XI Physics | Dot Product Solved Numericals

Q.15 Find the angle between A=2i + 2j – k and B=6i – 3j +2k. Q.16 Find the projection of the vector A= i – 2j + k on to the direction of vector B=4i – 4j +7k Q. 17 Find the angles alpha, beta and gamma which the vector A=3i – 6j +2k makes with the positive x,y,z axis respectively. Q.18 Find the work done in moving an object along a vector r=3i – 2j +5k is the applied force is F=2i – j – k. Q.19 Find the work done by a force 30,000 N in moving an object through a distance of 45 m when :(a) force is in the direction of motion;(b) force makes an angle of 40 degrees to the direction of motion.Find the rate at which the force is working at a time when the velocity is 2 m/s. Q.20 Two vectors A and B such that |A|=3, |B|=4 and A. B=-5 , find (a) The angle between A and B. (b) The Length |A+B| and |A-B| (c) The angle between (A+B) and (A-B)

X Lecture No.3 | Physical Quantities

Do you know ?

1. How long is One meter?
2. One Second is how long duration?
3. One kilogram is the mass of which object?
4. Which quantities are measured in Physics and Which are not measured in Physics?

This Lecture Includes:

1. Physical and Non Physical Quantities
2. Basic / Fundamental Quantities
3. Derived Quantities with examples
4. Conversions of length , mass and time.

XI Lecture No.3 | Significant Figures

Lecture Includes: XI Chapter No.1 : The Scope of Physics 1. Introduction to Significant Figures 2. Rules for counting Significant Figures 3. Examples of Significant Figures Previous Lectures: XI Lecture No.1|Introduction,History and Muslim Contribution: https://youtu.be/4tmmM3Hg5EU XI Lecture 2| Dimensions : https://youtu.be/svzZ2fUS__8

XI Physics | Solved Numericals | Chapter No.1 | The Scope of Physics

XI Physics Chapter No.1 : The Scope of Physics

1. Find the area of rectangular plate having length (21.3 +/- 0.2 ) cm and width (9.80 +/-0.10) cm.

2.Calculate (a) the circumference of circle of radius 3.5 cm. and (b) area of circle of radius 4.65 cm.

3.Prove that S=vit + 1/2 at^2 is correct dimensionally.

4.Suppose displace of a particle is related to a time according to a expression S= ct^3. what are the dimensions of constant c ?

5.Estimate the number of liters of gasoline used by all Pakistan's car each year.

XI Lecture 2| Dimensions

Chapter No.1 : The Scope of Physics

Topic :Dimension

Lecture contents:

1. Definition of Dimension

2.Dimension of Physical Quantities

3.Equation verification by using dimensions and Exam Questions

XI Lecture No.1 The Scope of Physics

Chapter No.1 : The Scope of Physics

Lecture contains;

1. Introduction and definition of Physics

2. Brief History of Physics

3. Contribution of Islamic World towards Science

Interpret velocity-time graph for constant direction and understand significance of area under velocity-time graph.

VELOCITY-TIME GRAPH FOR CONSTANT DIRECTION:

When an object is moving with a constant velocity, the line on the graph is horizontal. When an object is moving with a steadily increasing velocity, or a steadily decreasing velocity, the line on the graph is straight, but sloped. The diagram shows some typical lines on a velocity-time graph.

The steeper the line, the more rapidly the velocity of the object is changing. The blue line is steeper than the red line because it represents an object that is increasing in velocity much more quickly than the one represented by the red line.

Notice that the part of the red line between 7 and 10 seconds is a line sloping downwards (with a negative gradient). This represents an object that is steadily slowing down.

AREA UNDER VELOCITY-TIME GRAPH:

            Study this velocity-time graph.

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-gray triangle and the dark-gray rectangle.

1.    Area of light-gray triangle

o    The width of the triangle is 4 seconds and the height is 8 metres per second. To find the area, you use the equation:

o    area of triangle = ¹⁄₂ × base × height

o    so the area of the light-gray triangle is ¹⁄₂ × 8 × 4 = 16 m.

2.    Area of dark-gray rectangle

o    The width of the rectangle is 6 seconds and the height is 8 metres per second. So the area is 8 × 6 = 48 m.

