Physics NCERT Exemplar Solutions Class 12th Chapter Ten Wave Optics

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Explanation- T₂P= T₂O+OP= D+x

T₁P= T₁O-OP= D-x

S₁P $\frac{\mathit{\lambda }}{2}$ = $\sqrt[]{{{(S}_1{T}_1)}^2+{\left(P{T}_1\right)}^2}=\sqrt[]{D^2+{(d-x)}^2}$

S₂P= $\sqrt[]{{{(S}_2{T}_2)}^2+{\left(P{T}_2\right)}^2}=\sqrt[]{D^2+{(d+x)}^2}$

The minima will occur when S₂P- S₁P= (2n-1) $\frac{\mathit{\lambda }}{2}$

$\sqrt[]{D^2+{(d+x)}^2}$ - $\sqrt[]{D^2+{(d-x)}^2}$ = $\frac{\mathit{\lambda }}{2}$

$x=D$

[D²+4D²]^1/2-[D²+0]^1/2= $\frac{\mathit{\lambda }}{2}$

[5D²]^1/2- [D²]^1/2= $\frac{\mathit{\lambda }}{2}$

$\sqrt[]{5}$ D-D= $\frac{\mathit{\lambda }}{2}$

$\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{l}\mathrm{v}\mathrm{i}\mathrm{n}\mathrm{g}$ we get D= 0.404 $\mathit{\lambda }$

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Explanation- When angle of incidence is equal to Brewster’s angle, the transmitted light is unpolarised and reflected light is plane polarised.

Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented By arrows.

Polarisation by reflection occurs when the angle of incidence Is the Brewster’s angle

So tani_b = $\frac{{\mu }_1}{{\mu }_2}$  where $\mu$ ₂< $\mu$ ₁

When the light rays travels in such a medium, the critical angle is

Sini_c= $\frac{{\mu }_2}{{\mu }_1}$

Where $\mu$ ₂< $\mu$ ₁

As tani_b > sini_c for large angles i_bc

Thus the polarisation by reflection occurs definitely.

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Explanation- resolving power =1/d = $\frac{2sin\beta }{1.22\lambda }$  = d_min= $\frac{1.22\lambda }{2sin\beta }$  where $\lambda$  is the wavelength of light

So d_min= $\frac{1.22\times 5500\times {10}^{-10}}{2sin\beta }$

$\lambda$ = $\frac{12.27}{\sqrt[]{V}}$ = 0.12 $\times$ 10^-9m

$\frac{d\text{'}min}{dmin}$ = $\frac{0.12\times {10}^{-9}}{5500\times {10}^{-10}}$ = 0.2 $\times$ 10^-3

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Explanation- When we are considering a point source of sound wave. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of

wavefront is in accordance with Huygen's principle.

So, Huygen's principle is valid for longitudinal sound waves also.

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Explanation- The point image I₁, due to L₁ is at the focal point. Now, due to the converging lens L₂, let final image formed is I which is point image, hence the wavefront for this image will be of spherical symmetry.

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Explanation- We know that the sun is at very large distance from the earth. Assuming sun as spherical, it can be considered as point source situated at infinity. Due to the large distance the radius of wavefront can be considered as large (infinity) and hence, wavefront is almost plane.

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Explanation- As we know that the frequencies of sound waves lie between 20 Hz to 20 kHz so that their wavelength ranges between 15 m to 15 mm. The diffraction occur if the wavelength of waves is nearly equal to slit width.

As the wavelength of light waves is 7000 ×10^-10 m to 4000 $\times$ 10^-10m. The slit width is very near to the wavelength of sound waves as compared to light waves. Thus, the diffraction of sound waves is more evident in daily life than that of light waves.

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Explanation- angular resolution of the eye? = 5.8 x 10^-4

linear distance between two dots is l= 2.54/300=0.84 $\times$ 10^-2cm

? = l/z

z=l/? = $\frac{0.84\times {10}^{-2}}{5.8\times {10}^{-4}}$ = 14.5cm

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Explanation- As per the given question, monochromatic light emerging from polaroid (I) is plane polarised. When polaroid (II) is placed infront of this polaroid (I), and rotated till no light passes through polaroid (II), then (I) and (II) are set in crossed positions, i.e., pass axes of I and II are at 90°.

