Speed varies exponentially with height 'h'.

\(T=2 \pi \sqrt{\frac{\ell}{g}}\)
\(\text { when a bus is moving } g^{\prime} \sqrt{g^{2}+a^{2}}\)
\(\therefore T \propto \frac{1}{\sqrt{g^{2}+a^{2}}}\)

As it moves with respect to x axis
\(\Delta p_{x}=-m u, \Delta p_{y}=0\)

\(\text { Rigidity modulus }=\frac{1}{3} \times \text { Young's modulus }\)
\(\text { Poisson ratio }=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

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

For freely falling body, velocity while the body, hit the ground \(v=\sqrt{2 g h}\)
\(v_{1}=\sqrt{2 g h_{1}} \text { and } v_{2}=\sqrt{2 g h_{2}} \)
\(\therefore \frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m_{1} \sqrt{h_{1}}}{m_{2} \sqrt{h_{2}}}=\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

\(\text { Torque }=\text { Force } \times \mathrm{r}\)
\(\text { Dimension of torque }\)
\(=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)
\(\text { Energy }=\text { Force } \times \text { displacement }\)
\(\text { Dimension of Energy }\)
\(=\left[\mathrm{MLT}^{-2}\right][\mathrm{L}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\)

Assuming the orbit of the Moon to be circular, the energy of Moon is given by,
\({ E }_{ m }=-\frac { { GM }_{ E }{ M }_{ E } }{ { 2R }_{ m } } \)
where M_E is the mass of Earth 6.02\(\times\)10²⁴ kg; M_m is the mass of Moon 7.35\(\times\)10²² kg; and R_m is the distance between the Moon and the center of the Earth 3.84\(\times\)10⁵ km
G = 6.67\(\times\)10^-11 N m² kg².
\({ E }_{ m }=-\frac { { 6.67\times 10 }^{ -11 }\times 6.02\times { 10 }^{ 24 }\times 7.35\times { 10 }^{ 22 } }{ { 2 }\times 3.84\times { 10 }^{ 5 }\times { 10 }^{ 3 } } \)
\({ E }_{ m }=-38.42\times { 10 }^{ -19 }\times { 10 }^{ 46 }\)
\({ E }_{ m }=-38.42\times { 10 }^{ 46 }\ Joule\)
The negative energy implies that the Moon is bound to the Earth. Same method can be used to prove that the energy of the Earth is also negative.

(i) The value of g varies with height has
g a\(\frac { 1 }{ { \left( R+h \right) }^{ 2 } } \ or\ g\ a\ \frac { 1 }{ { r }^{ 2 } } \)
Thus the graph of g versus V is the parabolic curve AB

(ii) The value of g varies with depth d as
\(g=g\left( 1-\frac { d }{ R } \right) \)i.e g\(\alpha \)(R-d)
Thus the graph of g versus depth d is the straight line AB.
 

As Boltzmann's constant,
\({ K }_{ B }=\frac { R }{ N } \therefore R={ K }_{ B }N\)
Now, PV = nRT = nK_BNT
The number of molecules in the room
\(=nN=\frac { PV }{ { TK }_{ B } } \)
\(=\frac { 1.013\times { 10 }^{ 5 }\times 30 }{ 303\times 1.38\times { 10 }^{ -23 } } \)
\(=\frac { 1.013\times 30\times { 10 }^{ 5 }\times { 10 }^{ 23 } }{ 303\times 1.38 } \)
\(=\frac { 1.013\times 30\times \times { 10 }^{ 28 } }{ 418.14 } \)
\(=\frac { 30.39\times { 10 }^{ 28 } }{ 418.4 } \)
= 0.0726\(\times\)10²⁸
= 7.26\(\times\)10²⁶

(i) While skating a person presses the snow down ward with his weight i.e increases pressure on snow below the skates.
(ii) Due to it, the melting point of ice becomes lower & ice melts into water, which acts as lubricant for skates.
(iii) As the skater crosses the & now, the pressure on water is released & it is again converted into ice.