3.    Area under the whole graph

o    The area of the light-gray triangle plus the area of the dark-gray rectangle is:

o    16 + 48 = 64 m

o    This is the total area under the distance-time graph. This area represents the distance covered.

Summary

·         The gradient of a velocity-time graph represents the acceleration

The area under a velocity-time graph represents the distance covered

Derivation of Gravitational Potential Energy (without using Calculus)

The potential energy of a body at height “h” from centre of earth w.r.t. a point at which the gravitational field is zero i.e. a point which has no potential is called absolute gravitational potential energy.

Where R_E is the radius of earth.The minus sign indicates that the potential energy is “negative” at any finite distance that is the potential energy is zero at infinity and decreases as the separation distance decreases. This is due to the fact that the gravitational force acting on the particle by earth is attractive. As the particle moves in from infinity the Work is positive which means U is negative.

Download

Q. Explain Principle, Purpose and Features of Nuclear Reactors

NUCLEAR REACTOR

DEFINITION:

Nuclear reactor is a device which is used to perform controlled fission reaction and extract the hidden energy from the nucleus.

PURPOSE:

Following are some main purpose of the nuclear reactor.
1- To convert controlled fission reaction to controlled chain reaction.
2- To produce large amount of energy.
3- To produce radioisotopes.

PRINCIPLE:

It is used on the principle of nuclear fission a process in which heavy nucleus breaks in to lighter nuclei with the emission of large amount of energy”.

FEATURES OF NUCLEAR REACTOR:

Following are the general features, which are common or almost nuclear reactor.

1. NUCLEAR FUEL:

A Material consisting of the fissionable isotopes is called reactor fuel that may be used in a reactor. For example (Uranium 92U235).

2. MODERATORS:

In the nuclear fission process at least one or more neutrons are produce per fission. To reduce the energy of neutron some suitable material is required which are known as moderators. The ordinary water (light water) an attractive moderator because of its supply at low cost. Heavy water is the best material. Other moderator is graphite beryllium and its oxides and organic compound.

3. COOLANTS:

Huge amount of heat is generated in reactor core as a result of fission take place in the
unclear fuel. To remove this large amount of heat, material are required which are called coolant. These materials are circulated though the core in order to absorb heat and transfer it to the outside of the core.

4. CONTROL MATERIAL:

In order to control nuclear fission in a reactor, suitable neutron absorbing material is required to be placed in the core region. The control material should be such that it does become radioactive by neutron capture. Cadmium and boron are good control materials.

5. SHIELDING:

During nuclear reactions neutrons and gamma rays emits out and becomes hazardous in the vicinity of the reactor, therefore proper shielding material is always required. A shielding material used for such protection is called the biological shielding because its purpose is to protect health. Generally a layer of concrete about 6 to 8 feet thickness has been found to absorb both neutrons and gamma rays. For rector operating at high power thermal shielding is required. A few cm of iron or steel very close to the core of the reactor provide the required thermal shield.

WORK-ENERGY RELATION

WORK ENERGY EQUATION

DERIVATION:

            Let us consider a body of mass “m” is placed at point A at a height h from the surface of earth.At this point the body possesses gravitational potential energy equal to mgh w.r.t point C lying on the ground.

            Now consider a point B at a distance x below the point A during downward motion of body.At this stage the height of the body becomes (h-x).

so, potential energy at point B becomes,

                                                P.E=mg(h-x)

As we know that potential energy at point B is less than the potential energy at point A,i.e.

                                                mg(h-x) < mgh

            or                                 mgh – mgx <mgh

            The loss in potential energy at point B is mgx.

           

The Kinetic Energy at point A is equal to zero because the body is at rest.During its downward motion its velocity increases ,so its kinetic energy also increases.If there is no air friction then the loss of P.E is equal to the gain in K.E, means P.E is converted into K.E.

            When the body reaches at point C its P.E becomes zero which means all of its P.E is converted into K.E,so we can write as

Loss in P.E = Gain in K.E

Practically there is always a force of friction which opposes the downward motion of the body.Let if friction f is present in this case then some amount of P.E is lost in work done against friction.Now, the modified equation can be written as,

                        Loss in P.E = Gain in K.E + Work done against friction

                                    mgx= 1/2 mv² + fx

or                                 1/2 mv² = mgx – fx

In terms of “h”

                                      1/2 mv² = mgh – fh

            The above equation is known as “Work Energy Equation”.

