Physics

We know

Energy on any orbit is given by,

$E_n=\frac{-13.6}{n^2}{z}^2$

${}_{}\frac{}{}$  $\Rightarrow E_1=\frac{-13.6}{12}*9$ [n1 = 1] (1)

For 3rd orbit

$$ $E_3=\frac{-13.6}{9}*$   [n2 = 3]

E3 = 13.6ev

$\Delta E=E_3-E_1$

= 13.6 – (13.6 * 9)

$\Delta E=8*13.6ev=photonenergy$

$8*13.6ev=\frac{hc}{\lambda }$

$\Rightarrow 8*13.6ev=\frac{1242ev}{\lambda }nm$

$\lambda =11.415*10^9$

= 114.15 * 1010m

3d = 0.6mm

D = 80cm

= 800mm

Path difference is given by

BP – Andhra Pradesh = Dx

$\frac{}{}$ $\Rightarrow \frac{dy}{D}=\left(2n+1\right)\frac{\lambda }{2}$   [for Dark fringe at P]

n = 0, for first dark fringe

$\frac{dy}{D}=\frac{\lambda }{2}$

$\lambda =\frac{2d}{D}y$

$$ $=\frac{2d}{D}*\frac{d}{2}[y=\frac{d}{2},Given$  first dark fringe is observed on the screen directly opposite to one of the slits]

$\lambda =\frac{2*0.6*0.6}{800*2}$

$\lambda =450mm$

for lamp,   $P=\frac{v^2}{R}$

$\Rightarrow R=\frac{25*25}{5}=125\Omega$              

$l_{max}=\frac{v}{R}=\frac{25}{125}=\frac{1}{5}A$    

$\Rightarrow I_{max}=\frac{220}{125+R}=\frac{1}{5}$

$\Rightarrow 1100=125+R$        

R = 975  $\Omega$

T₂ = 200 k

$T_1=527°C+273=800K$      

w = 12000 kJ = 12 * 10⁶ J

Q₁ =?

  $x=1-\frac{T_2}{T_1}=\frac{w}{Q_1}=1-\frac{200}{800}=\frac{12*10^6}{Q_1}$              

$\Rightarrow \frac{3}{4}=\frac{12*10^6}{Q_1}$      

$\Rightarrow Q_1=16*10^{6}J$        

$V_T=2000c{m}^3=2000*10^{-6}{m}^3\left(Totalvolumeofmixture\right)$

T = 300 K

${}_{}{}^{}{}^{}$  $P_T=100kPa=100*10^{3}Pa=10^5$ Pa

mT = 8.3 J/K1 mol1

Now, using Ideal gas equation

$P_{T}V=\left(x_1+x_2\right)RT$

$\Rightarrow 10^5*200*10^{-6}=\left(x_1+x_2\right)*8.3*300$

$\frac{x_1}{x_2}=\frac{0.06}{0.02}=3:1$

Truth table for Input & Output

Now, Truth Table for option B.

Given absorb energy = 10.2 eV

$E_f=\frac{-13.6}{n^2}=-3.4$

$n^2=4$

n = 2

Given I = 2r

${\mu }_1=1\left(Air\right)$

${\mu }_2=\sqrt[]{2n}$

by, Snell's law

$\mu sini={\mu }_{2}sinr$

$i=2co{s}^{-1}\left[\sqrt[]{\frac{n}{2}}\right]$

$V_0=\frac{1}{5}C$

$V_0=\frac{C}{5},V_S=0$

$f=f_0\left[\frac{C+V_0}{C+V_s}\right]$

V0 observer velocity

C sound velocity

Vs source velocity

$=\frac{\frac{6}{5}{f}_0-f_0}{f_0}*100$

$=\frac{1}{5}*100$

% change in frequency = 20%

$x=Asin\omega t$

Angle travelled by while moving from A to B

$\theta =30°=\frac{\pi °}{6}$

$\theta =\omega t$

$\frac{\pi }{6}=\omega *3\left[t=3sec\right]$

$\omega =\frac{\pi }{18}$

$\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi *18}{\pi }=36sec$

T = 36 sec