\(l=\lambda \mathrm{f}=8.3 \mathrm{~Hz} \)
\(v=332 \mathrm{~m} / \mathrm{s} \quad \lambda=4 l \)
\(\lambda=\frac{v}{f}=\frac{332}{83}=4 \mathrm{~m} \)
\(l=\frac{\lambda}{4}=\frac{4 l m}{4}=1.0 \mathrm{~m} \)

\(l =0.9 \)
\(\mathrm{~T} =2 \pi \frac{l}{g} \)
\(T^{\prime} =2 \pi \frac{l_{X}}{g_{X}} \)
\(\mathrm{~g} \mathrm{x} =\mathrm{ng} \)
\(\therefore \mathrm{x} =0.9 \mathrm{n} \)

\(v_{m s}=1.73 \sqrt{\frac{k T}{m}}\)
\(\text { AT Increased by } 10 \text { times }\)
\(v_{\mathrm{ms}} \propto \Delta T\)
\(\text { RMS speed increases by } 10 \text { times. }\)

During the thermal equilibrium with surroundings internal energy is increased and heat energy will be increased.
\(\Delta U>0 Q>0\)

\(\text { Rigidity modulus }=\frac{1}{3} \times \text { Young's modulus }\)
\(\text { Poisson ratio }=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

\(\text {Potential energy } P . E=\alpha-\frac{\beta}{2} x^{2}\)
P.E = Work = Fx
\(P=\alpha-\frac{\beta}{2} x^{2}\)
\(\text {Force }=\frac{d p}{d x}=\frac{d}{d x}\left(\alpha-\frac{\beta}{2} x^{2}\right)\)
\(=0-\frac{\beta}{2} \times 2 x =-\beta x \)

\(T =2 \pi \sqrt{\frac{l}{g}} \)
\(g =4 \pi^{2} l / T^{2} \)
\(\frac{d g}{g} =\frac{d l}{l}-\frac{2 d T}{T} \)
\(\frac{d g}{g} \% =\left(\frac{d l}{l}\right) \%-2\left(\frac{d T}{T}\right) \% \)
\(=1 \%-2 \times(-3 \%) \)
\(=1+6=7 \% \)

A satellite of mass M to move in a circular orbit, centripetal force must be acting on the satellite. This centripetal force is provided by the Earth's gravitational force.
\(\frac { { mv }^{ 2 } }{ \left( { R }_{ E }+h \right) } =\frac { { GMM }_{ E } }{ { \left( { R }_{ e }+h \right) }^{ 2 } } \)
\({ v }^{ 2 }=\frac { { GM }_{ E } }{ \left( { R }_{ E }+h \right) } \)
\(v=\sqrt { \frac { { GM }_{ E } }{ \left( { R }_{ E }+h \right) } } \)
As h increases, the speed of the satellite decreases.

Here,
\({ V }_{ rms }=\left[ \frac { 3PV }{ M } \right] ^{ \frac { 1 }{ 2 } }=\left[ \frac { 3RT }{ M } \right] ^{ \frac { 1 }{ 2 } }\)
Let rms speed of oxygen is (V _rms)₁ and of helium is (V_rms)₂ is equal at temperature T₁ and T₂ respectively. \(\frac { { \left( { V }_{ rms } \right) }_{ 1 } }{ { \left( { V }_{ rms } \right) }_{ 2 } } =\sqrt { \frac { { M }_{ 2 }{ T }_{ 1 } }{ { M }_{ 1 }{ T }_{ 2 } } } \)
Atomic mass of Helium M₂ = 4
Atomic mass of Oxygen M₁ = 32
Temperature of helium -10^oC = 273 + (-10^oC)
= 263K
\(\left[ \frac { 4{ T }_{ 1 } }{ 32\times 263 } \right] =1\)
\({ T }_{ 1 }=\frac { 32\times 263 }{ 4 } =\frac { 8416 }{ 4 } =2104K\)
\(\therefore\)T₁ = 2104 K

Yes, during adiabatic compression the temperature
of a gas increases while no heat is given to it.
In adiabatic compression, dQ = 0
\(\therefore\) From first law of thermodynamics
dU = dQ - dW
dU = -dW
dU =- dW
In compression work is done on the gas i.e, work done is negative. Therefore,
dU = positive
Hence, internal energy of the gas increases due to which its temperature increases.

