We first find when and where the two balls collide. Let them collide at an instant f seconds after they start their respective motion. Clearly the two balls are at the same height above the ground at this instant.

The height of the first ball after t seconds = 49 t - 1/2 x 9.8t² = 4.9 t (10 - t)
Also the height of the second ball after t seconds = 98 - downward distance moved by it in t seconds.
= \(98-\frac{1}{2}\times 9.8 t^{2}=4.9(20-t^{2})\)
∴ \(4.9 t (100-t)=4.9(20-t^{2}) \)
or \(10t-t^{2}=20-t^{2}\ or \ t=2s\)
The balls thus collide two seconds after the start of their motion. Their velocities at this instant are
First ball: \(v_{1}=(49-9.8 \times 2) m/s\)
= 29.4 m/ s directed upwards
Second ball: v₂ = \((0+9.8 \times 2)\) m/s
= 19.6 m/s directed downwards
if v is the velocity of the combined mass of the two balls after they stick together folluwing 
their collision, we have, by principle of conservation of momentum.
\(200\times v=100 \times 29.4 -100 \times 19.6\)
∴ v = 4.9 m/s
The 'combined mass' thus moves upward, after collision with a velocity of 4.9 m/s. Its height above the ground at this instant is (considering the position of either of the two balls before collision)
\((98-\frac{1}{2}\times 9.8 \times 2^{2})m= (98-19.6)m = 78.4 m\)
We can now find the time t' taken by the 'combined mass' of the two balls to fall to ground.
We have for this 'combined mass',
u = 4.9 m/s ,s = -78.4 m, a = - g = -9.8 ms^-2
ஃ -78.4= 4.9t^'+1/2 (-9.8)t^'2
or t^'2-t^'-16=0
ஃ \(t^{'}=\frac{1\pm \sqrt{1+64}}{2} = \frac{1\pm 8.06}{2}\)
= 4.532 s (leaving out the negative solution)
The 'combined mass' thus takes 4.53 s to fall to the ground. Since the balls collided
2 s after they started their motion, their total time of flight is (2 + 4.53) s = 6.53 s.

Three forms of energy possessed by a flowing fluid are as follows:
1. Pressure energy:
Let an ideal fluid of density P be contained in a rectangular vessel, provided with a small side tube at a depth h_o below the free surface of fluid in the vessel. At the level of side tube, pressure of fluid along the axis of side tube
p = h₀pg
If we want to introduce more fluid into the vessel at this very pressure, we can force it through the side tube by doing work on the piston, If 'A' be the cross-section area of the piston, then force acting on the piston F = PA.
\(\therefore\) Work done in moving the piston through a small distance \(\Delta r\) will be
\(\Delta W=F\Delta x=PA\Delta x=P\Delta V\)
As a result of motion of piston, the mass of the fluid forced in the vessel
\(\Delta m=\rho A\Delta x=\rho\Delta V\)
The work done is stored up in the liquid in the form of its pressure energy.
\(\therefore\) Pressure energy of liquid per unit mass = Work done per unit mass
 \(=\frac{\Delta W}{\Delta m}=\frac{P\Delta V}{\rho\Delta V}=\frac{p}{\rho}\)
and pressure energy per unit volume = p.
2. Gravitational potential energy:
Let at any stage of its flow a fluid element of mass "m' be situated at a height 'h' from the reference line (generally taken to be earth's surface), then its gravitational potential energy in that position is mgh.
\(\therefore\) Gravitational potential energy per unit mass = \(\frac{mgh}{m}=gh\)
and gravitational potential energy per unit volume = pgh.
3. Kinetic energy:
Let at any stage of its flow, a fluid element of mass 'm' be moving with a speed 'v', then the kinetic energy of this fluid element is \(\frac{1}{2}v^2\)
\(\therefore\) Kinetic energy per unit mass = \(\frac{\frac{1}{2}mv^2}{m}=\frac{1}{2}v^2\)
and kinetic energy per unit volume = \(\frac{1}{2}\rho v^2\)
These three forms of energy possessed by a flowing fluid are mutually convertible from one form to another.

