1 amu =1.66 x 10^-27 kg
∴ 1 kg = (1/1.66 x 10^-27) amu = 0.6 x 10²⁷ amu

\(KE=\frac { 1 }{ 2 } mv^{ 2 }\text{ i.e.,dimensional formula of KE is} [ML^{ 2 }T^{ -2 }]\) 
\(Pressure=\frac { Force }{ Area } =\frac { [MLT^{ -2 }] }{ [L^{ 2 }] } =[ML^{ -1 }T^{ -2 }]\)

An object can be considered as a point object if the distance travelled by it is very large than its size.

The graph shows that 'with the passage of time, total path length first increases and then decreases.
The path length always increases or remains constant with passage of time and it does not decrease with time as shown in figure. Thus, this graph is not possible.

When the particle is moving with a constant velocity, there is no change in velocity with time and hence, its acceleration is zero.

Given, Mass of the object (m) = 500 g = 0.5 kg
Speed of the object (v) = 25 m/s
(i) Impulse imparted to the object
= change in the momentum = mv - mu
= m ( v - u ) = 0.5 ( 25 - 0) = 12.5 N-s
(ii) Velocity of the object after rebounding = -\(\frac { 25 }{ 2 } \) m/s
 v' = -12.5 m/s
\(\therefore \)  Change in momentum = m ( v' - v )
= 0.5 ( - 12.5 - 25 ) = -18.75 N-s

we know kinetic energy
\( K=\frac{p^2}{2 m} \)
\( K^{\prime}=\frac{p^{\prime} 2}{2 m}=\frac{(2 p)^2}{2 m}=4\left(\frac{p^2}{2 m}\right)=4 K \)
\( \frac{\Delta K}{K}=\frac{K^{\prime}-K}{K}=\frac{4 K-K}{K}=3 \)
\(\%\left(\frac{\Delta K}{K}\right)=3 \times 100=300 \%\)

Angular momentum of a body is measured with respect to certain origin.

So, a body in translatory motion can have angular momentum.
It will be zero, if origin lies on the line of motion of particle.

The gravitation force between satellite and the earth provides the centripetal force required by the satellite to move in a circular orbit.The satellite orbits around earth at such a higher height where air friction is neglible.

The value of g is more at poles than at the equator. Therefore, a body weighs more at poles than at equator.

Here,  original length, L = 2.5 m
Strain = \(\frac { \Delta L }{ L } \) = 0.012 % = \(\frac { 0.012 }{ 100 } \) 
 \(\Delta L\) = Strain x L
\(\Delta L\) = extension  = \(\frac { 0.012 }{ 100 } \times L\) 
= \(\frac { 0.012\times 2.5 }{ 100 } \) = 3 x 10^-4 m
= 0.3 mm

The viscosity of liquid decreases with the increase in temperature. Therefore, viscosity of hot liquid is less than that of cold liquid. Due to this, hot liquid moves faster than the cold liquid.

An ice box is made of double wall and the space in between the walls is filled with some non-conducting material to provide heat insulation, so that the loss of heat can be minimized.

The purpose of a coolant is to absorb maximum heat with least rise in its own temperature. This is possible only if specific heat is high because Q = mc \(\Delta \)T. For a given value of m and Q, the rise in temperature \(\Delta \)T will be small if c is large. This will prevent different parts of the nuclear reactor from getting too hot.

When a gas expands, it does work on the surroundings. This work is done on the expense of internal energy and that is why its internal energy and so its temperature decreases.

A reversible process is never possible in nature because of dissipative forces and condition for a quasi-static process is not practically possible.

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

Three important characteristics of an SHM are amplitude, time period (or frequency) and phase

The ball comes with large momentum after being hit by the batsman. When the player takes catch it causes large impulse on his palms which may hurt the cricketer. When he moves his hands backward the time of contact of ball and hand is increased so the force is reduced.