KINETIC MOLECULAR THEORY AND EFFECT OF PRESSURE ON GASES

DOWNLOAD AS PDF: https://drive.google.com/open?id=0B1t3t3Mn8__CVGJ4Y1FQaXZCV3M

SIGNIFICANT FIGURES AND RULES FOR COUNTING SIGNIFICANT FIGURES

SIGNIFICANT FIGURES

            “All the accurately known digits in a value and the first doubtful digit are known as significant figures.” In the measurement of any physical quantity the number of digits about which we are sure are called significant figures All physical measurements involve some degree of inaccuracy due to human error. instrumental error or due to both and therefore the knowledge of precision of a measurement is very important. A significant figure is that which is known to be reasonably reliable. The last figure being reasonably correct guarantees the certainty of the preceding figures.

RULES FOR COUNTING SIGNIFICANT FIGURES:

(i) In whole number values. all the digits except zeros at the right side m recognized as significant figures.

(ii) In decimal number values the zeros at the right side of the number are counted as significant figures but the zeros at the left side are not taken as significant figures.

(iii) Power or exponents to a certain base are not taken as significant figures.

(iv) In addition and subtraction process, the result should be rounded off to contain as many as decimal places as contained in the value of least number of decimal place.

(v) In multiplication and division process. the result should be rounded off to contain as many as significant figures as contained in the factor of least significant figures.

FOR EXAMPLE:

S.NO	Value	No. of significant figures
1	0.00045	2(4,5)
2	1.2000	5(1,2,0,0,0)
3	505	3(5,0,5)
4	34000	2(3,4)
5	6.67 x 1032	3(6,6,7)

Measure the uncertainity in the derived quantity

Download

The Cathode Ray Oscilloscope

Introduction 

The following should give the student some familiarisation with the function and uses of the cathode ray oscilloscope (C.R.O.).

Consider a simple sine wave electrical signal from some source as in Fig. 1a. If we can arrange things so that this sinusoidal voltage is applied to two horizontal conducting plates then in the region between these plates, the electric field will be alternating with period T seconds. It will increase in strength to a maximum, decrease to zero, turn over, and increase in the opposite direction to an equal maximum, then decrease to zero again, in each period of time T.

Now, if there is a beam of charged particles (electrons) streaming between these horizontal plates, the oscillating electric field there will bend the beam first up, then down, then back to the undeflected position in each time period T. Further, if the beam strikes a plate of material which fluoresces, one would see a spot of light on this plate (screen) which moves vertically up and down with period T.

 Now consider a set of vertical plates, also straddling the electron beam. An electric field applied to these plates will deflect the beam horizontally by an amount proportional to the voltage applied across the plates. If, connected to these plates we have a circuit which generates a linear ramp voltage, periodically, with the same period T, as in Fig. 1b, then the spot on the screen would be forced to start at the left side and move linearly in time across the screen, reaching its maximum travel to the right at time T. The spot would disappear then, and instantaneously reappear back where it started at the left hand side. This “sweep” would repeat every time T. You may wonder how one can easily produce a sawtooth wave at exactly the frequency of the input signal (or if you don’t, you should!). The answer is simple; you use the input signal to trigger the ramp (ie. to start it at its lowest voltage) every time the input voltage reaches a particular value going in a particular direction (ie. increasing or decreasing). In this way, if the input is periodic, then the sawtooth will have the same period, by definition.


The first set of plates, driven with the sinusoidally varying voltages will produce on the screen a spot travelling up and down in simple harmonic motion. If fast enough, it will appear as a solid vertical line. (Figure 2a) The second set of plates, driven with the ramp signal, if fast enough, produces a horizontal line on the screen, as in Figure 2b.

A combination of both sets of plates, one with a sinusoidal driving voltage of period T, the other with a ramp period of period T, will produce on the screen a picture like Figure 2c, (if the two circuits are synchronised so that they start as drawn on the voltage vs. time graphs).

