Let M be the total mass of a system supposed to be concentrated at the centre of mass whose position vector is" then in the presence of an external force \(\vec { F } \) we have
  \(\vec { F } =M\frac { { d }^{ 2 }\vec { r } }{ { dt }^{ 2 } } =M\frac { d }{ dt } \left( \frac { \vec { dr } }{ dt } \right) =M\frac { d }{ dt } ({ \vec { v } }_{ cm })\)
However for an isolated system, force F = 0 and hence, we have
\(M\frac { d }{ dt } ({ \vec { \upsilon } }_{ cm })=0\quad or\frac { d }{ dt } ({ \vec { \upsilon } }_{ cm })or{ \vec { \upsilon } }_{ cm }=a\quad constant\)
It means that for an isolated system the centre of mass moves with a uniform velocity along a straight line path.

The torque required to produce a unit twist in a solid shaft of radius r is given by
\(\tau =\frac { \pi \eta { r }^{ 4 } }{ 2l } \)    ....(1)
where η is the ~odulus of rigidity of the material and I is the length of the shaft. The torque required to produce a unit twist in a hollow shaft of inner and outer radii r_i and r₀ is given by
\({ \tau }^{ ' }=\frac { \pi \eta ({ r }_{ 0 }^{ 4 }-{ r }_{ i }^{ 4 }) }{ 2l } =\frac { \pi \eta ({ r }_{ 0 }^{ 2 }-{ r }_{ i }^{ 4 })({ r }_{ 0 }^{ 2 }+{ r }_{ i }^{ 2 }) }{ 2l } \)     .....(2)
Dividing (2) by (I), we get
\(\frac { { \tau }^{ ' } }{ \tau } =\frac { ({ r }_{ 0 }^{ 2 }-{ r }_{ i }^{ 2 })({ r }_{ 0 }^{ 2 }+{ r }_{ i }^{ 2 }) }{ { r }^{ 4 } } \)      ......(3)
Since the two shafts are made of the same material and the amounts of material are equal,
∴ \({ \pi r }^{ 2 }l=\pi ({ r }_{ 0 }^{ 2 }-{ r }_{ i }^{ 2 })l\quad \)  or r² = \({ r }_{ 0 }^{ 2 }-{ r }_{ i }^{ 2 }\)
From (3), \(\frac { { \tau }^{ ' } }{ \tau } =\frac { { r }_{ 0 }^{ 2 }+{ r }_{ i }^{ 2 } }{ r^{ 2 } } \)  or \(\frac { { \tau }^{ ' } }{ \tau } >1\) or \(\tau '>\tau \).

The given equation of stationary wave is \(y=4 cos \frac{\pi}{3} sin 40 \ \pi t\)
or \(y=2 \times 2 cos \frac{2\pi x}{6} sin \frac{2x(120)t}{6}\)   --- (i)
We know that \(y=2a cos \frac{2\pi x}{\lambda} sin \frac{2x vt}{\lambda}\)   --- (ii)
By comparing tow equations, we get 
a = 2 cm,λ = 6 cm and v = 220 cm /sec.
The component waves are 
\(y_{1}= a sin \frac{2\pi}{\lambda}(vt-x)\)
and \(y_{2}= a sin \frac{2\pi}{\lambda}(vt+x)\)
Distance between two adjacent nodes = \(\frac{\lambda}{2}=\frac{6}{2}=3 cm\).
Particle velocity \(\frac{dy}{dt} = 4 cos\frac{\pi}{x}cos(40\pi t).40 \pi\)
\(= 160 \ cos \frac{\pi x}{3} \ cos 40\pi t\)
=\(160 \ \pi \ cos \frac{\pi x}{3} cos (40\pi \times \frac{1}{8}) = 160 \pi \ \ [∵ cos \pi=cos 5\pi = -1]\)
Hence, particle velocity = 160 cm/sec.

Here, \(\rho\) = 0.09 kg m^-3 
At S.T.P., Pressure P = 1.01 x 10⁵ pa.
According to kinetic theory of gases.
\(p=\frac { 1 }{ 3 } \rho { C }^{ 2 }\quad or\quad C=\sqrt { \frac { 3p }{ \rho } } \)
\(=\sqrt { \frac { 3\times 1.01\times { 10 }^{ 5 } }{ 0.09 } } =1837.5\quad { ms }^{ -1 }\)
Volume occupied by one mole of hydrogen at S.T.P. = 22.4 liters = 22.4 x 10^-3 m³
\(\therefore \) Mass of hydrogen, M = volume x density
= 22.4 x 10^-3 x 0.09
= 2.016 x 10^-3kg
Average K.E/mole =\(\frac { 1 }{ 2 } \)MC2
=\(\frac { 1 }{ 2 } \) x (2.016 x 10^-3) x (1837.5)²
= 3403.4 J

(a) Volume of wind flowing per second = Av
Mass of wind flowing per second = Avp
Mass of air passing in second = Avpt
(b) Kinetic energy of air = \(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } (Avpt){ v }^{ 2 }=\frac { 1 }{ 2 } { av }^{ 3 }pt\)
(c) Electrical energy produced = \(\frac { 25 }{ 100 } \times \frac { 1 }{ 2 } { Av }^{ 3 }pt=\frac { { Av }^{ 3 }pt }{ 8 } \)
Electrical power = \(\frac { A{ v }^{ 3 }pt }{ 8t } =\frac { { Av }^{ 3 }p }{ 8 } \)
Now, A = 30 m², v = 36 kmh^-1 = \(36\times \frac { 5 }{ 18 } { ms }^{ -1 }10{ ms }^{ -1 }\)
and p = 1.2 kg ms^-1
\(\therefore\) Electrical power = \(\frac { 30\times 10\times 10\times 10\times 1.2 }{ 8 } W=4500\) 
W = 4.5 KW.

