The given equation is similar to
\(\mathrm{y} =\mathrm{A} \sin \left(\omega t-\frac{x}{\lambda}\right) \)
\(\omega=300, \ \mathrm{f} =\frac{300}{2 \pi}=\frac{150}{\pi} \)
\(\text { Speed } \mathrm{v} =\mathrm{f} \lambda \)
\(=\frac{150}{\pi} \times \pi \)
\(=150 \mathrm{~ms}^{-1} \)

Time period is the time taken by particle is return to B = 4 x 3 = 12s

Mean free path is independent of temperature and pressure

Pressure is independent of internal energy.

Moment of Inertia = MK²
K-Radius of gyration
During heating K would be increased.
Hence moment of inertia will increase

\(\text { Force } \overrightarrow{\mathbf{F}}=(2 i+\vec{j}) N\)
\(\text { Displacement } d=(5 \vec{i}+3 \vec{j})-(3 \vec{j}+\vec{k})\)
\(=(5 i-k) m\)
\(\text { Work done } W=F . d\)
\(=(2 \vec{i}+\vec{j})(5 i-k)\)
\(=10-0-0=10 J \)

Velocity and acceleration are in the same direction: So speed increases.

\(\text { Velocity of light } c=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)
\(c=\left(\mu_{0} \varepsilon_{0}\right)^{-\frac{1}{2}}\)
\(\text { Hence dimension }\left(\mu_{0} \varepsilon_{0}\right)^{-\frac{1}{2}} \text { is that of velocity. }\)

Unit of a (Planck's constant) - Js
Unit of c (Velocity of light) - ms^-1
Unit of G (Gravitational Constant) - \(\frac{\mathrm{Nm}^{2}}{\mathrm{Kg}^{2}}\)
\(\therefore \text { Unit of } \frac{h c}{G} \text { is }=\frac{J s \times m s^{-1}}{N m^{2} / k g^{2}} \)
\(=\frac{N m s \times m s^{-1} \times k g^{2}}{N m^{2}}[J=N m] =\mathrm{kg}^{2}\)

When two or more waves. superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a point is observed. This phenomenon is known as beats. The number of amplitude maxima per second is called beat frequency. If we have two sources, then their difference in frequency gives the beat frequency.
Number of beats per second
n = |f₁ - f₂| per second

In 1827, Robert Brown, a botanist reported that grains of pollen suspended in a liquid moves randomly from one place to other. The random (Zig - Zag path) motion of pollen suspended in a liquid is called Brownian motion. In fact we can observe the dust particle in water moving in random directions. This discovery puzzled scientists for long time. There were a lot of explanations for pollen or dust to move in random directions were found adequate. After a systematic study, Wiener and Gouy proposed that Brownian motion is to the bombardment of suspended particles by bombardment of suspended particles by molecules of the surrounding fluid. But during 19⁺⁺⁺ century people did not accept that every matter is made up of small atoms or molecules. In the year 1905, Einstein gave systematic theory of Brownian motion based on kinetic theory and he deduced the average size of molecules.
According to kinetic theory any particle suspended in a liquid or gas is continuously bombarded from all the directions so that the mean free path is almost negligible. This leads to the motion of the particles in a random and zig-zag manner as shown in Figure. But when we put our hand in water it causes no random motion because the mass of our hand is so large that the momentum transferred. by the molecular collision is not enough to move our hand.

Factors affecting Brownian Motion:
(i) Brownian motion increases with increasing temperature.
(ii) Brownian motion decreases with bigger particle size, high viscosity and density of the liquid (or) gas.

Limitations of Dimensional analysis: 
(i) This method gives no information about the dimensionless constants in the formula like 1, 2,................ \(\pi\), e, etc.
(ii) This method cannot decide whether the given quantity is a vector or a scalar.
(iii) This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
(iv) It cannot be applied to an equation involving more than three physical quantities.
(v) It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation.
For example, using dimensional analysis, s = ut + 1/3 at² is dimensionally correct whereas the correct relation is s = ut+1/2 at².

The weight of an object is defined as the downward force whose magnitude W is equal to the upward force that must be applied to the object to hold it at rest or at constant velocity relative to the Earth.

