Here m = 1 kg, k = 50 Nm^-1, A = 10 cm = 0.10 m. y = 5 cm = 0.05 m
Kinetic energy: \({ E }_{ K }=\cfrac { 1 }{ 2 } K\left( { A }^{ 2 }-{ Y }^{ 2 } \right) =\cfrac { 1 }{ 2 } \times 50\left[ \left( 0.10 \right) ^{ 2 }-\left( 0.05 \right) ^{ 2 } \right] =0.1875J\)
Potential energy, \({ E }_{ P }=\cfrac { 1 }{ 2 } { ky }^{ 2 }=\cfrac { 1 }{ 2 } \times 50\times \left( 0.05 \right) ^{ 2 }=0.0625J\)

Total energy, E = E_K + E_P = 0.1875 + 0.0625 = 0.25

\({ T }_{ 1 }=3=2\pi \sqrt { \cfrac { m }{ k } } \) 
\({ T }_{ 2 }=5=2\pi \sqrt { \cfrac { m+1 }{ k } } \)
\(\cfrac { { T }_{ 1 } }{ { T }_{ 2 } } =\cfrac { 3 }{ 5 } =\sqrt { \cfrac { m }{ m+1 } } \)
\(\cfrac { 9 }{ 25 } =\cfrac { m }{ m+1 } \Rightarrow 9m+9=25m\)
\(m=\cfrac { 9 }{ 16 } kg\)

For the stationary observer \({ f }_{ 1 }=\cfrac { v }{ v-{ v }_{ g } } \times f\)
\(\therefore { f }_{ 1 }=\cfrac { 340 }{ 340-34 } \times f_{ 2 }=\cfrac { 340 }{ 340-17 } \times f\)
 Hence,\(\cfrac { { f }_{ 2 } }{ { f }_{ 1 } } =\cfrac { 340-17 }{ 340-34 } =\cfrac { 19 }{ 18 } \Rightarrow f_{ 1 }:{ f }_{ 2 }=19:18\)

From rms speed, \({ v }_{ rms }=\sqrt { \cfrac { 3RT }{ M } } \)
\({ V }_{ O_{ 322 } }=\sqrt { \cfrac { 3RT{ O }_{ 2 } }{ { M }_{ { O }_{ 2 } } } } ;v_{ { H }_{ 2 } }=\sqrt { \cfrac { 3RT{ H }_{ 2 } }{ { M }_{ { H }_{ 2 } } } } \)
\(\cfrac { { V }_{ { O }_{ 2 } } }{ { V }_{ H_{ 2 } } } =\sqrt { \cfrac { { T }_{ O_{ 2 } } }{ { T }_{ { H }_{ 2 } } } ,\cfrac { { M }_{ { H }_{ 2 } } }{ { M }_{ { O }_{ 2 } } } } \)
\({ v }_{ { O }_{ 2 } }=\left( \sqrt { \cfrac { 1200 }{ 300 } \times \cfrac { 2 }{ 32 } } \right) \times 1930=\cfrac { 1930 }{ 2 } \)
\({ v }_{ { O }_{ 2 } }=965{ ms }^{ -1 }\)

From Mayer's equation, Cp - Cv = R
C_P-C_v=4150
\(\cfrac { { C }_{ P } }{ { C }_{ V } } =1.4\Rightarrow { C }_{ P }=1.4{ C }_{ V }\)
\(1.4{ C }_{ v }-{ C }_{ V }=4150\)
\(\therefore { C }_{ V }=\cfrac { 4150 }{ 0.4 } \Rightarrow { C }_{ V }=10375J/kg\)

It is a sphere drawn around a particular molecule as centre and molecular range as radius.

Young's modulus, Y = \(\frac{W}{A}\times \frac{L}{l}\)
∴ elongation, I = \(\frac{WL}{Ay}\)
on either side of wire, tension is w but length is \(\frac{l}{2}\)
∴ elongation produced along either side = \(\frac{l}{2}\)mm
Total elongation produced = \(\frac{l}{2}\)+\(\frac{l}{2}\)=1 mm

(i) The phase of a vibrating particle at any instant completely specifies the state of the particle.
(ii) It expresses the position and direction of motion of the particle at that instant with respect to its mean position.
(iii) where ωt+ φ₀ = φ is called the phase of the vibrating particle. y =A sin (ωt +φ₀ )
(iv) The displacement acquired by an oscilleting body when time t = 0 s (initial time), the phase φ = φ0 is called epoch (initial phase) where φ0 is called the angle of epoch.

(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)

The Brownian motion increases
(i) with the decrease in single of the suspended particle
(ii) with the increases in temperature of the fluid
(iii) with the decrease indensity of the fluid
(iv) with the decrease in viscosity of the fluid

In transverse wave motion, the constituents of the medium oscillate or vibrate about their mean positions in a direction perpendicular to the direction of propagation (direction of energy transfer) of waves.
Example: light (electromagnetic waves)

Case(a):
Pulley is fixed rigidly here.
When the mass displace by y and the spring will also stretch by y.
Hence, F = T = ky
\(T=2\pi \sqrt { \frac { m }{ k } } \)
Case(b):
Mass displace by y, pulley also displaces by y.
T = 4ky.
\(T=2\pi \sqrt { \frac { m }{ 4k } } \)

Pascal's law states that, if the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.

(i) Implies that the satellite is bound to the Earth and it cannot escape from the Earth
(ii) As h approaches ∝ the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth's gravity and is not bound to Earth at large distances.

