First harmonic of a closed organ pipe
\(\mathrm{L}_{\mathrm{c}}=\frac{\lambda}{4}\)
Third harmonic of an open organ pipe
\(\mathrm{L}_{\mathrm{o}} =\frac{3 \lambda}{2} \)
\(\frac{L_{c}}{L_{o}} =\frac{\lambda}{4} \times \frac{2}{3 \lambda} \)
\(\frac{L_{c}}{L_{o}} =, \frac{1}{6} \)

Escape speed v_e = \(\sqrt 2g R\)
if g' = 4g 
then \(v'_e\) = \(\sqrt (4g) R\)
= \(\sqrt 2g R \) \(\times\)2
= 2v_e

Moment of inertia of a disc
\(\mathrm{I}_{1}=\frac{M R^{2}}{2}\)
\(\text { Mass of small disc }=\frac{M}{\pi R^{2}} \times \pi \times\left(\frac{R}{2}\right)^{2}\)
\(=\frac{M}{\pi R^{2}} \times \frac{\pi R^{2}}{4}=\frac{M}{4}\)
By the theorem of parallel axis, the moment of inertia of the small disc. About an axis passing through 0 is
\(I_{2} =\frac{1}{2} \times \frac{M}{4}\left(\frac{R}{2}\right)^{2}+\frac{M}{4}\left(\frac{R}{2}\right)^{2} \)
\(=\frac{M}{8} \times \frac{R^{2}}{4}+\frac{M}{4} \times \frac{R^{2}}{4} \)
\(=\frac{M R^{2}}{32}+\frac{M R^{2}}{16}=\frac{M R^{2}+2 M R^{2}}{32} \)
\(I_{2} =\frac{3 M R^{2}}{32} \)
Moment of inertia of the remaining part is I= I₁ - I₂
\(=\frac{M R^{2}}{2}-\frac{3 M R^{2}}{32} \)
\(=\frac{16 M R^{2}-3 M R^{2}}{32}=\frac{13 M R^{2}}{32}\)
\(I =\frac{13 M R^{2}}{32} \)

Mass, length are scalars. Acceleration is a vector but magnitude of acceleration is a scalar

Here,
y(x, t) = 5.0 sin (48t + 0.0264x + \(\pi\over 6\))
The general equation of a plane progressive wave is,
v(x,t) = a Sin\(\left[{2\pi\over \lambda}(vt+x)+\phi\right]\)
It is observed that the given equation represent a travelling waveform right to left.
Velocity \(V={48\over 0.0264}=1818.18cms^{-1}, r=5cm\)
(a) Amplitude and frequency: 
Amplitude,
\({2\pi\over \lambda}=0.0264\)
or
\(\lambda={2\pi\over 0.0264}cm={2\times3.14\over 0.0264}={6.28\over 0.0264}=237.8cm\)
frequency,
From the equation
v = ⋋v,
\(v={v\over \lambda}={1818.18\over 2\pi}\times0.0264\)
\(={1818.18\over 2\times3.14}\times0.0264\)
= 289.51 x 0.0264
= 7.64 Hz
b) To find least distance between two successive crests in the wave.
\(\lambda={2\pi\over 0.0264 }={2\times3.14\over 0.0264}={6.28\over0.0264}\)
= 237.8 em = 2.38 m
(c) When \(x={\lambda\over 4}\)
\(\phi={2\pi\over \lambda}\times{\lambda\over 4}={\pi\over 2}rad\)

S.No	Isothermal	Adiabatic
1.	Temperature remains constant \(\triangle\)T =0	Heat content remains constant \(\triangle\)Q
2	Walls of the container is perfectly conductivity.	All walls and piston are perfectly an insulating.
3.	The changes occur slowly i.e. slow process.	The changes occur suddenly i.e. a fast process.
4.	Internal energy remains constant, i.e \(\triangle\)U = 0	internal energy changes \(\triangle\)U \(\neq \) 0
5.	Pv = constant	\({ P }_{ v }^{ \gamma }\) constant
6	Slope of isothermal curve on pv dig. \(\frac { -P }{ v } =\frac { dp }{ dv } \)	Slope is \(\frac { -\gamma p }{ v } \) \(\gamma >1\) Slope of adiabatic greater than isothermal.

