Rate of heat production
\(=\frac{\text { Work done }}{\text { Timetaken }}\)
Terminal velocity
\(v=\frac{2}{a} \frac{r^{2}(f-\sigma)}{\eta} g \)
\(v \propto r^{2} \)
\(\text { Work } \propto \text { Force } \times \text { distance }\)
\(\propto r^{2} \times r \Rightarrow 2^{5}\)

Acceleration of the solid sphere while rolling down without slipping
\(a_{1}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}}\)
Acceleration developed while slipping down \(a_{2}=g \sin \theta\)
\(\text { Required ratio } \frac{a_{1}}{a_{2}}=\frac{g \sin \theta}{1+\frac{k^{2}}{r^{2}}} / g \sin \theta\)
\(\frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{k^{2}}{r^{2}}}\)
\(\text { For a solid sphere } \frac{k^{2}}{r^{2}}=\frac{2}{5}\)
\(\therefore \text { Ratio of accelerations } \frac{a_{1}}{a_{2}}=\frac{1}{1+\frac{2}{5}}\)
\(=\frac{1}{5+\frac{2}{5}}=\frac{1}{\frac{7}{5}}=\frac{5}{7}\)
\(\therefore a_{1}: a_{2}=5: 7 \)

\(K \cdot E=\frac{1}{2} I \omega^{2} ; \quad L=I \omega ; \quad K \cdot E=\frac{2^{2}}{2^{2}} \)
\(\therefore K \cdot E \alpha L^{2} \quad E_{1} \alpha L_{1}^{2} \quad E_{2} \alpha L_{2}^{2}\)
\(\frac{E_{1}}{E_{2}}=\left(\frac{L_{1}}{L_{2}}\right)^{2} \)
\(\text { Here } E_{1}=E \quad E_{2}=\frac{E}{2} \)
\(L_{1}=L \quad \quad L_{2}=? \)
\(\frac{E}{\frac{E}{2}}=\left(\frac{L}{L_{2}}\right)^{2} \)
\(\frac{2 E}{E}=\left(\frac{L}{L_{2}}\right)^{2}\left(\frac{L_{1}}{L_{2}}\right)^{2}=2 \)
\(\therefore \frac{L}{L_{2}}=\sqrt{2} \)
\(L_{2}=\frac{L}{\sqrt{2}} \)

Using law of conservation of momentum,
\(2 m v =\sqrt{m^{2} v^{2}+m^{2} v^{2}} \)
\(=\sqrt{2 m^{2} v^{2}} \)
\(v =\frac{\sqrt{2} m v}{2 m}=\frac{v}{\sqrt{2}} \)
Energy released in explosion = \(2 \times \frac{1}{2} m v^{2} +\frac{1}{2} \times 2 m \times\left(\frac{v^{2}}{\sqrt{2}}\right)^{2} \)
\(=m v^{2}+m \times \frac{v^{2}}{2} \)
\(=\frac{3}{2} m v^{2} \)

Initially velocity has maximum value and at maximum height velocity becomes zero. After that the velocity becomes negative

F = kv²
Dimensional of k
\(=\frac{\text { Dimension of } \mathrm{F}}{\text { Dimension of }(v)^{2}}\)
\(=\frac{\mathrm{MLT}^{-2}}{\left(\mathrm{LT}^{-1}\right)^{2}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^{2} \mathrm{~T}^{-2}} \)
\(=\left[\mathrm{ML}^{-1-2} \mathrm{~T}^{-2+2}\right] \)
Dimension of proportionality constant \(=\left[\mathrm{ML}^{-1} \mathrm{~T}^{0}\right]\)

\(T =2 \pi \sqrt{\frac{l}{g}} \)
\(g =4 \pi^{2} l / T^{2} \)
\(\frac{d g}{g} =\frac{d l}{l}-\frac{2 d T}{T} \)
\(\frac{d g}{g} \% =\left(\frac{d l}{l}\right) \%-2\left(\frac{d T}{T}\right) \% \)
\(=1 \%-2 \times(-3 \%) \)
\(=1+6=7 \% \)

