f = 120Hz
Frequency of the 4th string is obtained from harmonic series
120,119, 118 and 117 Hz
(1)  (2)    (3)           (4)
\(\therefore\) f₄ = 117 Hz

Carbon dioxide
\(v_{r m s}=1.73 \sqrt{\frac{k T}{m}}\)

\(\lambda_{m} T =\mathrm{b} \)
\(\therefore T =\frac{2.898 \times 10^{-3}}{350 \times 10^{-9}}\)
\(\mathrm{~T} =0.00828 \times 10^{6} \)
\(=8280 \mathrm{~K} \)

\(-Q>0 \quad W>0\)
\(\Delta Q=\Delta U+\Delta W\)
\(\therefore \Delta Q \propto \Delta W\)
\(\text { If } \Delta Q>0 \text { then } \Delta W>0\)

For the wire 1
Young's modulus is less
\(\text { Young's modulus } Y \propto \frac{F}{A}\)
\(\therefore \text { Area is more }\)
\(\therefore \text { Wire } 1 \text { is thickest }\)

Radius =R
\(v_{1}^{2}-v_{2}^{2}=4 g R\)
\(\text { Tension } T_{2}=\frac{m v_{2}^{2}}{R_{2}}-m g\)
\(\text {To find minimum speed, let } T_{2}=0\)
\(0 =\frac{m v_{2}^{2}}{R}-m g \)
\(\frac{m v^{2}}{R} =m g \)
\(v_{2}^{2}=R g \ v_{2} =\sqrt{g R} \)
\(\text { sub (2) in the eqn (1) we get }\)
\(v_{1}^{2}-(\sqrt{g R})^{2} =4 g R \)
\(v_{1}^{2}-g R =4 g R \)
\(v_{1}^{2} =4 g R+g R \)
\(=5 g R \)
\(v_{1} =\sqrt{5 g R} \)

Use thumb rule

For a free falling body, initial vertical downward velocity = 0
For a body thrown horizontally with a velocity, initial vertical downward velocity = 0
Since both having same initial vertical downward velocity they cover equal vertical distance at any instant

A collection of forces is said to be concurrent, if the lines of forces act at a common point.

The moment of inertia of the disk about a perpendicular axis passing through its centre is
I= \(\frac{1}{2}\)mr²
From the parallel axis theorem, the moment of inertia of the disk about the pivot point is
I'=I+Md²=\(\frac{2.5\times0.6\times0.6}{2}\)+ 2.5 x 0.30 x 0.30
=\(\frac{0.9}{2}\)+0.225
= 0.45 + 0.225
=0.675kgm²
The angular frequency of small amplitude oscillations of a compound pendulum is given by
w=\(\sqrt { \frac { Mgd }{ I' } } =\sqrt { \frac { 2.5\times 9.8\times 0.30 }{ 0.675 } } \)
=\(\sqrt { \frac { 7.53 }{ 0.675 } } =\sqrt { \frac { 0.675 }{ 10.9 } } \)=3.3 rad/s

Latent heat capacity of a substance is defined as the amount of heat energy required to change the state of a unit mass of the material
Unit is J kg^-1.

Stoke's force F = 6πη αv
where a - radius of the sphere
v - velocity of the sphere and
η -  coefficient of viscosity of the liquid

The gravitational field at a point which is at a distance r from m, is defined as the gravitational force experienced by unit mass placed at that point.
It's unit is N/kg or ms^-2.

Examples of torque in day-to-day life are
1. The opening and closing of a door about the hinges and
2. Turning of a nut using a wrench.

Velocity:
Velocity is equal to the rate of change of position vector with respect to time. 
It is a vector quantity \(\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}\)
Speed:
The magnitude of velocity is called speed and is given by \(v= \sqrt{v^2_x+v^2_y+v^2_z}\). It is a positive scalar.

(i) Displacement is the difference between the final and initial positions of the object in a given interval of time. It can also be defined as the shortest distance between these two positions of the object and its direction is from the initial to final position of the object, during the given interval of time. It is a vector quantity.
(ii) Distance is the actual path length travelled by an object in the given interval of time during the motion. It is a positive scalar quantity.