T₂ = 27°C = 27 + 273
= 300K
η = 40%, T₂₂ = ?
From \(η = 1-{T_2\over T_1}\)
\({T_2\over T_1}=1-η=1-{40\over 100}={60\over 100}={3\over 5}\)
\(T_1={5\over 3}T_2={5\over 3}\times300=500K\)
Increase in efficiency 10% of 40 = 4%
∴ New efficiency 'Y]' 40 + 4 = 44%
Let T₁ be the new temperature of the source
As \(η'=1-{T_2\over T_1'}\)
\({T_2\over T_1'}=1-η'=1-{44\over 100}={56\over 100}\)
\(T_1'={100\over 56}T_2={100\over 56}\times300=535.7K\)
∴ Increase in temp. of source
= 535.7 - 500 = 35.7 K

Force experienced by box, F = ma = 40 x 2 = 80 N
Frictional force F_f = \(\mu\)mg = 0.15 x 40 x 10 = 60 N .
Net force = F - F_f = 80 - 60 = 20 N.
Backward acceleration produced in the box, n = \({20\over 4}({Net \ force \over m})\)
\(\Rightarrow a=0.5ms^{-2}\)
If t is time taken by the box to travel s = 5 metre and fall off the truck, then from
\(S=ut+{1\over2}at^2\)
\(5=0\times +{1\over2}\times 0.5 t^2\)
\(t=\sqrt{5\times2\over 0.5}=4.47 s.\)
If the truck travels a distance x during this time, then again from
\(S=ut+{1\over2}at^2\)
x = 0 x 4.47 + \(1\over2\) x 2 (4.47)²
= 19.98 m.

(a) (i) Given, function is
\(\sin { \omega t } -\cos { \omega t } \) = \(\sin { \omega t } -\sin { \left( \frac { \pi }{ 2 } -\omega t \right) } \) 
\(=2\sin { \left( \frac { \omega t+\frac { \pi }{ 2 } -\omega t }{ 2 } \right) } .\sin { \left( \frac { \omega t-\frac { \pi }{ 2 } +\omega t }{ 2 } \right) } \) 
Given function \(=2\sin { \left( \frac { \pi }{ 4 } \right) } .\sin { \left( \omega t-\frac { \pi }{ 4 } \right) } \)  
\(=\sqrt { 2 } \sin { \left( \omega t-\frac { \pi }{ 4 } \right) } \) 
This function represents a simple harmonic motion having period of \(T=\frac { 2\pi }{ \omega } \) and a phase angle \(\left( -\frac { \pi }{ 4 } \right) \) or \(\left( -\frac { 7\pi }{ 4 } \right) \).
(b) Periodic, but not SHM
The given function is:
sin3ωt = 143sin ωt - sin3 ωt
The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.(c) 3cos(\(\frac{\pi}{4}\)-2ωt) = 3 cos(2ωt -\(\frac{\pi}{4}\)) [∵ cos (- θ) = cos θ]
Clearly it represents SHM and its time period is 2π/2ω
(d) cos ωt + cos 3 ωt + cos 5 ωt. It represents the periodic but not S.H.M. Its time period is 2π/2ω
(e) \({ e }^{ -{ \omega }^{ 2 }{ t }^{ 2 } }\) It is an exponential function which never repeats itself. Therefore it represents non-periodic motion.
(f) 1 + ωt + ω²t²  also represents non periodic motion.

