Physics

4.21:

(a) Velocity , $\stackrel{?}{v}$ = 10.0 ? m/s

Acceleration, $\stackrel{?}{a}$ = (8.0 ? + 2.0 ?) m s-2

We know $\stackrel{?}{a}$ = $\frac{d\stackrel{?}{v}}{dt}$ = 8.0 ? + 2.0 ?

$\stackrel{?}{dv}$ = (8.0 ? + 2.0 ?)dt

Integrating both sides we get $\stackrel{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\stackrel{?}{u}$ ,

Where, $\stackrel{?}{u}=$ velocity vector of the particle at t =0

$\stackrel{?}{v}=$ velocity vector of the particle at time t

But $\stackrel{?}{v}$ = $\frac{\underset{dr}{\to }}{dt}$

$\stackrel{?}{dr}$ = $\stackrel{?}{v}$ dt

= (8.0t ? + 2.0t ? + $\stackrel{?}{u}$ )dt

Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

$\stackrel{?}{r}=$ $\stackrel{?}{u}$ t + ½ $*$ 8.0 t2 ? + ½ $*$ 2.0 $*$ t2 ? = $\stackrel{?}{u}$ t + 4.0 t2 ? + t2 ?

Substituting the value of $\stackrel{?}{u}$ , we get

$\stackrel{?}{r}=$ ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed as

x ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we get

x = 4.0 t2 and y = 10.0t + t2

t = $\sqrt{x/4}$

(a) When x = 16m, t = 2 s, y = 24m

(b) Velocity of the particle

$\stackrel{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\stackrel{?}{u}$

At t = 2 s,

$\stackrel{?}{v}$ (t) = 8.0 $*$ 2 ? + 2.0 $*2$ ? + 10 $*$ ? = 16 ? + 14 ?

The magnitude of $\stackrel{?}{v}$ (t) is given by

$\left|\mathrm{}\stackrel{?}{v}\right|$ = ( 162 + 142)1/2 = 21.26 m/s

According to Chapter 10 Physics Class 12, the Huygens' Principle in wave optics states that every point on a wave front spreads out in all directions at the speed of the wave, and these act as a source of secondary wavelets. According to this principle, all new wave front is the tangent to these secondary wavelets. The principle holds significance when it comes to explaining phenomena like refraction and reflection of light. It is instrumental in understanding the behavior of light in various media and provides a geometric method to determine the propagation of wave fronts. The Huygens' Principle lays the foundation for the wave theory of light.

13.5 Volume of the air bubble $,$ $V_1$ = 1.0 ${cm}^3$ = 1 $*{10}^{-6}$ $m^3$

Height achieved by bubble, d = 40 m

Temperature at a depth of 40 m, $T_1$ = 12 $?$ = 285 K

Temperature at the surface of the lake, $T_2$ = 35 $?$ = 308 K

The pressure at the surface of the lake, $P_2$ = 1 atm = 1.013 $*{10}^5$ Pa

The pressure at 40m depth: $P_1$ = 1 atm + d $?g$

Where $?$ is the density of water = ${10}^3$ kg/ $m^3$

G = acceleration due to gravity = 9.8 m/ $s^2$

Hence $P_1$ = 1.013 $*{10}^5+(40*{10}^3*9.8)$ = 493300 Pa

From the relation $\frac{P_1{V}_1}{T_1}$ = $\frac{P_2{V}_2}{T_2}$ , where $V_2$ is the volume of bubble when it reaches the surface, we get, $V_2$ = $\frac{P_1{V}_1{T}_2}{T_1{P}_2}$ = $\frac{493300*1*{10}^{-6}*308}{285*1.013\mathrm{}*{10}^5}$ $=$ 5.263 $*$ ${10}^{-6}{m}^3=5.263{cm}^3$

13.4 Volume of oxygen, $V_1$ = 30 litres = 30 $*{10}^{-3}{m}^3$

Gauge pressure, $P_1$ = 15 atm = 15 $*1.013*{10}^5$ Pa

Temperature, $T_1$ = 27 $?$ = 300 K

Universal gas constant, R = 8.314 J/mole/K

Let the initial number of moles of oxygen gas cylinder be $n_1$

From the gas equation, we get $P_1{V}_1=n_1\mathrm{R}{T}_1$

$n_1=\frac{P_1{V}_1}{\mathrm{R}{T}_1}$ = $\frac{15\mathrm{}*1.013*{10}^5*30*{10}^{-3}}{8.314*300}$ = 18.276

But $n_1$ = $\frac{m_1}{M}$ , where $m_1$ = initial mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_1$ = $n_1$ M = 18.276 $*32=584.84$ g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.

Hence, Volume, $V_2$ = 30 litres = 30 $*{10}^{-3}{m}^3$

Gauge pressure, $P_2$ = 11 atm = 11 $*1.013*{10}^5$ Pa

Temperature, $T_2$ = 17 $?$ = 290 K

Let the final number of moles of oxygen left in the cylinder be $n_2$

From the gas equation, we get $P_2{V}_2=n_2\mathrm{R}{T}_2$

$n_2=\frac{P_2{V}_2}{\mathrm{R}2}$ = $\frac{11\mathrm{}*1.013*{10}^5*30*{10}^{-3}}{8.314*290}$ = 13.864

But $n_2$ = $\frac{m_2}{M}$ , where $m_2$ = remaining mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_2$ = $n_2$ M = 13.864 $*32=443.65$ g

So the mass of oxygen taken out = $m_1-m_2$ = 141.19 gm = 0.141 kg