Process 1 to 2 = increase in volume. So heat must be added.
Process 2 to 3 = Volume remains constant. Increase in temperature. The given heat is used to increase the internal energy.
Process 3 to 1 = Pressure remains constant. Volume and Temperature are reduced. Heat flows out of the system. It is an isobaric compression where the work is done on the system.
In the process marked as 1-2, the gas undergoes isothermal 'expansion. It receiver certain amount of heat from the outside of spends all this heat in doing work the internal energy of the gas remains unchanged.
In the process marked as 2-3, the gas is heated isochorically (at constant volume). Since its volume does not change, no work is done. The internal energy of the gas is increased only due to the heat transferred to the gas from the outside.
In the process exhibited by 3-1, the gas is compressed isobarically (at constant pressure) of its temperature drops work is done on the gas, but its internal energy is reduced. This means the gas intensively gives up heat to the medium.

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.
Application of pascal's law
A practical application of Pascal's law is the hydraulic lift which is used to lift a heavy load with a small force. It is a force multiplier. It consists of two cylinders A and B connected to each other by a horizontal pipe, filled with a liquid.

They are fitted with frictionless pistons of cross sectional areas A₁ and A₂ (A₂ > A₁). Suppose a downward force F is applied on the smaller piston, the pressure of the liquid under this piston increases to\(\left(\right. where, \left.P=\frac{F_{1}}{A_{1}}\right).\) But according to Pascal's law, this increased pressure P is transmitted undiminished in all directions. So a pressure is exerted on piston B. Upward force on piston B is
\(\mathrm{F}_{2}=\mathrm{P} \times \mathrm{A}_{2}=\frac{F_{1}}{A_{1}} \times A_{2} \Rightarrow \mathrm{F}_{2}=\frac{A_{2}}{A_{1}} \times F_{1}\)
Hence by changing the force on the smaller piston A, the force on the piston B has been increased by the factor \(\frac{A_{2}}{A_{1}}\) and this factor is called the mechanical advantage of the lift.

\(F=-\eta A\frac { dv }{ dx } \)
n magnitude, \(\eta =\frac { F }{ A } \frac { dv }{ dx } \)
\(=\frac { \left( 2.5N \right) \quad \left( 2.5\times { 10 }^{ -3 }m \right) }{ \left( 2.5\times { 10 }^{ -4 }{ m }^{ 2 } \right) \left( 3\times { 10 }^{ -2 }{ ms }^{ -1 } \right) } \)
= 0.083\(\times\)10³ Nm^-2s

The Earth to Sun distance \(\mathrm{R}_{\mathrm{E}}=150 \times 10^{6} \mathrm{~km}=150 \times 10^{9} \mathrm{~m}\)
Mass of the Earth \( \mathrm{M}_{E}=5.9 \times 10^{24} \mathrm{~kg}\)
Mass of the Sun \(\mathrm{M}_{\mathrm{s}}=1.9 \times 10^{30} \mathrm{~kg}\)
Gravitational energy of the Earth and Sun
\(\mathrm{U} =-\frac{G m_{1} m_{2}}{r}  \)
\(\mathrm{U} =-\frac{G m_{E} m_{s}}{R_{E}}  \)
\(=-\left[\frac{6.67 \times 10^{-11} \times 5.9 \times 10^{24} \times 1.9 \times 10^{30}}{150 \times 10^{9}}\right]  \)
\(=-\left[\frac{6.67 \times 5.9 \times 1.9}{150} \times 10^{-11+24+30-9}\right]  \)
\( =-\left[\frac{74.7707}{150} \times 10^{34}\right]  \)
\(=-0.49847 \times 10^{34}  \)
\(\mathrm{U}=-49.847 \times 10^{32} \mathrm{~J}\)

From the safe turn condition the speed of the vehicle (v) must be less than or equal to \(\sqrt { { \mu }_{ s }rg } \)
v ≤ \(\sqrt { { \mu }_{ s }rg } \)
\(\sqrt { { \mu }_{ s }rg } \)=\(\sqrt { 0.81\times 10\times 10 } \)=9 ms^-1
For car C, \(\sqrt { { \mu }_{ s }rg } \) is less than v.
The speed of car A, B and Care 7 m s^-1, 8 m s^-1 and 10 m s^-1 respectively. The cars A and B will have safe turns. But the car C has speed 10 m s^-1 while it turns which exceeds the safe turning speed. Hence, the car C will skid.

