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

T² = C (R + h)³
\(\therefore T \alpha (R_E)^\frac{3}{2}\)
R_E = 2R_E
\(\therefore T \alpha (2R_E)^\frac{3}{2}\)
Time period increases by \(2^\frac{2}{3}\)= 2\(\sqrt 2\) 
No. of days in a year = (365.4) \(\times\)2\(\sqrt 2\)
= 1032

\(F=-\frac{d u}{d x} \quad F(x) =k x+a x^{3} \)
\(d u =-F d x \)
\(u(x) =-\int_{0}^{x}\left(-k x+a x^{3}\right) d x \)
\(=\int_{0}^{x} k x d x-a \int_{0}^{x} x^{3} d x \)
\(=\frac{k x^{2}}{2}-\frac{a x^{4}}{2} \)
\(U(x) =\frac{x^{2}}{2}\left(k-\frac{a x^{2}}{2}\right) \)
\(u(x)=0 \text { at } x=0 \text { and }\)
\(U(x) =0 ; k-\frac{a x^{2}}{2}=0 \)
\(=\frac{a}{2} x^{2}=-k \)
\(x^{2} =\frac{2 k}{a} \)
\(\therefore x =\sqrt{\frac{2 k}{a}} \)
\(\text { Clearly } u(x)=0 \text { at } x=0 \text { and }\)
\(x=\sqrt{\frac{2 k}{a}}\)
\(\text { For } x>\sqrt{\frac{2 k}{a}} U(x) \text { will be negative. } \)
\(\text { At } x=0 ; F=\frac{-d u}{d x}=0\)
(i.e.,) Slope of V - x graph is zero at x = 0
Hence the most appropriate answer is d.

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

Longitudinal strain can be classified into two types:
(i) Tensile strain: If the length is increased from its natural length then it is known as tensile strain.
(ii) Compressive strain: If the length is decreased from its natural length then it is known as compressive strain.

Weight of a body is defined as the gravitational force with which a body is attracted towards the centre of the earth. Hence the weight of a body is given by W = mg (or) \(\vec { w } =m\vec { g } \)

For a closed organ Pipe

\(\ell=\frac{\lambda}{4} \)
\(\therefore \lambda=4 \ell\)
For a open organ pipe \(L=\frac{\lambda}{2} \quad \therefore \lambda=2 L\)
Fundamental frequency of a closed pipe
\(f_{c} =250 \mathrm{~Hz}  \)
\(f_{0} =\frac{V}{\lambda}  \)
\(=\frac{V}{2 L}\)
Fundamental frequency of open organ pipe
\(f_{o} =2\left(\frac{V}{4 L}\right)  \)
\(=2 \times f_{c}  \)
\(=2 \times 250=500 \mathrm{~Hz}\)
∴ Frequency of open organ pipe =500 Hz.

An echo is a repetition of sound produced by the reflection of sound waves fiom a wall, mountain or other obstructing surfaces. The speed of sound in air at 20De is 344 ms^-1. If we shout at a wall which is at 344 m away, then the sound will take 1 second to reach the wall. After reflection, the sound will take one more second to reach us. Therefore, we hear the echo after two seconds. Scientists have estimated that we can hear two sounds properly if the time gap or time interval between each sound is \((\frac{1}{10})^{th}\) of a second (persistence of hearing) i.e., 0.1 s. Then,
Velocity = \(\frac{Distance\ travelled}{time\ taken}=\frac{2d}{r}\)
2d = 344\(\times\)0.1 = 34.4 m
d = 17.2 m
The minimum distance from a sound reflecting wall to hear an echo at 2^oCe is 17.2 meter.

