpendicular to the surface of each loop at all times. If the magnetic field is changed from 0.05T to 0.06T in 0.1s, find the average emf induced in the coil during this time.


P.No.15.3:- A coil of wire has 10 loops. Each loop has an area of  0.0015 meter spuare. A magnetic field is perpendicular to the surface of each loop at all times. If the magnetic field is changed from 0.05T to 0.06T in 0.1s, find the average emf induced in the coil during this time.


P.No.15.2:- The flux density B in a region between the pole faces of a horse shoe magnet is 0.5 Wb per meter squaredirected vertically downward. Find the emf induced in a straight wire 5.0cm long, perpendicular to B when it is moved in a direction at an angle of 60 degree with the horizontal with a speed of 100cm/s?




Problem No.15.1:- An emf of 0.45V is induced between the ends of a metal bar moving through a magnetic field of 0.22T. What field strength would be needed to produce an emf of 1.5V between the ends of the bar, assuming that all other factors remain the same?


Problem No.15.1:- An emf of 0.45V is induced between the ends of a metal bar moving through a magnetic field of 0.22T. What field strength would be needed to produce an emf of 1.5V between the ends of the bar, assuming that all other factors remain the same?
Solution:-
Given Data:-





Sunday, 16 December 2012


P.No.   4.6:- 100m³ of water is pumped from a reservoir into a tank, 10m higher than the reservoir, in 20 minutes. Find
       (a)  The increase in P.E
       (b)  The power delivered by the pump.
Solution:-
Given data:-
                 Volume of water   = V =     100m³
                  Height                  = h =      10m
                   Time taken          = t =      20min   =   20 × 60s   = 1200s
To determine:-
                          Potential energy =   P.E =?
                           Power                 =   P    =?
Calculation:-
 ( we know that work done is equal to change in energy it may be kinetic or potential but  in this case only P.E is involved. So,)
                 Work done = Increase in P.E     
                                    = mgh
                To determine mass ‘m’ we first take
                     Density = mass/volume
   Or,                 ρ = m/V

Q.No.6.12:- In an orbiting space station, would the blood pressure in major arteries in the leg ever be greater than the blood pressure in major arteries in the neck?


Q.No.6.12:- In an orbiting space station, would the blood pressure in major arteries in the leg ever be greater than the blood pressure in major arteries in the neck?
Ans:- Pressure is just like a force so to overcome a force we should exert a force which is  equal but opposite. In case of orbiting space station where is the condition of weightlessness, so the blood pressure will be equal in all parts of body

Q.No.6.11:- For which position will the maximum blood pressure in the body have the smallest value: (a) Standing up right (b) Sitting (c) Lying horizontally (d) Standing on one’s head?


Q.No.6.11:- For which position will the maximum blood pressure in the body have the smallest value: (a) Standing up right (b) Sitting (c) Lying horizontally (d) Standing on one’s head?
Ans:- When a man lying horizontally then in this case the heart will work quiet small. Consider the lying position, in this case all the parts are in level with the heart. When the blood circulates then every part of blood vessels make an angle of 90 degree with the earth surface. And the cos90 is zero(0) so the work done by heart against gravity  becomes zero( or approximately negligible).

Q.No. 6.10:- Explain the working of carburetors of a motorcar using Bernoulli’s equation.


Q.No. 6.10:- Explain the working of carburetors of a motorcar using Bernoulli’s equation.

Ans:- The carburetor of a car engine uses a Venturi duct to feed the correct mix of air and petrol to the cylinders. Air is drawn through the duct and along a pipe to the cylinders. A tiny inlet at the side of the duct is fed with petrol. The air through the duct moves very fast, creating low pressure in the duct, which draws petrol vapors into the air stream.
Extra Elucidation
    A venturi is a tube with a convex taper, (one end wider than the other). As air enters the wider end it's squeezed into the narrower section of the tube, lowering the air's pressure. The area of lowest pressure is just past the narrowest point and is called the depression. This has always seemed
counterintuitive to me, but Bernoulli's Principle outlines the fluid dynamics
involved in this effect. This lowered pressure, or comparative vacuum is separate from the engine vacuum. A variable venturi varies the venturi diameter at the depression 
by raising or lowering an obstruction. This obstruction is called a slide. On a CV the slide 
is called a piston or diaphragm valve.


Q.No. 6.9:- Explain how the swing is produced in a fast moving cricket ball?


Q.No. 6.9:- Explain how the swing is produced in a fast moving cricket

ball?
Ans:-  When a fast moving cricket ball moves in such a way that it spins as well as moves forward, then the air deflects due to the spin and unsmooth surface of ball the stream lines of air becomes closer to each other and results the increase of air speed and low pressure. On one side of ball there exists low pressure while on other side high. This change in pressure will cause the ball to lift. This will give an extra curvature to the ball known as swing, which deceives the batsman.

Extra Elucidation
Often Bernoulli's principle is used to explain the topspin effect, as the difference in speed between ball surface and air is greater on the top of the ball. For example, if the air flowing past the bottom of the ball is moving faster than the air flowing past the top then Bernoulli's principle implies that the pressure on the surfaces of the ball will be lower below than above. In other words, since there is more air friction occurring on the top surface of the ball compared to the bottom, this differential causes a greater pressure to be applied on the top of the ball, resulting in the ball being pushed down.
In ball sports, topspin is a property of a ball that rotates as if rolling in the same direction as it is moving. Topspin on a shot imparts a downward force that causes the ball to drop, due to its interaction with the air. It can be generated by hitting the ball with an up-and-forward swing, with the racquet facing below the direction it is moving. A topspin shot is the opposite of the slice; topspin itself is the opposite of backspin


Q.No. 6.8:- Two rowboats moving parallel in the same direction are pulled towards each other. Explain.