a. Expression for Potential Energy For the simple harmonic motion, the force and the displacement are related by Hooke's law
\(\vec { F } =-k\vec { r } \)
(i) Since force is a vector quantity, in three dimensions it has three components. Further, the force in the above equation is a conservative force field; such a force can be derived from a scalar function which has only one component. In one dimensional case
F = -kx   .....(i)
(ii) As we have discussed in unit 4 of volume I, the work done by the conservative force field is independent of path. The potential energy U can be calculated from the following expression.
F = \(\frac { dU }{ dx } \)   .......(2)
Comparing (1) and (2). we get
-\(\frac { dU }{ dx } \) = -kx
dU = kxdx
(iii) This work done by the force F during a small displacement dx stores as potential energy
U(x)=\(\int _{ 0 }^{ x }{ kx'dx=\frac { 1 }{ 2 } (x')^{ 2 }{ |_{ 0 }^{ x } } } =\frac { 1 }{ 2 } kx^{ 2 }\)   ....(3)
From equation \(\sqrt { \frac { k }{ m } } \) , we can substitute the value of force constant k=ω² in equation (3)
where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation y =A sin ωt,
we get x =A sin ωt
U(t)=\(\frac { 1 }{ 2 } mv_{ x }^{ 2 }=\frac { 1 }{ 2 } m\left( \frac { dx }{ dty } \right) ^{ 2 }\)   ......(4)
This variation of U is shown below.

Variation of potential energy with time t
b. Expression for Kinetic Energy
KE = \(\frac { 1 }{ 2 } mv_{ x }^{ 2 }=\frac { 1 }{ 2 } m\left( \frac { dx }{ dy } \right) ^{ 2 }\)
(i) Since the particle is executing simple harmonic motion, from equation
y =A sin ωt
x =A sin ωt
Therefore, velocity is
vx =\(\frac { dx }{ dt } \)Aω cosωt
\(A\omega \sqrt { 1-\left( \frac { x }{ A } \right) ^{ 2 } } \)
vx = \(\omega \sqrt { { A }^{ 2 }-{ x }^{ 2 } } \)     ....(5)
Hence
KE = \(\frac { 1 }{ 2 } mv_{ x }^{ 2 }=\frac { 1 }{ 2 } m{ \omega }^{ 2 }({ A }^{ 2 }-{ x }^{ 2 })\)    ...(6)
KE = \(\frac { 1 }{ 2 } m{ \omega }^{ 2 }A^{ 2 }cos^{ 2 }\omega t\)     ....(7)
This variation with time is shown below.

c. Expression for Total Energy
(i) Total energy is the sum of kinetic energy and potential energy
E = KE+U   ..............(8)
E = \(\frac { 1 }{ 2 } m{ \omega }^{ 2 }=\frac { 1 }{ 2 } m{ \omega }^{ 2 }({ A }^{ 2 }-{ x }^{ 2 })\)
Hence excelling x² term,
E = \(\frac { 1 }{ 2 } m{ \omega }^{ 2 }A^{ 2 }\) = constant   .....(9)
(ii) Alternatively, from equation (4), and equation (7), we get the total energy as
E =\(\frac { 1 }{ 2 } m{ \omega }^{ 2 }A^{ 2 }sin^{ 2 }\omega t+\frac { 1 }{ 2 } m{ \omega }^{ 2 }A^{ 2 }cos^{ 2 }\omega t\)
= \(\frac { 1 }{ 2 } m{ \omega }^{ 2 }A^{ 2 }(sin^{ 2 }\omega t+cos^{ 2 }\omega t)\)
(iii) From trigonometry identity,
sin²ωt+cos2ωt_=1
E = \(\frac { 1 }{ 2 } m{ \omega }^{ 2 }A^{ 2 }\) = constant
which gives the law of conservation of total energy. This is depicted.