When a piece of wood is pressed and released,
\(\mathrm{F}=\mathrm{ma}, \quad \mathrm{m} =\text { volume } \times \text { density }=\mathrm{A} \times \rho  \)
\(\text { Change in force } =\mathrm{mg}=\mathrm{A} \times \rho \mathrm{g}  \)
\(\therefore \text { Acceleration a } =\frac{F}{m}  \)
\(a =\left(\frac{A \rho g}{m}\right) x  \)   .....(1)
For SHM, \(a =\omega^{2} x\)   .....(2)
From equation (1) & (2) we get
\(\omega^{2}=\frac{A \rho g}{m} \quad \therefore \omega=\sqrt{\frac{A \rho g}{m}}\)
Time period \(\mathrm{T}=\sqrt{\frac{2 \pi}{\omega}} \quad \therefore \mathrm{T}=2 \pi \sqrt{\frac{m}{A \rho g}}\)

\(\text {The gravitational field }=-\frac{G m}{r^{2}} \hat{r} \)
\(\frac{G m}{x^{2}}=\frac{G \times 4 m}{(r-x)^{2}} \)
\(\frac{m}{x^{2}}=\frac{4 m}{(r-x)^{2}} \)
\(\frac{1}{x^{2}}=\frac{4}{(r-x)^{2}} \)
\(\frac{1}{x}=\frac{2}{r-x}\)
r - x = 2x
\(\mathrm{r}=3 x \quad \therefore x=\frac{r}{3} \)
\(\text {Gravitational potential } \mathrm{V}=-\frac{G m}{r}=-\frac{9 \mathrm{Gm}}{r}\)

The vehicle's engine has to do work against resistive force and make vehicle to move with an acceleration. Therefore, power delivered by the vehicle engine is
P (resistive force + mass x acceleration) (velocity)
\(P=\overrightarrow { { F }_{ -tot } } \overrightarrow { v } =\left( { F }_{ resistance }+F \right) \overrightarrow { v } \)
\(P=\overrightarrow { { F }_{ tot } } .\overrightarrow { v } =\left( { F }_{ resistance }+ma \right) \overrightarrow { v } \)
= (500 N + (1250 kg) \(\times\) (0.2 ms^-2)) (30 ms^-1) = 22.5 kW

\(|\vec A|=\sqrt {3^2+4^2}=5\)
also \(|\vec B|=\sqrt {7^2+24^2}=25\)
desired vector \(=|\vec B|\hat A=25\times\frac{3\hat i+4\hat j}{5}=5(3\hat i+4\hat j)=15\hat i+20\hat j\).

(a) I_xx' = 4 \(\times\) (0.3)² + 1 \(\times\) (0.8)² = 1 kgm² 
(b) I_xx'= 4 \(\times\) (3)² + 2 \(\times\) (2)² + 3 \(\times\) (4)² = 92 kgm²

Relation between angular momentum and rotational K.E.
Angular momentum, L = IW.
Rotational K.E., EK = \(\frac{1}{2}\)
\(2I={ (I\omega ) }^{ 2 }\)
\({ x }_{ k }=\frac { { L }^{ 2 } }{ 2I } \)
\(\sqrt { { E }_{ k } } =\frac { L }{ \sqrt { 2I } } \)
\(\frac { \sqrt { { E }_{ k } } }{ L } =\frac { I }{ \sqrt { 2I } } =constant\)
If I = 1, graph will be hyperbola.

(i) Accidental discoveries and well-analysed research outcome in the laboratory based on intuitive thinking and prediction.
(ii) For example, magnetism was accidentally observed but the reason for this strange behaviour of magnets was later analysed theoretically.
(iii) This analysis revealed the underlying phenomena of magnetism. With this knowledge, artificial magnets were prepared in the laboratories.