Here, heat abs or bed \(=\mathrm{Q}_1=6 \times 10^5 \mathrm{cal}\).
Initial temperature \(=\mathrm{T}_1=227^{\circ} \mathrm{C}=227+273=500 \mathrm{~K}\).
Final temperature \(=\mathrm{T}_2=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}\).
As, for Carnot engine;
\( \frac{Q_2}{Q_1}=\frac{T_2}{T_1} \)
\(Q_2=Q_1 \frac{T_2}{T_1} \)
\( \mathrm{Q}_2=\frac{400}{500} \times 6 \times 10^5 \)
\( \mathrm{Q}_2=4.8 \times 10^5 \mathrm{cal} \)
\( \mathrm{Q}_2=\text { Final heat emitted } \)
\( \text { As } \mathrm{w}=\mathrm{Q}_1-\mathrm{Q}_2=6 \times 10^5-4.8 \times 10^5 \)
\(=1.2 \times 10^5 \mathrm{cal} \)
\( \text { Work }=\mathrm{w}=1.2 \times 10^5 \times 4.2 \mathrm{~J} \)
\(\text { Dore }=5.04 \times 10^5 \mathrm{~J} \)

For neon Triple point T = 24.57 K
T_C = T(K) - 273.15
= 24.57 - 273.15 = - 248.58⁰C
T_F = 9/5T_C + 32= 9/5 x (- 248.58) + 32 = -415.44⁰F
For carbon dioxide Triple point, T = 216.55 K
T_C = T(K) - 273.15 = 216.55 - 273.15 = -56.6⁰C
T_F = 9/5T_C + 32 = 9/5x (-56.6) + 32 = 69.88⁰C

(a) Given, radius of artery    ( r ) = 2 \(\times\) 10^{- 3} m
\(\therefore \) Diameter of artery   D = 2r = 4 \(\times\) 10^-3m
Density of whole blood ( \(\rho\) ) = 1.06 \(\times\) 10³ kg / m³
Coefficient of viscosity of blood ( \(\eta\) ) = 2.084 \(\times\)10³ Pa - s
For laminar flow, maximum value of Reynold's number
R_e = 2000
Critical velocity  ( v_c ) = \(\frac{R_e\eta}{\rho D} \)
= \(\frac{2000\times2.084\times10^3}{1.06\times10^3\times4\times10^{-3}} \)
= 9.83 \(\times\)10⁵m / s
(b) Flow rate of blood = Volume of blood flowing per second
= Av_c
= \(\pi\)r2\(\times\)vc
= 3.14\(\times\)(2\(\times\)10^-3)²\(\times\)9.83\(\times\)105
= 12.35 m³/s

According to Kepler's third law of planetary motion,
\({ T }^{ 2 }\ \alpha \ { r }^{ 3 }\)
\(\\ Thus,\ \frac { { T }_{ S }^{ 2 } }{ { T }_{ E }^{ 2 } } =\frac { { r }_{ S }^{ 3 } }{ { r }_{ E }^{ 3 } } \ or \left( \frac { { T }_{ S } }{ { T }_{ E } } \right) ^{ 2 }=\left( \frac { { r }_{ S } }{ { r }_{ E } } \right) ^{ 3 }\)
\(\\ \left( \frac { { r }_{ S } }{ { r }_{ E } } \right) =\left( \frac { { T }_{ S } }{ { T }_{ E } } \right) ^{ 2/3 }\\\)
\( \\ { r }_{ S }=\left( \frac { { T }_{ S } }{ { T }_{ E } } \right) ^{ 2/3 }\times { r }_{ E }\)
\(\\ As,\quad \left( \frac { { T }_{ S } }{ { T }_{ E } } \right) =29.5\)
\(\\ { r }_{ E }=1.5\times 10^{ 8 }km\)
\(\\ { r }_{ S }=(29.5)^{ 2/3 }(1.5\times 10^{ 8 }km)\)
\(=14.3\times 10^{ 8 }km\) 

\(\because \)  Mass of the helicopter, m₁ = 1000 kg
Mass of the crew and the passengers, m₂ = 300 kg
Acceleration of the helicopter, a = 15 m/ s²
Acceleration  due to gravity, g = 10 m/s²
(i) Let R, be the reaction applied by the floor on the crew and the passengers.