We know that angular momentum \(\vec { l } \) of a particle having position vector \(\vec { r } \) and momentum \(\vec { p } \) is given by
\(\vec { l } \) = \(\vec { r } \)\(\times\)\(\vec { p } \)
But, \(\vec { r } \)= [x\(\vec { i} \)+ y\(\vec { j} \)+ z\(\vec { k} \)] where x,y,z are the components of
\(\vec { r } \)and\(\vec { p } \) = [p_x\(\vec { i} \)+ p_y\(\vec { j} \)+ p_z\(\vec { k} \)]
∴ \(\vec { l } \) = \(\vec { r } \)\(\times\)\(\vec { p } \) [x\(\vec { i} \)+ y\(\vec { j} \)+ z\(\vec { k} \)]\(\times\)[p_x\(\vec { i} \)+ p_y\(\vec { j} \)+ p_z\(\vec { k} \)]
or (lx\(\vec { i} \)+ ly\(\vec { j} \)+ lz\(\vec { k} \)) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ x & y & z \\ { p }_{ x } & { p }_{ y } & { p }_{ z } \end{matrix} \right| \)
 =(yp_z - zp_y)\(\vec { i} \) +(zp_x - xp_z)\(\vec { j} \)+(xp_y - yp_x)\(\vec { k} \)
From this relation, we conclude that 
lx= yp_z - zp_y, l_y= zp_x - xp_z and l_z = xp_y - yp_x
If the given particle moves only in the x - y plane, then z = 0 and p_z = 0 and hence,
\(\vec { l } \)=(xp_y-yp_x)\(\hat { k } \) ,which is only the .z-comonent of \(\vec { l } \)
It means that for a particle moving only in the x - y plane, the angular momentum has only the z-component

As in the dispersive medium wavelengths are different, hence the velocities, therefore, the shape of the pulse gets distorted.

Here h = 1000 m and ρ = 1.03 x 10 ³ kg m^-3
(a) From Eq. P₂ −P₁ = ρgh, absolute pressure
P = P_a + ρgh
= 1.01 x 10⁵ Pa
= + 1.03 x 10³ kg m^–3 x 10 m s^–2 x 1000 m
= 104.01 x 10⁵ Pa
≈ 104 atm
(b) Gauge pressure is P −P_a = ρgh = P_g
P _g = 1.03 x 10³ kg m^–3 x 10 ms² x 1000 m
= 10³ x 10⁵ Pa
≈ 103 atm
(c) The pressure outside the submarine is P = P_a + ρgh and the pressure inside it is P_a . Hence, the net pressure acting on the window is gauge pressure, P _g = ρgh. Since the area of the window is A = 0.04 m2 , the force acting on it is 
F = P _g A = 10³ x 10⁵ Pa x 0.04 m² = 4.12 x 10⁵ N

Let F be the tensile force applied. Since, each rivet shares the stretching force equally, so the shearing force on each rivet = F/2.
If A is the area of each rivet, then shearing stress on each rivet= \(\frac { F }{ 2A } \).Now, maximum shearing stress on each strip = 3.0\(\times \) 10⁸ Nm^-2
i.e. \(\frac { { F }_{ max } }{ 2A } \) = 3.0\(\times \) 10⁸ Nm^-2
where, F_max is maximum tangential force. 
or  F_max =  3.0\(\times \) 10⁸\(\times \) 2A = 6.0 \(\times \)10⁸\(\times \) \(\pi { r }^{ 2 }\) 
\(\therefore \)   r = 0.1 cm 
\(\Rightarrow \)    0.1 \(\times \)10^-2  
r = 1\(\times \)10^-3 m
\(\Rightarrow \) F_max = 6.0\(\times \) 10⁸\(\times \)\(​​\frac { 22 }{ 7 } \)\(\times \) (1\(\times \)10^-3)²
= 1885 N

Force, F = 10 kN = 1 x 10⁴ N
Radius, r = 2 mm = 2 x 10^-3 m
Area, A = \(\pi \) r² = \(\pi \) x ( 2 x 10^-3 )²
= 12.56 x 10^-6 m²
Stress = \(\frac { Force }{ Area } \) = \(\frac { 1\times { 10 }^{ 4 }N }{ 12.56\times { 10 }^{ -6 }{ m }^{ 2 } } \) 
= 0.0796 x 10¹⁰
= 7.96 x 10⁸ N / m²

Acceleration due to gravity is independent of the mass of the body.