A & B be the two given spheres & 0 = mid point
If E₁ & E₂ are the gravitational fields
\({ E }_{ 1 }=\frac { G\times 100 }{ { \left( 0.5 \right) }^{ 2 } } { E }_{ 2 }=\frac { G\times 100 }{ { \left( 0.5 \right) }^{ 2 } } \)
E₁ & E₂ are  equal & opposite, then resultant gravitational field at O is zero.
V - gravitational potential at O due to A & B 
\(V=-\frac { D.100 }{ 0.5 } -\frac { G.100 }{ 0.5 } =-\frac { 2G.100 }{ 0.5 } \)
\(=\frac { -200\times { 6.67 }^{ - }{ 10 }^{ -11 } }{ 0.5 } =-2.7\times { 10 }^{ -8 }J/kg\)

The situation in diagram
Let A & B be a rod of length 2m suppose a weight w is hung at C at a distance x from A. Let T₁ & T₂ be the tension in the steel & brass rods.
(i) stress in steel wire = \(\frac{T_{1}}{A_{1}}\)
stress in brass wire = \(\frac{T_{2}}{A_{2}}\) as both the stresses are equal, so
\(\frac{T_{1}}{A_{1}}\)=\(\frac{T_{2}}{A_{2}}\) or \(\frac{T_{1}}{T_{2}}=\frac{A_{1}}{A_{2}}\)

\(= \frac{0.1}{0.2}=\frac{1}{2}\)
Now moments about C are equal system is in equilibrium
\(T_{1}=T_{2}(2-x) \ or \ \frac{T_{1}}{T_{2}}=\frac{2-x}{x}\)
\(\frac{1}{2}=\frac{2-x}{x}\)
x = 4-2x
3x = 4 or \(x=\frac{4}{3}=1.33 m\)
(ii) Now \(y= \frac{stress}{stram}\)
∴ strain = stress/ y
strain in steel wire = \(\frac{T_{1}/A_{1}}{y}\)
strain in brass wire = \(\frac{T_{2}/A_{2}}{y_{2}}\)
Now, \(\frac{T_{1}}{A_{1}y_{1}}=\frac{T_{2}}{A_{2}y_{2}}\)
\(\frac{T_{1}}{T_{2}}=\frac{A_{1}y_{1}}{A_{2}y_{2}}\)
=\(\frac{0.1 cm^{2}\times 20\times 10^{10}Nm^{-2}}{0.2 m^{2}\times 10\times 10^{10}Nm^{-2}}=1\)
again, T₁x = T₂(2-x)
\(1=\frac{2-x}{2}\) 
x = 2 - x
2x = 2 or x = 1m

(i) Thermal radiations are EM waves having wavelength range 1 mm to 100 mm. It is also called IR waves
(ii) They travel in straight lines.
(iii) It obeys the laws of reflection and refraction
(iv) It shows the phenomenon of interference; diffraction and polarisation.
(v) It produces heat when they fall on the substance

Using equation 
v = \(\omega \sqrt { { A }^{ 2 }-{ x }^{ 2 } } \Rightarrow { v }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }^{ 2 } \right) \)
Therefore, at position x₁,
\({ v }_{ 1 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) \)   .................(1)
Similarly, at position x₂,
\({ v }_{ 2 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) \)  ...................(2)
Subtrating (2) from (1), we get
\({ v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x_1 }^{ 2 } \right) -{ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) \)
\(= { \omega }^{ 2 }\left( { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) \)
\({ \omega }=\sqrt { \frac { { v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 } }{ { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } } } \Rightarrow T=2\pi \sqrt { \frac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 } }-{ { v }_{ 2 }^{ 2 } } } } \) ....................(3)
Dividing (1) and (2), we get
\(\frac { { v }_{ 1 }^{ 2 } }{ { v }_{ 2 }^{ 2 } } =\frac { { \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) }{ { \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) } \Rightarrow A=\sqrt { \frac { { { v }_{ 1 }^{ 2 }x }_{ 2 }^{ 2 }-{ v }_{ 2 }^{ 2 }{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 } }-{ { v }_{ 2 }^{ 2 } } } } \) .................(4)
Dividing equation (3) and equation (4), we have
\(\frac { T }{ A } =2\pi \sqrt { \frac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 }x }_{ 2 }^{ 2 }-{ { v }_{ 2 }^{ 2 }x }_{ 1 }^{ 2 } } } \)