Observer is stationary
(i) Source (train) is moving towards an observer:
Apparent frequency due to train arriving station is
\({ f }_{ in }=\frac { f }{ \left( 1-\frac { { v }_{ s } }{ v } \right) } =\frac { 240 }{ \left( 1-\frac { 8 }{ 330 } \right) } =246Hz\)
(ii) Source (train) is moving away form an observer:
Apparent frequency due to train leaving station is
\({ f }_{ out }=\frac { f }{ \left( 1-\frac { { v }_{ s } }{ v } \right) } =\frac { 240 }{ \left( 1-\frac { 8 }{ 330 } \right) } =234Hz\)
So the number of beats = | f_in -f_out| = (246-234) = 12

Here \(\omega=200\)rad s^-1, \(\tau\)-=180 Nm
∵ Power P = \(\tau \omega=180 \times 200\) = 36,000 W = 36 kW. 

When a body is pulled on a rough, horizontal surface with constant velocity. Work is done on the body but K.E. remains unchanged.

(i) Basic laws of electricity and magnetism led to the discovery of wireless communication technology which has shrunk the world with effective communication over large distances.
(ii) The launching of satellite into space has revolutionized the concept of communication.
(iii) Microelectronics, lasers, computers, superconductivity and nuclear energy have comprehensively changed the thinking and living style of human beings.

When a track goes round a level curve, the force of friction between the tyres and the road provides the necessary centripetal force. If the frictional force, which acts as centripetal force and keeps the body moving along the circular road is not enough to provide the necessary centripetal force, the car will skid. In order to avoid skidding while going round a curved path the outer edge of the road is raised above the level of the inner edge. This is known as banking of curved roads or tracks.
Let the surface of the road makes angle ፀ with horizontal surface. Then the normal force makes the same angle d with the vertical. When the car takes a turn, there are two forces acting on the car:
(a) Gravitational force mg (downwards)
(b) Normal force N (perpendicular to surface).
We can resolve the normal force into two components. N cos ፀ and N sin ፀ as shown in Figure. The component N cos ፀ balances the downward gravitational force 'mg' and component N sin ፀ will, provide the necessary centripetal acceleration. By using Newton second law
N cos ፀ = mg .....(1)
N sin ፀ = \(\frac { { mv }^{ 2 } }{ r } \)  ....(2)
Eq (2) / (1) ⇒ tan ፀ = \(\frac { { v }^{ 2 } }{ rg } \)
v= \(\sqrt { rg\tan\theta } \)
Need Banking of Tracks
1. The banking angle ፀ and radius of curvature of the road or track determines the safe speed of the car at the turning.
2. If the speed of car exceeds this safe speed, then it starts to skid outward but frictional force comes into effect and provides an additional centripetal force to prevent the outward skidding.
3. At the same time, if the speed of the car is little lesser than safe speed, it starts to skid inward and frictional force comes into effect, which reduces centripetal force to prevent inward skidding. 
4. However if the speed of the vehicle is sufficiently greater than the correct speed, then frictional force cannot stop the car from skidding.

Angular displacement: While a particle is revolving around a point in a circular path, the angle described by the particle about the axis of rotation (or at the centre of the circle) in a given time is called angular displacement. Its unit is radian.
Angular velocity: The rate of change of angular displacement is called angular velocity. Its unit is rad s^-1
\(\omega =\frac { d\theta }{ dt } \)

We can distinguish between longitudinal and transverse ways by performing polarization experiments. Transverse waves can be polorised while longitudinal waves cannot be poloriused.