1. Law of orbits
Each planet moves around the Sun in an elliptical orbit with the Sun at one of the foci.
2. Law of area
The radial vector (line joining the Sun to a planet) sweeps equal areas in equal intervals of time.
3. Law of period
The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi major axis of the ellipse. It can be written as :
\(T^{2} \propto a^{3}  \)
\(\frac{T^{2}}{a^{3}}=\text { constant. }\)

Force is a vector and magnitude of the vector is represented by the length of the vector. Here \(\vec { { F }_{ 1 } } \) has greater length compared to other two. So \(\vec { { F }_{ 1 } } \) is largest of the three.

The centre of mass of a body is defined as a point where the entire mass of the body appears to be concentrated.

Characteristics of elastic collision are
1. Total momentum remains conserved
2. Total kinetic energy remains conserved.
3. In elastic collision conservative forces are involved. Hence total kinetic energy is conserved.
4. In elastic collision, mechanical energy is not dissipated.
Characteristics of inelastic collision are
1. Total momentum is conserved.
2. Total kinetic energy is not conserved.
3. Forces involved are non-conservative forces
4. Mechanical energy is dissipated into heat, light, sound etc.

The surface area of parachute is very large. And when it descends downwards, the air provides resistance to it and so it descends slowly.

The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the applied force.
\(F =\frac{\vec{d} p}{d t}=\frac{d(m \vec{v})}{d t}=m \frac{d \vec{v}}{d t}=m \vec{a}\left(\text { as } a=\frac{d \vec{v}}{d t}\right) \)
\(\vec{F} =m \vec{a} \Rightarrow \text { Force }=\text { mass } \times \text { acceleration }\)

The acceleration of a particle at any instant 'r' is equal to the rate of change of velocity.
Acceleration \(\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}\). It is a vector quantity.

(i) A quantity which is described by both its magnitude and direction is called a vector quantity.
(ii) Geometrically, a vector is a directed line segment
Examples: Force, velocity displacement, acceleration, position vector, linear momentum and angular momentum.

There are three types of elastic modulus.
(a) Young's modulus
(b) Rigidity modulus (or Shear modulus)
(c) Bulk modulus
(a) Young's modulus:
When a wire is stretched or compressed, then the ratio between tensile stress (or compressive stress) and tensile strain (or compressive strain) is defined as Young's modulus. Young modulus of a material =\(\frac{Tensile \ stress \ or \ compressive \ stress}{Tensile \ strain\ or\ compressive\ strain}\)
\(Y=\frac{\sigma_{t}}{\epsilon_{t}} \ or \ Y=\frac{\sigma_{c}}{\epsilon_{c}}\)
The unit for Young modulus has the same unit of stress because, strain has no unit. So, S.I. unit of Young modulus is Nm^-2 or pascal.
(b) Bulk modulus:
Bulk modulus is defined as the ratio of volume stress to the volume strain.
Bulk modulus, K = \(\frac{Normal\ (perpendicular)\ stress\ or\ pressure}{Volume \ strain}\)
The normal stress or pressure is
\(\sigma_{n}=\frac{F_{n}}{\Delta A}=\Delta p\)
The volume strain is \(\epsilon_{v}= \frac{\Delta V}{V}\)
Therefore, Bulk modulus is
\(K= - \frac{\sigma_{n}}{\epsilon_{v}}= - \frac{\Delta p}{\frac{\Delta V}{V}}\)
The negative sign in the equation means that when pressure is applied on the body, its volume decreases. Further, the equation implies that a material can be easily compressed if it has a small value of bulk modulus. In other words, bulk modulus measures the resistance of solids to change in their volume.
(c) The rigidity modulus or shear modulus:
The rigidity modulus is defined as Rigidity modulus or Shear modulus,
\(\eta_{R}=\frac{shearing \ stress}{angle\ of \ shear \ or \ shearing \ strain}\)
The shearing stress is \(\sigma _{s}=\frac{trangential \ force}{area\ over\ which\ it\ is\ applied}=\frac{F_{t}}{\Delta A}\)
The angle of shear or shearing strain
\(\epsilon_{s}=\frac{x}{h}=\theta\)
Therefore, Rigidity modulus is
\(\eta = \frac{\sigma_{s}}{\epsilon_{s}}=\frac{\frac{F_{t}}{\Delta A}}{\frac{x}{h}}=\frac{\frac{F_{t}}{\Delta A}}{\theta}\)
Further, the equation implies, that a material can be easily twisted if it has small value of rigidity modulus.

 As the distance between two masses increases, the strength of the force tends to decrease because of inverse dependence on r². Physically it implies that the planet Uranus experiences less gravitational force from the Sun than the Earth since Uranus is at larger distance from the Sun compared to the Earth.