(a) For t < 0. No force as Particles is at rest.
For t > 4s, No force again particle comes at rest.
For 0 < t < 4s, as slope of OA is constant so velocity constant i.e., a = 0, so force must be zero.
(b) Impulse = change in momentum
I = m(v-u) = 4(0-0.75) = 3 kg ms^-1
Impulse at t = 4s
I = m(v-u) = 4(0-0.75) = -3 kg ms^-1

Life of the man = 100 years
100 years includes 76 normal years and 24 leap years
Total no of days = 76\(\times\)365 + 24\(\times\)366 = 36524 days
Number of seconds = 36524\(\times\)24\(\times\)3600 = 3.155\(\times\)10° second
\(\text{Number of hearts beats}=\frac{Total\ no\ of\ seconds}{Time\ period\ of\ heart\ beat}=\frac{3.155\times10^9}{0.8s}=3.94\times10^9\)

Relation between power and velocity
The work done by a force \(\overrightarrow{\mathbf{F}}\) for a displacement \(d \vec{r}\) is
\(W=\int \overrightarrow{\mathrm{F}} \cdot d \vec{r}\)  ......(1)
Left hand side of the equation (1) can be written as
\(W=\int d W=\int \frac{d W}{d t} d t\)
(multiplied and divided by dt) (2)
Since, velocity is \(\vec{v}=\frac{d \vec{r}}{d t} ; \overrightarrow{d r}=\vec{v} d t.\) Right hand side of the equation (1) can be written as
\(\int \overrightarrow{\mathrm{F}} \cdot d \vec{r}=\int\left(\overrightarrow{\mathrm{F}}, \frac{d \vec{r}}{d t}\right) d t=\int(\overrightarrow{\mathrm{F}} \cdot \vec{v}) d t\left[v=\frac{d \vec{r}}{d t}\right] \ldots \ldots\) (3)
Substituting equation (2) and equation (3) in equation (1), we get
\(\int \frac{d W}{d t} d t=\int(\overrightarrow{\mathrm{F}} \cdot \vec{v}) d t\)
Or
\(\int\left(\frac{d W}{d t}-\overrightarrow{\mathrm{F}} \cdot \vec{v}\right) d t=0\)
This relation is true for any arbitrary value of dt. This implies that the term within the bracket must be equal to zero, i.e.,
\(\frac{d W}{d t}-\overrightarrow{\mathrm{F}} \cdot \vec{v} =0  \)
\(\frac{d W}{d t} =\overrightarrow{\mathbf{F}} \vec{v}\)
Examples: Motors, Engines and Automobiles
A vehicle of mass 1250 kg is driven with an acceleration 0.2 ms^-2 along a straight level road against an external resistive force 500 N. Calculate the power delivered by the vehicle's engine if the velocity of the vehicle is 30 ms^-1.
Solution
The vehicle's engine has to do work against resistive force and make vehicle to move with an acceleration. Therefore, power delivered by the vehicle engine is
\(P =(\text { resistive force }+\text { mass } \times \text { acceleration }) \text { (velocity) }  \)
\(P =\overrightarrow{\mathbf{F}}_{-\mathrm{ma}} \vec{v}=\left(F_{\text {reistie }}+F\right) \vec{v}  \)
\(P =\overrightarrow{\mathbf{F}}_{\text {tot }} \vec{v}=\left(F_{\text {reithie }}+m a\right) \vec{v}  \)
\(=500 \mathrm{~N}+\left((1250 \mathrm{~kg}) \times\left(0.2 \mathrm{~ms}^{-2}\right)\right)\left(30 \mathrm{~ms}^{-1}\right)=22.5 \mathrm{kw}\)

(i) Consider a rigid body capable of rotating about an axis AB as shown in Figure. Let the force F act at a point P on the rigid body.
(ii) The force F may not be on the plane ABP. The origin O at any random point on the axis AB is taken.