Consider the above diagram where a third polaroid (III) is placed between polaroid (I) and polaroid II.

When a third polaroid (III) is placed in between (I) and (II), no light will emerge from (II), if pass axis of (III) is parallel to pass axis of (I) or (II). In all other cases, light will emerge from (II), as pass axis of (II)will no longer be at 90° to the pass axis of (III).

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Explanation- when polariser is not used

A=A_perp+A_||

letA₁= asinwt and A₂=asin(wt+ $\varnothing$ )

now superposition principle for perpendicular polariser

A_R= asinwt+ asin(wt+ $\varnothing$ )

A_R=a(2cos $\varnothing /2$ sin(wt+ $\varnothing$ ))

A_R=2acos $\varnothing /2$  sin(wt+ $\varnothing$ )

This eqn is also same for parallel polariser

A_R=2acos $\varnothing /2$  sin(wt+ $\varnothing$ )

And we know that intensity is directly proportional to square of amplitude

(A_R)²= (A_perp)²+(A_||)²

So resultant intensity is

I=4(a)²cos^{2 $\varnothing /2\frac{1}{T}{\int }_0^T{sin}^2(wt+\varnothing )$} dt + 4(a)²cos^{2 $\varnothing /2\frac{1}{T}{\int }_0^T{sin}^2(wt+\varnothing )$} dt

I= 8(a)²cos^{2 $\varnothing /2$} (1/2)                                     (after integrating sin value we got 1/2)

I= 4(a)²cos^{2 $\varnothing /2$}

So I= 4(a)²

But when there is a polariser

Perpendicular polariser is blocked

I=2(a)²cos^{2 $\varnothing /2\frac{1}{T}{\int }_0^T{sin}^2(wt+\varnothing )$} dt + (a)^{2 $\frac{1}{T}{\int }_0^T{sin}^2\left(wt\right)$}

I=2(a)²cos^{2 $\varnothing /2+$} 1/2(a)²

When $\varnothing =0$

I_max=2(a)²+1/2(a)²

I_max=5/2(a)²

I_max=5/2(I/4)=5I/8

At first minima it becomes I/8

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Explanation- as the refractive index of the class $\mu$ , the path difference will be calculated as $?x$ =2dsin $\theta$ +( $\mu -1$ )L

For principal maxima ,(path difference is zero)

2dsin $\theta$ ₀+( $\mu -1$ )L=0

Sin $\theta$ ₀= - $\frac{L(\mu -1)}{2d}$ = $\frac{-L\left(0.5\right)}{2d}$

Sin $\theta$ ₀=-1/16

OP=Dtan $\theta$ ₀= Dsin $\theta$ ₀=-D/16

For pat $?$ h difference $?\frac{\lambda }{2}$

2dsin $\theta$ ₁+0.5L= $?\frac{\lambda }{2}$

Sin $\theta$ ₁= $\frac{?\frac{\lambda }{2}-0.5L}{2d}$ = $\frac{?\frac{\lambda }{2}-\frac{d}{8}}{2d}$

= $\frac{\frac{\lambda }{2}-\frac{\lambda }{8}}{2\lambda }$ = $?$ 1/4 -1/16

So two possible values $\frac{1}{4}-\frac{1}{16}=\frac{3}{16}$  and- $\frac{1}{4}-\frac{1}{16}$ = $-\frac{5}{16}$

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Explanation-consider the disturbance at the receiver R₁ which is at a distance d from B

Y_A= acos(wt)  and path difference is $\frac{\lambda }{2}$  hence phase difference is $\pi$ .

