Imporatnt questions for law of motion for 11 class

Question.1. Which one out of four fundamental forces is the weakest force between two protons at a                       distance of 1 fermi .
Answer       Gravitational

Question .2. According to newtons first law of motion a body moving with a uniform speed along a                        straight line should continue moving. In pratice , a body in motion stops after some time .
                    Explain with reason .
Answer =      The force of friction act as the external retarding force . It bring the body to rest.

Question.3 Why a horse cannot pull a cart and run in empty space?
Answer =  To pull the cart a horse pushes the ground in backward direction with feet . The reaction                        from the ground makes the cart to move forward . In an empty space the horse cannot
                  push in backward direction and get a forward reaction . That is why a horce cannot pull a
                  cart and run in the empty space.

Question 4  Why do we fall forward when a moving bus suddenly brakes . Explain the law required ?
Answer =    It is because of inertia of uniform motion . When the speeding bus stops suddenly the
                  lower part of the body in contact with the seat of the bus comes to rest where as the upper                    part of the body till tends to remain in motion.

Question 5 According to newton third law every force is a accompanied by an equal and opposite
                force . How can a moment ever take place ?
Answer = Since action and reaction do not act on same body,they do not cancel each other. Therefore
              a body may move either under the effect of the action of force or the other body may move
                 under the effect of reaction of it .

Question 6 A man riding horse is caught in a marshy ground and to his utter disgust finds himself and
                  the horse sinking into the ground . In the effort to pull the horse out the marshy ground ;
                  sitting on the back of the horse , he starts pulling the horse by the hair on its neck .Will the
                  man come out of the situation .Give explanation for your answer in brief
Answer= The man and the horse will not get out the marshy ground.The reason is that when the man
                 applies force on the horse by pulling the hair on its neck the applied force is an can get out
                  of it only if some external force i.e.m  a force outside the man-horse system acts . It is
                  because internal forces cannot bring about a change in the state of a system.

question of phyiscs

A sample of gas occupies 1.50 litre at 25°c if the temperature is raised 60°c what is the new volume of gas at constant pressure.

solution 

expalnation by charles law

v1/t1=v2/t2

where,

 are the initial volume and temperature of the gas.

 are the final volume and temperature of the gas.

given we have

v1=1.50liter

t1=25degree celcius =(25+273)k=298k

v2=?

t2=60degree celcius=(60+273)k=333k

Putting values in above equation, we get:

1.50 liter / 298k = v2 / 333

we have,

v2=1.68liter

Wave Optics

Revision Notes on Wave Optics 

Huygens Principle:- Wave-front  of a wave, at any instant , is defined as the locus of all the particles in the medium which are being disturbed at the same instant of time and are in the same phase of vibration.

(a) Each point on a wave front acts as a source of new disturbance and emits its own set of spherical waves called secondary wavelets. The secondary wavelets travel in all directions with the velocity of light so long as they move in the same medium.

(b) The envelope or the locus of these wavelets in the forward direction gives the position of new wave front at any subsequent time.

• Determination of Phase Difference:-

The phase difference between two waves at a point will depend upon

(a) The difference in path lengths of the two waves from their respective sources.

(b) The refractive index of the medium

(c) Initial phase difference between the source if any.

(d) Reflections, if any, in the path followed  by waves.

• Reflection of plane wave at plane surface (Laws of reflection):-

?(a) The incident ray, the reflected ray and normal to the reflecting surface at the point of incidence, all lie in one plane and that plane is perpendicular to the reflecting surface.

(b) The angle of incidence is equal to the angle of reflection.

So, ∠i = ∠r

This signifies angle of incidence is equal to the angle of reflection.

• Refraction of light:-

Refraction is the phenomena by virtue of which a wave going from one medium to another undergoes a change in velocity.

(a) The sine of the angle between the incident ray and the normal bears a constant ratio to the sine of the angle between refracted ray and the normal.

sin i/sin r = v1/v2 = ¹µ2 = constant

Here, v1 and v2 are the velocities of sound in first and second medium respectively.¹µ2 is the refractive index of the second medium with respect to first.

(b) The incident ray, the refracted ray and the normal to the refracting surface lie in the same plane.

