The amplitude of the velocity wave is same in all four cases

\(n \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3}\)
\(\therefore \text { Energy }=3 V T\left(\frac{1}{r}-\frac{1}{R}\right) \text { is released }\)

Time period T = \(\frac{2\pi}{\sqrt GM_E} (R_E+ h)^\frac{3}{2}\)
\(\therefore\) It is independemt of mass

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

For freely falling body, velocity while the body, hit the ground \(v=\sqrt{2 g h}\)
\(v_{1}=\sqrt{2 g h_{1}} \text { and } v_{2}=\sqrt{2 g h_{2}} \)
\(\therefore \frac{m_{1} v_{1}}{m_{2} v_{2}}=\frac{m_{1} \sqrt{h_{1}}}{m_{2} \sqrt{h_{2}}}=\frac{m_{1}}{m_{2}} \sqrt{\frac{h_{1}}{h_{2}}}\)

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 \(x={\lambda\over 4}\)
\(\phi={2\pi\over \lambda}\times{\lambda\over 4}={\pi\over 2}rad\)

A molecule of mass m moving with a velocity \(\vec{v}\) having components \(\left(v_{x}, v_{y}, v_{z}\right)\) hits the right side wall. Since we have assumed that the collision is elastic, the particle rebounds with same speed and its x-component is reversed. The components of velocity of the molecule after collision are \(\left(-v_{x}, v_{y}, v_{z}\right)\)
The x-component of momentum of the molecule before collision = mv_x
The x-component of momentum of the molecule after collision = mv_x
The change in momentum of the molecule in x direction
= Final momentum - initial momentum 
= \(-\mathrm{mv}_{\mathrm{x}}-\mathrm{mv}_{\mathrm{x}} \)
= \(-2 \mathrm{mv}_{\mathrm{x}}\)
According to law of conservation of linear momentum, the change in momentum of the wall \(=2 \mathrm{mv}_{\mathrm{x}}\)
The number of molecules hitting the right side wall in a small interval of time ∆t is calculated as follows.
The molecules within the distance of v_x∆t  from the right side wall and moving towards the right will hit the wall in the time interval ∆t. The number of
molecules that will hit the right side wall in a time interval ∆t is equal to the product of volume \(\left(\mathrm{Av}_{x} \Delta t\right)\)and number density of the molecules (n). Here A is area of the wall and n is number of molecules per unit volume \(\left(\frac{N}{V}\right)\). We have assumed that the number density is the same throughout the cube.
Not all the n molecules will move to the right, therefore on an average only half of the n molecules move to the right and the other half moves towards left side. The number of molecules that hit the right side wall in a time interval
\(\Delta t=\frac{n}{2} A v_{x} \Delta t\)   .....(1)
In the same interval of time ∆t, the total momentum transferred by the molecules.
\(\Delta \dot{p} =\frac{n}{2} A v_{x} \Delta t \times 2 m v_{x}  \)
\(=A v_{x}^{2} m n \Delta t\)   ....(2)
From Newton's second law, the change in momentum in a small interval of time gives rise to force.
The force exerted by the molecules on the wall (in magnitude)

\(\mathrm{F} =\frac{\Delta p}{\Delta t} \)
\(=n m A v_{x}^{2} \)   ...(3)
Pressure, P = force divided by the area of the wall.

\(\mathrm{P} =\frac{F}{A}  \)
\(=n m v_{x}{ }^{2}\)
Since all the molecules are moving completely in random manner, they do not have same speed. So we can replace the term v_x² by the average \(\overline{v_{x}^{2}}\)
\(\mathrm{P}=\frac{F}{A}=n m v_{x}^{2}  \)
\(\mathrm{P}=n m \overline{v_{x}^{2}}\)
Since the gas is assumed to move in random direction, it has no preferred direction of motion. (the effect of gravity on the molecules is neglected). It implies that the molecule has same average speed in all the three direction. So.\( \overline{v_{x}^{2}}=\overline{v_{y}^{2}}=\overline{v_{x}^{2}}\).
The mean square speed is written as
\(\overline{v^{2}}=\overline{v_{\dot{x}}^{2}}+\overline{v_{y}^{2}}+\overline{v_{z}^{2}}=\overline{3 v_{x}^{2}}  \)
\(\overline{v_{x}^{2}}=\frac{1}{3} \overline{v^{2}}  \)
\(\mathrm{P}=n m \overline{v_{x}^{2}}  \)
\(\mathrm{P}=\frac{1}{3} n m \overline{v^{2}} \text { or } P \frac{1}{3} \frac{N}{V} m \overline{v^{2}} \quad \text { as }\left[n=\frac{N}{V}\right]\)

Hooke's law states that within the elastic limit, the strain produced in a body is directly proportional to the stress applied.
It can be verified in a simple way by stretching a thin straight wire (stretches like spring) of length L and uniform cross-sectional area A suspended from a fixed point O. A pan and a pointer are attached at the free end of the wire as shown in Figure. The extension produced on the wire is measured using a vernier scale arrangement. The experiment shows that for a given load, the corresponding stretching force is F and the elongation produced on the wire is ΔL. It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and ΔL on the Y-axis. This graph is a straight line passing through the origin as shown in Figure.