R₁- m₂g = m₂a
or
R₁= m₂g + m₂a = m₂(g+a)
= 300(10 + 15) = 7500N(upward direction)
(ii) Action of the rotor of the helicopter on the surrounding air
= ( m₁ + m₂ ) g + ( m₁ + m₂ ) a
= ( m₁ + m₂ ) ( g + a )
= ( 1000 + 300 ) x (10 + 15 )
= 1300 x 25 = 32500 N
Force (action ) of the rotor of the helicopter on the surrounding air = 32500 N (downward)

When the boat is anchored in the harbour, the flag flutters along the N-E direction. It shows that the velocity of wind is along the North-East direction. When the boat starts moving, the flag will flutter along the direction of relative velocity of wind W.r.t. boat. Let v_wb be the relative velocity of wind W.r.t.boat and \(\beta \) be the angle between v_wb and v_w.
Then,    v_wb= v_w + (-v_b) 

Here,  |v_w| = 72 km/h and |-v_b| = 15 km/h
Angle between v_w and - v_b is \(135°\)  i.e. \(\theta =135°\) Then, 
\(\tan { \beta } =\frac { 51\sin { 135° } }{ 72+51\cos { 135° } } =\frac { 51\sin { 45° } }{ 72+51(-\cos { 45°) } } \)
=\(\frac { 51\times (1/\sqrt { 2 } ) }{ 72-51(1/\sqrt { 2 } ) } \) = 1.0039
\(\therefore \)  \(\beta =\tan ^{ -1 }{ (1.0039)=45.1° } \)
Angle w.r.t. East direction = \(45.1°\)-\(45°\)=\(0.1°\)
It means the flag will flutter almost due East.    

Let t be the time taken by the ball to reach the highest point where height from ground be s.
Taking vertical upward motion of the ball, we have u =- 29.4ms^-, a = 9.8ms^-, v = 0, s = S,t = ?
As, v²-u² = 2 as
0 -(-29.4)² = 2 x 9.8 x S
or \(S=\frac { -{ (29.4) }^{ 2 } }{ 2\times 9.8 } =-44.1m\)
Here, negative sign shows that the distance is covered in upward direction.
As, v = u + at
\(\therefore \quad 0=-29.4+9.8\times t\quad or\quad t=\frac { 29.4 }{ 9.8 } =3s\)
It means time of ascent = 3s
When an object moves under the effect of gravity alone, the time of ascent is always equal to the time of descent.
Therefore, total time after which the ball returns to the player's hand = 3 + 3 = 6s.

Imagine you along with your friends are in a spaceship which is moving towards Mars. The body of the speceship is made of a specially designed matter which become more harder as its temperature increases. nuclear power plants in spaceship.
Two of them work alternatively and third is for emergency. The speed of the spaceship is very high and all of you are very happy. The energy produced in power plants are converted into electric energy which runs the motors of the spaceship.
You along with your friends reach safely on Mars, collects data, takes photographs and then returns to the Earth. In return journey is working and due to overheating. its efficiency is decreasing continuously.
You and your friends try to reduce the temperature of the plant and try to repair the fuse of the other power plants.
Finally, fuse of one other plant is repaired and start to work before the first plant crosses the danger limit of an excess of temperature. Finally, you and your friends return safely on Earth.

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

As E ∝ A², the energy of harmonic oscillator will became 4 times, its original value when its amplitude is doubled.

The radial component of a force does not contribute towards torque.

Since, S \(\alpha\) v², therefore motor car will cover a distance four times longer than before.

No, as it is valid only when v_s < v . When v_s >v ,shock waves are produced.

Whenever a mass collides and gets stuck to the other mass, the collision is said to be inelastic.

Density of a solid decreases as it is heated (i.e., as its temperature rises) as per following relation: p / = p(1 - \(\gamma \).\(\triangle\).T), where p = density of given solid at temperature T and p' = density of given solid at temperature (T + \(\triangle\).T).

Compressibility is defined as the reciprocal of bulk modulus (B) i.e.,
K = \(\frac{1}{B}\).

Change in pressure will be more than 3 times the initial pressure.

The theory with which scientists try to unify all forces is called unified field theory.

Oxygen is a diatomic gas.
Number of moles of O₂ gas
\(=\frac { Atomic\ wt. }{ Molecular\ wt. } =\frac { 8 }{ 32 } \)
\(\\ =\frac { 1 }{ 4 } =0.25\)
\(\\ \therefore \ Energy\ associated\ with\ 1\ mole\ of\ oxygen\)
\(\\ U=\frac { 5 }{ 2 } RT\)
\(\\ \therefore \ Internal\ enreyg\ of\ 8g\ of\ oxygen=0.25\times \frac { 5 }{ 2 } \times 8.31\times 273=1417.9J\)

Surface energy of first soap bubble 
= Surface tension \(\times\) Surface area
\(=2\times 4\pi { R }_{ 1 }^{ 2 }S=8\pi { R }_{ 1 }^{ 2 }S\)
Surface energy of second soap bubble\(=8\pi { R }_{ 1 }^{ 2 }S\)
Let the radius of the new soap bubble is R SO, the surface energy of new bubble \(=8\pi { R }_{ 1 }^{ 2 }S\)
By the law of conservation of energy
\(8\pi { R }^{ 2 }S=8\pi { R }_{ 1 }^{ 2 }S+8\pi { R }_{ 1 }^{ 2 }S\)
\(\\ { R }^{ 2 }={ R }_{ 1 }^{ 2 }+{ R }_{ 2 }^{ 2 }=36+64\)
\(\\ { R }^{ 2 }=100{ cm }^{ 2 }\Rightarrow R=10\ cm\) 