Periodic motion is any motion which repeats itself in a fixed time internal.
Example:
Hands in pendulum clock, swing of a cradle, the revolution of the Earth around the Sun, waxing and waning of Moon, etc.
Non periodic motion is any motion which does not repeat itself after a regular interval of time.
Example: Occurrence of Earth quake, eruption of volcano, etc.

Using equation 
v = \(\omega \sqrt { { A }^{ 2 }-{ x }^{ 2 } } \Rightarrow { v }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }^{ 2 } \right) \)
Therefore, at position x₁,
\({ v }_{ 1 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) \)   .................(1)
Similarly, at position x₂,
\({ v }_{ 2 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) \)  ...................(2)
Subtrating (2) from (1), we get
\({ v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 }={ \omega }^{ 2 }\left( { A }^{ 2 }-{ x_1 }^{ 2 } \right) -{ \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) \)
\(= { \omega }^{ 2 }\left( { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) \)
\({ \omega }=\sqrt { \frac { { v }_{ 1 }^{ 2 }-{ v }_{ 2 }^{ 2 } }{ { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } } } \Rightarrow T=2\pi \sqrt { \frac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 } }-{ { v }_{ 2 }^{ 2 } } } } \) ....................(3)
Dividing (1) and (2), we get
\(\frac { { v }_{ 1 }^{ 2 } }{ { v }_{ 2 }^{ 2 } } =\frac { { \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 1 }^{ 2 } \right) }{ { \omega }^{ 2 }\left( { A }^{ 2 }-{ x }_{ 2 }^{ 2 } \right) } \Rightarrow A=\sqrt { \frac { { { v }_{ 1 }^{ 2 }x }_{ 2 }^{ 2 }-{ v }_{ 2 }^{ 2 }{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 } }-{ { v }_{ 2 }^{ 2 } } } } \) .................(4)
Dividing equation (3) and equation (4), we have
\(\frac { T }{ A } =2\pi \sqrt { \frac { { x }_{ 2 }^{ 2 }-{ x }_{ 1 }^{ 2 } }{ { { v }_{ 1 }^{ 2 }x }_{ 2 }^{ 2 }-{ { v }_{ 2 }^{ 2 }x }_{ 1 }^{ 2 } } } \)

P₁ = 1 atmospheric Pressure
V₁ = V
V₂ = V/8
T₁ = 32 + 273
= 305
T₂ = ? 
\(\gamma =1.4  \)
\(T_{2} V_{2}^{\gamma-1} =T_{1} V_{1}^{\gamma-1}  \)
\(\frac{T_{1}}{T_{2}} =\left(\frac{V_{2}}{V_{1}}\right)^{\gamma-1}  \)
\(\frac{T_{1}}{T_{2}} =\left(\frac{V / 8}{V}\right)^{1.4-1}  \)
\(=\left(\frac{1}{8}\right)^{0.4} \)
\(\frac{305}{T_{2}} =\frac{1}{8^{0.4}}\)
\(\therefore T_{2} =305 \times 8^{0.4}  \)
\(=400^{\circ} \mathrm{C}\)
Temperature of the compressed air
\(T \cong 400^{\circ} \mathrm{C}\)
 

Whenever we analyze the motion of objects in rotating frames we must take into account the centrifugal force. Even though we treat the Earth as an inertial frame, it is not exactly correct because the Earth spins about its own axis. So when an object is on the surface of the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.