The problem here may be stated as
Find x when v = 70.0 m/s
It contains the single unknown x, as well as aand v, which are known with u = 0, v_x² = 2axx
Solving for x, we obtain
x = \( \frac{v^2}{2a} \)
= \(\frac{(70.00 m/s)^2}{2(4.00 m/s^2)}\)
= 613 m

Period of a revolution of a satellite is the time taken by the satellite to complete one revolution round the earth. It is denoted by T.
\(\therefore T=\frac { Circumference\ of\ circular\ orbit }{ Orbital\ velocity } \) 
or    \(T=\frac { 2\pi r }{ { v }_{ o } } \)
or     \(T=\frac { 2\pi (R+h) }{ { v }_{ o } } \quad \quad \quad \quad \quad [\therefore r=R+H]\)
or     \(T=2\pi (R+h)\sqrt { \frac { R+h }{ GM } } \left[ \because \quad { v }_{ o }=\sqrt { \frac { GM }{ R+h } } \right] \)
or     \(T=2\pi \sqrt { \frac { (R+h)^{ 2 } }{ GM } } \)     
Also,  \(T=2\pi \sqrt { \frac { (R+h)^{ 2 }(R+h) }{ GM } } \)
or       \(T=2\pi \sqrt { \frac { (R+h)^{ 3 } }{ gR^{ 2 } } } \)
 \(\because \quad \quad g{ R }^{ 2 }=GM\)  
\(\therefore T=2\pi \sqrt { \frac { (R+h)^{ 2 } }{ gR^{ 2 } } } \)   

(a) The pressure of liquid column is given by \(p=h\rho g\), where h is depth, \(\rho \)i s density and g is acceleration due to gravity.
Therefore, pressure of liquid column increases with depth. The height of blood column in human body is more at feet than at the brain. Therefore, the blood pressure in humans is greater at the feet than the brain.
(b) The density of air is maximum near the surface of the earth and decreases rapidly with height. At a height of 6 km, the density of air decreases to nearly half its value at the seal level. Beyond 6km height, the density of air decreases very slowly with height.Hence, the atmospheric pressure at a height of about 6km decreases to nearly half of its value at the sea level.
(c) When force is applied on a liquid, the pressure is transmitted equally in all directions inside the liquid. Therefore, hydrostatic pressure has no fixed direction and hence, it is a scalar quantity.

Consider (say \(\mu \) mole) an ideal gas, which is undergoing an adiabatic expansion. Let the gas expands by an infinitesimally small volume dV, at pressure p, then the  infinitesimally small work done given by
dW = pdV
The net work done from an initial volume V₁ is given by
\(W=\int _{ { v }_{ 1 } }^{ { v }_{ 2 } }{ pdV } \)
For an adiabatic process, \(p{ V }^{ \gamma }=constant=K\)
\(\\ p=\frac { K }{ { V }^{ \gamma } } =K{ V }^{ -\gamma }\)
\(\\ \therefore \ W=\int _{ { v }_{ 1 } }^{ { v }_{ 2 } }{ (K{ V }^{ -\gamma })dV } =k\left[ \frac { { V }^{ -\gamma +1 } }{ -\gamma +1 } \right] _{ { v }_{ 1 } }^{ { v }_{ 2 } }\)
\(\\ =\frac { K{ V }_{ 2 }^{ -\gamma +1 }-K{ V }_{ 1 }^{ -\gamma +1 } }{ (1-\gamma ) } \)
\(\\ For\ an\ adiabatic\ process,\)
\( K={ p }_{ 1 }{ V }_{ 1 }^{ \gamma }={ p }_{ 2 }{ V }_{ 2 }^{ \gamma }\)
\(\\ \Rightarrow W=\frac { { { p }_{ 2 }{ V }_{ 2 }^{ \gamma }.{ V }_{ 2 }^{ -\gamma +1 }-{ p }_{ 1 }{ V }_{ 1 }^{ \gamma }.{ V }_{ 2 }^{ -\gamma +1 } } }{ (1-\gamma ) } \)
\(\\ =\frac { 1 }{ (1-\gamma ) } ({ p }_{ 2 }{ V }_{ 2 }-{ p }_{ 1 }{ V }_{ 1 })\)
\(\\ For\ an\ ideal\ gas,\ { p }_{ 1 }{ V }_{ 1 }=\mu R{ T }_{ 1 }\ and\ { p }_{ 2 }{ V }_{ 2 }=\mu R{ T }_{ 2 }.\ So,\ we\ have\)
\(\\ W=\frac { 1 }{ (1-\gamma ) } [\mu R{ T }_{ 2 }-\mu R{ T }_{ 1 }]=\frac { \mu R }{ (1-\gamma ) } [{ T }_{ 1 }-{ T }_{ 2 }]\)

(i) During driving, temperature of the gas increases while its volume remains constant. So, according to Charles' law, at constant V, p \(\propto \)  T. Therefore, pressure of gas increases.
(ii) This is because in a harbor town, the relative humidity is more than in a desert town. Hence, the climate of a harbour town is without extremes of hot and cold.