(iii) The torque of the force \(\overset { \rightarrow }{ F } \) about O is \(\overset { \rightarrow }{ \tau } =\overset { \rightarrow }{ r } \times \overset { \rightarrow }{ F } \) . The component of the torque along the axis is the torque of \(\overset { \rightarrow }{ F } \) about the axis. To find it, we should first find the vector \(\overset { \rightarrow }{ \tau } =\overset { \rightarrow }{ r } \times \overset { \rightarrow }{ F } \) and then find the angle φ between ፒ and AB. (Remember here, \(\overset { \rightarrow }{ F } \) is not on the plane ABP). The torque about AB is the parallel component of the torque along AB, which is \(\left| \overset { \rightarrow }{ r } \times \overset { \rightarrow }{ F } \right| \) cos\(\phi \) . And the torque perpendicular to the axis AB is  \(\left| \overset { \rightarrow }{ r } \times \overset { \rightarrow }{ F } \right| \)sin\(\phi \).

The gravitational force exerted by Earth on the mass m near the surface of the Earth is given by
\(\overrightarrow { F } =-\frac { { GMM }_{ e } }{ { R }_{ e }^{ 2 } } \hat { r } \)
Now equating Gravitational force to Newton's second law,
\(ma=-\frac { { GMM }_{ e } }{ { R }_{ e }^{ 2 } } \hat { r } \)
Hence, acceleration is,
\(\overrightarrow { a } =-\frac { { GM }_{ e } }{ { R }_{ e }^{ 2 } } \hat { r } \)
The acceleration experienced by the object near the surface of the Earth due to its gravity is called acceleration due to gravity. It is denoted by the symbol g. The magnitude of acceleration due to gravity is
\(\left| g \right| =\frac { { GM }_{ e } }{ { R }_{ e }^{ 2 } } \hat { r } \)
It is to be noted that the acceleration experienced by any object is independent of its mass. The value of g depends only on the mass and radius of the Earth.

Here ρ = 7.8 g/cm³ ; rw = 10 cm/s
ηw = 8.5\(\times\)10^-4 ρa.s
Pg=1.2g/cm³ ; ηg =13.2 ρa s; Vg=?
Terminal velocity V = \(\frac{2r^{2}(\rho-\rho_{0})g}{9\eta}\)
\(v \propto \frac{(\rho-\rho_{0})}{\eta}\)
When ball falls in water, then
\(V_{w} \propto \frac{(\rho-\rho_{w})}{\eta_{w}}\)
When ball falls in glycerine
\(V_{g} \propto \frac{(\rho-\rho_{g})}{\eta_{g}}\)
\(\frac{V_{g}}{V_{w}}=(\frac{\rho-\rho_{g}}{\rho-\rho_{w}})\frac{\eta_{w}}{\eta_{g}}\)
\(V_{g}={V_{w}}=(\frac{\rho-\rho_{g}}{\rho-\rho_{w}})\frac{\eta_{w}}{\eta_{g}}\)
= 10\((\frac{7.8-1.2}{7.8-1})\times \frac{8.5 \times 10^{-4}}{13.2}\)
= 6.25 \(\times\)10^-4 cm/s

Sound reflects from a harder flat surface, is called as specular reflection.
Specular reflection is observed only when the wavelength of the source is smaller than dimensions of the reflecting surface, as well as smaller than surface irregularities.
When the sound waves hit the plane wall, they bounce off in a manner similar to that of light. Suppose a loudspeaker is kept at an angle with respect to a wall (plane surface), then the waves coming from the source (assumed to be a point source) can be treated as spherical wave fronts (say, compressions moving like a spherical wave front). Therefore, the reflected wave front on the plane surface is also spherical, such that its centre. of curvature (which lies on the other side of plane surface) can be treated as the image of the sound source (virtual or imaginary loud speaker) which can be assumed to be at a position behind the plane surface.