(a) Suppose a particle of mass m executes SHM of time period T. The displacement of the particle of any instant A is given by
\(\mathrm{y} =\mathrm{A} \sin \omega t  \) .....(1)
\(\text {Velocity } \mathrm{v} =\frac{d y}{d t}=\frac{d}{d t}\)
\((\mathrm{A} \sin \omega t) =\omega A \cos \omega t\)
\( \therefore \text {Kinetic energy } E_{k} =\frac{1}{2} m v^{2}=\frac{1}{2} m \omega^{2} \cdot A^{2} \cos ^{2} \omega t \)   ......(2)
Potential energy \(\mathrm{E}_{\mathrm{p}}=\frac{1}{2} m \omega^{2} y^{2}\)
\(\mathrm{E}_{p}=\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t\)   .....(3)
Average K.E for one period of oscillation is
\(E_{k_{\alpha v}}=\frac{1}{T} \int_{0}^{T} E_{k} d t \)
\(=\frac{1}{T} \int \frac{1}{2} m \omega^{2} A^{2} \cos ^{2} \omega t d t \)
\(=\frac{1}{2 T} m \omega^{2} A^{2} \int_{0}^{T} \frac{(1+\cos 2 \omega t)}{2} d t \)
\(E_{k_{o v}}=\frac{1}{4 T} m \omega^{2} A^{2}\left[t+\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T}=\frac{1}{4 T} m \omega^{2} A^{2}(T) \)
\(E_{k_{\alpha v}}=\frac{1}{4 T} m \omega^{2} A^{2}\)  .....(4)
Average potential energy over a period of oscillation is
\(E_{p_{\omega}} =\frac{1}{T} \int_{0}^{T} E_{p} d t=\frac{1}{T} \int_{0}^{T} \frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t d t  \)
\(=\frac{1}{2 T} m \omega^{2} A^{2} \int_{0}^{T}\left(\frac{1-\cos 2 \omega t}{2}\right) d t  \)
\(=\frac{1}{4 T} m \omega^{2} A^{2}\left[t-\frac{\sin 2 \omega t}{2 \omega}\right]_{0}^{T}=\frac{1}{4 T} m \omega^{2} A^{2} C_{1}  \)
\(E_{p_{a v}} =\frac{1}{4 T} m \omega^{2} A^{2} \)   .....(5)
clearly from equation (4) and (5)
\(E_{k_{\mathrm{ow}}}=E_{p_{\mathrm{wo}}}\)
Average value of K.E = Average value of P.E
(b) Total energy
\(\text {T.E } =\frac{1}{2} m \omega^{2} y^{2}+\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right)  \)
\(\text {But y } =\mathrm{A} \sin \omega t  \)
\(\text {T.E } =\frac{1}{2} m \omega^{2} A^{2} \sin ^{2} \omega t+\frac{1}{2} m \omega^{2} A^{2} \cos ^{2} \omega t \)
\(=\frac{1}{2} m \omega^{2} A^{2}\left(\sin ^{2} \omega t+\cos ^{2} \omega t\right)\)
\(But \sin ^{2} \omega t+\cos ^{2} \omega t=1\)
Equation (6) becomes
\(\text {T.E }=\frac{1}{2} m \omega^{2} A^{2}(1)=\frac{1}{2} m \omega^{2} A^{2}\)
Average potential energy = Average kinetic energy
\(=\frac{1}{2} \text { (Total energy) }\)

Frequency of simple harmonic motion is defined as the number of oscillations produced by the Particle.

For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change.

All naturally occuring processes are irreversible. Here we give some interesting examples.
(a) When we open a gas bottle, the gas molecules slowly spread into the entire room. These gas molecules can never get back in to the bottle. 

(b) Suppose one drop of an ink is dropped in water, the ink droplet slowly spreads in the water. It is impossible to get the ink droplet back.
(c) When an object falls from some height, as soon as it hits the earth it comes to rest. All the kinetic energy of the object is converted to kinetic energy of molecules of the earth surface, molecules of the object and small amount goes as sound energy. The spreaded kinetic energy to the molecules never collected back and object never goes up by itself. 
Note that according to first law of thermodynamics all the above processes are possible in both directions. But second law of thermodynamics forbids The processes to occur in the reverse direction. The second law of thermodynamics is one of the very important laws of nature. It controls the way the natural processes occur.

Pascal's law states that, if the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.

Angle of inclination θ = 45⁰
∴ Coefficient of friction μ= tan θ = tan45⁰=1

This paradox arises due to wrong application of Newton's second and third laws. Before applying Newton's laws, we should decide 'what is the system?'. Once we identify the 'system', then it is possible to identify all the forces acting on the system. We should not consider the force exerted by the system. If there is an unbalanced force acting on the system, then it should have acceleration in the direction of the resultant force. By following these steps we will analyse the horse and cart motion.
If we decide on the cart+horse as a 'system', then we should not consider the force exerted by the horse on the cart or the force exerted by cart on the horse. Both are internal forces acting on each other. According to Newton's third law, total internal force acting on the system is zero and it cannot accelerate the system. The acceleration of the system is caused by some external force. In this case, the force exerted by the road on the system is the external force acting on the system. It is wrong to conclude that the total force acting on the system (cart+horse) is zero without including all the forces acting on the system. The road is pushing the horse and cart forward with acceleration. As there is an external- force acting on the system, Newton's second law has to be applied and not Newton's third law. The following figures illustrates this.