Thus the wave R₁ because of B

Y_B= acos(wt- $\pi$ )= - acoswt here path difference is $\lambda$  and hence phase difference is $2\pi$

Thus R₁ because of C

Y_c= acos(wt-2 $\pi$ )= acoswt

(i)let the signal picked up at R₂ from B be Y_B= a₁cos(wt)

The path difference between signal at D and that B is $\frac{\lambda }{2}$

Y_D= -a₁cos(wt)

The path difference between signal at A and that atB is

$\sqrt[]{d^2+{\left(\frac{\lambda }{2}\right)}^2}$ -d = d( ${1+\frac{{\lambda }^2}{4D^2})}^{1/2}$ -d = $\frac{{\lambda }^2}{8d^2}$

$asd?\lambda$  therefore path difference os 0

$phasedifference=\frac{2\pi }{\gamma }\frac{{\lambda }^2}{8d^2}$

_{$Y$ A}=a₁cos(wt- $\varnothing$ )

_{$Y$ C}=a₁cos(wt- $\varnothing$ )

Therefore signal picked up by R₂ is

Y_A+Y_b+Y_c+Y_d= y = 2a₁cos(wt- $\varnothing$ )

|y|² = 4a₁²cos²(wt- $\varnothing$ )

=2a₁². Thus R₁ picks up the larger signal.

(ii) if B is swiched off

R₁ picks up y = acoswt

^{IR1= $\frac{1}{2}$ a2}

R₂ picks up  y = a coswt

I_R2= $\frac{1}{2}$ a²

(iii) thus R₁ and R₂ pick up the same signal

If D is switched off

R₁ picks up by y = acoswt

I_R1= $\frac{1}{2}$ a²

Y= 3acoswt

I_R2=9a²/2

R₂ picks up larger signal compared to R₁

(iv) thus a signal at R₁ indicates B has been switched off and an enhanced signal at R₂ indicates D has been switched off.

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Explanation-All points with the same optical path length must have the same phase.

So – $\sqrt[]{{\mu }_r{\epsilon }_r}AE$   =BC-–  $\sqrt[]{{\mu }_r{\epsilon }_r}CD$

BC= $\sqrt[]{{\mu }_r{\epsilon }_r}$ (CD-AE)

BC>0, si must be greater than AD

But in other figure

–  $\sqrt[]{{\mu }_r{\epsilon }_r}AE=BC–\mathrm{ }\mathrm{ }\sqrt[]{{\mu }_r{\epsilon }_r}CD$

So BC= –  $\sqrt[]{{\mu }_r{\epsilon }_r}\left(CD-AE\right)$

But clearly here BE is less than zero

To proving snells law we know that

BC=ACsin $\theta$ and CD-AE=ACsin $\theta$

So n= sini/sinr

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Explanation- refractive index = 1.38 refractive index = 1.5

^{$\lambda =5500A$ 0}

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.

This is partly reflected at the film-glass interface and a part transmitted. A part of the

reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r 1. Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence, rays r 1 and r2 shall dominate the behaviour. If incident light is to be transmitted through the lens, r 1 and r2 should interfere destructively. Both the reflections at A and D are from lower to higher refractive index and hence, there is no phase change on reflection. The optical path difference between r₂ and r₁ is

n(AD+CD)-AB

if d is the thickness of the film , then AD=CD=d/cosr

AB=AC sini

AC/2= d tanr

AC= 2dtanr

Hence AB= 2d tanr sini

Thus the optical path difference = $\frac{2nd}{cosr}$ -2dtanrsini

= 2. $\frac{sinid}{sinrcosr}$ - 2d $\frac{sinr}{cosr}sini$

= 2dsin{ $\frac{1-{sin}^{2}r}{sinrcosr}$ }

= 2ndcosr

For path difference $\frac{\lambda }{2}$

2ndcosr= $\frac{\lambda }{2}$

ndcosr= $\frac{\lambda }{4}$

for photographic lenses, the sources are normally in vertical plane

i=r=0⁰

ndcos0= $\frac{\lambda }{4}$

d= $\frac{\lambda }{4n}$ =5500/4 $\times 1.38$ = 1000A⁰

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Answer- (c)

Explanation-Consider the diagram the light beam incident from air to the glass slab at Brewster's angle (ip ). The incident ray is unpolarised and is represented by dot (.).