 If the ramp period is now doubled, so that the spot sweeps across the screen in a time 2/T, on the sweep left to right the sinusoidal voltage will complete two full cycles and the picture on the screen will look like Figure 2d. Hopefully, this introduction will have presented an idea of how the C.R.O. functions in displaying a.c. (ie. time-varying) signals on the screen.

The C.R.O. in Detail 

The main part of the C.R.O. is a highly evacuated glass tube housing parts which generates a beam of electrons, accelerates them, shapes them into a narrow beam, and provides external connections to the sets of plates described above for changing the direction of the beam. The main elements of the C.R.O. tube are shown in Figure 3.

1. K, an indirectly heated cathode which provides a source of electrons for the beam by “boiling” them out of the cathode.
2. P, the anode (or plate) which is circular with a small central hole. The potential of P creates an electric field which accelerates the electrons, some of which emerge from the hole as a fine beam. This beam lies along the central axis of the tube.
3. G, the grid. Controlling the potential of the grid controls the number of electrons for the beam, and hence the intensity of the spot on the screen where the beam hits.
4. F, the focusing cylinder. This aids in concentrating the electron beam into a thin straight line much as a lens operates in optics.
5. X, Y, deflection plate pairs. The X plates are used for deflecting the beam left to right (the x direction) by means of the “ramp” voltage. The Y plates are used for deflection of the beam in the vertical direction. Voltages on the X and Y sets of plates determine where the beam will strike the screen and cause a spot of light.
6. S, the screen. This is coated on the inside with a material which fluoresces with green light (usually) where the electrons are striking.

What is Elastic Potential Energy?

Elastic Potential Energy

Any object than can be deformed (have its shaped changed) and then return to its original shape can store elastic potential energy.

• We’re still talking about potential energy, since it is stored energy until the object is allowed to
bounce back.
• “Elastic” does not refer to just things like elastic bands…other materials that would be referred
to as elastic would be
• pole vaulter’s pole
• springs

You learned in Physics that Hooke’s Law is…
F = kx
F = force (N)
k = spring constant for that object (N/m)
x = amount of expansion or compression (m)
We can use this formula to figure out a formula for the energy stored in the spring.
• Remember that W = F d

• We might be tempted to just shove the formula for Hooke’s Law into this formula to get
W = kxd = kx2
, but this is wrong!
• You have to take into account that the force is not constant as the object returns to its original
shape… it’s at a maximum when it is deformed the most, and is zero when the object is not
deformed.
• Let’s graph Force vs Distance of Expansion for a spring that was stretched and we are now
letting go of it…

But this is really just a Force vs Displacement Graph like the ones we just looked at a couple of
sections back! To figure out the energy of the spring we can just figure out the work it does by looking at the area under the graph.

Area = ½ bh
 = ½ F x
 = ½ (kx) x
Area = ½ kx2 = W
So the work done by the spring (and then energy it stored) can be calculated using…
Ee = ½ kx2
Ee = eleastic potential energy (J)
k = spring constant (N/m)
x = amount of expansion or compression [deformation] (m)


Example 1: Determine how much energy a spring with a spring constant of 15 N/m stores if it is
stretched by 1.6m.
Ee = ½ kx2
 = ½ (15N/m) (1.6 m)
2
Ee = 19 J

Young's Double Slit Experiment

This is a classic example of interference effects in light waves. Two light rays pass through two slits, separated by a distance d and strike a screen a distance, L , from the slits, as in Fig. 22.10.


Figure 22.10: Double slit diffraction



If d < < L then the difference in path length r1 - r2 travelled by the two rays is approximately:

r1 - r2 dsin

where is approximately equal to the angle that the rays make relative to a perpendicular line joining the slits to the screen.

If the rays were in phase when they passed through the slits, then the condition for constructive interference at the screen is:

dsin = m ,m = 1, 2,...

whereas the condition for destructive interference at the screen is:

dsin = (m + ) ,m = 1, 2,...