(iv) Thus the amplitude of simple harmonic oscillator, can be expressed in terms of total energy.
A =\(\sqrt { \frac { 2E }{ m{ \omega }^{ 2 } } } =\sqrt { \frac { 2E }{ k } } \) .

(i) All the molecules of a gas are identical, elastic spheres.
(ii) The molecules of different gases are different.
(iii) The number of molecules in a gas is very large and the average separation between them is larger than size of the gas molecules.
(iv) The molecules of a gas are in a state of continuous random motion.
(v) The molecules collide with one another and also with the walls of the container.
(vi) These collisions are perfectly elastic so that there is no loss of kinetic energy during collisions.
(vii) Between two successive collisions, a molecule moves with uniform velocity.
(viii) The molecules do not exert any force of attraction or repulsion on each other except during collision. The molecules do not possess any potential energy and the energy is wholly kinetic.
(ix) The collisions are instantaneous. The time spent by a molecule in each collision is very small compared to the time elapsed between two consecutive collisions.
(x) These molecules obey Newton's laws of motion even though they move randomly.

The specific heat of one mole of a monoatomic gas C_V = \(\frac{3}{2}\)R
For \(\mu\)₁ mole , C_V = \(\frac{3}{2}\)\(\mu\)₁R  C_p = \(\frac{5}{2}\)\(\mu\)₁ R
The specific heat of one mole of a diatomic gas
C_v = \(\frac{5}{2}\)R
For μ₂ mole, C_V = \(\frac{5}{2}\)μ₂ R C_P = \(\frac{7}{2}\)μ₂ R
The specific heat of the mixture at constant volume C_V = \(\frac{3}{2}\)\(\mu\)₁R +\(\frac{5}{2}\)\(\mu\)₂ R
The specific heat of the mixture at constant pressure C_P = \(\frac{5}{2}\)\(\mu\)₁ R + = \(\frac{7}{2}\)\(\mu\)₂ R
The adiabatic exponent \(\gamma =\frac { { C }_{ p } }{ { C }_{ V } } =\frac { 5{ \mu }_{ 1 }+{ 7\mu }_{ 2 } }{ 3{ \mu }_{ 1 }+{ 5\mu }_{ 2 } } \)

(1) Excess of pressure inside air bubble in a liquid.
Consider an air bubble of radius R inside a liquid having surface tension T as shown in Figure. Let P₁ and P₂ be the pressures outside and inside the air bubble, respectively. Now, the, excess pressure inside the air bubble is
\(\Delta P=P_{1}-P_{2}.\)