(i) Physical quantities are classified into two types. There are fundamental and derived quantities.
(ii) Fundamental or base quantities are quantities which cannot be expressed in terms of any other physical quantities.
(iii) These are length, mass, time, electric current, temperature, luminous intensity and amount of substance.
(iv) Quantities that can be expressed in terms of fundamental quantities are called derived quantities.
(v) For example, area, volume, velocity, acceleration, force.

The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

Consider a container of gas with volume V, pressure P and temperature T. If we add sand particles one by one slowly on the top of the piston, the piston will move inward very slowly. This can be taken as almost a quasi-static process. It is shown in the figure

Sand particles added slowly- quasi-static process

Escape speed is the minimum speed of an object thrown vertically up such that it escapes the Earth's gravity and would never come back.

m = 2 kg, d = 10 m, F_ext = 20 N, \(\mu\)_k = 0.9.
when an object is in motion on he horizontal surface, it experiences two forces.
(a) External force, F_ext = 20 N
(b) Kinetic friction,
f_k = \(\mu\)_k mg = 0.9 \(\times\) (2) \(\times\) 10 = 18N
The work done by the external force W_ext = Fd = 20 x 10 = 200J
The work done by the force of kinetic friction W_k = f_kd = (-18) \(\times\) 10 = -180 J. Here the negative sign implies that the force of kinetic friction is opposite to the direction of displacement.
The total work done on the object W_total = W_ext + W_k = 200 J - 180 J = 20 J.
Since the friction is a non-conservative force, out of 200 J given by the external force, the 180 J is lost and it can not be recovered.

The moment of inertia of a disc about an axis passing through the center of mass and perpendicular to the disc is, \(I=\frac{1}{2}MR^{2}\)
In terms of radius of gyration, I = MK²
Hence, \(Mk^{2}=\frac{1}{2}MR^{2} ; K^{2}=\frac{1}{2}R^{2}\)
\(K=\frac{1}{\sqrt{2}}R\ or\ K=\frac{1}{1.414}R\ or\ K=(0.707)R\)
From the case of a rod and also a disc, we can conclude that the radius of gyration of the rigid body is always a geometrical feature like length, breadth, radius or their combinations with a positive numerical value multiplied to it.

Velocity of the particle,
\(\vec { v } =\frac { d\vec { r } }{ dt } =\frac { d }{ dt } (3t)\hat { i } +\frac { d }{ dt } (5{ t }^{ 2 })\hat { j } +\frac { d }{ dt } (7)\hat { k } \)
\(\frac { d\vec { r } }{ dt } =3\hat { i } +10t\hat { j } \)
Acceleration of the particle
\(\frac { d\vec { r } }{ dt } =\frac { { d }\vec { v } }{ dt } =\frac { { d }^{ 2 }\vec { r } }{ dt^{ 2 } } =10\hat { j } \)
Here, the particle has acceleration only along positive y direction. According to Newton's second law, net force must also act along positive y direction. In addition, the particle has constant velocity in positive x direction and no velocity in z direction. Hence, there are no net force along x or z direction.

one

Dimension formula for 
 \(\frac { 1 }{ 2 } \)mv² = [M][LT^-1]² = [ML²T^-2]
Dimension formula for 
mgh = [M][LT^-2][L] = [ML²T^-2]
[ML²T^-2] = [ML²T^-2]
Both sides are dimensionally the same, hence the equations \(\frac { 1 }{ 2 } \)mv² = mgh is dimensionally correct

Scalar is a property which can be described only by magnitude.
Examples:
Distance, mass, temperature, speed and energy.

(a) Closed organ pipes:
Closed organ is a pipe with one end closed and the other end open. If one of a pipe is closed, the wave reflected at this closed end is 180^o out of phase with the incoming wave. Thus there is no displacement of the

particles at the closed end. Therefore, nodes are formed at the closed end and anti-nodes are formed at open end.
Let us consider the simplest mode of vibration of the air column called the fundamental mode. Anti-node is formed at the open end and node at closed end. From the figure let L be the length of the tube and λ be the wavelength of the wave produced. For the fundamental mode of vibration, we have,
\(L={\lambda_1\over4}(or)\lambda_1=4L\)
The frequency of the note emitted is
\(f_1={v\over \lambda_1}={v\over 4L}\)
which is called the fundamental note.
The frequencies higher than fundamental frequency can be produced by blowing air strongly at open end. Such frequencies are called overtones.
The figure (2) shows the second mode of vibration having two nodes and two anti-nodes

4L = 3λ₂ 
\(L={3⋋_2\over 2}or\ ⋋_2={4L\over 3}\)
The frequency for this
\(f_1={v\over \lambda_2}={3v\over 4L}=3f_1\)
is called first over tone, since here, the frequency is three times the fundamental frequency it is called third harmonic.
The Figure (3) shows third mode of vibration having three nodes and three anti-nodes.

