Physics

According to question, we can write

 10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$                        

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$  

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$  

$\frac{dT}{dt}=-k\left(T-T_0\right)$  

$\frac{dT}{\left(T-T_0\right)}=-kdt$  

$\Rightarrow k=-\frac{1}{6}\mathcal{l}n\left(\frac{2}{3}\right)$             - (1)

Now $\frac{dT}{dt}=-k\left(T-T_0\right)$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

$Q=\Delta \upsilon +w\Rightarrow nC\Delta T=n{C}_v\Delta T=\frac{nC\Delta T}{4}=\frac{3}{4}nC\Delta T=n{C}_v\Delta T$

$C=\frac{4}{3}{C}_v=\frac{4}{3}*\frac{3}{2}R=2R$

given N = 1000 turns

A = 1m²

$\omega =1rev/sec$

$=1*2\pi rad/sec$

$V=\frac{-d{?}_B}{dt}=\frac{-d}{dt}\left(NBAcos\omega t\right)$  

=  $140*\frac{22}{7}$  

= 20 * 22

= 440 volts

$d=4\sqrt[]{3}$ cm (Lateral shift)

By Snell's law

${\mu }_{air}sin60°={\mu }_{g}sin\theta$

θ = 30

$\Rightarrow 1*\frac{\sqrt[]{3}}{2}=\sqrt[]{3}sin\theta$  

$sin\theta =\frac{1}{2}$  

t = 12 cm

$forA,t_{\frac{1}{2}}=4sec$  

$=m{A}_0{e}^{-}\frac{0.693}{4}*16$

$m_A=m_{A0}{e}^{-4*0.693}$    -(1)

for B,

for B,  $t_{\frac{1}{2}}=8sec$  

$m_B=m_{B_0}{e}^{-2*0.693}$               -(2)

$\frac{m_A}{m_B}=\frac{m_{AO}}{m_{BO}}\frac{e^{-4}*0.693}{e^{-2}*0.693}$

$\frac{m_A}{m_B}=\frac{25}{100}=\frac{x}{100}x=25$

For maxima = y = (2n + 1) $\frac{\lambda }{2}\frac{D}{a}$

For 1^st maxima for l₁ wavelength (n = 1)

$y_1=\frac{3{\lambda }_1}{2}\frac{D}{a}$          - (1)

First maxima for l₂ wavelength

$y_2=\frac{3}{2}{\lambda }_2\frac{D}{a}$          - (2)

$\Rightarrow y_2-y_1=\frac{3}{2}\frac{D}{a}\left[{\lambda }_2-{\lambda }_1\right]$

$=\frac{3}{2}*\frac{2*5*10^{-9}}{0.5*10^{-3}}$

$=\frac{3}{10^{-3}}*10^{-8}$

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

$l_{rms}=\frac{l_0}{\sqrt[]{2}}$  

$V_{rms}=l_{rms}z$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(X_L\right)$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(2\pi *50*200*10^{-3}\right)$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(20\pi \right)$

$l_0=\frac{220\sqrt[]{2}}{20\pi }$

$l_0=\frac{11\sqrt[]{2}}{\pi }=\frac{\sqrt[]{a}}{\pi }$

a = 121 * 2

a = 242

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_B=\frac{V_B}{I_B}=\frac{10*10^{-3}}{10*10^{-6}}=10^3\Omega$  

$A_v=\left(\frac{\Delta l_C}{\Delta l_B}\right)*\left(\frac{R_C}{R_B}\right)$

=  $\frac{1.5*10^{-3}}{10*10^{-6}}*\frac{5*10^3}{10^3}$

A_v = 750