In order to find the excess pressure inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get,
(i) The force due to surface tension acting towards right around the rim of length \(2 \pi \mathrm{R} \ is \ \mathrm{F}_{\mathrm{T}}=2 \pi \mathrm{RT}\)
(ii) The force due to outside pressure P₁ is to the right acting across a cross sectional area of \(\pi \mathrm{R}^{2} \  is \ F_{P_{1}}=P_{1} \pi R^{2}\)
(iii) The force due to pressure P₂ inside the bubble, acting to the left is \(F_{P_{2}}=P_{2} \pi R^{2}\).
As the air bubble is in equilibrium under the action of these forces, \(F_{P_{2}}=F_{T}+F_{P_{1}}\)
\(\mathrm{P}_{2} \pi \mathrm{R}^{2} =2 \pi \mathrm{RT}+\mathrm{P}_{1} \pi \mathrm{R}^{2}  \)
\(\Rightarrow\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right) \pi \mathrm{R}^{2} =2 \pi \mathrm{RT}\)
Excess pressure is \(\Delta \mathrm{P}=P_{2}-P_{1}=\frac{2 T}{R}\)
(2) Excess pressure inside a soap bubble
Consider a soap bubble of radius R and the surface tension of the soap bubble be T. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and other outside the bubble. Hence, the force on the soap bubble due to surface tension is \(2 \times 2 \pi\) RT. The various forces acting on the soap bubble are,
(i) Force due to surface tension \(F_{T}=4 \pi R T\) towards right
(ii) Force due to outside pressure, \(F_{P_{1}}=P_{1} \pi R^{2}\) towards right
(iii) Force due to inside pressure, \(F_{P_{2}}=P_{2} \pi R^{2}\) towards left
As the bubble is in equilibrium, \(F_{P_{2}}=F_{T}+F_{P_{1}}\)
\(\mathrm{P}_{2} \pi \mathrm{R}^{2} =4 \pi \mathrm{RT}+\mathrm{P}_{1} \pi \mathrm{R}^{2}  \)
\(\Rightarrow\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right) =\pi \mathrm{R}^{2}=4 \pi \mathrm{RT} \pi \mathrm{R}^{2}  \)
\(\text {Excess pressure is } \Delta \mathrm{P} =\mathrm{P}_{2}-\mathrm{P}_{1}=\frac{4 T}{R}\)
(3) Excess pressure inside the liquid drop
Consider a liquid drop of radius R and the surface tension of the liquid is T. 
The various forces acting on the liquid drop are,
(i) Force due to surface tension \(F_{T}=2 \pi R T\) towards right
(ii) Force due to outside pressure, \(F_{P_{1}}=P_{1} \pi R^{2}\) towards right
(iii) Force due to inside pressure, \(F_{P_{2}}=P_{2} \pi R^{2}\) towards left
As the bubble is in equilibrium, \(F_{P_{2}}=F_{T}+F_{P_{1}}\)
\(\mathrm{P}_{2} \pi \mathrm{R}^{2} =2 \pi \mathrm{RT}+\mathrm{P}_{1} \pi \mathrm{R}^{2}  \)
\(\Rightarrow\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right) \pi \mathrm{R}^{2}=2 \pi \mathrm{RT} \)
\(\text {Excess pressure is } \Delta \mathrm{P} =\mathrm{P}_{2}-\mathrm{P}_{1}=\frac{2 T}{R}\)

Consider the Earth and mass system, with r, the distance between the mass m and the Earth's centre. Then the gravitational potential energy,
\(\mathrm{U}=-\frac{G M_{e} m}{r}\) ......(1)
Here r = R_e + h, where R_e is the radius of the Earth. h is the height above the Earth's surface.
\(\mathrm{U}_{\mathrm{c}}=-G \frac{M_{e} m}{\left(R_{e}+h\right)}\) .......(2)
If h << R_e, equation can be modified as
\(\mathrm{U} =-G \frac{M_{e} m}{\mathrm{R}_{e}\left(1+h / R_{e}\right)}  \)
\(\mathrm{U} =-G \frac{M_{e} m}{\mathrm{R}_{e}}\left(1+h / R_{e}\right)^{-1}\)  .......(3)
By using Binomial expansion and neglecting the higher order terms, we get
\(\mathrm{U}=-G \frac{M_{e} m}{\mathrm{R}_{e}}\left(1-\frac{h}{R_{e}}\right)\)   ........(4)
We know that, for a mass m on the Earth's surface,
\(G \frac{M_{e} m}{\mathrm{R}_{e}}=m g \mathrm{R}_{e}\)    ........(5)
Substituting equation (5) in (4) we get
\(\mathrm{U}=-\mathrm{mg} \mathrm{R}_{e}+\mathrm{mgh}\)   .......(6)
It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h₁ to h₂ then the potential energy at h₁ is
\(\mathrm{U}\left(\mathrm{h}_{1}\right)=-\mathrm{mg} \mathrm{} \mathrm{R}_{\mathrm{e}}+\mathrm{mgh}_{1}\)  .......(7)
and the potential energy at h₂ is
\(\mathrm{U}\left(\mathrm{h}_{2}\right)=-\mathrm{mg} \mathrm{} \mathrm{R}_{\mathrm{e}}+\mathrm{mgh}_{2}\)  .....(8)
The potential energy difference between h₁ and h₂ is
\(U\left(h_{2}\right)-U\left(h_{1}\right)=m g\left(h_{1}-h_{2}\right)\) .......(9)