Therefore,
ΔL = (slope)F
Multiplying and dividing by volume,
V = AL,
F (slope) = \(\frac{AL}{AL} \Delta L\)
Rearranging, we get
\(\frac{F}{A}=[\frac{L}{A(Slope)}]\frac{\Delta L}{L}\)
Therefore, \(\frac{F}{A} \alpha [\frac{\Delta L}{L}]\)
Comparing with equations stress and strain \(\sigma=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}, \varepsilon=\frac{\text { Change in size }}{\text { Original size }}=\frac{\Delta l}{l}\),
we get volume strain, \(\varepsilon_{v}=\frac{\Delta V}{V}\) equation as
\(\sigma \propto \varepsilon\)
i.e., the stress is proportional to the strain in the elastic limit.

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)

Force \(\overrightarrow { F } =mg\left( -\hat { j } \right) =-mg\hat { j } \)
Displacement vector \(d\vec{r}\) = dx\(\hat { i } \) + dy \(\hat { j } \)
(As the displacement is in two dimension; unit vectors \(\hat { j } \) and \(\hat { i } \)  are used)
(a) Since the motion is only vertical, horizontal displacement component dx s zero. Hence, work done by the force along path 1 (of distance h).
\({ W }_{ push\ \ 1 }=\int _{ A }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dy\hat{j})=-mg\int _{ 0 }^{ h }{ dy=-mgh } } } \)
Total work done for path 2 is
\({ W }_{ push\ \ 2 }=\int _{ A }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ C }{ \overrightarrow { F } .d\overrightarrow { r } +\int _{ C }^{ D }{ \overrightarrow { F } .d\overrightarrow { r } +\int _{ D }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } } } } } \)
But \(\int _{ A }^{ C }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dx\hat { i } )=0 } } \)
\(\int _{ A }^{ D }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ C }^{ D }{ (-mg\hat { j } ).(dy\hat { j } )=-mg\int _{ 0 }^{ h }{ dy } =-mgh } } \)
\(\int _{ D }^{ B }{ \overrightarrow { F } .d\overrightarrow { r } =\int _{ A }^{ B }{ (-mg\hat { j } ).(dx\hat { i } )=0 } } \)
Therefore, the total work done by the force along the path 2 is
\({ W }_{ push\ \ 2 }=\int _{ A }^{ B }{ \vec { F } } .d\overrightarrow { r } =-mgh\)
Note that the work done by the conservative force is independent of the path.

For both persons, the Kinematic equations are the same, with u=0,a_c =g and a_moon  =\(g\over 6\)
a_e = g and a_m =\(g\over 6\)
For a person on earth, V_earth = \(\sqrt{2gh}=\sqrt{2\times 10\times 100}\)
Hence, V_earth =\(\sqrt{2000} ms^{-1}\) gives the velocity at the ground, on earth.
Similarly, for a person on the moon,
 V_earth =\(\sqrt{2gh\over6}={\sqrt{2000}\over \sqrt{6}}ms^{-1}\)
The person on earth reaches ground with greater velocity than the person on the moon.

According to parallel axis theorem,
I = I₀ + mx²
=\(\frac { M{ l }^{ 2 } }{ 12 } +M\left( \frac { l }{ 2 } \right) ^{ 2 }\)
=\(\frac { { Ml }^{ 2 } }{ 12 } +\frac { Ml^{ 2 } }{ 4 } ={ Ml }^{ 2 }\left[ \frac { 1 }{ 12 } +\frac { 1 }{ 4 } \right] \)
=\({ Ml }^{ 2 }\left[ \frac { 1 }{ 3 } \right] \)
I=\(\frac { { Ml }^{ 2 } }{ 3 } \) .

Work done by a constant force:
(i) When a constant force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation,
dW= (F cos\(\theta\) ) dr
(ii) The total work done in producing a displacement from initial position r_i to final position r_f is,
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dw\)
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } \left( F\cos { \theta } \right) dr=\left( F\cos { \theta } \right) \)\(\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dr=\left( F\cos { \theta } \right) \left( { r }_{ f }-{ r }_{ i } \right) \)
(iii) The graphical representation of the work done by a constant force is shown in Figure. The area under the graph shows the work done by the constant force.
              