Reflection of so d ' through the curved surface.
The behaviour of sound is different when it is reflected from different surfaces-convex or concave or plane. The sound reflected from a convex surface is spread out and so it is easily attenuated and weakened. Whereas, if it is reflected from the concave surface it will converge at a point and this can be easily amplified. The parabolic reflector (curved reflector) which is used to focus the sound precisely to a point is used in designing the parabolic mics which are known as high directional microphones. We know that any surface (smooth or rough) can absorb sound.

In general our interest our interest is to find how many gas molecules have the range of speed from v to v + dv. This is given by Maxwell's speed distribution function.
\({ N }_{ v }=4\pi N{ \left( \frac { m }{ 2\pi KT } \right) }^{ \frac { 3 }{ 2 } }{ v }^{ 2 }{ e }^{ \frac { { mv }^{ 2 } }{ 2KT } }\)   ....(1)
The above expression is graphically shown as follows

From the figure it is clear that, for a given temperature the number of molecules having lower speed increases parabolically but decreases exponentially after reaching most probable speed. The rms speed, average speed and most probable speed are indicated in the figure. It can be seen that the rms speed is greatest among the three. To Know the number of molecules in the range of speed between \(50 \mathrm{~m} \mathrm{~s}^{-1} \ and \ 60 \mathrm{~m}\mathrm{s}^{-1}\), we need to integrate \(\int_{50}^{60} 4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{\frac{m v^{2}}{2 k T}} d v=\mathrm{N}\left(50\right.\ to \ \left.60 \mathrm{~ms}^{-1}\right)\). In general the number of molecules within the range of speed v and v + dv is given by
\(\int_{v}^{v+d v} 4 \pi N\left(\frac{m}{2 \pi k T}\right)^{\frac{3}{2}} v^{2} e^{\frac{m v^{2}}{2 k T}} d v=N(v \text { to } v+d v) \)
The exact integration is beyond the scope of the book. But we can infer the behaviour of gas molecules from the graph.
(i) The area under the graph will give the total number of gas molecules in the system.
(ii) Figure shows the speed distribution graph for two different temperatures. As temperature increases, the peak of the curve is shifted to the right. It implies that the average speed of each molecule will increase. But the area under each graph is same since it represents the total number of gas molecules.

The specific heat of one mole of a monoatomic gas C_V = \(\frac{3}{2}\)R
For \(\mu\)₁ mole , C_V = \(\frac{3}{2}\)\(\mu\)₁R  C_p = \(\frac{5}{2}\)\(\mu\)₁ R
The specific heat of one mole of a diatomic gas
C_v = \(\frac{5}{2}\)R
For μ₂ mole, C_V = \(\frac{5}{2}\)μ₂ R C_P = \(\frac{7}{2}\)μ₂ R
The specific heat of the mixture at constant volume C_V = \(\frac{3}{2}\)\(\mu\)₁R +\(\frac{5}{2}\)\(\mu\)₂ R
The specific heat of the mixture at constant pressure C_P = \(\frac{5}{2}\)\(\mu\)₁ R + = \(\frac{7}{2}\)\(\mu\)₂ R
The adiabatic exponent \(\gamma =\frac { { C }_{ p } }{ { C }_{ V } } =\frac { 5{ \mu }_{ 1 }+{ 7\mu }_{ 2 } }{ 3{ \mu }_{ 1 }+{ 5\mu }_{ 2 } } \)