Initially, \({ E }_{t }=\frac { -GMm }{ 4R_E } \)
While finally,  \({ E }_{ f }=\frac { -GMm }{ 8R_E } \)
Change in the total energy is given by
\(\triangle E={ E }_{ f }-{ E }_{ i }=\frac { -GM_Em }{ 8R_E } \frac { mR_E }{ 8 } \)
\(∆E =\frac { gmR }{ 8 } =\frac { 9.81\times 400\times 6.37\times { 10 }^{ 6 } }{ 8 } \)
\(\\ \triangle E=3.13\times { 10 }^{ 9 }J\)
The kinetic energy is reduced and it mimics ∆E, namely
\(\therefore \triangle K={ K }_{ f }-{ K }_{ i }=-3.13\times { 10 }^{ 3 }J\)
The change in potential energy is twice the change in the total energy, namely
∆V = V_f – V_i = – 6.25 × 10⁹ J

(i) False, internal as well as external forces can change the kinetic energy.Again forces of conservative may change the potential energy of a system.
(ii) False, for non-conservative force the work done over a closed loop is not zero.
(iii) It is usually true but not always true.As an exmple in the exlposion of a cracker final kinetic energy is greater than the initial kinetic energy. Again final kinetic energy of gun-bullet system after firing is more than initial kinetic energy before collision.

The nuclear decay may be represented as follows
Fe⁵⁷\(\rightarrow\) Fe⁵⁷ + hv (\(\gamma \)-ray photon)
According to de-Broglie hypothesis, momentum of a photon of energy y E is 
\(p=\frac { E }{ c } =\frac { 14.4\times 1.6\times { 10 }^{ -16 }J }{ 3\times { 1 }0^{ 8 }{ ms }^{ -1 } } \)
\(\\ p=\ 7.68\ \times { 10 }^{ -24 }kg { ms }^{ -1 }\)
By conservation of momentum, the momentum of daughter nucleus, p=momentum of \(\gamma \) -ray photon
\(=d\ 7.68 \ \times { 10 }^{ -24 }kg \ { ms }^{ -1 }\)
The recoil energy of the nucleus will be
\(k= \ \frac { { p }^{ 2 } }{ 2m } =\frac { ({ 7.68 \ \times { 10 }^{ -24 }) }^{ 2 } }{ 2\times \ 56.935 \ \times 1.66 \times { 10 }^{ -27 } }\)
\( \\ =0.32\times { 10 }^{ -21 }\)
\(\\ =\frac { 0.312\times { 10 }^{ -21 } }{ 1.6\times { 10 }^{ -16 } } keV\)
\(K=1.95{ \times 10 }^{ -16 }keV\)

When a man is standing on a weighing scale, it will read the normal reaction R as apparent weight.
Given , mass of man (m) = 70 kg
In each case the weighing scale will read the reaction R, i.e. the apparent weight.
(i) As lift is moving upward with a uniform speed, therefore, its acceleration a = 0
\(\therefore \)  Normal reaction w = R = mg = 70 x 10 N = 700 N
w acts vertically downwards and R acts vertically upwards.
\(\therefore \)  Reading on weighing scale = \(\frac { 700 }{ 10 } \) = 70 kg
(ii) Acceleration of the lift, a = 5 m/ s² (\(\downarrow \) )
\(\therefore \)  Normal reaction, R = m (g - a) = 70 ( 10 - 5) N
= 70 x 5N = 350 N
\(\therefore \)  Reading on weighing scale = \(\frac { 350\quad N }{ 10\quad m/{ s }^{ 2 } } \) = 35 kg
(iii) Acceleration of the lift, a = 5 m/s² (\(\uparrow \) ) 
\(\therefore\)  Normal reaction R = m ( g + a )
= 70 ( 10 + 5 ) = 1050 N
\(\therefore\)   Reading on weighing scale = \(\frac { 1050N }{ 10m/{ s }^{ 2 } } \) = 105 kg
(iv) Acceleration of the lift when it is falling freely under gravity
a = g (\(\downarrow\) )
\(\therefore\)  Normal reaction, R = m ( g - g ) = 0
\(\therefore\)  Reading on weighing scale = 0