(i) Torque is defined as the moment of the external applied force about a point or axis of rotation.
(ii) \(\overset { \rightarrow }{ \tau } =\overset { \rightarrow }{ r } \times \overset { \rightarrow }{ F } \)
where, \(\overset { \rightarrow }{ r } \) is the position vector of the point where the force \(\overset { \rightarrow }{ F } \) is acting on the body as shown in Figure.

(iii) Here, the product of \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ F } \) is called the vector product or cross product. The vector product of two vectors results in another vector that is perpendicular to both the vectors. Hence, torque (\(\overset { \rightarrow }{ \tau } \)) is a vector quantity.
(iv) Torque has a magnitude (r F sinፀ) and direction perpendicular to \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ F } \). Its unit is N m.
\(\overset { \rightarrow }{ \tau } =(r\ F\sin\theta )\hat { n } \)
(v) Here, \(\theta\) is the angle between \(\overset { \rightarrow }{ r } \) and \(\overset { \rightarrow }{ F } \) and \(\hat { n } \) is the unit vector in the direction of \(\overset { \rightarrow }{ \tau } \).

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.

i. Every object continues to be in the state of rest or of uniform motion unless there is external force acting on it.
ii. The force acting on an object is equal to the rate of change of its momentum.
iii. For every action there is an equal and opposite reaction.
Discussion:
i) Newton's laws are vector laws. The equation \(\vec{F}=\mathrm{m} \vec{a}\) can be written in cartesian coordinates as
\(\mathrm{F}_{x} \hat{\mathrm{i}}+\mathrm{f}_{\mathrm{y}} \hat{\mathrm{j}}+\mathrm{F}_{z} \hat{\mathrm{k}}=\operatorname{ma}_{x} \hat{\mathrm{i}}+\operatorname{ma}_{\mathrm{y}} \hat{\mathrm{j}}+\operatorname{ma}_{\mathrm{z}} \hat{\mathrm{k}}\)
On comparing both sides,
\(\mathrm{F}_{x}=m \mathrm{~m}_{x}  \)
\(\mathrm{F}_{\mathrm{y}}=m \mathrm{ma}_{\mathrm{y}}  \)
\(\mathrm{F}_{\mathrm{z}}=m \mathrm{ma}_{\mathrm{z}}\)
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along n direction. In the same way, F₂ can not affect a_y and a_n
ii) The acceleration experienced by the body at a time t depends on the force which acts on the body at that instant of time. Thus, \(\vec{F}(\mathrm{t})=\mathrm{m} \vec{a}(\mathrm{t})\)
when a bowler throws the ball to a batsman the acceleration of the ball is determined by the gravitational and air frictional forces and not by the speed which it is thrown.
iii) The direction of force may be different from the direction of force. The following motions are possible.
a) Force and motion in the same direction.
Example: Falling of an apple from the tree.
b) Force and motion are not in the same direction.
Example: The Moon experiences a force towards the Earth but Moon moves in the elliptical path.
c) Force and motion are in the opposite direction.
Example: The motion of a body thrown vertically upward.
d) Zero net force, but there is motion.
Example: The falling of rain drops.
4) If multiple forces \(\vec{F}_{1}, \vec{F}_{2}, \vec{F}_{3} \ldots\) act on the same body then total force is equal to the vectorial sum of the individual forces \(\vec{F}_{\text {net }}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}+\ldots\)
5) Newton's second law can be written in the second derivative of position vector as  \(\vec{F}=\mathrm{m} \frac{\mathrm{d}^{2}\overrightarrow{\mathrm{r}}}{\mathrm{dt}^{2}}\). It means that whenever the second derivative of position vector is not zero, there must be a force acting on the body.
6) If no force acts on the body then \(m \frac{d \vec{v}}{\mathrm{dt}}=0\). It implies that \(\vec{v}\) is constant. Thus second law is consistent with the first law even though they are independent to each other.
7) Newton's second law is cause and effect relation force is the cause and acceleration is effect.