• Interference:- The modification in the distribution of light energy obtained by the superposition of two or more waves is called interference.
• Principle of superposition:- It states that a number of waves travelling, simultaneously, in a medium behave independent of each other and the net displacement of the particle, at any instant, is equal to the sum of the individual displacements due to all the waves.
• Displacement equation:- y =  R sin 2π/λ (vt+x/2)
• Amplitude:- R = 2a cos πx/λ
• Intensity:- I = K4a² cos² (πx/λ)    [I = KR²]
• Maxima:- A point having maximum intensity is called maxima.

x = 2n (λ/2)

A point will be a maxima if the two waves reaching there have a path difference of even multiple of λ/2.

Imax = 4Ka² = 4i   (Here, i = Ka²)

• Minima:- A point having minimum intensity is called a minima.

x = (2n+1) (λ/2)

A point will be a minima if the two waves reaching there have a path difference of odd multiple of λ/2.

Imin = K. 4a²×0 = 0

• Condition for constructive interference:-

Path difference = (2n)λ/2

Phase difference = (2n)π

• Condition for destructive interference:-

Path difference = (2n+1)λ/2

Phase difference = (2n+1)π

• Coherent Sources:- Coherent sources are the sources  which either have no phase difference or have a constant difference of phase between them.
• Conditions for interference:-

(a) The two sources should emit, continuously, waves of same wavelength or frequency.

(b) The amplitudes of the two waves should be either or nearly equal

(c) The two sources should be narrow.

(d) The sources should be close to each other.

(e) The two sources should be coherent one.

• Young’s double slit experiment:-

Path difference, x = yd/D

Maxima, y = nλD/d

Here, n = 0,1,2,3….

Minima, y = (2n+1) λD/d

Here, n = 0,1,2,3….

• Fringe Width:- It is the distance between two consecutive bright and dark fringes.

β = λD/d

• Displacement of fringes due to the introduction of a thin transparent medium:-

(a) Shift for a particular order of fringes:-

?y = (β/λ) (µ-1)t

(b) Shift across a particular point of observation:-

µ = (mλ/t) +1

• Lloyd’s single mirror:-

?λ = β .2a/D

• Power of lens:- P = 100/f
• Magnifying power or magnification of a simple microscope:- M= 1+(D/f)
• Magnifying power or magnification of a compound microscope:-

M= L/f0 (1+D/fe)

Here, f0 is the focal length of the objective, fe is the focal length of the eyepiece and L is the length of the microscope tube.

• Magnification of astronomical telescope in normal adjustment:-

M = f0/fe

• Magnification of astronomical telescope, when the final image is formed at the distance of distinct vision:-

M = (f0/fe) [(fe+D)/D]

• Magnifying power M of Galileo’s telescope:-

M = focal length of objective/focal length of eye lens = F/f

• Diffraction:- Diffraction is the bending  or spreading of waves that encounter an object ( a barrier or an opening) in their path.

(a) In Fresnel class of diffraction, the source and/or screen are at a finite distance from the aperture.

(b) In Fraunhofer class of diffraction, the source and screen are at infinite distance from the diffracting aperture. Fraunhofer is a special case of Fresnel diffraction.

If Im represents the intensity at O, its value at P is

Iθ = Im (sinα/α)²

Here, α = ?/2 = πa sinθ/λ

A minimum occurs when, sin α = 0 and α is not equal to zero.

so  α = nπ,       n = 1, 2, 3...

So, πa sinθ/λ = nπ

Or, a sinθ = nλ

Angular width of central maxima of diffraction pattern = 2θ1 = 2 sin^-1(λ/a)

[ θ1 gives the angular position of first minima]

• Polarization: - Polarization of two interfering wave must be same state of polarization or two source of light should be un polarized.
• Brewster Law:-

?According to this law when un polarized light is incident at polarizing angle (i) on an interface separating a rarer medium from a denser medium, of refractive index m as shown in Fig., below such that,

 µ = tan i

Then light reflected in the rarer medium is completely polarized. Reflected and refractive rays are perpendicular to each other.

• Reduction in Intensity:- Intensity of polarized light is 50% of that of the un polarized light, i.e.,

 Ip = Iu/2

Here, Ip = Intensity of polarized light.

Iu = Intensity of un polarized light.

           For notes of thermal property: 1

           For notes on laws of motion  :2

 Revision Notes on Gravitation and Projectile: 3
 Revision Notes on Circle:4
 Straight lines revision notes:5
 Hydrocarbons revison notes :6
 Some basic concepts of chemstry revision notes:7
 SI units  : 8
 M.s dhoni great captian: 9
 Virat kholi is a great cricketer:10

             Revision Notes on Rotational Motion:11

written by arpan ruhil