Mass of the man (m₁) is, m₁= 50 kg
Mass of the boat (m₂) is, m₂ = 300 kg
With respect to a stationary observer:
The man moves with a velocity, v₁ = 2 m s^-1 and the boat moves with a velocity v₂ (which is to be found)
(i) To determine the velocity of the boat with respect to a stationary observer on land. As there is no external force acting on the system, the man and boat move due to the friction, which is an internal force in the boat-man system. Hence, the velocity of the center of mass is zero (V_CM= 0). Using equation,
\(\overrightarrow{{v}}_{CM}=\frac{\sum { m_1v_1} }{\sum{m_1}}=\frac{m_1v_1+m_2v_2}{m_1+m_2}\)
0 = \(\frac{\sum{m_1v_1}}{\sum{m_1}}=\frac{m_1v_1+m_2v_2}{m_1+v_2}\)
0 = m₁v₁ + m₂v₂ - m₂v₂ = m₁v₁
-m₂v₂ = m₁v₁
\(v_2=-\frac{m_1}{m_2}v_1\)
\(v_2= -\frac{50}{300}\times 2= -\frac{100}{300}\)
v₂ = -0.33 ms^-1
The negative sign in the answer implies that the boat moves in a direction opposite to that of the walking man on the boat to a stationary observer on land.
(ii) To determine the velocity of the boat with respect to the walking man: We can find the relative velocity as
v₂₁ = v₂ - v₁
where, v₂₁ is the relative velocity of the boat with respect to the walking man.
v₂₁ = (-0.33)-(2)
v₂₁ = -2.33 ms^-1
The negative sign in the answer implies that the boat appears to move in the opposite direction to the man walking in the boat.

Let us take, m₁= 3 kg and m₂ = 5 kg
(i) To find center of mass from the origin: The point masses are at positions, x₁ = 4 m, x₂ = 8m from the origin along X-axis.

The centre of mass x_CM can be obtained using equation 5.4.
\({ x }_{ CM }=\frac { { m }_{ 1 }{ x }_{ 1 }+{ m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
\({ x }_{ CM }=\frac { \left( 3\times 4 \right) +\left( 5\times 8 \right) }{ 3+5 } \)
\({ x }_{ CM }=\frac { 12+40 }{ 8 } =\frac { 52 }{ 8 } =6.5m\)
(ii) To find the center of mass from 3 kg mass: The origin is shifted to 3 kg mass along X-axis. The position of 3 kg point mass is zero (x₁ = 0) and the position of 5 kg point mass is 4 m from the shifted origin (x₂ = 4 m).
\({ x }_{ CM }=\frac { \left( 3\times 0 \right) +\left( 5\times 4 \right) }{ 3+5 } \)
\({ x }_{ CM }=\frac { 0+20 }{ 8 } =\frac { 20 }{ 8 } =2.5m\)

The center of mass is located 2.5 m from 3 kg point mass (and 1.5 m from the 5 kg point mass) on x-axis.
1. This result shows that the center of mass is located closer to larger mass.
2. If the origin is shifted to the center of mass, then the principle of moments holds good m₁x₁ = m₂x₂, 3\(\times\)2.5 = 5\(\times\)1.5; 7.5 = 7.5.
When we compare case
(i) with case
(ii), the x_CM = 2.5 m from 3 kg mass could also be obtained by subtracting 4 m (the position of 3 kg mass) from 6.5 m, where the center of mass was located in case (i).

Note: Before applying Newton's laws, the following steps have to be followed:
1. Choose a suitable inertial coordinate system to analyse the problem. For most of the cases we can take Earth as an inertial coordinate system.
2. Identify the system to which Newton's laws need to be applied. The system can be a single object or more than one object.
3. Draw the free body diagram.
4. Once the forces acting on the system are identified, and the free body diagram is drawn, apply Newton's second law. In the left hand side of the equation, write the forces acting on the system in vector notation and equate it to the right hand side of equation which is the product of mass and acceleration. Here, acceleration should also be in vector notation.
5. If acceleration is given, the force can be calculated. If the force is given, acceleration can be calculated.