The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression v²= u² + 2as, the dimensions of v², u² and 2 as are the same and equal to [L²T^-2].
This method is used to
(i) Convert a physical quantity from one system of units to another.
(ii) Check the dimensional correctness of a given physical equation.
(iii) Establish relations among various physical quantities.
(i) To convert a physical quantity from one system of units to another: This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant. i.e, n₁[u₁] = constant (or) n, n₁[u₁ ] = n₂[u₂].
Consider a physical quantity which has dimension 'a' in mass, 'b' in length and 'c' in time.
If the fundamental units in one system are M₁, L₁ and T₁ and the other system are M₂, L_2, and T₂ respectively, then we can write, n₁ [M₁^a L₁^b T₁^c] = n₂ [ M ₂^a  L₂^b T₂^c]
We have thus converted the numerical value of physical quantity from one system of units into the other system.
Example: Convert 76 cm of mercury pressure into Nm^-2 using the method of dimensions.
Solution: In cgs system 76 cm of mercury pressure =76\(\times\)13.6\(\times\)980 dyne cm^-2 
The dimensional formula of pressure P is [ML^-1T^-2]
\(P_{1}\left[M_{1}^{a} L_{1}^{b} T_{1}^{c}\right]=P_{2}\left[M_{2}^{a} L_{2}^{b} T_{2}^{c}\right]\)
We have
\(P_{2} =\left[\frac{\mathrm{M}_{1}}{\mathrm{M}_{2}}\right]^{a}\left[\frac{\mathrm{L}_{1}}{\mathrm{~L}_{2}}\right]^{b}\left[\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right]^{c} \)
\(M_{1} =1 \mathrm{~g}, \mathrm{M}_{2}=1 \mathrm{~kg}\)
\(L_{1}=1 \mathrm{~cm}, \mathrm{~L}_{2}=1 \mathrm{~m} \)
\(T_{1}=1 \mathrm{~s}, T_{2}=1 \mathrm{~s}\)
So a=1, b=1 and c=-2
Then
\(P_{2} =76 \times 13.6 \times 980\left[\frac{\mathrm{g}}{1 \mathrm{~kg}}\right]^{1}\left[\frac{\mathrm{cm}}{1 \mathrm{~m}}\right]^{-1}\left[\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right]^{-2} \)
\(=76 \times 13.6 \times 980\left[\frac{10^{-3} \mathrm{~kg}}{1 \mathrm{~kg}}\right]^{1}\left[\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~m}}\right]^{-1}\left[\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right]^{-2} \)
\(=76 \times 13.6 \times 980 \times\left[10^{-3}\right] \times 10^{2} \)
\(P_{2} =1.01 \times 10^{5} \mathrm{Nm}^{-2}\)
(ii) To check the dimensional correctness of a given physical equation:
Example: The equation \(1\over 2\) mv² = mgh can be checked by using this method as follows.
Solution: Dimensional formula for
\(\boxed{{1\over 2}mv^2=[M][LT^{-1}]^2=[ML^2T^{-2}]}\)
Dimensional formula for
\(\boxed {mgh=[M][LT^{-2}][L]=[ML^{2}T^{-2}] \\ [ML^{2}T^{2}]=[ML^{2}T^{-2}]}\)
Both sides are dimensionally the same, hence the equations\(1\over 2\) mv² = mgh is dimensionally correct.
(iii) To establish the relation among various physical quantities:
 If the physical quantity Q depends upon the quantities Q₁, Q₂ and Q₃ ie. Q is proportional to Q₁, Q₂ and Q₃.
Then,
\(Q \alpha Q_{1}^{a} Q_{2}^{b} Q_{3}^{c}  \)
\(Q=k Q_{1}^{a} Q_{2}^{b} Q_{3}^{c}\)
where k is a dimensionless constant. When the dimensional formula of Q₁, Q₂ and Q₃ are substituted, then according to the principle of homogeneity, the powers of M, L, T are made equal on both sides of the equation. From this, we get the values of a, b, c.
Example:
Obtain an expression for the time period T of a simple pendulum. The time period T depend upon (i) mass 'm' of the bob (ii) length 'l' of the pendulum and (iii) acceleration due to gravity g at the place where the pendulum is suspended. (Constant k=2π ) i.e
Solution:
\(\boxed{T \alpha m^a l^b g^c \\ T=k.m^al^bg^c}\)
Here k is the dimensionless constant. Rewriting the above equation with dimensions.
\(\boxed{[T^1]=[M^a][L^b][LT^{-2}]^c\\ [M^oL^oT^1]=[M^aL^{b+c}T^{-2c}]}\)
Comparing the powers of M, L and T on both sides, a = 0, b + C = 0, -2c = 1
Solving for a, b and c a = 0, b = 1/2, and c = -1/2
From the above equation
T = k. m^o l^1/2 g-^1/2
T=\(k{1\over g}^{1\over 2}=k\sqrt{1\over g}\)
Experimentally k = 2\(\pi\) , hence \(T=2\pi \sqrt{l\over g}\)