Consider an object moving in a straight line with uniform or constant acceleration 'a'. Let u be the velocity of the object at time I = 0, and v be velocity of the body at a later time t.
Velocity - time relation
(i) The acceleration of the body at any instant is given by the first derivative of the velocity with respect to time, \(a={{dv}\over{dt}}or\ dv=a.dt\)
Integrating both sides with the condition that as time changes from 0 to I, the velocity changes from u to v. For the constant acceleration,
\(\int _{ u }^{ v }{ dv } =\int _{ 0 }^{ v }{ a\ dt } =a\int _{ u }^{ v }{ dt } \Rightarrow{[v]}^{v}_{u}=a{[t]}_{0}^{t}\)
v - u = a (or) v = u + at
Displacement - time relation
(ii) The velocity of the body is given by the first derivative of the displacement with respect to time.
\(v={{ds}\over{dt}}\) or ds = vdr
and since v = u + at,
We get ds = (u + at) dt
Assume that initially at time 1=0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time - independent, we have
\(\int _{ 0 }^{ s }{ ds } =\int _{ 0 }^{ t }{ u\ dt } +\int _{ 0 }^{ t }{ at } \ dt\) (or) s = ut + \({{1}\over{2}}{at}^{2}\)
Velocity - displacement relation
(iii) The acceleration is given by the first derivative of velocity with respect to time.
\(a={{dv}\over{dt}}={{dv}\over{ds}}{{ds}\over{dt}}={{dv}\over{ds}}v\)
[since dsl dt = v] where s is distance traversed]
This is rewritten as a \(={{1}\over{2}}{{{dv}^{2}}\over{ds}}\)
or ds \(={{1}\over{2a}}d({v}^{2})\)
Integrating the above equation, using the fact when the velocity changes from u² to v², displacement changes from 0 to s, we get
\(\int _{ 0 }^{ s }{ ds } =\int _{ u }^{ v }{{{1}\over{}2a} (v^2) } \)
\(\therefore s\ ={{1}\over{2a}}(v^2-u^2)\)
\(\therefore v^2=u^2+2as\) ............(3)
We can also derive the displacement's f in terms of initial velocity 'u' and final velocity v.
From equation 1, we can write
at = v - u
Substitute this in equation 2, we get
\(s=ut+{{1}\over{2}}(v-u)t\)
\(s={{(u+v)t}\over{2}}\)  .....................(4)
The equations 1, 2, 3 and 4 are called kinematic equations of motion, and have a wide variety of practical applications.
Kinematic equations
v = u+at
\(s=ut+{{1}\over{2}}{at}^{2}\)
v²=u²+2as
\(s={{(u+v)t}\over{2}}\)

The phase of a harmonic is a quantity that gives complete information of the wave at any time and at any position.

We know that \(Y=\cfrac { Fl }{ { \pi r }^{ 2 }\Delta l } \Rightarrow \cfrac { Fl }{ { \pi r }^{ 2 }Y } \)
Therefore, for the two wires,
\(\left( \Delta l \right) _{ 1 }=\cfrac { { F }_{ 1 }{ l }_{ 1 } }{ { \pi r }^{ 2 }\Delta l } \Rightarrow \Delta l=\cfrac { F_{ 2 }l_{ 2 } }{ { \pi r }^{ 2 }Y } \)
The ratio of their elongations
\(\cfrac { \left( \Delta l \right) _{ 1 } }{ \left( \Delta l \right) _{ 2 } } =\cfrac { { F }_{ 1 }{ l }_{ 1 }{ r }_{ 2 }^{ 2 } }{ { F }_{ 2 }{ l }_{ 2 }{ r }_{ 1 }^{ 2 } } =\cfrac { 40 }{ 80 } \times \cfrac { 1 }{ 2 } \times \left( \cfrac { 2 }{ 1 } \right) ^{ 2 }\)
\(\cfrac { \left( \Delta l \right) _{ 1 } }{ \left( \Delta l \right) _{ 2 } } =\cfrac { 1 }{ 1 } \)
\(\left( \Delta l \right) _{ 1 }:\left( \Delta l \right) _{ 2 }=1:1\)