We have 4L= 5⋋₃
\(L={5\lambda_3\over4}\ or\ \lambda_3={4L\over 5}\)
The frequency
\(f_3={v\over \lambda _3}={5v\over 4L}=5f_1\)
is called second over tone, and since n = 5 here, this is called fifth harmonic. Hence, the closed organ pipe has only odd harmonics and frequency of the n^th harmonic is f _n = (2n + 1)f₁. Therefore, the frequencies of harmonics are in the ratio
f₁:f₂:f₃:f₄:.... = 1:3:5:7:....
(b) Open organ pipe: 
Consider the picture of a flute, shown in Figure. It is a pipe with both the ends open. At both open ends, anti-nodes are formed. Let us consider the simplest mode of vibration of the air column called fundamental mode. Since anti-nodes are formed at the open end, a node is formed at the mid-point of the pipe

From Figure (5), if L be the length of the tube, the wavelength of the wave produced is given by
\(L={\lambda_1\over 2}or\ {\lambda_1=2L}\)
The frequency of the note emitted is
\(f_1={v\over \lambda_1}={v\over 2L}\)
which is called the fundamental note.
The frequencies higher than fundamental frequency can be produced by blowing air strongly at one of the open ends. Such frequencies are called overtones.

The Figure (6) shows the second mode of vibration in open pipes. It has two mode of vibration in open pipes. It has two nodes and three anti-nodes, and therefore,
L= ⋋₂ or ⋋₂= L
The frequency
\(f_2={v\over \lambda_2}={v\over L}=2\times{v\over 2L}=2f_1\)
is called first over tone. Since n = 2 here, it is called the second harmonic.

The Figure (7) above shows the third mode of vibration having three nodes and four anti-nodes
\(L={3\over2}\lambda,\ or\ \lambda_3={2L\over 3}\)
The frequency
\(f_3={v\over \lambda_3}={3v\over 2L}=3f_1\)
is called second over tone. Since n = 3 here, it is called the third harmonic. 
Hence, the open organ pipe has all the harmonics and frequency of n^th harmonic is f _n = nf₁, Therefore, the frequencies of harmonics are in the ratio
f₁ : f₂ : f₃ : f₄ :... = 1 : 2 : 3 : 4 : ....

S.No	Simple Harmonic Motion	Angular Harmonic Motion
1.	The displacement of the particle is measured in terms of linear displacement \(\vec { r } \).	The displacement of the particle is measured in terms of angular displacement  \(\vec { \theta } \)(also known as angle of twist).
2.	Acceleration of the particle is \(\vec { a } =-{ \omega }^{ 2 }\vec { r } \)	Angular acceleration of the particle is \(\vec { \alpha } =-{ \omega }^{ 2 }\vec { \theta } \)
3.	Force, \(\vec { F } =m\vec { a } \) where m is called mass of the particle.	Torque, \(\vec { \tau } =I\vec { \propto } \) where I is called moment of inertia of a body.
4.	The restoring force \(\vec { F } =-k\vec { r } \), where k is restoring force constant.	The restoring torque \(\vec { \tau } =-k\vec { \theta } \), where the symbol k (kappa) is called restoring torsion constant It depends on the property of a particular torsion fiber.
5.	Angular frequency \(\omega =\sqrt { \frac { k }{ m } } \)rads-1	Angular frequency, \(\omega =\sqrt { \frac { k }{ I } } \) rad s-1

(i) All the molecules of a gas are identical, elastic spheres.
(ii) The molecules of different gases are different.
(iii) The number of molecules in a gas is very large and the average separation between them is larger than size of the gas molecules.
(iv) The molecules of a gas are in a state of continuous random motion.
(v) The molecules collide with one another and also with the walls of the container.
(vi) These collisions are perfectly elastic so that there is no loss of kinetic energy during collisions.
(vii) Between two successive collisions, a molecule moves with uniform velocity.
(viii) The molecules do not exert any force of attraction or repulsion on each other except during collision. The molecules do not possess any potential energy and the energy is wholly kinetic.
(ix) The collisions are instantaneous. The time spent by a molecule in each collision is very small compared to the time elapsed between two consecutive collisions.
(x) These molecules obey Newton's laws of motion even though they move randomly.