The reflected light is plane polarised represented by arrows.

As the emergent ray is unpolarised, hence intensity cannot be zero when passes through polaroid.

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Answer- ( a)

Explanation- given width of slit is 10⁴10^-10m= 10^-6

Wavelength of sunlight varies from 4000A⁰ to 8000A⁰

As the width of slit is comparable to that of wavelength, hence diffraction occurs with maxima at centre. So, at the centre all colours appear i.e., mixing of colours form white patch at the centre.

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Answer- a

Explanation- due to refraction from the glass medium there is a phase change of $\pi$

$?t$ = $\frac{o{p}^{\text{'}\text{'}}}{v}$ = $\frac{\frac{d}{cosr}}{\frac{c}{n}}=\frac{nd}{ccosr}$

According to snells law n= sin $i$ /sinr

Cosr = $\sqrt[]{1-{sin}^{2}r}=\sqrt[]{1-\frac{{sin}^2\theta }{n^2}}$

$?t$ = $\frac{nd}{c({\frac{{1-sin}^2\theta }{n^2})}^{1/2}}=\frac{n^{2}d}{c}$ ( $({\frac{{1-sin}^2\theta }{n^2})}^{-1/2}$ )

Phase difference = $?\varnothing$ = $\frac{2\pi }{T}\times ?t$ = $\frac{2\pi nd}{\lambda }$ ( $({\frac{{1-sin}^2\theta }{n^2})}^{-1/2}$ )

So net difference = $?\varnothing +\pi$ = $\frac{4\pi d}{\lambda }$ ( $1-\frac{1}{n^2}{sin}^2\theta$ )-1/2+ $\pi$

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Answer – (c)

Explanation- For the interference pattern to be formed on the screen, the sources should be coherent and emits lights of same frequency and wavelength. In a Young's double-slit experiment, when one of the holes is covered by a red filter and another by a blue filter. In this case due to filteration only red and blue lights are present. In YDSE monochromatic light is used for the formation of fringes on the screen. Hence, in this case there shall be no interference fringes.

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Answer- d

Explanation- According to question, there is a hole at point P₂. From Huygen's principle, wave will propagates from the sources S₁ and S₂. Each point on the screen will acts as secondary sources of wavelets. Now, there is a hole at point P₂ (minima). The hole will act as a source of fresh light for the slits S₃ and S₄.

Therefore, there will be a regular two slit pattern on the second screen

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Answer- (a, b, d)

Explanation-Consider the pattern of the intensity shown in the figure

(i) As intensities of all successive minima is zero, hence we can say that two sources S₁ and S₂ are having same intensities.

(ii) As width of the successive maxima (pulses) increases in continuous manner, we can say that the path difference (x) or phase difference varies in continuous manner.

(iii) We are using monochromatic light in YDSE to avoid overlapping and to have very clear pattern on the screen.

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Answer – (b, d)

Explanation- We know that wavelength of sunlight ranges from 4000 Å to 8000 Å.

Clearly, wavelength λ < width of the slit.

Hence, light is diffracted from the hole. Due to diffraction from the slight the image formed On the screen will be different from the geometrical image.

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Answer- (a, b)

Explanation- (a) When a decreases w increases. So, size decreases.

(b) Now, light energy is distributed over a small area and intensity∝1/Area  is decreasing so intensity increases

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Answer- (a, b)

Explanation- Due to the point source light propagates in all directions symmetrically and hence, wavefront will be spherical as shown in the diagram.

If power of the source is P, then intensity of the source will be I= p/4 $\pi r$ ²

 where, r is radius of the wavefront at anytime.

Pressure × time $=\frac{Ft}{A}=\left[ML{T}^{-2}\right]\left[T\right]\left[L^{-2}\right]=\left[M{L}^{-1}{T}^{-1}\right]$

Coefficient of viscosity,  $\eta =\left[M{L}^{-1}{T}^{-1}\right]$

$F_{viscous}=-\eta A\frac{\Delta v}{\Delta y}$

a = k²rt²

$\Rightarrow \frac{v^2}{r}=k^{2}r{t}^2$                         

             

$\Rightarrow v=krt$

$a_t=\frac{dv}{dt}=kr$                

⇒ tangential force, F_t = ma_t = mkr

$\therefore Powerdelivered,P=F_{t}v=\left(mkr\right)\left(krt\right)=m{k}^2{r}^{2}t$                

Note → Power delivered by centripetal force will be zero.