Acceleration due to gravity at the equator is
\(g_{ e }=g-R\omega ^{ 2 }\)
\(\\ mg_{ e }=mg-mR\omega ^{ 2 }\ or\)
\(\\ \frac { 3 }{ 5 } mg=mg-mR\omega ^{ 2 }\ \left[ \because mg_{ e }=\frac { 3 }{ 5 } mg \right] \)
\(\\ \therefore \omega =\sqrt { \frac { 2g }{ 5R } } ={ \sqrt { \frac { 2\times 9.8 }{ 5\times 6400\times 10^{ 3 } } } =7.8\times 10^{ -4 } }rad/s\) 

Centre of mass and centre of gravity are two different concepts. But if g goes not vary from one part of body to other than CG and CM coincides.
So, when vertical height of the object is very small compared to a radius of earth, we call object small, otherwise, we call it extended. In above context, building and pond are small objects and a deep lake and a mountain are large extend objects.

Rohit and suresh were very alert and courageous. They also have presence of mind.

Given, A = 20 N, B = 20 N, R = 30 N, \(\theta \) = ?
 \(R=\sqrt { { A }^{ 2 }+{ B }^{ 2 }+2AB\cos { \theta } }\)
\( \\ 30=\sqrt { { 2 }0^{ 2 }+{ 2 }0^{ 2 }+2\times 20\times 20\cos { \theta } } \)
\(\\ \cos { \theta } =\frac { { 30 }^{ 2 }-{ 2 }0^{ 2 }-{ 2 }0^{ 2 } }{ 2\times { 2 }0^{ 2 } } =0.125\)
\( \Rightarrow \quad \theta ={ 82 }^{ 0 }{ 49 }^{ ' }\) 

From the above question, we draw the following figure.

Total path length = AB + BC = 360 + 240 = 600 m

In 1 teragram = 10¹²g
In 1 metric ton = 10³kg = 10³ x 10³ = 10⁶g
Number of metric tons are in teragram
\(\frac { { 10 }^{ 12 }g }{ { 10 }^{ 6 }g } ={ 10 }^{ 6 }\)

One

v = \(\frac{dx}{dt}\)
Since v = \(\alpha\sqrt{x}\)
we have \( \frac{dx}{dt}=\alpha\sqrt{x} \) or \(\frac{dx}{\sqrt{x}}=\alpha dt\)
Intefrating from \( t=0(x=0) to \ t=t(x=x)\)
we have \( \overset { x }{ \underset { 0 }{ \int } } { x }^{ -1/2 }dx=\alpha \overset { t }{ \underset { 0 }{ \int } } dt\)
∴ \({ \left| \frac { { x }^{ 1/2 } }{ 1/2 } \right| }_{ 0 }^{ x }=\alpha t\)
or \( x = \frac{\alpha^2t^2}{4}\)
The time dependence of the velocity is obtained by differentiating both sides of this relation w.r.t. time t. Thus
\(v=\frac{dv}{dt}=\frac{\alpha.2t}{4}=\frac{\alpha^2}{2}t\)
The velocity x of the particle is thus increasing in direct proportion to time.
Similarly, the time dependence of acceleration is obtained by differentiating both sides of this relation w.r.t. 't', Thus
\( a=\frac{dv}{dt}=\frac{\alpha^2}{2}\)
The particle is thus moving with a constant acceleration.
To find the average velocity over the first s metre, we assume that the time taken to cover this distance is T. Using
\( x=\frac{\alpha^2t^2}{4}\)
we get s= \(\frac{\alpha^2T^2}{4} or \ T=\frac{2\sqrt{s}}{\alpha}\)
The average velocity v_av \((=s/T)\) is, therefore
\( v_{av}=(\frac{\alpha}{2}\sqrt{s}) \).

The radius of gyration of a body about the axis of rotation of a body is the point at which the weighed mass of the body acts. It is also equal to the square root of moment of inertia of all particles of the body about the axis of rotation divided by the total mass of the body i.e.,
\(k=\sqrt {\frac{m_1r_1^2+m_2r_2^2+m_3r_3^2....+m_nr_n^2}{m_1+m_2+m_3...+m_n}}\)
\(\sqrt {\frac{\sum mR}{M}}=\sqrt {\frac{I}{M}}\)
Its dimensions are those of length and its is measured in metre in SI units. The moment of a body is a quantity which comes in rotational motion and plays same role in rotational motion as does mas in translational motion. Thus a body continues of rotate or be at rest in the absence of any external torque. This is similar to the law of inertia in translational motion. This aspect is used in over creasing the dead points in the engines and crankshafts. Similarly, the kinetic energy of rotation is dependent on the moment of inertia of the body in the same manner the kinetic energy of translation of motion depends, on the mass of the body. For a given angular velocity (\(\omega\)) kinetic energy of rotation \(\propto\) I. If equal torques are applied I₁ and I₂ the their angular acceleration are inversely proportional to the moments of inertia of the bodies.
\(\frac{I_1\alpha_1}{I_2\alpha_2}=1\)
\(\frac{\alpha_1}{\alpha_2}=\frac{I_2}{I_1}\)
\(\alpha\propto\frac{1}{I}\)
Similarly if two bodies have same angular acceleration, then the moments of inertia are directly proportional to the torque applied on then.
\(\frac{\tau_1}{\tau_2}=\frac{I_1}{I_2}\ or \alpha_1=\alpha_2\)
The linear momentum of a body depends on its mass and velocity. If two bodies have same velocity, then their moments are proportional to their masses i.e.,
\(\frac{\rho_1}{\rho_2}=\frac{m_1}{m_2}\)
Similarly for angular momentum
\(\frac{L_1}{L_2}=\frac{I_1}{I_2}\)
Thus, moment of inertia of a body plays same role in the rotational motion as does mass in translational motion.
The moment of inertia determines the amount of torque to be applied to produce desired angular acceleration.