Newton's law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.
\(\frac{d Q}{d t} \propto-\left(\mathrm{T}-\mathrm{T}_{s}\right)\)   ...(1)
The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,
T = Temperature of the object
T_s = Temperature of the surrounding

From the graph in figure it is clear that the rate of cooling is high initially and decreases with falling temperature.
Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,
dQ = msdT ........(2)
(iv) Dividing both sides of equation (2) by dt
\(\frac { dQ }{ dt } =\frac { msdT }{ dt } \)..........(3)
From Newton's law of cooling
\(\frac { dQ }{ dt } \propto -(T-{ T }_{ s })\)
\(\frac { dQ }{ dt } =-\alpha (T-{ T }_{ s })\)   ...(4)
Where a is some positive constant.
From equation (2) and (4)
-\(\alpha\) (T - T_s) = \(md\frac { dt }{ dt } \)
\(\frac { dt }{ T-{ T }_{ s } } =\frac { a }{ ms } dt\)   ....(5)
Integrating equation (5) on both sides,
\(\int _{ 0 }^{ \infty }{ \frac { dt }{ T-{ T }_{ a } } =-\int _{ 0 }^{ 1 }{ \frac { a }{ ms } dt } } \)
ln (T - T_s) = \(-\frac { a }{ ms } t+{ b }_{ 1 }\)
Where b₁ is the constant of integration taking exponential both sides, we get
\(\mathrm{T} =T_{s}+b_{2} e^{-\frac{a}{m s} t} \)
here \(\mathrm{~b}_{2} =e^{b_{1}}=\text { constant }\)

Hooke's law states that within the elastic limit, the strain produced in a body is directly proportional to the stress applied.
It can be verified in a simple way by stretching a thin straight wire (stretches like spring) of length L and uniform cross-sectional area A suspended from a fixed point O. A pan and a pointer are attached at the free end of the wire as shown in Figure. The extension produced on the wire is measured using a vernier scale arrangement. The experiment shows that for a given load, the corresponding stretching force is F and the elongation produced on the wire is ΔL. It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and ΔL on the Y-axis. This graph is a straight line passing through the origin as shown in Figure.


Therefore,
ΔL = (slope)F
Multiplying and dividing by volume,
V = AL,
F (slope) = \(\frac{AL}{AL} \Delta L\)
Rearranging, we get
\(\frac{F}{A}=[\frac{L}{A(Slope)}]\frac{\Delta L}{L}\)
Therefore, \(\frac{F}{A} \alpha [\frac{\Delta L}{L}]\)
Comparing with equations stress and strain \(\sigma=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}, \varepsilon=\frac{\text { Change in size }}{\text { Original size }}=\frac{\Delta l}{l}\),
we get volume strain, \(\varepsilon_{v}=\frac{\Delta V}{V}\) equation as
\(\sigma \propto \varepsilon\)
i.e., the stress is proportional to the strain in the elastic limit.

Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed Vi' the initial total energy of the object is
E_i = \(\frac { 1 }{ 2 } { Mv }_{ i }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)   ............(1) 
where, M_E is the mass of the Earth and R_E- the radius of the Earth. The term \(\frac { GMM_{ E } }{ R_{ E } } \) is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∝) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise Kinetic energy can be non-zero.
E_f = 0
According to the law of energy conservation,
E_i = E_f  .............(2)
Substituting (1) in (2) we get,
\(\frac { 1 }{ 2 } { Mv }_{ i }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)   =0
\(\frac { 1 }{ 2 } { Mv }_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)  = 0  .............(3)
Consider the escape speed, the minimum speed required by an object to escape Earth's gravitational field, hence replace v_i with v_e. i.e.,
\(\frac { 1 }{ 2 } { Mv }_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)
\(v_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } .\frac { 2 }{ M } \)
\(v_{ e }^{ 2 }=\frac { 2G{ M }_{ E } }{ { R }_{ E } } \)  ..............(4)
Using g  = \(\frac { G{ M }_{ E } }{ { R }_{ e } } \)  ..............(5)
\(v_{ e }^{ 2 }=2g{ R }_{ E }\)
\({ v }_{ e }=\sqrt { 2g{ R }_{ E } } \)   .................(6)