(i) The moon  has small gravitational; force and hence the escape velocity is small .As the moon is in tyhe proximity of the earth as seen from the sun, the moon has the same amount of heat per unit area  as that of the earth , The air molecules have l;arge range of speeds.
Even though the rms speed of the air molecules is smaller than the escape velocity on the moon, a significant number of  molecules have speed greater than escape velocity and they escape.
Now, rest of the molecules arrange the speed distribution for the equilibrium temperature. Again, a significant number of molecules escape as their speeds exceed escape sppeed. Hence, over a long time the moon has lost most of its atmosphere.
(ii) As the molecules move higher , their potential energy increases and hence kinetic energy decreases and hence temperature reduces.
At greater height, more volume is available and gas expands and hencde some cooling takes place.

The fundamental frequency of a string fixed at both ends is given by
\(v=\frac{1}{2 L} \sqrt{\frac{F}{\mu}}\)
As F and mu are fixed, \(\frac{v_1}{v_2}=\frac{L_2}{L_1}$ or , $L_2=\frac{v_1}{v_2} L_1=\frac{124 \mathrm{~Hz}}{186 \mathrm{~Hz}}(90 \mathrm{~cm})=60 \mathrm{~cm}\).
Thus, the string should be pressed at 60 cm from an end.

Here, K 1200 Nm^-1; m = 3.0 kg, a = 2.0 cm = 0.02 m
(i) Frequency, \(v=\frac { 1 }{ T } =\frac { 1 }{ 2\pi } \sqrt { \frac { k }{ m } } =\frac { 1 }{ 2\times 3.14 } \sqrt { \frac { 1200 }{ 3 } } =3.2{ s }^{ -1 }\)
(ii) Acceleration, \(A={ \omega }^{ 2 }\quad y=\frac { k }{ m } y\)
Acceleration will be maximum when y is maximum i.e., y = a
∴ max. acceleration, \({ A }_{ max }=\frac { ka }{ m } =\frac { 1200\times 0.02 }{ 3 } =8{ ms }^{ -2 }\)
(iii) Max. speed of the mass will be when it is passing through mean position
\({ V }_{ max }=a\omega =a\sqrt { \frac { k }{ m } } =0.02\times \sqrt { \frac { 1200 }{ 3 } } =0.4{ ms }^{ -1 }\) 

The lower layer of earth's atmosphere reflects infra-red radiations from earth back to the surface of the earth. So the heat radiation received by the earth from the sun during del) time are trapped by the atmosphere. Therefore, if the earth did not have atmosphere, its surface would become too cold to tolerate

If M be the mass of the sun and R be the distance of the earth from the sun, then escape velocity,
\({ v }_{ e }=\sqrt { \frac { 2GM }{ R } } =\sqrt { \frac { 2\times 6.67\times 10^{ -11 }\times 2\times 10^{ 30 } }{ 1.5\times { 10 }^{ 11 } } } { ms }^{ -1 }\)
 \( =\sqrt { \frac { 4\times 6.67 }{ 1.5 } } \times { 10 }^{ 4 }ms^{ -1 }=4.217\times { 10 }^{ 4 }ms^{ -1 }\)
\({ v }_{ e }=42.17\ kms^{ -1 }\ [1km=1000m]\)
\(\therefore \) The escape speed for solar system is \(42.17\quad kms^{ -1 }\) 