Given, Avogadro number N = 6.02 x 10²³
Boltzmann's constant k = 1.38 x 10^-23 joule/molecule K
Kelvin temperature T = 27 + 273 = 300 K.
(a) An oxygen molecule has three degrees of freedom with respect of translation. Hence the average kinetic energy of translation of molecule
\(3\times \frac { 1 }{ 2 } kT=\frac { 3 }{ 2 } kT\)
(\(\because \) Average kinetic energy of a gas molecule per degree of freedom is \(\frac { 1 }{ 2 } \) kT.)
\(=3\times \frac { 1 }{ 2 } (1.38\times { 10 }^{ -23 })\times 300\)
= 10.35 x 10^-21 joule/molecule 
One gm molecule of oxygen contains N molecules. Hence the total kinetic energy of 1 gm molecule of the gas
\(=N\times \frac { 5 }{ 2 } kT\)
= (6.02 x 10²³) x 10.35 x 10^-21
= 6231 joule/mole

For calorimetry and its principle, see text
Heat is lost by stearn in getting condensed and gained by water, ice and the calorimeter.
Let the calorimeter originally contain x grams of ice and (200- x) grams of water , we then have
Heat gained by calorimeter = \(\frac { 100 }{ 1000 } \) x 0.42 x 10³ x (50 - 0)
= 2100 J
Heat gained by ice = \(\frac { x }{ 1000 } \) {3.36 x 10⁵ + 4.2x10³ (50 - 0) }
= x {336 + 210} = 546 x J
Heat gained by water  =\(\frac { \left( 200-x \right) }{ 1000 } \) x 4.2 10³ x (50 - 0)
 = (42000 - 210 x) J 
Heat lost by stearn = \(\frac { (330-200-100) }{ 1000 } \) x [22.5 x 10⁵ + 4.2 x 10³ x (100 - 50)]
= 30 (2250 + 210) = 73800 J
Heat gained = Heat lost
546 x + 2100 + 42000 - 210x = 73800
Mass of ice in the original mixture = 88.4 g
Mass of water in the original mixture = 200 - 88.4 = 111.6 g
Ratio of ice present to water present = 88.4 : 111.6
= 1 : 1.26

For Poisson's ratio, see text.
Poisson's ratio
where r is the radius of the wire and l its length.
Volume of the wire before expansion V₁
Volume of the wire after expansion
V₂ = \(\pi\)(r-Δr)²(l+Δl)
If the volume is to remain unchanged during expansion, we require that
V₁ = V₂, i.e.,
\(\pi\)r²l = \(\pi\)[r²-2r Δr + (Δr)²](l+Δl)
= \(\pi\)r²(l+Δl)-2\(\pi\)r Δrl-2r Δr Δl \(\pi\) + (Δr)²(l+Δl)\(\pi\)
or 2\(\pi\)r Δrl =\(\pi\)r² Δl + terms containing the product of Δr and Δl, which can be neglected.
Hence \(\frac { \triangle r/r }{ \triangle l/l } =\frac { 1 }{ 2 } \)
or σ = 0.5
Thus the volume of the wire does not change if the Poisson's ratio of the material of the wire is 0.5.

The centripetal force required by a satellite of mass m revolving in a circular orbit of radius r with a speed v is supplied by the gravitational force extended by the planet of mass M on the satellite. Thus
\(\frac{Mv^2}{r}=G\frac{Mm}{r^2}\)
\(v=\sqrt {\frac{Gm}{r}}\)
The period of revolution of the satellite is
\(T=\frac{2\pi r}{v}=2\pi\sqrt {\frac{r^3}{GM}}\)
For satellite S₁ let T=T₁,r=r₁,v=v₁
Then \(T_1^2=\frac{4\pi^2r_1^3}{GM}\)
For satellite S₂ , \(T_2^2=\frac{4\pi^2r_2^2}{GM}\therefore \frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}\)
\(r_2=r_1(\frac{T_2}{T_1})^{\frac{2}{3}}\)
\(=10^4(\frac{8}{1})^{\frac{2}{3}}\)
\(=4\times10^4\ km\)
\(v_1=\frac{2\pi r_1}{T_1}\)
\(=\frac{2\pi\times10^4}{1}=2\pi\times10^4\ km/hr\)
\(v_2=\frac{2\pi r_2}{T_2}\)
\(=\frac{2\pi\times4\times10^4}{8}\)
\(=\pi\times10^4\ km/hour\)
Velocity of S₂ relative to \(S_1=v_2-c_1=\theta_r(say)\)
\(v_r=(\pi\times10^4-2\pi\times10^4)\ km/hr\)
\(=-\pi\times10^4\ km/hour\)
Let r₂-r₁ = r
The angular velocity of S₂ relative to S₁ is given by
\(\omega=\frac{v_r}{r}=\frac{\pi\times10^4}{(4-1)10^4}\ rad/hour\)
\(\omega=\frac{\pi}{3}\ rad/hour\)