If we consider the horse as the 'system', then there are three forces acting on the horse.
(i) Downward gravitational force (m_hg)
(ii) Force exerted by the road (F_r)
(iii) Backward force exerted by the cart (F_c)
It is shown in the following figure.

F_r - Force exerted by the road on the horse
F_c - Force exerted by the cart on the horse
F^丄_r-Perpendicular component of F_r=N
F^||_r-Parallel component of Fr which is reason for forward movement.
The force exerted by the road can be resolved into parallel and perpendicular components. The perpendicular component balances the downward gravitational force. There is parallel component along the forward direction. It is greater than the backward force (F_c). So there is net force along the forward direction which causes the forward movement of the horse.
If we take the cart as the system, then there are three forces acting on the cart.
(i) Downward gravitational force (m_cg)
(ii) Force exerted by the road (F_r)
(iii) Force exerted by the horse (F_h)
It is shown in the figure.

The force exerted by the road (\(\vec { { F }_{ r } } \) ) can be resolved into parallel and perpendicular components. The perpendicular component cancels the downward gravity (m_cg). Parallel component acts backwards and the force exerted by the horse (\(\vec { { F }_{ h } } \) ) acts forward. Force (\(\vec { { F }_{ h } } \) ) is greater than the parallel component acting in the opposite direction. So there is an overall unbalanced force in the forward direction which causes the cart to accelerate forward.
If we take the cart+horse as a system, then there are two forces acting on the system.
(i) Downward gravitational force (m_h + m_c)g
(ii) The force exerted by the road (F_r) on the system.
It is shown in the following figure.

(iii) In this case the force exerted by the road (F_r) on the system (cart+horse) is resolved in to parallel and perpendicular components. The perpendicular component is the normal force which cancels the downward gravitational force (m_h + m_c)g. The parallel component of the force is not balanced, hence the system (cart+horse) accelerates and moves forward due to this force.

Lifting a table from the floor by two persons. Pushing the car by two persons.

Electrical energy is measured in the unit called kilowatt hour (kWh).
1 electrical unit = 1 kWh = 1\(\times\)(10³ W)\(\times\)(3600 s)
1 electrical unit = 3600\(\times\)10³ Ws
1 electrical unit = 3.6\(\times\)10⁶J
1 kWh =3.6\(\times\)10⁶J

Sprinkled water takes circular motion.
The centripetal acceleration \(=\frac{v^2}{r}\). The centripetal acceleration is provided by the acceleration due to gravity.
\(\therefore g=\frac { v^2 }{ r } \ or \ r=\frac { { v }^{ 2 } }{ g } \)
∴ Area around the fountain that gets wet = πr²
\(=\pi { \left( \frac { { r }^{ 2 } }{ g } \right) }^{ 2 }=\frac { { \pi r }^{ 4 } }{ { g }^{ 2 } } \)

1 light year is distance travelled by light in vacuum in one year. 1 Light Year = 9.467\(\times\)10¹⁵ m.

Number of molecule in 0.014 kg of nitrogen
\(n=\frac { 0.014\times { 10 }^{ 3 } }{ 27 } \)=0.0005\(\times\)103
\(=0.5=\frac { 1 }{ 2 } \)mole.
Molecular specific heat
\({ C }_{ V }=\frac { 5 }{ 2 } \times R\)
\({ C }_{ V }=\frac { 5 }{ 2 } R=\frac { 5 }{ 2 } \times 2\)
\(=\frac { 10 }{ 2 } =5\)cal / molecule k
[\(\therefore\) Nitrogen diatom molecule S_o=\(\frac{5}{2}\)]
Ratio of volume  \(\frac { { V }_{ 2 } }{ { V }_{ 1 } } =\sqrt { \frac { { T }_{ 2 } }{ { T }_{ 1 } } , } { T }_{ 2 }=4{ T }_{ 1 }\)
\(\Delta\) T=T₂-T₁=4T₁-T₁=3T₁
= 3\(\times\)300=900K
\(\Delta\)Q = nC_V\(\Delta\)T= \(\frac{1}{2}\)\(\times\)5\(\times\)900
\(\therefore\)Heat transfer \(\Delta\)Q = 2250 cal

Consider a pipe AB of varying cross sectional area a₁ and a₂ such that a₁ > a_2. A non-viscous and incompressible liquid flows steadily through the pipe, with velocities v₁ and v₂ in area a₁ and a₂, respectively as shown in Figure.