This centrifugal force is given by mω²R'.
 R' =  R cos λ       ....(1)
where λ  is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
 a_c = ω²R' cos λ = ω²R cos² λ
since R' = R cos λ
Therefore,
g' = g -ω²R cos² λ     ...(2)
From the expression (2), we can infer that at equator, λ  = 0; g' = g - ω²R. The acceleration due to gravity is minimum. At poles λ  = 90; g' = g, it is maximum. At the equator, g' is minimum.

kWh is a bigger unit of energy.
\(\frac { 1kwh }{ 1eV } =\frac { 3.6\times { 10 }^{ 6 }J }{ 1.6\times { 10 }^{ -19 }J } =2.25\times10^{25}\)

(x, y) = (0.54 m, 0.36 m)

Minimum inaccuracy = Vernier constant
 = 1 MSD -1 VS.D
=1 MSD-\(\frac { 49 }{ 50 } \) MSD
= \(\frac { 1 }{ 50 } \)(0.5 mm)= 0.01 mm

Initial momentum of the object P_i = 10 \(\times\)15 = 150 kgm s^-1
Final momentum of the object p_f = 0
Δp = 150 - 0 = 150 kg ms^-1
(a) Impulse J = Δp = 150 N s and Average force \({ F }_{ avg }=\frac { \triangle p }{ \triangle t } =\frac { 150 }{ 0.03 } \)= 5000N
b) Impulse J = Δp = 150 N s and Average force F_avg =\(\frac { 150 }{ 10 } \)= 15N

(a) v_x - remains constant
(b) v_y - decreases and increases
(c) a - remains downward
(d) r - varies

Static friction: The opposing force that comes into play when one body tends to move over the surface of another, but the actual motion has yet not started is called static friction.
Limiting friction: If the applied force is increased the force of static friction also increases. If the applied force exceeds a certain (maximum) value, the body starts moving. This maximum value of static friction up to which body does not move is called limiting friction.
Kinetic or dynamic friction: If the applied force is increased further and sets the body in motion, the friction opposing the motion is called kinetic friction.
We can reduce friction
(1) By polishing.
(2) By lubrication.
(3) By proper selection of material.
(4) By streamlining the shape of the body.
(5) By using ball bearing.

An echo is a repetition of sound produced by the reflection of sound waves from a wall, mountain or other obstructing surfaces.

Given y₁ = 5 sin(240\(\pi\)t) and y₂ = 4 sin(244\(\pi\)t)
Comparing with y = A sin(2\(\pi\) f₁t), we get
2\(\pi\)f₁ = 240π \(\Rightarrow\) f₁ = 120Hz
2\(\pi\)f₂ = 244π \(\Rightarrow\) f₂ = 122Hz
The number of beats produced is | f₁ − f₂| = |120 − 122| = |− 2|=2 beats per sec

The rise or fall of a liquid in a narrow tube is called capillarity or capillary action.

Let the particle of mass m cross a horizontal distance x in time t.

Angular momentum \(\overrightarrow{L}=\int{\overrightarrow{\tau}}{dt}\)
But \(\overrightarrow{\tau}=\overrightarrow{r}\times \overrightarrow{F}\)
\(\overrightarrow{r}=x\hat { i } +y\hat { j} \) and \(\hat { F }=-mg \hat { j } \)
\(\therefore \overrightarrow { \tau }=(x\hat { i }+y\hat { j } ) \times (-mg \hat { j } )\)
\(\overrightarrow { \tau } =-mgx(\hat { i } \times \hat { j } )=-mgx\hat { k } \)
\(\overrightarrow { L } = -mg \int{(xdt)}\hat {k} =-gv\ cos\theta (\int t dt)\hat { k } \)
Let initial time t = 0 and final time t = t_f
\(\overrightarrow {L}=-mg \cos \theta \left( ^{ t }\int _{ 0 }^{ f }{ tdt } \right) \hat k = -\frac{1}{2}mgv\ cos \theta\ t^{2}_{f} \hat{k}\)
Negative sign indicates, \(\overrightarrow {L}\) point inwards. 

 
As shown in the Figure, the positive x-axis is taken as east direction, positive y-direction is taken as north. After the trip, the person reaches the point P whose position vector given by
\(\vec r=-2\hat i-3\hat j\)
The displacement direction is south west.

Dimension formula for 
 \(\frac { 1 }{ 2 } \)mv² = [M][LT^-1]² = [ML²T^-2]
Dimension formula for 
mgh = [M][LT^-2][L] = [ML²T^-2]
[ML²T^-2] = [ML²T^-2]
Both sides are dimensionally the same, hence the equations \(\frac { 1 }{ 2 } \)mv² = mgh is dimensionally correct

The energy is dissipated in the form of heat. The heat energy so produced is not available for work.