\(According\ to\ the\ equation\ of\ continutity,\ we\ get\)
\(\\ { a }_{ 1 }{ v }_{ 1 }={ a }_{ 2 }{ v }_{ 2 }\)
\(\\ \Rightarrow \ \pi { r }_{ 1 }{ v }_{ 1 }=\pi { r }_{ 2 }{ v }_{ 2 }\)
\(\\ \therefore { v }_{ 2 }={ \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }\ { v }_{ 1 }\)
\(\\ Given,\ { r }_{ 1 }=\frac { 4 }{ 2 } =2cm=2\times { 10 }^{ -2 }m\)
\({ r }_{ 2 }=\frac { 6 }{ 3 } =3cm\ =3\times { 10 }^{ -2 }m\)
\( { v }_{ 1 }=5m/s\)
\({ v }_{ 2 }={ \left( \frac { 2 }{ 3 } \right) }^{ 2 }\times 5=2.22m/s\)
\(\\ Now,\ applying\ the\ Bernoulli's\ theorem\)
\(\\ { p }_{ 1 }+\frac { 1 }{ 2 } { \rho { v } }_{ 1 }^{ 2 }={ p }_{ 2 }\frac { 1 }{ 2 } { \rho v }_{ 2 }^{ 2 }\)
\({ p }_{ 2 }={ p }_{ 1 }+\frac { 1 }{ 2 } \rho ({ { v } }_{ 1 }^{ 2 }+{ v }_{ 2 }^{ 2 })\)
\(=3.0\times { 10 }^{ 4 }+\frac { 1 }{ 2 } \times { 10 }^{ 3 }({ 5 }^{ 2 }-{ 2.22 }^{ 2 })\)
\(=3\times { 10 }^{ 4 }+\frac { 1 }{ 2 } \times { 10 }^{ 4 }+500\times 20.08\)
\( { p }_{ 2 }=4\times { 10 }^{ 4 }N/{ m }^{ 2 }\) 

We know that C_p - C_v = R
and \({C_P\over C_v}=γ\)
From eqn. (ii) C_p = γC_v and sustituting this value in (i),
We have \(⋎C_v-C_v=R⇒C_v={R\over (⋎-1)}\)
\(C_v=γ.C_v={γR\over(γ-1)}\)

The velocity required to put a satellite into its orbit around the earth is called orbital velocity. A satellite moves around the earth in its orbit with orbital velocity 
Consider a satellite of mass m revolving around the earth in an orbit of radius R + h, where R is the radius of earth and h is the height of satellite from the surface of earth.
Let v_o be the orbital velocity of the satellite. The gravitational force between the satellite and the earth provides the necessary centripetal force to the satellite to move in a circular path around the earth. i.e., Gravitational force = Centripetal force
\(\frac{GMm}{(R+h)^2}=\frac{mv_0^2}{(R+h)}\Rightarrow v_0^2=\frac{GM}{(R+h)}\)
\(v_0=[\frac{GM}{(R+h)}]^{\frac{1}{2}}\)
\(\frac{GM}{R^2}=g\ or\ GM=gR^2\)
\(\therefore v_0=[\frac{gR^2}{(R+h)}]^{\frac{1}{2}}\)
If satellite is very close to the surface of earth, then (R + h) = R
Hence \(v_0=\sqrt {gR}.\)

Let at any instant, the displacement of a particle executing SHM is y, mass of the particle 'm',
We know, displacement, y = a sin ωt
velocity, \(v={dy\over dt}={d\over dt}a\ sin\ ωt\)
or v= a ω cos ωt
kinetic energy = \({1\over 2}mv^2\)
\(E_k={1\over 2}m(a\ ω\ cos\ ωt)^2={1\over 2}ma^2ω^2cos^2ωt\)
or \(E_k={1\over 2}ma^2ω^2(1-sin^2ωt)\) 
or \(E_k={1\over 2}ma^2ω^2(1-{y^2\over a^2})\)
or \(E_k={1\over 2}mω^2(a^2-y^2)\)
For Potential Energy:
As velocity, v = aω cos ωt
acceleration = \({d\over dt}(v)=(aω\ cos\ ωt)\)
=-aω² sin ωt = -ω²(a sin ωt) = -ω²y
Restoring force at any instant, F = - ω² y. m = - mω²y
The negative sign indicates that the restoring force is always directed towards the mean position. In order to maintain the particle at displacement y, a force In ω²y acting away from the mean position has to be applied on the particle. Let dW be the work done by the applied force to displace a given particle through a distance dy away from the mean position
Then, dW = mω²y dy
Let W be the total work done in increasing the displacement from O to y.
Thus, \(W=\int_0^ymω^2y dy = mω^2\int_0^yydy\)
or \(W=mω^2\left[ y^2\over 2\right]_0^4=mω^2\left({y^2\over 2}-0\right)={1\over 2}mω^2y^2\)
This work done is stored in the particle as potential
\(∴\ \ E_P={1\over 2}mω^2y^2\)
Total energy, E = E_k + E_p
\(⇒\ E={1\over 2}mω^2(a^2-y^2)+{1\over 2}mω^2y^2\)
or \(E={1\over 2}mω^2a^2-{1\over 2}mω^2y^2+{1\over 2}mω^2y^2\)
or \(E={1\over 2}mω^2a^2\)