The points of constructive interference will appear as bright bands on the screen and the points of destructive interference will appear as dark bands. These dark and bright spots are called interference fringes. Note:
In the case that y , the distance from the interference fringe to the point of the screen opposite the center of the slits (see Fig.22.10) is much less than L ( y < <L ), one can use the approximate formula:

sin y/Lso that the formulas specifying the y - coordinates of the bright and dark spots, respectively are:

y Bm = brightspots

y Dm = darkspotsThe spacing between the dark spots is

y =

If d < < L then the spacing between the interference can be large even when the wavelength of the light is very small (as in the case of visible light). This give a method for (indirectly) measuring the wavelength of light. 

The above formulas assume that the slit width is very small compared to the wavelength of light, so that the slits behave essentially like point sources of light.

Electromagnetic Induction:C.R.Q's /Questions

15.1 Does the induced emf in a circuit depend on the resistance of the circuit?does the induced current depend on the resistance of the circuit?

Ans. The Induced emf in a coil depends upon the rate of change of flux through it (E=-Nt) .Hence its value does not depend upon the resistance of the coil.But the induced current that flows through a coil is equal to I=E/R and it’s value depends on the resistance of the coil.If , resistance increases then the current flowing through the coil decreases.Because the product of I and R must remains constant.i.e. I x R = Constant.

15.2 A square loop of wire i moving through a uniform magnetic field.The normal to the loop is oriented parallel to the magnetic field.Is emf induced in th loop?Give a reason for your answer.

Ans.No, induced emf will not be produced because there is no change of flux linking to the loop.i.e.t=0 .So according to the relation (E=-Nt) ,E=0.If the square loop is being rotated in magnetic field in such a way that the loop is cutting the magnetic field lines due to its motion,then emf will induce in the coil.

15.3 A light metallic ring is released from above into a vertical bar magnet.Viewed for above ,does the current flow clockwise or anticlockwise in the ring?

Ans.According to Faraday’s law of electromagnetic induction an induced emf and hence induced current will be produced in the metallic ring.According to Lenz law , the current in the ring should flow in such a direction as to oppose the cause producing it.So, the induced current in the coil must produce magnetic field which opposes the motion of ring towards bar magnet.The side of the ring facing magnet will be North Pole of the induced magnetic field.Right hand rule shows that the magnetic field will produce in this manner only when the current will flow in clockwise direction in ring.

15.4 What is the direction of the current through resistor R in the figure.When switch S is (a) closed (b) opened.

Ans.(a) When the switch is closed ,the current in the coil increases from zero to maximum steady value;during during this interval magnetic flux in the second coil from zero to max. and induced current will flow in it.The side of the current carrying coil facing the other coil becomes North pole.So,to oppose N pole , the current in the other coil must flow anti clockwise.Hence current in R flow from left to right according to the figure.

(b) however, when the switch is opened, the current in the coil decreases from max. to zero and flux linked to the other coil also decreases and induced current is produced in the reverse direction.So, the current will flow from Right to Left (clockwise) according to the figure.

15.5 Does the induced emf always act to decrease the magnetic flux through a circuit?

Ans.No, the induced emf does not act as to decrease magnetic flux through a circuit.According to Lenz law , the current in the ring should flow in such a direction as to oppose the cause producing it.If an induced emf appears in a circuit  due to decreasing magnetic flux linking that circuit then induced current flowing through the circuit will produce its own magnetic field that oppose the decrease of the magnetic field.In other word it is increasing the magnetic flux through a circuit.

15.6 When the switch in the circuit is closed a current is established in the coil and the metal ring jumps upward.Why?Describe what would happen to the ring is battery polarity were reversed?

Ans.From establishment of current, induced magnetic flux will be produced in the  cylinder. From Lenz’s law an opposing emf in the ring will be produced. The face of the ring opposite to coil develops similar pole of magnet and experiences repulsion, which makes it to jump upward.

The ring will jump upward in the same manner, if the battery polarity is reversed. The same process will happen as mentioned above.

15.7 The Fig. Shows a coil of wire in the xy plane with a magnetic field directed along the y- axis. Around which of the three coordinate axes should the coil be rotated in order to generate an emf and a current in the coil?

Ans.The coil must be rotated about x-axis to get change of magnetic flux and induced current through it.

15.8 How would you position a flat loop of wire in a changing magnetic field so that there

is no emf induced in the loop?

Ans.If the flat loop of wire is parallel to the field. When the coil is held parallel to the direction of B, then the angle between vector area A and B will be 90o

.