By following the above steps:
We fix the inertial coordinate system. on the ground as shown in the figure.
The forces acting on the mango are
(i) Gravitational force exerted by the Earth on the mango acting downward along negative y-axis.
(ii) Tension (in the cord attached to the mango) acts upward along positive y-axis.
The free body diagram for the mango is shown in the figure


\(\bar { { F }_{ g } } =mg(-\hat { j } )=-mg\hat { j } \)
Here, mg is the magnitude of the gravitational force and \((\hat { -j } )\) represents the unit vector in negative y-direction.
\(\vec { T } =T\hat { j } \)
Here T is the magnitude of the tension force and \((\hat { j } )\) represents the unit vector in positive y direction.
\(\vec { F_{ net } } ={ F }_{ g }+\vec { T } =-mg\hat { j } +T\hat { j } =(T-mg)\hat { j } \)
From Newton's second law \(\vec { F_{ net } } =m\vec { a } \)

Since the mango is at rest with respect to us (inertial coordinate system) the acceleration is zero \((\vec { a } =0)\)
So, \(\vec { F_{ net } } =m\vec { a } \) = 0
\((T-mg)\hat { j } \) = 0
By comparing the components on both sides of the above equation, we get T - mg = 0. So the tension force acting on the mango is given by T = mg.
Mass of the mango m = 400 g and g = 9.8 ms^-2
Tension acting on the mango is T = 0.4\(\times\)9.8 = 3.92 N.

By following the law of triangular addition, the resultant vector is given by \(\vec R\) = \(\vec A\) + \(\vec B\) as illustrated below.
The magnitude of the resultant vector \(\vec R\) is given by
\(R=|\vec R|=\sqrt{5^2+7^2+2\times 5\times 7\cos 60^o}\)
\(R=\sqrt{25+49+\frac{70\times 1}{2}}=\sqrt{109}\) units

The angle \(\alpha\) between \(\vec R\) and \(\vec A\) is given by
\(\tan\alpha=\frac{B\sin\theta}{A+B\cos\theta}\)
\(\tan\alpha=\frac{7\times\sin60^o}{5+7\cos60^o}=\frac{7\sqrt{3}}{10+7}=\frac{7\sqrt{3}}{17}\) = 0.713
\(\therefore\alpha=35^o\)

By the principle of homogeneity, a / V² is of the dimensions of pressure and b is of the dimensions of volume.
[a] = [pressure] [V²] = [ML⁻¹T⁻²] [L⁶]
= [ML⁵T^-2]
[b] = [V] = L³

Mass of ball(m) = \(\frac { 35 }{ 100 } =0.035kg\)
A ball hits by a stumps at an angle (\(\theta\)₁) = 30°
Ball rebounds at an .angle (\(\theta\)₂) = 60°
Change in momentum along horizontal direction
= -mu cos30° - (mu cos30⁰)
= -2 mu cos30°
= -2\(​​\times\)0.035\(​​\times\)20\(​​\times\)cos30°
= -2 \(​​\times\)0.035\(​​\times\)20\(​​\times\)\(\sqrt { \frac { 3 }{ 2 } } \)  \(\quad \left[ \because cos{ 30 }^{ 0 }=\sqrt { \frac { 3 }{ 2 } } \right] \)
= \(-1.4\times \frac { \sqrt { 3 } }{ 2 } =1.21\quad kgms^{ -1 }\)
The impulse received by a ball j = 1.2 kg ms^-1

(i) Instantaneous acceleration or acceleration of a particle at time 't' is given by the ratio of change in velocity over \(\Delta t\), as \(\Delta t\) approaches zero.
Acceleration \(\overrightarrow { a } =\lim _{ \Delta t\rightarrow 0 }{ \frac { \Delta \overrightarrow { v } }{ \Delta t } =\frac { d\overrightarrow { v } }{ dt } } \)
(ii) In other words, the acceleration of the particle at an instant t is equal to rate of change of velocity.
(iii) Acceleration is a vector quantity. Its SI unit is ms^-2 and its dimensional formula is M^oL¹T^-2
(iv) Acceleration is positive if its velocity is increasing, and is negative if the velocity is decreasing. The negative acceleration is called retardation or deceleration.

3.08 x 10¹⁶m equivalent to 1 parsec
1 metre is equivalent to \({{1}\over{3.08\times{10}^{16}}}\)
= 0.324\(\times\)10^-16
= 3.24\(\times\)10^-17 m
In one metre 3.24\(\times\)10^-17 parsec are present.