\(\\ U=\cfrac { -GMm }{ R } =\cfrac { -6.67\times { 10 }^{ -11 }\times 100\times 10\times { 10 }^{ -3 } }{ 10\times { 10 }^{ -2 } } \)
U = 6.67\(\times\)10^-10 J
So, the amount of work done to take the particle upto infinite will be 6.67\(\times\)10^-10 J

(i) The amount of heat required to raise the temperature of one mole of a substance by 1K Dr 1°C at constant volume is called molar specific heat capacity at constant volume (C_v).
(ii) If a pressure is kept constant, it is called molar specific heat capacity at constant pressure (C_p)

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.

m = 2 kg, d = 10 m, F_ext = 20 N, \(\mu\)_k = 0.9.
when an object is in motion on he horizontal surface, it experiences two forces.
(a) External force, F_ext = 20 N
(b) Kinetic friction,
f_k = \(\mu\)_k mg = 0.9 \(\times\) (2) \(\times\) 10 = 18N
The work done by the external force W_ext = Fd = 20 x 10 = 200J
The work done by the force of kinetic friction W_k = f_kd = (-18) \(\times\) 10 = -180 J. Here the negative sign implies that the force of kinetic friction is opposite to the direction of displacement.
The total work done on the object W_total = W_ext + W_k = 200 J - 180 J = 20 J.
Since the friction is a non-conservative force, out of 200 J given by the external force, the 180 J is lost and it can not be recovered.

Relative velocity of A with respect to B, \( { v } _{ AB }\)= 80 km h^-1 due east
Thus to a passenger in train B, the train A will appear to move east with a velocity of 80 km h^-1.The relative velocity of B with respect to A, V_BA = 80 km h^-1 due west.
To a passenger in train A, the train B will appear to move westwards with a velocity of 80 km h^-1

Meter scale is longer and larger than a match stick. Meter scale's centre of gravity is higher but match stick has centre of gravity much lower as compared to scale. Higher the centre of gravity easier it is to balance.

Units used to measure extremely large distances are,
(i) Astronomical Unit:
It is the mean distances of the earth from the sun. 1 AU = 1.496\(\times\)10¹¹m.
(ii) Lightyear:
It is the distance travelled by light in vacuum in one year. 1 Iy = 9.46\(\times\)10¹⁵m
(iii) Parallactic second:
It is the distance at which an arc of length 1 astronomical unit subtends an angle of 1 second of arc.1 parsec = 3.084\(\times\)10¹⁶m = 3.261y.

According to Weber-Fechner's law, "loudness (L) is proportional to the logarithm of the actual intensity (1) measured with an accurate nonhuman instrument". This means that
L ∝ 1n I|
L = k  1n I
where k is a constant, which depends on the unit of measurement. The difference between two loudnesses, L₁ and L_o measures the relative loudness between two precisely measured intensities and is called as sound intensity level. Mathematically, sound intensity level is
ΔL = L₁-L_o = k In I₁- k In I₀ = k In \(\left[I_1\over I_0\right]\)
If k = 1, then sound intensity level is measured in bel, Therefore,
\(ΔL=In\)\(\left[I_1\over I_0\right]\)bel
However, this to express smaller unit, decibel. Thus, [decibel = \(1\over10\)bel] by multiplying and dividing by 10
\(ΔL=10\left(In\left[I_1\over I_0\right]\right){1\over 10}bel\)
\(ΔL=10In\left[I_1\over I_0\right]\)decibel with k = 10
For practical purposes,
\(ΔL=10log_{10}\left[I_1\over I_0\right]\)decibel.