 $x=4sin\left(\frac{\pi }{2}-\omega t\right)$ - (i)

$y=4sin\left(\omega t\right)$ - (ii)

From (i) and (ii) cos2 $\omega t+si{n}^2\omega t={\left(\frac{x}{4}\right)}^2+{\left(\frac{y}{4}\right)}^2=1$

$\Rightarrow x^2+y^2={\left(4\right)}^2$

Use formula for M.I.

According to kepler’s third law of time period –

${\left(\frac{T_A}{T_B}\right)}^2={\left(\frac{r_A}{r_B}\right)}^3\Rightarrow \frac{r_A}{r_B}=4^{1/3}$

$\Rightarrow r_4^3=4$$r_{}^{}$$B_{}^{}$$3_{}^{}$

Gain in surface energy, $\Delta U=T\Delta A$  

from volume centenary, $\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$  

$\Rightarrow r=\frac{R}{4}$                                            

Initial surface area, Ai = 4pR²

final surface area, $A_f=64\times 4\pi r^2$  

$\therefore \Delta U=12\pi R^2.T=0.075\times 12\times 3.14\times 10^{-4}=2.8\times 10^{-4}J$          

In case of adiabatic process

$Workdone,W=\frac{\eta R\left(T_2-T_1\right)}{1-Y}$

$As,\Delta Q=\Delta U+W$

for adiabatic process :- $\Delta Q=0$

$\Rightarrow \Delta U=-W$

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

$E=\frac{kq}{2a^2}$

$E\text{'}=\frac{kq}{{\left(\sqrt[]{2}a\right)}^2}=\frac{kq}{2a^2}$

$E_{Net}=2Ecos45°+E\text{'}$

$=\frac{kq}{a^2}\left(\frac{1}{\sqrt[]{2}}+\frac{1}{2}\right)$

$E_0=2.25\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$ Speed of declromagnetic signal,  $v=\frac{E_0}{B_0}=\frac{2.25v/m}{1.5\times 10^{-8}T}$

$B_0=1.5\times 10^{-8}T\Rightarrow v=\frac{225}{150}\times 10^8=1.5\times 10^{8}m/s$

Time to reach each to the same

radar,  $t=\frac{2\times 3\times 10^{3}m}{1.5\times 10^{8}m/s}=4\times 10^{-5}S$

At minimum deviation –

$\delta =2i-A\Rightarrow i=\frac{\delta +A}{2},r=\frac{A}{2}$

Using snell’s low : - $sini=\mu =sinr$

$\Rightarrow \frac{\delta +A}{2}=\frac{\pi -A}{2}\Rightarrow \delta =\pi -2A$

Resolving power, = $\frac{d}{1.22\lambda }=\frac{24.4\times 10^{-2}}{1.22\times 2440\times 10^{-10}}$

${\lambda }_e=\frac{h}{P_e}=\frac{h}{\sqrt[]{2m_e{K}_e}}\Rightarrow {\lambda }_e^2=\frac{h^2}{2m_e{k}_e}\Rightarrow K_e=\frac{h^2}{2m_e{\lambda }_e^2}$    - (i)

${\lambda }_P=\frac{h}{P_P}=\frac{hc}{K_P}\Rightarrow K_P=\frac{hc}{{\lambda }_P}$     - (ii)

K_e = K_P

$\Rightarrow {\lambda }_P\alpha {\lambda }_e^2$           

X + p → Y + b

$Q=K_Y+K_b-K_P$

$Q+K_P=K_Y+K_b$                

$\Rightarrow Q+K_P>0$                 

From Truth table $Y=\overline{AB}$ .

Diode conducts current only in one direction.

Using factual data.

Using factual data.