Frequency of a vibrating body \(\gamma \alpha \frac{1}{l} \sqrt{\frac{F}{M}}\)
\( a\text { Dimension of frequency } =\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}  \)
\(\text {Dimension of length } =\mathrm{L}=\mathrm{M}^{0} \mathrm{LT}^{0}  \)
\(\text {Dimension of Force } =\mathrm{MLT}^{-2}  \)
\(\text {Dimension of Mass } =\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{0}  \)
\(\text {Frequency } \gamma =x  \)
\(\gamma =\mathrm{K}\left[\mathrm{F}^{x}\right][\mathrm{M}]^{y}[\mathrm{~L}]^{z}\)
Using dimensions we get
\(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}=\left[\mathrm{MLT}^{-2}\right]^{x}[\mathrm{M}]^{y}[\mathrm{~L}]^{z}  \)
\(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}=\mathrm{M}^{x+y} \mathrm{~L}^{x+z} \mathrm{~T}^{-2 x}\)
Comparing the powers we get
x + y = 0  z = -x = \(\frac{1}{2} \)
x + z = 0  y = -x = \(\frac{-1}{2} \)
-2 x = -1 
\(\therefore x =\frac{-1}{-2}=\frac{1}{2}  \)
\(\therefore \gamma =1 \times[\mathrm{F}]^{1 / 2}[\mathrm{M}]^{-1 / 2}[\mathrm{~L}]^{-1 / 2}  \)
\(\gamma =\frac{1}{l} \sqrt{\frac{F}{m}}\)

(i) According to law of conservation of angular momentum L = mur = constant.
∴ The comet moves faster when it is close to the sun and moves slower when it is farther away from the sun.
The speed of the comet does not remain constant.
(ii) As no external torque is acting on angular momentum \(\tau =\frac { dL }{ dt } =0\)
∴ Angular momentum of the comet remains constant.
(iii) Kinetic energy = \(\frac{1}{2}\)mv². As the linear speed of the comet changes, kinetic energy also changes. Potential energy also changes as the kinetic energy changes.
But Total energy remam constant throughout its orbit.

The Avogadro's number N_A is defined as the number of carbon atoms contained in exactly 12 g Of ¹²C.

From the ideal gas equation, \(V=\left( \frac { \mu R }{ P } \right) T\)
V-T graph is a straight line passing the origin.
The slope = \(\frac { \mu R }{ P } \)
The slope of V-T graph is inversely proportional to the pressure. If the slope is greater, lower is the pressure.
Here P₁ has larger slope than P₂. So P₂ > P₁.

\(\frac { \Delta L }{ L } ={ \alpha }_{ L }{ \Delta }_{ T }\)
\(\Delta L={ L\alpha }_{ L }{ \Delta }_{ T }\)
\(\Delta\)L = 10\(\times\)10^-6 \(\times\)300\(\times\)23 = 0.69 m=69 mm
Area Expansion
For a small change in temperatur ΔT the fractional change in area \(\left(ΔA\over A_0\right)\) substance is directly proportional to ΔT and it can be written as
\({ΔA\over A_0}=\alpha_AΔT\)
Therefore, \(\alpha_A={ΔA\over A_0ΔT}\)
Where, α_A = coefficient of area expansion.
ΔA = Change in area
A₀ = Original area
ΔT = Change in temperature
Volume Expansion
For a small change in temperature ΔT the fractional change in volume \(\left(ΔV\over V_0\right)\) of a substance is directly proportional to ΔT.
\({ΔV\over V_0}=\alpha_VΔT\)
Therefore, \(\alpha_V={ΔV\over V_0ΔT}\)
Where, α_V = coefficient of volume expansion.
ΔV = Change in volume
V₀ = Original volume
ΔT = Change in temperature
Unit of coeffi cient of linear, area and volumetric expansion of solids is ^oC^-1 or K^-1

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

\(\vec { A } +\vec { B } =\left( 5\hat { i } +7\hat { j } -4\hat { k } \right) +\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) =11\hat { i } +10\hat { j } -2\hat { k } \)
\(\vec { B } +\vec { A } =\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) +\left( 5\hat { i } +7\hat { j } -4\hat { k } \right) =\left( 6+5 \right) \hat { i } +\left( 3+7 \right) \hat { j } +\left( 2-4 \right) \hat { k } =11\hat { i } +10\hat { j } -2\hat { k } \)
\(\vec { A } -\vec { B } =\left( 5\hat { i } +7\hat { j } -4\hat { k } \right) -\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) =-\hat { i } +4\hat { j } -6\hat { k } \)
\(\vec { B } -\vec { A } =-\hat { i } +4\hat { j } -6\hat { k } \)
Note that the vector \(\vec A+\vec B\) and \(\vec B+\vec A \) are same and the vectors \(\vec A-\vec B\) and \(\vec B-\vec A \) are opposite to each other.