(i) This is because a body with large reflectivity is poor absorber of heat, and poor aborbers of heat are emitters.
(ii) When we touch a brass tumber on a chilly day, heat flows from our body to the tumbler quickly (as thermal conductivity of brass is very high) and as a result, it appears colder.
(iii) On the other hand, as the wood a bad conductor, heat does not flow to the wooden tray from our body, on touching it.

Time period of a simple harmonic oscillator
\(T=2\pi \sqrt { \frac { m }{ k } } \)
m - mass k - force constant

This law states that at constant temperature and pressure, equal volumes of all gases contain same number of molecules. For two different gases at the same temperature and pressure, according to kinetic theory of gases. We get
\(\mathrm{P}=\frac{1}{3} n m \overline{v^{2}} \ or \ \mathrm{P}=\frac{1}{3} \frac{N}{V} m \overline{v^{2}}\)
\(\mathrm{P}=\frac{1}{3} \frac{N_{1}}{V} m_{1} v_{1}^{2} \)
\(=\frac{1}{3} \frac{N_{2}}{V} m_{2} v_{2}^{2}\)   ......(1)
where \(\overline{v_{1}^{2}} \ and \ \overline{v_{2}^{2}}\) are the mean square speed for two gases and \(\mathrm{N}_{1} \ and \ \mathrm{N}_{2}\) are the number of gas molecules in two different gases.
At the same temperature, average kinetic energy per molecule is the same for two gases.
\(\frac{1}{2} m_{1} \overline{v_{1}^{2}}=\frac{1}{2} m_{2} \overline{v_{2}^{2}}\)   ........(2)
Dividing the equation (1) by (2) we get
N₁ = N₂

From the shadows it is revealed that Earth has spherical shape with bulging along equator and flat at poles.

Mass of a body m = 2 kg
Magnitude of the two force, F₁= 3N and F₂ =5N,
Angle (\(\theta\) ) = 90°
Resultant force on the body = 
F = \(\sqrt { { F }_{ 1 }^{ 2 }={ F }_{ 2 }^{ 2 }+2{ F }_{ 1 }{ F }_{ 2 }cos\theta } \)
= \(\sqrt { { 3 }^{ 2 }+{ 5 }^{ 2 }+2\times 3\times 5\times cos90° } \)
= \(\sqrt { 9+25+2\times 3\times 5\times 0 } \)   \(\left[ \because \cos90°=0 \right] \)
= \(\sqrt { 9+25 } \)
F = \(\sqrt { 34 } \)
Net acceleration experienced by the body
a =\(\frac { F }{ m } \)
= \(\frac { \sqrt { 34 } }{ 3.5 } \) = 1.66 ms^-2

(i) Consider a rigid body rotating about a fixed axis. A point the body rotating about an axis perpendicular to the plane of the page. A tangential force F is applied on the body.
(ii) It produces a small displacement ds on the body. The work done (dw) by the force is,
dw=Fds
(iii) As the distance ds, the angle of rotation d\(\theta\) and radius r are related by the expression
ds=r d\(\theta\)
The expression for work done now becomes,
dw=F ds; dw=F r d\(\theta\)
(iv) The term (Fr) is the torque ፒ produced by the force on the body.
dw=\(\tau\)d\(\theta\)
This expression gives the work done by the external torque ፒ, which acts on the body rotating about a fixed axis through an angle d\(\theta\).

1. Equal mass balancing each side so reading shows Zero.
2. Mass of string acting downward direction in the inclined plane.
mg sin θ = T
T = 2kg x 9.8m/s² x sin(30^o)
T = 9.8N