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) Vertical motion
Consider two blocks of masses m₁ and m₂ (m₁ > m₂) connected by a light and inextensible string that passes over a pulley as shown in fig. A little consideration will show, that the greater mass m₁ will move downwards, whereas the smaller one will move upwards. Since the string is inextensible, the upward acceleration of the mass m₂ will be equal to the downward acceleration of the mass m₁

Condition	Acceleration	Tension
	T - m2g = m2a .......(1) T - m1g = m1a   m1g - T = m1a .......(2) Adding equations (1) and (2), we get m1g - m2g = m1a + m2a (m1 - m2)g = (m1 + m2)a  .......(3) (vi) From equation (3), the acceleration of both the masses is \(a=\left( {{m_1-m_2}\over{m_1+m_2}} \right)g\) .......(4) If both the masses are equal (m1 = m2), from equation then a = 0	T - m2g = m2a .......(1)\(a=\left({m_1-m_2 \over m_1+m_2 } \right)g\)  .......(4) Substituting the equation (4) into(1) \(T-m_2g=m_2\left({m_1-m_2 \over m_1+m_2 } \right)g\) \(T=m_2 g\left(1+{m_1-m_2 \over m_1+m_2 } \right)\) \(T=g\left({2m_1m_2\over m_1+m_2} \right)\)

ii) Horizontal Motion: Mass m, is kept on a horizontal table and mass m, is hanging through a small pulley as shown in the figure. Assume that there is no friction on the surface. As both blocks are connected to the unstretchable string, if m, moves with an accelerating 'a' downward them m₂ move with the same acceleration 'a' horizontally.

Condition	Acceleration	Tension
(i) The forces acting on mass m2 are 1. Downward gravitational force (m2g) 2. Upward normal force (N) exerted by the surface 3. Horizontal tension (T) exerted by the string. (ii) The forces acting on mass m1 are 1. Downward gravitational force (m1g). 2. Tension (T) acting upwards	T-m1g= -m1a ...........(1) T = m2 a  .......(2) There is no acceleration along y-direction for m2 N - m2g = 0 ⇒ N= m2g .... (3) m2a-m1g= -m1a m2a+m1a= m1g \(a={m_1 \over m_1+m_2}g\)    ......(4)	Tension in the string can be obtained by substituting equation (4) in equation (2). \(T=m_2({m_1 \over m_1+m_2})g\) \(T=g({m_1m_2 \over m_1+m_2})\) This result has an important application in industries. The ropes used in conveyor belts (horizontal motion) work for longer duration than those of cranes and lifts (vertical motion).