 
According to triangle law of vector addition, OA + AB = OB
AB = OB -OA = v₂-v₁ = Change in velocity
|v₂-v₁| = AB= \(\sqrt { (OA)^{ 2 }+(O{ B) }^{ 2 } } =\sqrt { (5)^{ 2 }+{ (5) }^{ 2 } } =5\sqrt { 2 } \) m/s
Hence, average acceleration = \(\frac { \left| { v }_{ 2 }-{ v }_{ 1 } \right| }{ t } \quad =\frac { 5\sqrt { 2 } }{ 10 } \) = \(\frac { 1 }{ \sqrt { 2 } } \) m/s²
Along North-West direction.

\(\sqrt { { A }^{ 2 }+B^{ 2 }+AB } \)

(i) Mass vaccination is good as, it is used to make the socity free from the diseases like small pox.
(ii) Television for eradication of illiteracy and for mass communication of news and ideas is good as, it is a medium which is easily within the reach of common man and also they are very habitual to it.
(iii) Prenatal sex determination is bad because people are misusing it. Some of the people after determination of sex of child, think to abort. They do it especially with girl child.
(iv) Computer for increase in work efficiency is good as using the computer, a man can do much more work with greater efficiency and accuracy.
(v) Putting artificial satellite into orbits around the Earth is good for development as these satellites serve many purpose like remote sensing, weather foresting.
(vi) Development of nuclear weapons is bad as they can be used in mass destruction.
(vii) Development of new and powerful tool of chemical and biological warfare are bad, as they can also be used for mass destruction.
(viii) Purification of water for drinking purpose is good as we can save ourself from the diseases which we can have due to drinking of the water.
(ix) Plastic surgery is good as with the help of it a man or women can remove the skin defects occurring due to accident or some other reasons. It has some bad effects too but they are not very considerable.
(x) Cloning is good as far as animals are concerned. With the help of it, we can develop some species of animals which can be used to serve some specific purpose. But it is not good for human beings.

Three. Their resultant will be zero provided they can be represented by three sides of a triangle taken in order.

Curved roads are generally banked so as to help in providing centripetal force needed to balance the centrifugal force, arising due to circular motion on the curved road.

A Camot engine is an ideal heat engine from the following points of view:
(i) There is absolutely no friction between the walls of cylinder and the piston.
(ii) The working substance is an ideal gas. In a real engine, these conditions cannot be fulfilled and hence no heat engine working between the same two temperatures can have efficiency greater than that of camot engine

The speedometer measures the instantaneous speed.

According to Stoke's law, viscous force F = 6 \(\pi\eta rv\)
\(\therefore\) Dimensions of \(\eta\) are \(\frac{[F]}{[r][v]}=\frac{MLT^{-2}}{L\times LT^{-1}}[M^1 L^{-1}T^{-1}]\)

Geology.

In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.
No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

(i) No negative sign on RHS, hence, not SHM
(ii) Displacement on RHS is squared, hence not SHM
(iii) a = -10x follows the condition of SHM, acceleration α -displacement hence, SHM.
(iv) No negative sign on RHS and displacement appears as cubed, hence, not SHM

75 J / kg , for 1 mole

The air between the two walls of the thermo flask is evacuated. This prevents heat loss due to conduction and convention. The loss of heat due to radiation is minimized by silvering the inside surface of the double wall. As the loss of heat due to the three processes is minimized the tea remains hot for a long time.

The situation is shown in the diagram.
Now, Young's modulus (Y) = \(\frac { f }{ A } \times \frac { L }{ l } \)
  
For first wire, Y = \(\frac { f }{ \pi { r }^{ 2 } } \times \frac { L }{ l } \)      ....(i)
For second wire, Y = \(\frac { 2f }{ \pi { (2r) }^{ 2 } } \)\(\times \frac { 2L }{ l\prime } =\frac { f }{ \pi { r }^{ 2 } } \times \frac { L }{ l\prime } \)  ...(ii) 
For Eqs. (i) and (ii), \(\frac { f }{ \pi { r }^{ 2 } } \times \frac { L }{ l } \)= \(\frac { f }{ \pi { r }^{ 2 } } \times \frac { L }{ l\prime } \) 
\(​​\therefore\)    l = l'     [ \(​​\because\) both wires are of same material, hence, Young's modulus will be same].

