Physics

Across zener diode & RL

8 = I_LR_L

8 = $\left(20-10\right)R_{L\left(max\right)}$

$R_{Lmax}=\frac{8}{10}k\Omega$

At max current in loop (1)

10 = i_max R + 8

i_max = $\frac{2}{R}=\frac{2}{100}=0.02A$

= 20 mA

R_L (max) = $\frac{8}{10}k\Omega$

At minimum curre3nt through zener

i_max R_L (minimum) = 8

R_L (minimum) = $\frac{8}{20}k\Omega$

$\frac{R_{Lmax}}{R_{Lmin}}=2$

Given x₁ = 1.22mm

 x₂ = 1.23mm

x₃ = 1.19mm

&   x₄ = 1.20mm

$x_{mean}=\frac{x_1+x_2+x_3+x_4}{4}=\frac{1.22+1.19+1.20}{4}=1.21$

$\left|\Delta x_1\right|=\left|x_{mean}-x_1\right|=\left|1.21-1.22\right|=0.01$

$\left|\Delta x_2\right|=\left|x_{mean}-x_2\right|=\left|1.2\right|-\left|1.23\right|=0.02$

$\left|\Delta x_3\right|=\left|x_{mean}-x_3\right|=\left|1.21-1.19\right|=0.02$

$\left|{\Delta }_4\right|=\left|x_{mean}-x_4\right|=\left|1.21-1.20\right|=0.01$

${\left(\Delta x\right)}_{mean}=\frac{0.01+0.02+0.02+0.01}{4}$

$=\frac{0.06}{4}$

$\mathrm{\%}error=\frac{\Delta x_{mean}}{x_{mean}}=\frac{0.06}{4*1.21}*100$

$=\frac{600}{4*121}$

$=\frac{150}{121}\mathrm{\%}$

x = 150

$V_{AM}=10\left[1+0.4cos\left(2\pi *10^{4}t\right)\right]cos\left(2\pi *10^{7}t\right)$

Main wave frequency (carrier frequency)

Bandwidth = 2 fm

= $2*10^4{H}_2=20*10^3{H}_2=20K{H}_2$

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5*10^{10}}{2.0*10^{10}}=\frac{3}{4}$

According to question, we can write

$\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

According to question, we can write

$F_{CA}=F_{CB}=\frac{KqQ}{x^2+{\left(\frac{d}{2}\right)}^2}\Rightarrow F=2F_{CA}cos\theta =\frac{2KqQx}{{\left[x^2+{\left(\frac{d}{2}\right)}^2\right]}^{\frac{3}{2}}}$

For maxima of force

$\frac{dF}{dx}=0,so$

$x=\frac{d}{2\sqrt[]{2}}$

According to question, we can write

 $d=d_0\left(1+\alpha \Delta T\right)\Rightarrow 6.241=6.230\left(1+1.4*10^{-5}*\Delta T\right)$

$\Rightarrow T=126.18+27=153.18°C$

We know, N = $N_0{e}^{-\lambda t}$

Where, N Number of un-decayed Nuclei

No Initial No. Of Nuclei

N $e^{-\lambda t}$

$N=N_0{\overline{e}}^{\lambda t}$

Taking log both side log N = log N0 + loge $e^{-\lambda t}$

log N = log No $\lambda t$

Slope = $\lambda =\frac{1}{t_{av}}$

According to question, we can write

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

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

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

E Energy with which photons are incident on a metallic surface.

${\lambda }_1=3{\lambda }_2\Rightarrow {\lambda }_1>{\lambda }_2$

E $\alpha \frac{1}{\lambda },$ E₁ < E₂

We know E₁ = ${?}_0+k_1$

$\frac{hc}{{\lambda }_1}=?+k_1$

k₁ = $\frac{hc}{{\lambda }_1}-{?}_0----\left(i\right)$

& E₂ = ${?}_0+k_2$

$\frac{hc}{{\lambda }_2}={?}_0+k_2$

$\frac{3hc}{{\lambda }_1}=?+k_2$

k₂ = $\frac{3hc}{{\lambda }_1}-{?}_0$ (i)

from (1) & (2)

$k_z=3\left[k_1+{?}_0\right]-{?}_0=3k_1+2{?}_0$

k₂ > $3k_1\Rightarrow \raisebox{1ex}{$k_2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.>k_1$