(i) O to P, Average velocity = 20 ms^-1
(ii) O to P and back to Q
Average velocity = 10ms^-1
Average speed = 20 ms^-1

A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum remain constant.

19.4

(a) I_xx' = 4 \(\times\) (0.3)² + 1 \(\times\) (0.8)² = 1 kgm² 
(b) I_xx'= 4 \(\times\) (3)² + 2 \(\times\) (2)² + 3 \(\times\) (4)² = 92 kgm²

(i) Both charged particles shall be dissimilar charge. (i.e. positive and negative)
(ii) After collision the charged particles should stick together permanent.
(iii) They should move with common velocity.

Initial speed u = 5 ms^-1
Angle of projection = 30°; g = 9.8 ms^-2
Height_max=?
Range R = ?
Height, \({ h }_{ max }=\frac { { u }^{ 2 }{ sin }^{ 2 }\theta }{ 2g } ;R=\frac { { u }^{ 2 }sin2\theta }{ g } \)
\({ h }_{ max }=\frac { { 5 }^{ 2 }{ sin }^{ 2 }30° }{ 2\times 9.8 } =\frac { 25\times \frac { 1 }{ 4 } }{ 19.6 } =\frac { 25 }{ 19.6 } \times \frac { 1 }{ 4 } \)
h_max = 0.318 m
Range_, \(R=\frac { 25\quad sin60° }{ 9.8 } =\frac { 25\times \sqrt { 3 } }{ 19.6 } =\frac { 25\times 1.732 }{ 19.6 } \)
Range, R = 2.209 m. 

Radius of the circle r = 3.12 m
Area of the circle A = \(\pi\) r²
= 3.14\(\times\)3.12\(\times\)3.12
= 30.566016 m²
According to the rule of significant 
A = 30.6 m² 

Since the gravitational force is conservative; the total energy is conserved throughout the motion.

Final values of potential energy, kinetic energy and total energy are measured at the height h.
By law of conservation of energy, the initial and final total energies are the same.
\(\frac { 1 }{ 2 } { mv }_{ 0 }^{ 2 }=\frac { 1 }{ 2 } { mv }^{ 2 }+mgh\)
\({ v }_{ 0 }^{ 2 }={ v }^{ 2 }+2gh\)
\(v=\sqrt { { v }_{ 0 }^{ 2 }-2gh } \)

R₁ = 100 ± = 3\(\Omega\); R₂ = 150 ± 2\(\Omega\)
Equivalent resistance R =?
Equivalent resistance R = R₁+ R₂ = (100 ± 3) + (150 ± 2) = (100 + 150) ± (3 + 2)
R = (250 ± 5) \(\Omega\)

t₁ = 2.63s, t₂ = 2.56s,
t₃ = 2.42s, t₄ = 2.71s, t₅ = 2.80s
(i) T_m =\({t_1+t_2+t_3+t_4+t_5\over 5}={2.63+2.56+2.42+2.71+2.80\over 5}\)
T_m = 2.62s (Rounded off to 2^nd decimal place)
(ii) Absolute error
\(\triangle\)T = Tm - t
\(\triangle\)T_I = 2.62 - 2.63 = +0.01s
\(\triangle\)T₂ = 2.62 - 2.56 = +0.06s
\(\triangle\)T₃ = 2.62 - 2.42 = +0.20s
\(\triangle\)T₄ = 2.62 - 2.71 = +0.09s
\(\triangle\)T₅ = 2.62 - 2.80 = +0.18s
(iii) Mean absolute error = \({\sum |\triangle T_1|\over n}\)
\(\triangle\)T_m = \({0.01+0.06+0.20+0.09+0.18\over 5}\)
\(\triangle\)T_m​​​​=\({0.54\over 5}=0.108s=0.11s\) (Rounded off to 2^nd decimal place)
(iv) Relative error: S_T =\({\triangle T_m\over T_m}={0.11\over 2.62}=0.0419\)
S_T = 0.04
(v) Percentage error in T = 0.04\(\times\)100% = 4%
(vi) Time period of simple pendulum = T = (2.62 ± 4%)s

Frequency of a vibrating body \(\gamma \alpha \frac{1}{l} \sqrt{\frac{F}{M}}\)
\( a\text { Dimension of frequency } =\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}  \)
\(\text {Dimension of length } =\mathrm{L}=\mathrm{M}^{0} \mathrm{LT}^{0}  \)
\(\text {Dimension of Force } =\mathrm{MLT}^{-2}  \)
\(\text {Dimension of Mass } =\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{0}  \)
\(\text {Frequency } \gamma =x  \)
\(\gamma =\mathrm{K}\left[\mathrm{F}^{x}\right][\mathrm{M}]^{y}[\mathrm{~L}]^{z}\)
Using dimensions we get
\(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}=\left[\mathrm{MLT}^{-2}\right]^{x}[\mathrm{M}]^{y}[\mathrm{~L}]^{z}  \)
\(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}=\mathrm{M}^{x+y} \mathrm{~L}^{x+z} \mathrm{~T}^{-2 x}\)
Comparing the powers we get
x + y = 0  z = -x = \(\frac{1}{2} \)
x + z = 0  y = -x = \(\frac{-1}{2} \)
-2 x = -1 
\(\therefore x =\frac{-1}{-2}=\frac{1}{2}  \)
\(\therefore \gamma =1 \times[\mathrm{F}]^{1 / 2}[\mathrm{M}]^{-1 / 2}[\mathrm{~L}]^{-1 / 2}  \)
\(\gamma =\frac{1}{l} \sqrt{\frac{F}{m}}\)