The given displacement function is
x(t) = A cos (ωt + \(\phi \)) ...(i)
At t 0, x(0) = 1 cm. Also, ω = \(\pi\)s^-1
∴ 1 = A cos(\(\pi\)\(\times\)0+\(\phi \))
⇒ A cos\(\phi \) = 1 ....(ii)
Also, differentiating eqn. (i) w.r.t. 't'.
v = \(\frac{d}{dt}x(t)\)=-Aωsin(ωt + \(\phi \)) ....(iii)
Now at t = 0, v = ω
∴ from eqn. (iii), ω = -Aωsin(\(\pi\)\(\times\)0 + \(\phi \))
or A sin \(\phi \)=-1 ....(iv)
Squaring and adding eqns. (ii) and (iv).
A² cos²\(\phi \)+ A² sin²\(\phi \) = 1² + 1² or A =\(\sqrt2\)cm
Dividing eqns. (ii) and (iv),
\(\frac { Asin\phi }{ Acos\phi } =\frac { -1 }{ 1 } \therefore tan\phi =-1\Rightarrow \phi =\frac { 3\pi }{ 4 } \)
If instead we use the sine function, i.e.,
x = B sin (ωt+ α),then v =\(\frac{d}{dt}\)Bωcos(ωt +  α)
∴ At t = 0, using x = 1and v = ω, we get 1 = B sin(ω\(\times\)0+α)
or B sin α = 1 ...(v)
and ω = Bωcos(ω\(\times\)0+α) or Bcosα = 1 ...(vi)
Dividing (v) by (vi),
tanα = 1 or α = \(\frac{\pi}{4}\) or \(\frac{5\pi}{4}\)
Squaring (v) and (vi), we get
B² sin²α + B² cos² α= 1² + 1²
⇒ B =​ \(\sqrt2\) cm

(a) Let a drop of a liquid L be poured on a solid surface S placed in air A. If T_SL, T_LA and T_SA be the surface tensions corresponding to solid-liquid layer, liquid-air layer and solid-air layer respectively and \(\theta\) be the angle of contact between the liquid and solid, then
       
\(T_{LA}\cos\theta+T_{SL}=T_{SA}\)
\(\Rightarrow \cos\theta=\frac{T_{SA}-T_{SL}}{T_{LA}}\)
For the mercury-glass interface, T_SA < T_SL, Therefore, cos \(\theta\) is negative. Thus \(\theta\) is an obtuse angle. For the water-glass interface, T_SA > T_SL, Therefore cos \(\theta\) is positive. Thus, \(\theta\) is an acute angle.
(b) Water on a clean glass surface tends to spread out i.e., water wets glass because force of cohesion of water is much less than the force of adhesion due to glass. In case of mercury force of cohesion due to mercury molecules is quite strong as compared to adhesion force due to glass. Consequently, mercury does not wet glass and tends to form drops.
(c) Surface tension of liquid is the force acting per unit length on a line drawn tangentially to the liquid surface at rest. Since t; .is force is independent of the area of liquid surface therefore, surface tension is also independent of the area of the liquid surface.
(d) We know that the clothes have narrow pores or spaces which act as capillaries. Also, we know that the rise of liquid in a capillary tube is directly proportional to cos \(\theta\) (Here \(\theta\) is the angle of contact). As \(\theta\) is small for detergent, therefore cos \(\theta\) will be large. Due to this, the detergent will penetrate more in the narrow pores of the clothes.
(e) We know that any system tends to remain in a state of minimum energy. In the absence of any external force for a given volume of liquid its surface area and consequently. Surface energy is least for a spherical shape. It is due to this reason that a liquid drop, in the absence of an external force is spherical in shape.

495 Hz, 594 Hz, Yes

Given, m = 0.2kg, T = 8s
\(\omega =\frac { 2\pi }{ T } =\frac { 2\pi }{ 8 } =\frac { \pi }{ 4 } rad/s\)
When t  = 1s, v = 6m/s
\(v(t)=\omega A\cos { \omega t } \)
\(\\ 6=\frac { \pi }{ 4 } \times Acos\left( \frac { \pi }{ 4 } \times 1 \right) \)
\(\\ 6=\frac { \pi }{ 4 } \times A\times \frac { 1 }{ \sqrt { 2 } } \)
\( A=\frac { \sqrt [ 4 ]{ 2\times 6 } }{ \pi } =\frac { \sqrt [ 24 ]{ 2 } }{ \pi } m\)
The total energy of the body
E = \(\frac { 1 }{ 2 } m{ \omega }^{ 2 }{ A }^{ 2 }=\frac { 1 }{ 2 } \times 0.2\times \left( \frac { \pi }{ 4 } \right) ^{ 2 }\times \left( \frac { \sqrt [ 24 ]{ 2 } }{ \pi } \right) ^{ 2 }\)
\( =\frac { 230.4 }{ 23 } =7.2\ J\)
Potential energy, PE =E - KE =7.2-\(\frac { 1 }{ 2 } \)mv²
= \(7.2-\frac { 1 }{ 2 } \times 0.2\times (6{ ) }^{ 2 }=7.2-3.6\)
Potential energy = 3.6 J