Space also an orientation.We also have the frames of reference in space.Hence, orientational problem will affect the astronaut in space .

Given \({ m }_{ 1 }=1kg ,\ { v }_{ 1 }=(3i-2j)m/s,{ m }_{ 2 }=2 \ kg\ and \ { v }_{ 2 } = (4j-6k)m/s\) When two particles experience a perfectly inelastic collision. They stick together and move with a common velocity v given by 
\(v=\frac { { m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 1(3i-2j)+2(4j-6k) }{ 1+2 }\)
\( =(I+2j-4k)m/s\)
Speed of combined particle
\(v=\sqrt { { 1 }^{ 2 }+{ (2) }^{ 2 }+{ (-4 }^{ )2 }=\sqrt { 21 } m/s } \)

\(\text{Linear mass density }=\frac { Mass }{ Length } =\frac { \left[ M \right] }{ \left[ L \right] } =\left[ ML^{ -1 } \right] \)

\(\frac { \triangle x }{ x } =\frac { 0.001 }{ 0.004 } =0.25\)

(a) Force constant \(k=\frac { F }{ l } =\frac { mg }{ l } \)
Here m = 3.0 kg and elongation in length of spring l = 0.2 m
∴ Force constant k = \(\frac{3.0\times 9.8}{0.2}=174Nm^{-1}\)
(b) Period of oscillation \(T=2\pi \sqrt { \frac { m }{ k } } =2\times 3.14\times \sqrt { \frac { 3 }{ 147 } } =0.9s\) 

Here mass of bullet, m = 0.05 kg, initial velocity of bullet before firing u = 0, length of barrel of gun, moving through which the bullet is accelerated s = 1m, final muzzle velocity of bullet v = 400ms^-1.
(I) Using the relation v² - u² = 2as, we have
(400)² - (0)² = 2 x a x 1
\(\Rightarrow a={400\times 400\over 2\times 1}=8\times 10^4 ms^{-2}\)
(ii) Force F = ma = 0.05 x 8 x 10⁴ = 4000 N
(iii) Impulse given to the bullet by the gun, J = change in momentum of bullet
= m (v - u) = 0.05 x (400 - 0)
= 20 Ns.

(I) Curiosity, awareness, sharp mind and keen observation.
(ii) The astronauts and the satellite require the centripetal force to revolve around the earth. Their weight is used up in providing the necessary centripetal force. Hence an astronaut feels weightlessness in the space.

(a) v^'=2v.
It is possible is source is approaching the stationary listener i.e., v_s is +
As \(v^{'}=\frac{v-v_{L}}{v-v_{s}}V\)
∴ 2v =  \(\frac{v}{v-v_{s}}V\)
or 2 = \(\frac{v}{v-v_{s}}\)
or 2v - 2v_s = v
or 2v_s = v
or \(v_{s}=\frac{v}{2}\)
Therefore, source should approach the listener with half the velocity of sound propagation.
(b) \(v^{'}=\frac{v}{2}\)
It is possible if source is receding away from the stationary listener i.e., v_s is negative and v_L= 0
∴ \(v^{'}=\frac{v}{v+v_{s}}v\)
or \(\frac{v}{2}=\frac{v}{v+v_{s}}v \ or\ \frac{1}{2}=\frac{v}{v+v_{s}}\)
or v + v_s = 2v or v_s = v
Therefore, source should recede away from the listener with the velocity of sound propagation.