Q.No. 6.8:- Two rowboats moving parallel in the same direction are pulled towards each other. Explain.

Ans:- When two boats moving parallel in the same direction, the water between them is also dragged along with them. When water dragged then the speed of water will be fast, than the water on the other sides of the boats. Hence the pressure will be low where

the speed is high and vice versa.

    So, the high pressure at the other sides will pull the boats to the side where the pressure is low. Hence as a result the boats will pull toward each other.

Q.No. 6.7:- Identify the correct answer. What do you infer from Bernoulli’s theorem? (i)Where the speed of the fluid is high the pressure will be low. (ii)Where the speed of the fluid is high the pressure is also high. (iii)This theorem is valid only for turbulent flow of the liquid.


Q.No. 6.7:- Identify the correct answer. What do you infer from Bernoulli’s theorem?
(i)Where the speed of the fluid is high the pressure will be low.
(ii)Where the speed of the fluid is high the pressure is also high.
(iii)This theorem is valid only for turbulent flow of the liquid.
Ans:- The  correct statement is (i) where the speed of the fluid is high the pressure will be low.

Q.No. 6.6:- A person is standing near a fast moving train. Is there any danger that he will fall towards it?


Q.No. 6.6:- A person is standing near a fast moving train. Is there any danger that he will fall towards it?
Ans:- From the application of Bernoulli’s equation we have come to know that the place where the speed of  fluid is high there the pressure will be low and vice versa.
So, a person standing near a fast moving train will obviously in danger. Because when a train moving fast then speed of air near the train will be high and away from the person will be low. So this pressure difference will pull the man toward the fast moving train.

Q.No.6.5:- State Bernoulli’s relation for a liquid in motion and describe some of its applications.


Q.No.6.5:- State Bernoulli’s relation for a liquid in motion and describe some of its applications.

Ans:- Bernoulli’s equation is the fundamental  equation in fluid dynamics which states that the sum of pressure and kinetic and potential energies per unit volume is a steady flow of an incompressible and non viscous liquid has the same value.







Where” ρ” is the density of the fluid and “g” is the acceleration due to gravity.

Some of the applications of Bernoulli’s equation are

(i)                  Lift in an airplane.

(ii)                 Swing of a cricket ball

(iii)               Working of a filter pump

(iv)              Measurement of speed of liquid or gas flow through a pipe.

Q.No.6.4:- Explain the difference between laminar flow and turbulent flow.


Q.No.6.4:- Explain the difference between laminar flow and turbulent flow.

Ans:-  

  Laminar Flow:-  If every particle of the fluid that passes a point moves along the same path, as followed by particles, which passed that point earlier is called laminar flow.

Turbulent Flow:- The irregular or unsteady flow of the fluid is called turbulent flow. In this case the exact path of the particles of the fluid cannot by predicted as the velocity of the fluid may change abruptly.

Q.No.6.3:- Why fog droplets appear to the suspended in air?


Q.No.6.3:- Why fog droplets appear to the suspended in air?

Ans:-    Let suppose that the fog droplet is just like a small sphere. So when it flows through the air (which is fluid) then droplet will experience a retarding force of the magnitude 

As the droplet is very small so its radius is small and as a whole the drag force will be small. The drag force will be equal to its weight which is also very small. So the drop let will soon attains a terminal velocity which is



  

Q.No.6.2:- What is meant by drag force? What are the factors upon which drag force acting upon a small sphere of radius r, moving down through a liquid, depend?


Q.No.6.2:- What is meant by drag force? What are the factors upon which drag force acting upon a small sphere of radius r, moving down through a liquid, depend?

Ans:-   When an object moving through a fluid then it will experience a  retarding(opposing) force called drag force. This force increase as the speed of the object increases in the fluid. The drag force ‘F’ is given by ‘Stockes Law’




The factors upon which drag force depends are:

(i)                  Speed of the sphere

(ii)                 Radius of the sphere

(iii)               Coefficient of viscosity



Q.No.6.1:- Explain what do you understand by the term viscosity?

Q.No.6.1:- Explain what do you understand by the term viscosity?
Ans:-      From the microscopic study of flowing fluid we know that fluid flows when the layers of fluid slid over on another. But there exists another opposing force between the layers of fluid which wants to stop this sliding. This force is called the viscosity.
    In short we can say that viscosity is the opposing force (frictional effects between the relative motion of layers of fluid) between the layers of fluid which will slow down the flow of fluid or stop it if it is very large.

P.No. 4.6:- 100m³ of water is pumped from a reservoir into a tank, 10m higher than the reservoir, in 20 minutes. Find (a) The increase in P.E (b) The power delivered by the pump.

P.No.    4.6:- 100m³ of water is pumped from a reservoir into a tank, 10m higher than the reservoir, in 20 minutes. Find (a) The increase in P.E (b) The power delivered by the pump.  
Solution:-
Given data:-
                 Volume of water   = V =     100m³
                  Height                  = h =      10m
                   Time taken          = t =      20min   =   20 × 60s   = 1200s
To determine:-
                          Potential energy =   P.E =?
                           Power                 =   P    =?
Calculation:-
 ( we know that work done is equal to change in energy it may be kinetic or potential but  in this case only P.E is involved. So,)
                 Work done = Increase in P.E     
                                    = mgh
                To determine mass ‘m’ we first take
                     Density = mass/volume
   Or,                 ρ = m/V