 $\left|f_1-f_2\right|=\frac{40}{12}=\frac{10}{3}$

$\Rightarrow v\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)=\frac{10}{3}$

$\Rightarrow$$V$$=$$\frac{10}{3}$$\times$$\frac{{\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}$$=$$\frac{10}{3}$$\times$$\frac{4.08\times 4.16}{.08}$$=$$\frac{10}{3}$$\times$$\frac{408\times 4.16}{8}$$=$$7$$0$$7$$.$$2$$m$$/$

From conservation of Energy:

$mg\mathcal{l}\left(1-cos60°\right)=\frac{1}{2}m{v}^2$

$\Rightarrow v^2=2g\mathcal{l}\left(\frac{1}{2}\right)=g\mathcal{l}$

$=10\times \frac{25}{10}=5m/s$

$x_R=120?$

$x_L=?L=100*100*10^{?3}10?$

$x_C=\frac{1}{?c}=\frac{1}{100*100*10^{?6}}=100?$

$z=\sqrt[]{120^2+{\left(100?10\right)}^2}=\sqrt[]{120^2+90^2}$

= 150 $?$

$?32=\frac{160}{75}*t?t=\frac{32*75}{160}=15$

$\overrightarrow{L}=\overrightarrow{r}\times m\overrightarrow{v}$

= $\left(3\widehat{i}-\widehat{j}\right)\times \left(3\widehat{j}+\widehat{k}\right)$

$=9\widehat{k}+3\left(-\widehat{j}\right)-\widehat{i}$

$=-\widehat{i}-3\widehat{j}+9\widehat{k}$

$\left|\overrightarrow{L}\right|$$=$=√$\text{exception 25:}$$=$=√91

Using conservation of momentum:

60 × v = (60 + 120) × 2

⇒  v = 6m/s

Using newton’s law:

2mg – T = 4ma  - (1)

T – mg = ma       - (2)

From (1) & (2) :

$mg=5ma\Rightarrow a=\frac{g}{5}=2m/s^2$                

$T=mg+\frac{mg}{5}=\frac{6mg}{5}$                     

$U=n{c}_{v}T$

$\begin{array}{l}=2\times \frac{3}{2}\times R\times 300\\ =7479J\end{array}$      

 $r=\frac{mv}{qB}=\frac{\sqrt[]{2km}}{qB}$

= $\frac{\sqrt[]{2\times 5\times 10^3\times 1.6\times 10^{-19}\times 24\times 1.66\times 10^{-27}}}{1.6\times 10^{-19}\times \frac{1}{2}}$

= 9.975 × 102 cm

= 9.975 cm

C = 10^-6 f

f = 0.5 × 10³ = 500

$z=\sqrt[]{x_R^2+{\left(x_L-x_C\right)}^2}$                

for minimum impedance :

x_L = x_C

$\Rightarrow {\omega }^2=\frac{1}{LC}$

$\Rightarrow L=\frac{1}{{\pi }^2}=\frac{1}{10}\times 1000=100mH$                                

Using Ampere’s law for long hollow cylinder carrying current on its surface

B_in = 0

$B_{out}=\frac{{\mu }_{0}i}{2\pi r}$

${\overrightarrow{r}}_1=2\widehat{i}+4\widehat{j}-2\widehat{k},{\overrightarrow{r}}_2=\widehat{i}+2\widehat{j}+\alpha \widehat{k}$

Projection of ${\overrightarrow{r}}_1$ ${\stackrel{}{}}_{}$ on ${\overrightarrow{r}}_2$ ${\stackrel{}{}}_{}$ is zero $\Rightarrow {\overrightarrow{r}}_1.{\overrightarrow{r}}_2$ ${\stackrel{}{}}_{}{\stackrel{}{}}_{}$ = 0

2 + 8 - 2 = 0=> α= 5

t_1/2 = 2hr 30 min

Radiation intensity a disintegrations / sec (Activity)

As,   $A=\frac{A_0}{{2^t}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

For safe working, $A=\frac{A_0}{64}\Rightarrow \frac{A_0}{64}=\frac{A_0}{2^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$}\right.}}$  