Initially 1000g of water is at 5°C.
Let m gram of it be cooled to ice at O°c.
Heat released due to this = (m x 1 x 5) + (m x 80)
= 5 m + 80 m = 85 meal.
The heat required by (1000 - m) g of water at 5 "C to become steam at lOOoe
[(1000 - m) (100 - 5) + (1000 - m) 540] cal
(1000 - m) (95 + 540) cal
(1000 - 111) (635) cal
Now, 85m (1000 - 111) (635) or 720m = 635 x 1000
m = \(\frac { 635\times 1000 }{ 720 } =881.9g\)
Hence 881.9 g of water by turning into at ooe will supply heat to evaporate 118.1 g of water.

As is clear from Fig.
Change in pressure, Δp = EF = 5.0 - 2.0 = 3.0 atm = 3.0 x 10⁵ Nm^-2
Change in volume, ΔV=DF = 600 - 300 = 300 cc = 300 x 10^-6 m³
Work done by the gas from 0 to E to F = area of ΔDEF
\(W={1\over 2}\times DF\times EF\)
\(={1\over 2}\times (300 \times 10^{-6})x (3.0 \times 10^5) = 45 J\)

Here, R = 2m M = 100 kg
v = 20 cm/s = 0.2 m/s
Total energy of the hoop = \(\frac { 1 }{ 2 } { Mv }^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 }\)
= \(\frac { 1 }{ 2 } {Mv }^{ 2 }+\frac { 1 }{ 2 } (MR^{ 2 }){ \omega }^{ 2 }\)
=\(\frac { 1 }{ 2 } { Mv }^{ 2 }+\frac { 1 }{ 2 } { Mv }^{ 2 }={ Mv }^{ 2 }\)
Work required to stop the hoop = total energy of the hoop
W = Mv² = 100 (0.2)² = 4 Joule.

The microscopic domain of physics deals with the constitution and structure of matter at atomic and nuclear scale.
The quantum theory is currently accepted as the proper framework for explaining microscopic phenomena.

Given u = 126 km/h = 126 x \(\frac { 5 }{ 18 } \) rn/s = 35 m/s
S 200 m and v = 0
As  v² - u² = 2as
∴ 0- (35)² 2a x 200
\(\Rightarrow a=\frac { { -35 }^{ 2 } }{ 400 } =-3.06\quad m/{ s }^{ 2 }\)
Also v = u + at
\(\Rightarrow t=\frac { v-u }{ a } =\frac { 0{ -35 } }{ -3.06 } =11.4 s\)

Given, \({ T }_{ 1 }=300k,{ T }_{ 2 }=?\)
\(\\ { V }_{ 1 }=V,{ V }_{ 2 }=\frac { 4 }{ 3 } V\)
According to Charles law, we get
\(\frac { { V }_{ 2 } }{ { V }_{ 1 } } =\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
\(\\ \Rightarrow { T }_{ 2 }={ T }_{ 1 }\frac { { V }_{ 2 } }{ { V }_{ 1 } } =300\times \frac { 4 }{ 3 } =400\ k\)

Given for steel wireLength (l₁) = 4.7 m
Area of cross-section (A₁) = 3.0 x 10^-5m²
For copper wire
Length (l₂) =3.5 m
Area of crosssection (A₂) = 4.0 x 10^-5m²
Let F be the given load under which steel and copper wires be streched by the same amount \(\Delta l\)
Young's modulus \((Y)=\frac { F/A }{ \Delta l/l } =\frac { F\times l }{ A\times \Delta l } \)
For steel   \({ Y }_{ s }=\frac { F\times l_{ 1 } }{ { A }_{ 1 }\times { \Delta l } } \)
For copper \({ Y }_{ c }=\frac { F\times l_{ 2 } }{ { A }_{ 2 }\times { \Delta l } } \)
Dividing Eq (i)by Eq(ii) we get
\(\frac { { Y }_{ s } }{ { Y }_{ c } } =\frac { F\times { l }_{ 1 } }{ { A }_{ 1 }\times \Delta l } \times \frac { { A }_{ 2 }\times \Delta l }{ F\times { l }_{ 2 } }\)
\( \\=\frac { { l }_{ 1 } }{ { l }_{ 2 } } \times \frac { { A }_{ 2 } }{ { A }_{ 1 } } =\frac { 4.7 }{ 3.5 } \times \frac { 4.0\times { 10 }^{ -5 } }{ 3.0\times { 10 }^{ -5 } } \)
\( \frac { { Y }_{ s } }{ { Y }_{ c } } =\frac { 18.8 }{ 10.5 } =1.79=1.8\)