Work done by a variable force:
(i) When the component of a variable force F acts on a body, the small work done (dW) by the force in producing a small displacement dr is given by the relation
dw = (F cos\(\theta\)) dr
[F cos\(\theta\) is the component of the variable force F]
where, F and \(\theta\) are variables. The total work done for a displacement from initial position r_i to final position r_f is given by the relation, 
\(W=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } dw=\overset { { r }_{ f } }{ \underset { { r }_{ i } }{ \int } } F\cos { \theta } dr\)
(ii) A graphical representation of the work done by a variable force is shown in Figure. The area under the graph is the work done by the variable force.
                

The force which always opposes the relative motion between an object and the surface where it is placed is called frictional force. Frictional force always acts on the object parallel to the surface on which the object is placed. During the time of Newton and Galileo, frictional force was considered as one of the natural forces like gravitational force. But now it is understood that frictional force is the electromagnetic force between the atoms on the two surfaces. Because of this even a well polished surfaces have frictional force. Frictional force is independent of surface area but depends on the normal force acting on it.
Angle of friction:
The angle of friction is defined as the angle between the normal force (N) and the resultant force (R) of normal force and maximum friction force (f_s^max). from the figure, tan ፀ \(=\frac{f_s^{max}}{N}\)

But f_s^max = μs N where us is the coefficient of static friction.
∴ μs = tan ፀ
To show that angle of friction is equal to angle of repose

A jerk is given on a stretched string at time t = 0 s. Assume that the wave pulse created during this disturbance moves along positive x direction with constant speed v as shown in Figure (a).

Represent the shape of the wave pulse, mathematically as y = y(x, 0) = j(x) at time t = 0s. Assume that the shape of the wave pulse remains the same during the propagation. After some time t, the pulse moving towards the right and any point on it can be represented by x' (read it as x prime) as shown in Figure (b). Then
y(x, t) =j(x') =j(x - vt)
Similarly, if the wave pulse moves towards left with constant speed v, then y =j(x + vt). Both waves y =j(x + v!) and y =j(x - vt) will satisfy the following one dimensional differential equation known as the wave equation
\({∂^2y\over ∂x^2}={1\over4 v^2}{∂^2y\over ∂t^2}\)
where the symbol p represent partial derivative (read \({∂y\over ∂x}\)  as partial y by partial x). Not all the solutions satisfying this differential equation can represent waves, because any physical acceptable wave must take finite values for all values of x and t. But if the function represents a wave then it must satisfy the differential equation. Since, in one dimension (one independent variable), the partial derivative with respect to x is the same as total derivative in coordinate x, we write So it can be written as
\({d^2y\over dx^2}={1\over v^2}{d^2y\over dt^2}\)

(i) In the two wires, the wire A is more ductile. Because, the material of wire A have large plastic range of extension than the wire B.
(ii) In the two wires, the slope of graph in wire A is greater than the slope of graph in wire B, so wire A has greater young's modulus.
(iii) Wire A is stronger than wire B because it can withstand more load without breaking. For Wire A, the breakeven point is higher.
(iv) WireB is more brittle. Because the material of wire B have small range of extension than the wire A.

a. The displacement of the particle executing simple harmonic motion
y = A sin\(\omega t\)
Velocity of the particle is 
v = A \(\omega\) cos \(\omega t\) = A \(\omega\) sin\(\left( \omega t+\frac { \pi }{ 2 } \right) \)
The phase difference between displacement and velocity is \(\frac{\pi}{2}\)
b. The velocity of the particle is
v = A \(\omega\) cos \(\omega t\)
Acceleration of the particle is 
a = -A\({ \omega t }^{ 2 }\) sin \(\omega t\) = A \({ \omega }^{ 2 }\)cos \(\left( \omega t+\frac { \pi }{ 2 } \right) \)
The phase difference between velocity and acceleration is\(\frac{\pi}{2}\)
c. The displacement of the particle is
y = A sin\(\omega t\)
Acceleration of the particle is
a = − A \({ \omega }^{ 2 }\) sin ωt = A \({ \omega }^{ 2 }\) sin(\(\omega t\) + \(\pi\))
The phase difference between displacement and acceleration is \(\pi\) radian.

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

\(F-T =m_{1} a \ \& \ T=m_{2} a  \)
\(a =\frac{F}{m_{1}+m_{2}} \ \& \ T=\frac{m_{2} F}{m_{1}+m_{2}}  \)
\(T =\frac{10 k g \times 500 N}{(10 k g+15 k g)}  \)
\(T =\frac{5000}{25} N  \)
\(T =200 \mathrm{~N}\)

The concepts to form the integrand to find the moment of inertia are to be followed. Now, the origin is fixed to the left 'end of the rod and the limits are to be taken from 0 to l.