$\Rightarrow t=6t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=6\times 2.5hr$                            

= 15

${\lambda }_1=560nm,$ fringe width, B = 7.2 mm

$\Rightarrow B_1=\frac{D{\lambda }_1}{d}=7.2mm$

For ${\lambda }_2\to B_2=\frac{D{\lambda }_2}{d}=8.1mm$ ${}_{}{}_{}\frac{{}_{}}{}$

$\frac{B_2}{B_1}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{8.1}{7.2}\Rightarrow {\lambda }_2\frac{81}{72}{\lambda }_1$

$=\frac{9}{8}\times 560nm$

= 630 nm

 $\frac{{}_{}}{\text{exception 25:}}$ $i=\frac{{\mathcal{l}}_0}{\sqrt[]{2}}$  at ${\omega }_1$ ${}_{}$ = 212 rad/s $R=5\Omega$ $$

${}_{}$  ${\omega }_2$ = 232 rad/s $\Delta \omega ={\omega }_2-{\omega }_1=20rad/s$ ${}_{}{}_{}$

$P=\frac{P_{max}}{2}at{\omega }_{1}and{\omega }_2$

So, $\Delta \omega ={\omega }_2-{\omega }_1=\frac{R}{L}$ ${}_{}{}_{}\frac{}{}$

$\Rightarrow L=\frac{R}{\Delta \omega }=\frac{5}{20}=.25H$

= 250 mH

$R_{AB}=20\Omega$

L = 300 cm

Null point is at 60 cm from A, so

$20mV=\frac{4}{R+20}\times 20\times \left(\frac{60}{300}\right)\times 10^3$

$\Rightarrow 20=\frac{16}{R+20}\times 10^3$

$R=\frac{15600}{20}=780\Omega$

$P_1=1.2\times 10^{-30}C.m{P}_2=2.4\times 10^{-30}C.m$

$E_1=5\times 10^{4}N{C}^{-1}{E}_2=15\times 10^{4}N{C}^{-1}$

So, $\frac{{\tau }_1}{{\tau }_2}=\frac{P_1{E}_{1}sin90°}{P_2{E}_{2}sin90°}=\frac{\left(1.2\times 10^{-30}\right)\times \left(5\times 10^4\right)}{\left(2.4\times 10^{-30}\right)\times \left(15\times 10^4\right)}$

$=\frac{1}{2}\times \frac{1}{3}=\frac{1}{6}\Rightarrow x=6$

${\upsilon }_0=320Hz.v_{train}=36km/h=\frac{36\times 1000}{60\times 60}=10m/s$

${\upsilon }_{\left(reflectedSound\right)}={\upsilon }_0\left(\frac{v}{v-v_{train}}\right)$

$=320\left(\frac{330}{330-10}\right)=330Hz$

$\Rightarrow {\upsilon }_{echo}={\upsilon }_{reflected}\left(\frac{v+v_{train}}{V}\right)$

$=330\left(\frac{330+10}{330}\right)=340Hz.$

d₁ = 2mm = 2r

$\rho =1750kg/m^3$ coeff of viscosity H

  $\frac{d}{dt}\left(2r\right)=0.35cm/s$              

here,   $\left(\frac{4}{3}\pi r^3\right)\rho g=6\pi nrv$

$\Rightarrow \eta =\frac{4r^2\rho g}{3\times 6v}=\frac{2r^2\rho g}{9v}$

$\frac{350}{9\times 35}=\frac{10}{9}?1$  

loss in k = Gain in U spring

$\Rightarrow \frac{3E}{4}=\frac{1}{2}k{\left(\frac{25}{100}\right)}^2$

$\Rightarrow \frac{3E}{4}=\frac{k}{32}\Rightarrow k=\frac{3F}{4}\times 32=24E$                

$I_{OO\text{'}}=\left[\frac{1}{4}M{\left(\frac{a}{2}\right)}^2\right]\times 2+\left[\frac{5}{4}m{\left(\frac{a}{2}\right)}^2\right]\times 2$