Let m1 be the mass of fluid flowing through section A in time Δ_t, m₁ = (a₁v₁Δt) p
Let m2 be the mass of fluid flowing through section B in time Δt, m₂= (a₂v₂Δt) p
For an incompressible liquid, mass is conserved m₁ =m₂
a₁v₁Δtρ = a₂v₂Δtρ
a₁v₁ = a₂v₂ ⇒ av = constant
which is called the equation of continuity and it is a statement of conservation of mass in the flow of fluids.
In general, a v = constant, which means that the volume flux or flow rate remains constant throughout the pipe. In other words, the smaller the cross section, greater will be the velocity of the fluid.

F ∝ m^a v^b r^c ;
F = k m^a v^b r^c
where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k= 1, we have
[MLT^-2] = [M]^a [LT^-1]^b [L]^c
= [M^aL^bT^-bL^c]
[MLT^-2] = [M^aL^b+cT^-b]
Comparing the powers of M, L and T on both sides
 a = 1 ; b + c = 1; -b =-2
2 + c = 1 ; b = 2;
a = 1 b = 2 and c = -1
From the above equation we get
F = m^av^br^c
F =m¹v²r^-1
or F = \(\frac { m{ v }^{ 2 } }{ r } \)

For two non-zero vectors \(\vec A\) and \(\vec B\) which are inclined to each other at an angle θ, the difference \(\vec A-\vec B\) is obtained as follows. First obtain \(-\vec B\) as in figure (Subtraction of vectors). The angle between \(\vec A\) and \(-\vec B\) is 180-θ.The difference \(\vec A-\vec B\) is the same as the resultant of \(\vec A\)and \(-\vec B\). 

We can write \(\vec A-\vec B=\vec A+(-\vec B)\) and using the equation, we have \(\left| \vec { A } -\vec { B } \right| =\sqrt { { A }^{ 2 }+{ B }^{ 2 }+2ABcos\left( 180-\theta \right) } \)
Since, cos(180 -θ ) = -cosθ, we get \(\Rightarrow \left| \vec { A } -\vec { B } \right| =\sqrt { { A }^{ 2 }+{ B }^{ 2 }-2ABcos\theta } \)
Again from the figure, and using an equation similar to the equation
\(tan\ { \alpha }_{ 2 }=\frac { Bsin\left( 180^{ 0 }-\theta \right) }{ A+Bcos\left( 180^{ 0 }-\theta \right) } \)
But sin(180°- θ ) = sine hence we get
\(\Rightarrow tan\ { \alpha }_{ 2 }=\frac { Bsin\theta }{ A-Bcos\theta } \)

Kinetic energy of rotation = \(\frac { 1 }{ 2 } I{ \omega }^{ 2 }\)
=\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 2 } { mr }^{ 3 } \right) { \omega }^{ 2 }\)
=\(\frac { 1 }{ 4 } \times 20\times (0.25)^{ 2 }\times 100\times 100\)
=
= 3125 J

Mass of a car (m) = 2\(\times\)10³ kg
The coefficient of friction (μ)= 0.05
Speed (v) = 40 ms^-1
Displacement (s) = 2 km = 2000 m
Power, p = ?
Watt, w = ?
F = before of friction = umg
Power = F v = μmg\(\times\)v
= 0.05\(\times\)2\(\times\)10³\(\times\)10\(\times\)40
= 4\(\times\)10⁴ Watt
= 40 kW
Workdone, W = F.S = μmgs
= 0.05\(\times\)2\(\times\)10³ \(\times\)10\(\times\)2\(\times\)10³
= 2\(\times\)10⁶ J