The gravitational forces between two particles always constitute an action reaction pair. It implies that the gravitational force exerted by the Sun on the Earth is always towards the Sun. The reaction-force is exerted by the Earth on the Sun. The direction of this reaction force is towards Earth.
The torque experienced by the Earth due to the gravitational force of the Sum is zero given by
\(\vec { \tau } =\vec { r } \times \vec { F } =\vec { r } \times \left( -\frac { { GM }_{ s }{ M }_{ E } }{ { r }^{ 2 } } \hat { r } \right) =0\)
Since \(\vec { r } =r\hat { r } ,(\hat { r } \times \hat { r } )=0\)
So, \(\hat { \tau } =\frac { d\vec { L } }{ dt } =0\)
It implies that angular momentum \(\vec{L}\) is a constant vector. The angular momentum of the Earth about the Sun is constant throughout the motion. It is true for all the planets. In fact, this constancy of angular momentum leads to the Kepler's second law.
The expression \(\vec{F}=-\frac{G M_{1} M_{2}}{r^{2}} \hat{r}\) has one inherent assumption that both M₁ and M₂ are treated as point masses. When it is said that Earth orbits around the Sun due to Sun's gravitational force, we assumed Earth and Sun to be point masses. This assumption is a good approximation because the distance between the two bodies is very much larger than their diameters. For some irregular and extended objects separated by a small distance, we cannot directly use the equation. Instead, we have to invoke separate mathematical treatment which will be brought forth in higher classes.
However, this assumption about point masses holds even for small distance for one special case. To calculate force of attraction between a hollow sphere of mass M with uniform density and point mass m kept outside the hollow sphere, we can replace the hollow sphere of mass M as equivalent to a point mass M located at the center of the hollow sphere. The force of attraction between the hollow sphere of mass M and point mass m can be calculated by treating the hollow sphere also as another point the center of the hollow sphere. It is shown in the Figure.
There is also another interesting result. Consider a hollow sphere of mass M. If we place another object of mass 'm' inside this hollow sphere as in Figure, the force experienced by this mass 'm' will be zero.
The triumph of the law of gravitation is that it concludes that the mango that is falling down and the Moon orbiting the Earth are due to the same gravitational force.

(i) Parallel axis theorem states that the moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its center of mass and the product of the mass of the body and the square of the perpendicular distance between the two axes.
(ii) If I_C is the moment of inertia of the body of mass M about an axis passing through the center of mass, then the moment of inertia I about a parallel axis at a distance d from it is given by the relation,
I = I_C + Md²
(iii) Let us consider a rigid body as shown in Figure. Its moment of inertia about an axis AB passing through the center of mass is I_C DE is another axis parallel to AB at a perpendicular distance d from AB. The moment of inertia of the body about DE is I. We attempt to get an expression for I in terms of I_C For this, let us consider a point mass m on the body at position x from its center of mass.

(iv) The moment of inertia of the point mass about the axis DE is, m(x + d)². The moment of inertia I of the whole body about DE is the summation of the above expression.
\(I=\sum { m\left( x+d \right) ^{ 2 } } \)
This equation could further be written as,
\(I=\sum { m\left( { x }^{ 2 }+{ d }^{ 2 }+2xd \right) } \)
\(I=\sum { \left( { mx }^{ 2 }+m{ d }^{ 2 }+2dmx \right) } \)
\(I=\sum { { mx }^{ 2 }+\sum { m{ d }^{ 2 } } +2d\sum { mx } } \)
(v) Here, \(\sum { mx^{ 2 } } \) is the moment of inertia of the body about the center of mass. Hence, 
I_C = \(\sum { mx= } 0\) because, x can take positive and negative values with respect to the axis AB. The summation \(\left( \sum { mx } \right) \) will be zero.
Thus, I = I_c + \(\sum { md^{ 2 } } \) = I_C + \(\left( \sum { m } \right) d^{ 2 }\)
(vi) Here, \(\sum { m } \) is the entire mass M of the object \(\left( \sum { m=M } \right) \)
I = I_C + Md²
Hence the parallel axis theorem is proved.