Initial KE = \(\frac { 1 }{ 2 } { I }_{ i }{ \omega }_{ i }^{ 2 }=\frac { 1 }{ 2 } \times{ I }_{ i }\times(40{ ) }^{ 2 }\)
 \(=\frac { 1 }{ 2 } \times{ I }_{ i }\times1600=800{ I }_{ i }\)
Final KE = \(\frac { 1 }{ 2 } { I }_{ f }{ \omega }_{ f }^{ 2 }=\frac { 1 }{ 2 } \times\frac { 2 }{ 5 } { I }_{ i }\times(100{ ) }^{ 2 }\)
 \(=\frac { 1 }{ 2 } \times\frac { 2 }{ 5 } \times{ I }_{ i }\times100\times100=2000{ I }_{ i }\)
Clearly, final KE > initial KE. This energy is obtained from conversion of muscular work in KE. Muscular work has to be done in folding of arms.

(i) Area swept by blades of windmill = A and wind velocity = v|
\(\therefore \) Volume of air passing per unit time = Av
\(\therefore \) Mass of air passing per unit time = Avp
and mass of air passing in time t, M = Avpt
(ii) KE of said quantity of air , K = \(\frac { 1 }{ 2 } { Mv }^{ 2 }=\frac { 1 }{ 2 } { Aptv }^{ 3 }\)
(iii) 

If efficiency of windmill be 25%, then Output electrical power = 25% of input power
\(\frac { 25 }{ 100 } \times \frac { 1 }{ 2 } Ap{ v }^{ 3 }\)

As \(A={ 30m }^{ 2 },\ v=36\ km/h\ =36\times \frac { 5 }{ 18 } m/s\)
\(\\ =10m/s\ and\ p=1.2\ { kgm }^{ -3 }\)
\(\\ \therefore Output\ electrical\ power=\frac { 25 }{ 100 } \times \frac { 1 }{ 2 } \times 30\times 1.2\times ({ 10 })^{ 3 }\)
\(\\ =4500\quad W=4.5\ kW\ 1kW=1000W]\)

Given, R = 1000m
Horizontal range of the bullet fired at an angle \(\theta \) is
\(R=\frac { { u }^{ 2 }sin2\theta }{ g } \Rightarrow 1000=\frac { { u }^{ 2 }sin\theta cos\theta }{ g } ....(i)\)
Bullet is fired from the car moving with 36km/h
i.e.10m/s, then horizontal component of the velocity of = usin\(\theta \) + 10
Vertical component of the velocity of the bullet=usin​\(\theta \) 
Then, new range of the bullet is
\({ R }_{ 1 }=\frac { 2 }{ g } (usin\theta )(ucos\theta +10)\)
\(\\ =\frac { 2 }{ g } { u }^{ 2 }sin\theta cos\theta +\frac { 20 }{ g } usin\theta \Rightarrow { R }_{ 1 }=R+\frac { 20 }{ g } usin\theta\)
\( \\ \Rightarrow { R }_{ 1 }-R=\frac { 20 }{ g } usin\theta ....(ii)\)
\(\\ From\ Eq.(i),we \ have \ u=\sqrt { \frac { 1000\times g }{ 2sin\theta cos\theta } } .....(iii)|\)
\(\\ Now, \ substituting \ the \ value \ of \ u \ in \ Eq.(ii), \ we \ get\ \)
\(\\ { R }_{ 1 }-R=\frac { 20 }{ g } \sqrt { \frac { 1000\times g }{ 2sin\theta cos\theta } } sin\theta =20\sqrt { \frac { 500\times sin\theta }{ gcos\theta } }\)
\( \\ =20\sqrt { \frac { 500 }{ 9.8 } tan\theta } =142.9\sqrt { tan\theta } \) ​​​​