φB = B•A = BAcos90o= 0

15.9 In a certain region the earth’s magnetic field point vertically down. When a plane flies due north, which wingtip is positively charged?

Ans.[At the two magnetic poles, the direction of the earth’s magnetic field is vertical. At north magnetic pole it is downward into the ground, at south magnetic pole, it is upward out of the ground. Here on both places, the compass needle does not indicate any particular direction along the ground.]

Left wingtip will be positively charged. The electrons in the wing experience the magnetic force [ F = -e(vxB)] From R.H. rule, the electrons will move towards right, (the direction of conventional current is left). Due to it left wingtip (West side) will be positively charged.

15.10 Show that ε and ∆Φ / ∆t have the same units.

Ans.

15.11 When an electric motor, such as an electric drill, is being used, does it also act as a

generator? If so what is the consequences of this?

Ans.Yes it also acts as a generator. When the electric motor is running, due to rotation of its coil, an emf is induced in it. It is called back emf, which produces opposing current. It increases with speed of motor. This means that it also acts as a generator.

15.12 Can a D.C. motor be turned into a D.C. generator? What changes required to be done?

Ans. Yes a d.c. motor can be turned into a d.c. generator.To change it, needs some arrangement to rotate the armature. Disconnect the brushes of the commutator from d. c. supply and connect it with some external circuit.

15.13 Is it possible to change both the area of the loop and the magnetic field passing

through the loop and still not have an induced emf in the loop?

Ans. Yes, if the flux remains constant. From the equation; ∆φ = B•A ,B and A are inversely proportional to each other. If the area of the loop and magnetic field passing through the loop are changed in such a way to make product constant, then no induced emf will be produced.

Secondly, if plane of the coil is parallel to the magnetic field, changing in area and the field will not induce any emf in the loop.

15.14 Can an electric motor be used to drive an electric generator with the output from the generator being used to operate the motor?

Ans. No. An electric motor cannot be used to drive an electric generator. Perpetual

motion machine is not possible according to law of conservation of energy.

15.15 A suspended magnet is oscillating freely in horizontal plane. Oscillations are strongly damped when a metal plate is placed under the magnet. Explain why this occurs?

Ans. The metal plate produces an induced emf, due to oscillations in the suspended magnet.This induced emf produces current, which produces its own magnetic field that will oppose the motion of the suspended magnet. So oscillations are strongly damped.

Q.16 Four unmarked wires emerge from a transformer. What steps would you take to

determine the turns ratio?

Ans. Separate primary and secondary coils by ohmmeter. Connect primary coil with a.c. supply of known voltage Vp . measure the voltage induced Vs by voltmeter. Calculate turns ratio from; Vs / Vp = Ns / Np

15.17 a) Can a step-up transformer increase the power level? b) In a transformer, there is no transfer of charge from the primary to the secondary. How is, then the power transferred?

Ans. a) No. A step up transformer cannot increase the power level. As for ideal

case : power input = power out .It can increase or decrease voltage or current but power, P = VI, will remain same.

b) Due to induced emf, power is transferred. There is no transfer of charge, but the change of flux in one coil is linked with the other coil and emf is produced.

15.18 When the primary of a transformer is connected to a.c. mains the current in it

a) is very small if the secondary circuit is open, but

b) increases when the secondary circuit is closed. Explain these facts.

Ans. a) The output power is zero, if the secondary circuit is open, very small current is drawn by the primary coil from a.c. mains.

b) Output power will increase, when the secondary circuit is closed.

Power input = Power output , Greater current is needed in primary for equalizing power in the secondary coil.

The ratio of the dimensions of G to those of g is..... ?

The dimension of G can be found by the formula F=Gm₁m₂/r²

i.e. G=F r²/ m₁m₂      

Dimensions: [F]=MLT^-2         [r]=L               [m]=M

So [G] = MLT^-2 L²M^-2=M^-1L³T^-2

Now, the dimension of g is given by

[g]=LT^-2

Now, finding the ratio of G to g;

ð M^-1L³T^-2/ LT^-2

ð M^-1 L² 

Hence it is the ratio of dimensions of G to the dimensions of g.