Work-Kinetic Energy Theorem
Work and energy are equivalents. This is true in the case of kinetic energy also. To prove this, let us consider a body of mass m at rest on a frictionless horizontal surface.
The work (W) done by the constant force (F) for a displacement (s) in the same direction is,
W = Fs
The constant force is given by the equation,
F = ma
The third equation of motion can be written as,
\(v^{2} =u^{2}+2 a s  \)
\(a =\frac{v^{2}-u^{2}}{2 s}\)
Substituting for a in equation (2),
\(F=m\left(\frac{v^{2}-u^{2}}{2 s}\right)\)
Substituting equation (2), (1)
\(w=m\left(\frac{v^{2}}{2 s} s\right)-m\left(\frac{u^{2}}{2 s} s\right)  \)
\(w=\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}\)
The expression for kinetic energy:
The term \(\left(\frac{1}{2} m v^{2}\right)\) in the above equation is the kinetic energy of the body of mass (m) moving with velocity(v).
\(K E=\frac{1}{2} m v^{2}\)
Kinetic energy of the body is always positive. From equations (4) and (5)
\(\Delta K E =\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}  \)
\(\text {Thus, } W =\Delta K E\)
The expression on the right hand side (RHS) of equation (6) is the change in kinetic energy (\(\Delta\)KE) of the body.
This implies that the work done by the force on the body changes the kinetic energy of the body, This is called work-kinetic energy theorem.
The work-kinetic energy theorem implies the following.
1. If the work done by the force on the body is positive then its kinetic energy increases.
2. If the work done by the force on the body is negative then its kinetic energy decreases.
3. If there is no work done by the force on the body then there is no change in its kinetic energy, which means that the body has moved at constant speed provided its mass remains constant.

i) Systematic errors
Systematic errors are reproducible inaccuracies that are consistently in the same direction. These occur often due to a problem that persists throughout the experiment. Systematic errors can be classified as follows
1) Instrumental errors
When an instrument is not calibrated properly at the time of manufacture, instrumental errors may arise. If a measurement is made with a meter scale whose end is worn out, the result obtained will have errors. These errors can be corrected by choosing the instrument carefully.
2) Imperfections in experimental technique or procedure
These errors arise due to the limitations in the experimental arrangement. As an example, while performing experiments with a calorimeter, if there is no proper insulation, there will be radiation losses. This results in errors and to overcome these, necessary correction has to be applied.
3) Personal errors
These errors are due to individuals performing the experiment, may be due to incorrect initial setting up of the experiment or carelessness of the individual making the observation due to improper precautions.
4) Errors due to external causes
The change in the external conditions during an experiment can cause error in measurement. For example, changes in temperature, humidity or pressure during measurements may affect the result of the measurement.
5) Least count error
Least count is the smallest value that can be measured by the measuring instrument, and the error due to this measurement is least count error. The instrument's resolution hence is the cause of this error. Least count error can be reduced by using a high precision instrument for the measurement.
ii) Random errors
Random errors may arise due to random and unpredictable variations in experimental conditions like pressure, temperature, voltage supply etc. Errors may also be due to personal errors by the observer who performs the experiment. Random errors are sometimes called "chance error". When different landings are obtained by a person every time he occurs. For example, consider the case of the thickness of a wire measured using a screw gauge. The readings taken may be different for different trials. In this case, a large number of measurements are made and then the arithmetic mean is taken.
If n number of trial readings are taken in an experiment, and the readings are a₁, a₂, a₃,...............a_n. The arithmetic mean is
\( a_{m}=\frac{a_{1}+a_{2}+a_{3}+\ldots \ldots \ldots \ldots \ldots \ldots a_{n}}{n} \text { or } \)
\(a_{m}=\frac{1}{n} \sum_{i=1}^{t=n} a_{i}\)
Usually this arithmetic mean is taken as the best possible true value of the quantity.
Certain procedures to be followed to minimize experimental errors, along with examples are shown in Table.
Type of error Example How to minimize it Random error Suppose you measure the mass of a ring three times using the same balance and get slightly different values. 15.46 g, 15.42 g, 15.44 g Take more data. Random errors can be evaluated through statistical analysis and can be reduced by averaging over a large number of observations. Systematic error Suppose the cloth tape measure that you use to measure the length of an object has been stretched out from years of use. (As a result all of the length measurements are not correct). Systematic errors are difficult to detect and cannot be analysed statistically, because all of the data is in the same direction. (Either too high or too low) iii) Gross Error
The error caused due to the sheer carelessness of an observer is called gross error.
For example
(i) Reading an instrument without setting it properly.
(ii) Taking observations in a wrong manner without bothering about the sources of errors and the precautions.
(iii) Recording wrong observations.
(iv) Using wrong values of the observations in calculations.
These errors can be minimized only when an observer is careful and mentally alert.