(i) This problem is designed to give an idean about cooling by evaporation
\({ V }^{ 2 }_{ rms }=\frac { \underset { i }{ \Sigma { n }_{ i } } { v }_{ i }^{ 2 } }{ \Sigma { n }_{ i } }\) 
\(=\ \frac { 10\times \left( 200 \right) ^{ 2 }\times \left( 400 \right) ^{ 2 }+40\times \left( 600 \right) ^{ 2 }+20\times \left( 800 \right) ^{ 2 }+10\times \left( 1000 \right) ^{ 2 } }{ 100 }\)
\(=\frac { 10\times { 100 }^{ 2 }\times \left( 1\times 4+2\times 16+4\times 36+2\times 64+1\times 100 \right) }{ 100 } \)
\(1000\times \left( 4+32+144+128+100 \right)\)
\( \\ =408\times 1000\ { m }^{ 2 }/{ s }^{ 2 }\)
\(\\ \therefore \ { v }_{ rms }=639\ m/s\)
\(\\ \frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 3 }{ 2 } kT\)
\(\\ \therefore \ T=\frac { 1 }{ 3 } \frac { mv^{ 2 }_{ rms } }{ k } =\frac { 1 }{ 3 } \times \frac { 30\times { 10 }^{ -26 }\times 4.08\times { 10 }^{ 5 } }{ 1.38\times { 10 }^{ -23 } } \)
\(\\ =2.96\times { 10 }^{ 2 }=296\ K\)
(ii) This problem is designed to give an
 \({ v }_{ rms }^{ 2 }=\frac { 10\times \left( 200 \right) ^{ 2 }+20\times \left( 400 \right) ^{ 2 }+40\times \left( 600 \right) ^{ 2 }+20\times \left( 800 \right) ^{ 2 } }{ 90 } \) 
\(\\ =\frac { 10\times { 100 }^{ 2 }(1\times 4+2\times 16+4\times 36+2\times 64) }{ 90 } \)
\(\\ { v }_{ rms }^{ 2 }=10000\times \frac { 308 }{ 9 }\)
\( \\ { v }_{ rms }^{ 2 }=342\times 1000\ { m }^{ 2 }/{ s }^{ 2 }\)
\(\\ { v }_{ rms }=584\ m/s\)
\(T=\frac { 1 }{ 3 } \frac { m{ v }^{ 2 }_{ rms } }{ k } \)  
T = 248 K

\(For \ path \ I, \ \triangle Q=\triangle W+\triangle U\ \Rightarrow \ \triangle U=\triangle Q-\triangle W\)
\( =-80-(-120)=40\ J\)
\(\\ As,\ \triangle U=\ a\ state\ function,\ so\ d=e=f=g=40\ J\)
\(\\ Now\ for\ Path\ III,\ \triangle Q=\triangle W+\triangle U\)
\(\\ 90=c+40\ \Rightarrow \ c=50\ J\)
\(\\ Similarly,\ a=80J,\ b=0\)

(i) Aman loves nature and his helpful boy. He also has presence of mind as he thought of an excellent idea to help cartman.
(ii) \({ L }_{ 1 }=5.231m,{ L }_{ 2 }=5.243m,{ T }_{ 1 }=27 ^{0}C,{ T }_{ 2 }=?\)
As we know 
\(\therefore { T }_{ 2 }-{ T }_{ 1 }=\frac { { L }_{ 2 }-{ L }_{ 1 } }{ { L }_{ 1 }\times \alpha } \)
\(\\ { T }_{ 2 }=\frac { { L }_{ 2 }-{ L }_{ 1 } }{ { L }_{ 1 }\alpha } \)
\({ T }_{ 2 }=\frac { 5.243-5.231 }{ 5.231\times 1.2\times { 10 }^{ -5 } } +27\)
\(\\ =218 \ ^{0} C\)
(iii) Linear expansion of solids i.e increase in length of solid on heating.

(i) Sahil is very helpful and generous person.
(ii) People don't bother about the road accident victims because they probably think that they don't know the victim or they don't want to get involved in a police case.
(iii) Given, 
\(h=1.3m,\rho =1020kg/{ m }^{ 3 },p=h\rho g\)
\(p=1.3\times 1020\times 9.8\)
\(\\ r=\frac { D }{ 2 } =\frac { 0.36 }{ 2 } =0.18mm=0.8\times { 10 }^{ -3 }m\)
\(\\ l=3cm\ =\ 3\times { 10 }^{ -2 }m\)
\(\\ V=\frac { Total\ volume }{ Time } =\frac { 4.5\times { 10 }^{ -6 } }{ 60 } { m }^{ 3 }/s\)
\(Viscosity,\eta =\frac { \pi p{ r }^{ 4 } }{ 8Vl }\)
\(=\frac { 22 }{ 7 } \times \frac { (1.3\times 1020\times 9.8)\times (0.18\times { 10 }^{ -3 })^{ -4 } }{ 8\times \left( \frac { 4.5\times { 10 }^{ -6 } }{ 60 } \right) \times 3\times { 10 }^{ 2 } } \)
\(\\ \eta =0.00238\ Pa-s\) 