Time delay T = 5.6s
\(t={{T}\over{2}}={{5.6}\over{2}}=2.8\)
Speed of radio waves = speed of light (v) 3\(\times\)10⁸ m/s
Distance from the surveillance ship to enemy ship D = v \(\times\) t
= 3\(\times\)10⁸\(\times\)2.8
= 8.4\(\times\)10⁸m
Distance (D) = 8.4\(\times\)10⁸ km.

(i) According to Newton's third law when one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.
(ii) When you push a car from inside, the reaction force of your pushing is balanced out by your body moving backward and eventually the seat behind you pushes against to bring things to static equilibrium.

Using Graham's law of diffusion
\(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } \)
Squaring both side, we get.
\({ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }={ \left( \sqrt { \frac { { M }_{ 2 } }{ { M }_{ 1 } } } \right) }^{ 2 }\)
\({ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }=\frac { { M }_{ 2 } }{ { M }_{ 1 } } \)
\({ M }_{ 2 }={ M }_{ 1 }{ \left( \frac { { R }_{ 1 } }{ { R }_{ 2 } } \right) }^{ 2 }\)
\(={ \left( \frac { { 28.7 } }{ { 7.2 } } \right) }^{ 2 }\)
\(=\frac { 823.69 }{ 51.84 } =15.88\)
=16
The gas is identified as oxygen.

Entropy and second law of thermodynamics:
(i) We have seen in the equation that the quantity \(\cfrac { { O }_{ H } }{ { T }_{ H } } \) is equal to \(\cfrac { { O }_{ H } }{ { T }_{ H } } \) The quantity\(\cfrac { Q }{ T } \) is called entropy. It is a very important thermodynamic property of a system.
(ii) It is also a state variable. \(\cfrac { { O }_{ H } }{ { T }_{ H } } \) is the entropy received by the Carnot engine from hot reservoir and \(\cfrac { { Q }_{ L } }{ { T }_{ L } } \) is entropy given out by the Camot engine to the cold reservoir. For reversible engines (Chamot Engine) both entropies should be same, so that the change in entropy of the Camot engine in one cycle is zero.
(iii) This is proved in equation (8.66). But for all practical engines like diesel and petrol engines which are not reversible engines, they satisfy the relation\(\cfrac { { Q }_{ L } }{ { T }_{ L } } >\cfrac { { Q }_{ H } }{ T_{ H } } \) fact we, can reformulate the second law of thermodynamics as follows   
(iv) "For all the processes that occur in nature (irreversible process), the entropy always increases. For reversible process entropy will not change". Entropy determines the direction in which natural process should occur.
(v) Because entropy increases when heat flows from hot object to cold object. If heat were to. flow from a cold to a hot object, entropy will decrease leading to violation of second law thermodynamics
(vi) Entropy is also called 'measure of disorder'. All natural process occur such that the disorder should always increases.
(vii) Consider a bottle with a gas inside. When the gas molecules are inside the bottle it has less disorder. Once it spreads into the entire room it leads to more disorder.
(viii) In other words when the gas is inside the bottle the entropy is less and once the gas spreads into entire room, the entropy increases.
(ix) From the second law of thermodynamics, entropy always increases. If the air molecules go back in to the bottle, the entropy should decrease, which is not allowed by the second law of thermodynamics.
(x) The same explanation applies to a drop of ink diffusing into water. Once the drop of ink spreads, its entropy is increased. The diffused ink can never become a drop again. So the natural processes occur in such a way that entropy should increase for all irreversible proc       