(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

Radius (r) of moon orbit from the centre of earth
\(r =(384,000 \mathrm{~km}) \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} \)
= 384,000,000 m
Time Period (T) = (27 days ) \(\times \frac{24 \text { hours }}{1 \text { day }} \times \frac{60 \mathrm{~min}}{1 \mathrm{hr}} \times \frac{60 \mathrm{sec}}{1 \mathrm{~min}}\)
= 2,332,800 sec
\(velocity (v)=\frac{\text { circumference }}{\text { Time period }}=\frac{2 \pi \mathrm{r}}{\mathrm{T}}\)
\(=\frac{2 \pi \times 384,000,000 m}{2,332,800 \mathrm{~s}}=329 \pi \mathrm{m} / \mathrm{s}\)
Centripetal acceleration \(\left(a_{\perp}\right)=\frac{v^{2}}{r}=\frac{(329 \pi \mathrm{m} / \mathrm{s})^{2}}{384,000,000 \mathrm{~m}}\)
\(a_{\perp}=2.78 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2}\)
Tension due to Centripetal force
\(F_{c} =T=m a_{\perp}=7.34 \times 10^{22} \mathrm{~kg} \times 2.78 \times 10^{-3} \mathrm{~m} / \mathrm{s}^{2} \)
\(\mathrm{~T} =2.04052 \times 10^{20} N\)

\(|3\hat { i } -2\hat { j } +\sqrt { 3 } \hat { k } |=\sqrt { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ \sqrt { 3 } }^{ 2 } } \)
\(\sqrt { 9+4+3 } =\sqrt { 16 } =4\)

(i) The satellites orbiting the Earth have different time periods corresponding to different orbital radii. Orbital radius of a satellite if its time period is 24 hours is calculated below:
Kepler's third law is used to find the radius of the orbit.
T² = \(\frac { 4\pi ^{ 2 } }{ { GM }_{ E } } \) (R_E + h)³
(R_E + h)³ = \(\frac { { GM }_{ E }{ T }^{ 2 } }{ 4\pi ^{ 2 } } \)
R_E + h = \(\left( \frac { { GM }_{ E }{ T }^{ 2 } }{ 4\pi ^{ 2 } } \right) ^{ 1/3 }\)
(ii) Substituting for the time period (24 hrs = 86400 seconds), mass, and radius of the Earth, h turns out to be 36,000 km. Such Satellites are called "geo-stationary satellites", They appear to be stationary when seen from Earth.
India uses the INSAT group of satellites that are basically geo-stationary satellites for the purpose of telecommunication.
Another group of satellite which is placed at a distance of 500 to 800 km from the surface of the Earth orbits the Earth from north to south direction. This type of satellite that orbits Earth from North Pole to South Pole is called a polar satellite. The time period of a polar satellite is nearly 100 minutes and the satellite completes many revolutions in a day. A polar satetrlite covers a small strip of area from pole to pole during one revolution it covers a different strip of area since the Earth would have moved by a small angle. In this way polar satellites cover the entire surface area of the Earth.

Applying component of velocities,

The x - component of velocity is
usin \(\theta\) = vsin f   ......(1)
The magnitude of y-component of velocity is not same, therefore, using coefficient of restitution,
\(e=\frac{v \cos \varphi}{u \cos \theta}\)   ......(2)
Squaring (1) and (2) and adding we get
\(v^{2} \sin ^{2} \varphi=u^{2} \sin ^{2} \theta  \)
\(v^{2} \cos ^{2} \varphi=e^{2} u^{2} \cos ^{2} \theta\)
adding
\(v^{2}=u^{2} \sin ^{2} \theta+e^{2} u^{2} \cos ^{2} \theta\)
\(\therefore v^{2}=u^{2}\left[\sin ^{2} \theta+e^{2} \cos ^{2} \theta\right]\)
v=u\(\sqrt { { sin }^{ 2 }\theta +{ e }^{ 2 }{ cos }^{ 2 }\theta } \)