I =  \(\frac { M }{ l } \int _{ 0 }^{ l }{ { x }^{ 2 }dx } =\frac { M }{ l } \left[ \frac { { x }^{ 3 } }{ 3 } \right] _{ 0 }^{ l }=\frac { M }{ l } \left[ \frac { { l }^{ 3 } }{ 3 } \right] \)
I = \(\frac { 1 }{ 3 } \)Ml²

(i) If he stops his hands soon after catching the ball, the ball comes to rest very quickly. It means that the momentum of the ball is brought to rest very quickly.
(ii) So the average force acting on the body will be very large.
(iii) Due to this large average force, the hands will get hurt.
(iv) To avoid getting hurt, the player brings the ball to rest slowly

For closed organ pipe, 3rd harmonics = \(\cfrac { 3v }{ 4l } \)
For open organ pipe, fundamental frequency,=\(\cfrac { v }{ 2l } \)
Given,\(\cfrac { 3v }{ 4l } =\cfrac { v }{ 2l } \)
\(l'=\cfrac { 4l }{ 3\times 2 } =\cfrac { 4l }{ 6 } \)
l = 20 cm \(l'=\cfrac { 4\times 20 }{ 6 } \)
l'=13.33 cm

Conduction is the process of direct transfer of heat through matter due to temperature difference. When two objects are in direct contact with one another, heat will be transferred from the hotter object to the colder one. The objects which allow heat to travel easily through them are called conductors.

Solid, liquid, gas, plasma, Bose-Einstein condensates, quark-gluon plasmas and hot plasma these are the physical states of matter.

Geocentric Model of Solar System:
In the second century, Claudius Ptolemy, a famous Greco-Roman astronomer, developed a theory to explain the motion of celestial objects like the Sun, the Moon, Mars, Jupiter etc. This theory was called the geocentric model. According to the geocentric model, the Earth is at the center of the universe and all celestial objects including the Sun, the Moon, and other planets orbit the Earth.

a. Periodic motion
b. Non-periodic motion
c. Periodic motion

One mole of an ideal gas at S.T.P occupies a volume of \(22.4 \times 10^{-3} \mathrm{~m}^{3}\)
∴ Number of moles \(/ \mathrm{m}^{3}=\mathrm{n}=\frac{6.023 \times 10^{23}}{22.4 \times 10^{-3}}\)
\(\therefore n=2.69 \times 10^{25} \mathrm{moles} / \mathrm{m}^{3}\)
In terms of number of molecules and radius the mean free path at S.T.P can be written as
\(\lambda =\frac{1}{4 \pi \sqrt{2} r^{2} n}  \)
\(\text {Radius } r =\frac{3 \times 10^{-10}}{2}=1.5 \times 10^{-6} \mathrm{~m}\)
∴ Mean free path, \(\quad \lambda=\frac{1}{4 \times 3.14 \times 1.414 \times\left(1.5 \times 10^{-10}\right)^{2} \times 2.69 \times 10^{25}}\)
\(\lambda=0.0931 \times 10^{-6}=9.31 \times 10^{-8} \mathrm{~m}\)
Mean free path, \(\lambda \approx 9 \times 10^{-8} \mathrm{~m}\)

The average speed
\(\overset { - }{ v } =\frac { 2+3+4+5+5+5+6+6+7+9 }{ 10 } =5.2{ ms }^{ -1 }\)
To find the rms speed, first calculate the mean square speed \(\overset { - }{ { v }^{ 2 } } \)
\(\overset { - }{ { v }^{ 2 } } =\frac { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+{ 5 }^{ 2 }+6^{ 2 }+{ 6 }^{ 2 }+{ 7 }^{ 2 }+{ 9 }^{ 2 } }{ 10 } \)
= 30.6ms²s^-2
The rms speed
\({ v }_{ rms }\sqrt { \overset { - }{ { v }^{ 2 } } } =\sqrt { 30.6 } =5.53{ ms }^{ -1 }\)
The most probable speed is 5 m s^-1 because three of the particles have that speed.

When the water is heated from 30°C to 60°C, there is only a slight change in its volume. So we can treat this process as isochoric. In an isochoric process the work done by the system is zero. The given heat supplied is used to increase only the internal energy.
ΔU = Q = ms_v ΔT
The mass of water = 500 g = 0.5 kg
The change in temperature = 30K
The heat Q = 0.5\(\times\)4184\(\times\)30 = 62.76 kJ

x = 2 - 5t + 6t²
Velocity, v = \(\frac { dx }{ dt } =\frac { d }{ dt } \) (2-5t+6t²) or v = -5+12t
For initial velocity, t = 0
∴ Initial velocity = -5ms^-1
The negative sign implies that at t = 0 the velocity of the particle is along negative x direction.

One second is the duration of 9,192,631,770 periods of radiation corresponding to the transition between the two hyperfine levels of the ground state of Cesium-133 atom.

The weight of tree exerts a torque about the point where the cut is made. This causes rotation of the tree about the cut has to be made at say point A to weaken the tree trunk and to shift the centre of mass to the right and eventually move towards the ground on the right.