Scalar product:
Scalar product or dot product of two vectors in defined as the product of the magnitudes of both the vectors and the cosine of the angle between them.
If \(\vec{A}\) and \(\vec{B}\) are two vector having an angle \(\theta\) between them, then \(\vec{A} \vec{B}=A B \cos \theta\) where A and B are magnitudes of \(\vec{A}\) and \(\vec{B}\) .
Example: work, energy and electric flux.
Properties:
1. \(\vec{A}\).\(\vec{B}\)  is always a scalar. It is positive if \(\theta\)<90 and it is negative if \(90^{\circ}<\theta<180^{\circ}\)
2. When the vectors are parallel, \(\theta=0^{\circ} \ and \ \cos 0^{\circ}=1 \therefore(\vec{A} \cdot \vec{B})_{\text {mat }}=A B\).
3. When the vectors are anti-parallel, \(\theta=180^{\circ}\ and \ \cos 180^{\circ}=-1 \therefore(\vec{A} \cdot \vec{B})_{\min }=-A B\)
4. When the vectors are perpendicular to each other, \(\theta=90^{\circ}\ and \ \cos 90^{\circ}=0\therefore \vec{A} \vec{B}=0\).
5. Scalar product is commutative i.e, \(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A}\)
6. It obeys distributive law i.e., \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\)
7. Self dot product is given by \(\vec{A} \cdot \vec{A}=A A \cos \theta=A^{2}, \ here\ \theta=0^{\circ}\). The magnitude of the vector \(\vec{A}\ is \ (\vec{A})=A=\sqrt{\vec{A} \cdot \vec{A}}\)
8. In the case of orthogonal unit vectors \(\vec{i}, \vec{j} \ and \ \vec{k}\)
\(\hat{i} \hat{j}=\hat{j} \hat{j}=\hat{k} \cdot \hat{k}=1 \text { and } \)
\(\vec{i} \cdot \vec{j}=\hat{j} \hat{k}=\hat{k} \hat{i}=0\)
9. The angle between the vectors \(\theta=\cos ^{-1}\left[\frac{\vec{A} \cdot \vec{B}}{A B}\right]\)
10. In terms of components,
\(\vec{A} \cdot \vec{B} =\left(A_{x} \hat{i}+\mathrm{A}_{y} \hat{j}+A_{z} \hat{k}\right)\left(B_{x} \hat{i}+\mathrm{B}_{y} \hat{j}+B_{z} \hat{k}\right) \)
\(=A_{x} B_{x}+A_{y} B_{y}+A_{i} B_{z} \text {, with all other terms zero. }\)
The magnitude of A is given by \(|\vec{A}|=A=\sqrt{A_{x}^{2}+A_{y}^{2}+A_{2}^{2}}\) and \(|\vec{B}|=B=\sqrt{B_{x}^{2}+B_{y}^{2}+B_{2}^{2}}\) Vector product:
The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them.
If \(\vec{A}\) and \(\vec{B}\) are two vectors, then \(\vec{A} \times \vec{B}=\vec{C}=(A B \sin \theta) \hat{n}\).
The direction \(\hat{n}\ of \ \vec{A} \times \vec{B}\) is perpendicular to the plane containing the vectors \(\vec{A}\) and \(\vec{B}\) and is determined by the right hand screw rule or right hand thumb rule.
Example: Torque \(\tau=\vec{r} \times \vec{F}\) and Angular momentum \(\vec{L}=\vec{r} \times \vec{p}\)
Properties:
1. The resultant of the vector product is always another vector whose direction is perpendicular to the plane containing these two vectors \(\vec{A}\) and \(\vec{B}\) even though the vectors \(\vec{A}\) and \(\vec{B}\) may or may not be mutually orthogonal.
2. It is not commutative. \(\vec{A} \times \vec{B} \neq \vec{B} \times \vec{A}\). But \(\vec{A} \times \vec{B}=-[\vec{B} \times \vec{A}]\)
3. When the vectors \(\vec{A}\) and \(\vec{B}\) are orthogonal to each other the vector product will have maximum magnitude as \(\theta=90^{\circ}\  and \ \sin \theta=1\).
\((\vec{A} \times \vec{B})_{\max }=A B \hat{n}\)
4. The vector product of two non-zero vectors will be minimum when (sin \(\theta\))=0, i.e., \(\theta=0^{\circ} \ or \ 180^{\circ}(\vec{A} \times \vec{B})_{\min }=0\).
It means that the vector product of two non-zero vectors vanishes if the vectors are parallel or anti parallel.
5. The self-cross product is a null vector. \(\vec{A} \times \vec{A}=A A \sin 0^{\circ} \hat{n}=\overrightarrow{0}\)
6. The self-vector products of unit vectors are then zero \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0\).
7. In the case of orthogonal unit vectors, \(\vec{i}, \vec{j} \ and \ \hat{k}\)
\(\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i} \ and \ \hat{k} \times \hat{i}=\hat{j}\) and
\(\hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{j}=-\hat{i} \ and \ \hat{i} \times \hat{k}=-\hat{j}\)
8. In terms of components,
\(\vec{A} \times \vec{B}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ A_{x} & A_{y} & A_{z} \\ B_{x} & B_{y} & B_{z} \end{array}\right|=\begin{array}{r} +\hat{i}\left(A_{y} B_{z}-A_{z} B_{y}\right) \\ +\hat{j}\left(A_{z} B_{x}-A_{x} B_{z}\right) \\ +\hat{k}\left(A_{0} B_{y}-A_{z} B_{x}\right) \end{array}\)
9. If two vectors \(\vec{A}\) and \(\vec{B}\) form adjacent sides of a parallelogram, then magnitude \((\vec{A} \times \vec{B})\) is equal to the area of the parallelogram.
10. If two vectors \(\vec{A}\) and \(\vec{B}\) are represented by the two sides of a triangle taken in order, then the area of the triangle is equal to \(\frac{1}{2}|\vec{A} \times \vec{B}|\)

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.