When a body is stretched, work is done against the restoring force (internal force). This work done is stored in the body in the form of elastic energy. Consider a wire whose un-stretch length is L and area of cross section is A. Let a force produce an extension I and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy. Then, the work done by the force F is equal to the energy gained by the wire.
The work done in stretching the wire by dl, dW = Fdl
The total work done in stretching the wire from 0 to l is
\(W=\int^{l}_{o}F dl\)   --- (1)
From Young's modulus of elasticity,
\(\mathrm{Y}=\frac{F}{A} \times \frac{L}{l} \Rightarrow \frac{Y A l}{L}\)  ......(2)
Substituting equation (2) in equation (1), we get
\(\mathrm{W}=\int_{0}^{1} \frac{Y A l}{L} d l\)
Since, l is the dummy variable in the integration, we can change l to l' (not in limits), therefore
\(\mathrm{W}=\int_{0}^{l} \frac{Y A l^{\prime}}{L} d l^{\prime} =\frac{Y A}{L}\left(\frac{l^{\prime 2}}{2}\right)_{0}^{i}=\frac{Y A}{I} \frac{l^{2}}{2}=\frac{1}{2}\left(\frac{Y A l}{L}\right) l=\frac{1}{2} F l \)
\(W =\frac{1}{2} F l=\text { Elastic potential energy }\)
Energy per unit volume is called energy density,
\(u =\frac{\text { Elastic potential energy }}{\text { Volume }}=\frac{\frac{1}{2} F l}{A L}  \)
\(\frac{1}{2} \frac{F}{A} \frac{l}{L} =\frac{1}{2}(\text { Stress } \times \text { Strain })\)  ......(3)

Let a rigid body of moment of inertia I rotate with angular velocity ധ
The angular momentum of a rigid body is, L= I_ധ
The rotational kinetic energy of the rigid body is, KE =\(\frac{1}{2}\)I_ധ²
By multiplying the numerator and denominator of the above equation with I, we get a relation between L  and KE as,
\(KE=\frac { 1 }{ 2 } \frac { { I }^{ 2 }{ \omega }^{ 2 } }{ I } =\frac { 1 }{ 2 } \frac { { \left( I\omega \right) }^{ 2 } }{ I } \)
\(KE=\frac { { L }^{ 2 } }{ 2I } \)

Consider the velocities \(\overrightarrow { { V }_{ A } } \)  and \(\overrightarrow { { V }_{ B } } \)  at an angle \(\theta \)  between their directions.
The relative velocity of A with respect to B,
\(\overrightarrow { { V }_{ AB } } =-\overrightarrow { { V }_{ B } } -\overrightarrow { { V }_{ A } } \)
Then, the magnitude and direction of \(\overrightarrow { { V }_{ AB } } \) is given by V_AB= \({ v }_{ aB }=\sqrt { { v }_{ A }^{ 2 }+{ v }_{ B }^{ 2 }-2{ v }_{ A }{ v }_{ B }cos\theta } \) and tan \(\beta =\frac { { v }_{ B }sin\theta }{ { v }_{ B }-{ v }_{ A }cos\theta } \) (Here \(\beta\)  is angle between \(\overrightarrow { { V }_{ AB } } and\quad \overrightarrow { { V }_{ B } } \) ) 
(i) When \(\theta\)  = 0, the bodies move along parallel straight lines in the same direction, We have v_AB = (v_A - v_B) in the direction of \(\overrightarrow { { V }_{ A } } \) . Obviously v_BA= (V_B+ v_A) in the direction of \(\overrightarrow { { V }_{ B } } \).
(ii) When \(\theta\) = 180°, the bodies move along parallel straight lines in opposite directions, We have v_AB= (v_A + V_B) in the direction of \(\overrightarrow { { V }_{ A } } \) .Similarly v_BA= (V_B+ v_A) in the direction of \(\overrightarrow { { V }_{ B } } \).
(iii) If the two bodies are moving at right angles to each other, then \(\theta\) =90°. The magnitude of the relative velocity of A with respect to \(B={ v }_{ BA }=\sqrt { { v }_{ A }^{ 2 }+{ v }_{ B }^{ 2 } } \).

The gravitational potential energy (U) at some height his, equal to the amount of work required to take the object from the ground to that height h.
\(U=\int{\bar{F}_a.d\bar{r}}=\int_{0}^{h}|\bar{F}_a||d\bar{r}|\cos\ \theta\)
Since the displacement and the applied force are in the same upward direction, the angle between them, \(\theta = 0°.\)Hence, cos°=1 and \(|\bar{F}_a|=mg\) and \(|d\bar{r}|=dr.\)
\(U=mg\int_{0}^{4}dr\Rightarrow mg {[r]}_{0}^{h} =mgh\)