Average force exerted to the quarterback's femur
So    F_av x \(\Delta \) t = M ( v_f - v_i )
i.e.     Fav = \(\frac { M({ v }_{ f }-{ v }_{ i }) }{ \Delta t } \)  = \(\frac { 125(4-2.38) }{ 0.1 } \) 
= \(\frac { 125\times 1.62 }{ 0.1 } \) = 2.03 x 10³ N
Shearing stress = \(\frac { F }{ A } \) = \(\frac { 2.03\times { 10 }^{ 3 } }{ 5\times { 10 }^{ -4 } } \) 
= 4.06 x 10⁶ Pa

Kepler's second law establishes that the linear speed of a planet, when closer to the sun, is more than its linear speed when away from the sun. The planets revolve around the sun on account of the gravitational pull of the sun. Thus, sun is the source of energy. The law implies that when you are closer to the source of energy, your speed/progress is faster. And when you are far away from the source of energy, your speed/progress become slow. This is so true in everyday life.

In pulling a lawn mower, a force F is applied in upward direction, making an angle \(\theta \) with the horizontal [Fig]. Its vertical component in upward direction decreasing the effective weight of the mower.
 
In pushing a lawn mower, a force F is applied in downward direction, making an angle \(\theta \) with the horizontal [Fig]. Its vertical component is in downward direction increasing the effective weight of the mower. Therefore, it is easier to pull a lawn mower than to push it.

Given,F₁ = 5N,F₂ = 8N, m = 0.3 kg and acceleration a = ?.The resultant force in the x-direction exerted on the puck.
\(\Sigma { F }_{ x }={ F }_{ 1x }+{ F }_{ 2x }={ F }_{ 1 }\cos { 20° } +{ F }_{ 2 }\cos { 60° } \)
\(\\=(5N)(0.940)+(8N)(0.500)=8.70N\)
The resultant force in the y-direction exerted on the puck,
\(\Sigma { F }_{ y }={ F }_{ 1y }+{ F }_{ 2y }={ -F }_{ 1 }\sin { 20° } +{ F }_{ 2 }\sin { 60° } \)
\(\\ =(5N)(0.342)+(8N)(0.8666)=5.22N\) 
Now, we can use Newton's second law in component form to find the x and y-component of acceleration
 \({ a }_{ x }=\frac { \Sigma { F }_{ x } }{ m } =\frac { 8.70N }{ 0.3kg } =29.0m/{ s }^{ 2 }\) 
\(\\ { a }_{ y }=\frac { \Sigma { F }_{ y } }{ m } =\frac { 5.22N }{ 0.3kg } =17.4m/{ s }^{ 2 }\)
Magnitude,    \(a=\sqrt { { (29.0) }^{ 2 }+{ (17.4) }^{ 2 } } m/{ s }^{ 2 }=33.8m/{ s }^{ 2 }\) 
and its direction is
\(\theta =\tan ^{ -1 }{ ({ a }_{ y }/{ a }_{ x }) } =\tan ^{ -1 }{ (17.4/29.0)=31.0° } \) 
relative to the positive x-axis.

(i) \(T=\frac { { 2u }_{ 0 }\sin { \beta } }{ g\cos { \alpha } } \)
(ii) \(R=\frac { { 2u }_{ 0 }\sin { \beta } \cos { \left( \alpha +\beta \right) } }{ g\cos { ^{ 2 }\alpha } } \)
(iii) \(\beta =\frac { \pi }{ 4 } -\frac { \alpha }{ 2 } \)

The instrument whose least count is minimum, is called the most precise device. 
i) Number of divisions on vernier scale = 20
Main scale Division (MSD) = 1 mm
As 20 divisions on vernier scale will be equal to the 19 divisions on the vernier scale will be equal to the 19 division on main scale.
 ∴ vernier scale Division (VSD) \(=\frac { 19 }{ 20 } MSD\) 
Least count of vernier callipers
= 1MSD − 1VSD
\(\\=1MSD-\frac { 19 }{ 20 } MSD=\frac { 1 }{ 20 } MSD\)
\(\\ =\frac { 1 }{ 20 } mm=\frac { 1 }{ 200 } cm=0.005cm\)
ii) Pitch of screw guage = 1mm
Number of divisions on circular scale = 100
Least count of screw guage
\(=\frac { Pitch }{ Number\ of\ division\ on\ circular\ scale } \)   
\(\\ =\frac { 1 }{ 100 } mm=\frac { 1 }{ 1000 } cm=0.001cm\)
iii) Wavelength of light \((\lambda )\approx { 10 }^{ -7 }m\) 
= 10^-5cm = 0.00001 cm
\(\therefore \)  As the given optical instrument can measure length to within a wavelength of light, therefore, least count of the given optical instrument
= Wavelength of light
= 0.00001 cm
The least count is minimum for the given optical instrument. Therefore, the given optical instrument is the most precise.