Integration is basically an area finding process. For certain geometric shapes we can directly find the area. But for irregular shapes the process of integration is used.

To find the area of the irregular shaped curve given by fix), we divide the area into rectangular strips as shown in the figure. The area under the curve is approximately equal to sum of areas of each rectangular strip.
This is given by \(A\approx f\left( a \right) \triangle x+f\left( { x }_{ 1 } \right) \triangle x+f\left( { x }_{ 2 } \right) \triangle x+f\left( b \right) \triangle x\)
Where f(a) is the value of the function f(x) at x = a,f(x₁) is the value of f(x) for x = x₁ and so on.
As we increase the number of strips, the area evaluated becomes more accurate. If the area under the curve is divided into N strips, the area under the curve is given by
\(A=\sum _{ n=1 }^{ N }{ { f }_{ n }\left( { x }_{ n } \right) \triangle { x }_{ n } } \)
As the number of strips goes to infinity,\(N\rightarrow \infty \) the sum becomes an integral,
\(A=\int _{ a }^{ b }{ f\left( x \right) dx } \)   (Note: As \(N\rightarrow \infty ,\triangle x\rightarrow 0\))
The integration will give the total area under the curve f(x).

(i) The total kinetic energy (KE) can be written as the sum of kinetic energy due to translational motion (KE_TRANS) and kinetic energy due to rotational motion (KE_ROT) 
KE = KE_TRANS + KE_ROT
(ii) If the mass of the rolling object is M, the velocity of center of mass is v_CM its moment of inertia about center of mass is ICM and angular velocity is ω, then
KE = \(\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }+\frac { 1 }{ 2 } { I }_{ CM }\omega ^{ 2 }\)
(iii) With center of mass as reference:
The moment of inertia (lCM) of a rolling object  about the center of mass is,
I_CM= MK² and vCM= Rω. Here, K is radius of gyration.
\(KE=\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }+\frac { 1 }{ 2 } \left( MK^{ 2 } \right) \frac { v^{ 2 }_{ CM } }{ R^{ 2 } } \\ \\ \)
\(KE=\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }+\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }\left( \frac { { K }^{ 2 } }{ R^{ 2 } } \right) \)
\(KE=\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }\left( 1+\frac { K^{ 2 } }{ R^{ 2 } } \right) \)
(iv) With point of contact as reference: We can also arrive at the same expression by taking the momentary rotation happening with respect to the point of contact (another approach to rolling). if we take the point of contact as 0, then,
\(KE=\frac { 1 }{ 2 } { I }_{ 0 }\omega ^{ 2 }\)
Here, I_o is the moment of inertia of the object about the point of contact. By parallel axis theorem, I_o = I_CM+ MR². Further we can write, I_o = MK² + MR². With v_CM= Rω or 
\(\omega =\frac { { v }_{ CM } }{ R } \)
\(KE=\frac { 1 }{ 2 } \left( MK^{ 2 }+MR^{ 2 } \right) \frac { v^{ 2 }_{ CM } }{ { R }^{ 2 } } \)
\(KE=\frac { 1 }{ 2 } Mv^{ 2 }_{ CM }\left( 1+\frac { { K }^{ 2 } }{ R^{ 2 } } \right) \)
(vi) K.E. in pure rolling can be determined by anyone of the following two cases.
(a) The combination of translational motion and rotational motion about the center of mass. (or)
(b) The momentary rotational motion about the point of contact.