The equivalent resistance of a parallel combination
\(R'=\frac{R_1R_2}{R_1+R_2}=\frac{150\times220}{150+220}=\frac{33000}{370}=89.1\ Ohm\)
We know that, \(\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}\)
\(\frac { \triangle { R }^{ ' } }{ \left( { R }^{ ' } \right) ^{ 2 } } =\frac { \triangle { R }_{ 1 } }{ { R }_{ 1 }^{ 2 } } +\frac { \triangle { R }_{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
\(\triangle { R }^{ ' }=\left( { R }^{ ' } \right) ^{ 2 }\frac { \triangle { R }_{ 1 } }{ { R }_{ 1 }^{ 2 } } +\left( { R }^{ ' } \right) ^{ 2 }\frac { \triangle { R }_{ 2 } }{ { R }_{ 2 }^{ 2 } } =\left( \frac { { R }^{ ' } }{ { R }_{ 1 } } \right) ^{ 2 }\triangle { R }_{ 1 }+\left( \frac { { R }^{ ' } }{ { R }_{ 2 } } \right) ^{ 2 }\triangle { R }_{ 2 }\)
Substituting the value,
\(\triangle { R }^{ ' }=\left[ \frac { 89.1 }{ 150 } \right] ^{ 2 }\times 2+\left[ \frac { 89.1 }{ 220 } \right] ^{ 2 }\times 6=0.070+0.098=0.168\)
R' = 89.1 ± 0.168 Ohm.

(i) t = \(\theta\) to t = 3:
velocity at t = 0, u = 0
velocity at t = 3 see, v = 30 mls
So, from velocity v = u + at
Acceleration \(a={{u}\over{t}}={{30}\over{3}}=10{ms}^{-2}\)
So distance covered between 0 to 3 see
\(s=ut+{{1}\over{2}}{at}^{2}\)
initial velocity u = 0 and t = 3
\(s=ut+{{1}\over{2}}{at}^{2}\)
\(=0\times3+{{1}\over{2}}\times10{(3)}^{2}\)
\(=0+{{1}\over{2}}\times10\times9\)
\(={{1}\over{2}}\times90\)
s = 45 m
From 3 to 6 see, velocity is same 30 m/s.
So distance travelled, = 30\(\times\)3 = 60. m
Total distance covered between 0 to 6 sec.
= 30 + 60
= 90m
(ii) At t = 3 sec, u = 30 ms^-1, at t = 6 see,
v = 0 m/s .
So from, v = u + at
0= 30 + a\(\times\) 3
a =-10m/s²
So, distance covered from t = 3 to t = 6 see is,
v² = u² +2as
(0)² = (30)² - 2\(\times\)10\(\times\) s
\(s={{900}\over{20}}=45\ m\)
So total distance covered = 90 + 45
= 135 m

Momentum is conserved in all types of collision
Let m ➝ mass of the molecule
M ➝ mass of the wall.
kinetic energy after collision = \(\frac{1}{2}mv^{2}+\frac{1}{2}MV^{2}\)
=\(\frac{1}{2}m(350)^{2}+\frac{1}{2}M(0)^{2}\)
=\(6.1250\times10^{4}mJ\)
kinetic energy before collision = \(\frac{1}{2}m(350)^{2}\)
=\(\frac{1}{2}\times m\times122500\)
=6.1250\(\times\)10⁴ mJ
Therefore,
Kinetic energy before collision = kinetic energy after collision.

If there is a loss of kinetic energy during a collision, then it is called as an inelastic collision
In the case of inelastic collision,
(i) Total kinetic energy is not conserved.
(ii) Some or all of the forces involved are non-conservative.
(iii) A part of the mechanical energy is transformed into heat, sound, light etc.
Examples for inelastic collision:
(i) Collision between ball and floor
(ii) Collision between two vehicles
Examples for perfectly inelastic collision:
(i) Mud thrown on a wall and sticking to it
(ii) a man jumping into a moving trolley
(iii) a bullet fired into a wooden block and remaining embedded in it.