Class 12th

$\lambda =\frac{c}{v}=\frac{Q^3}{W^2}=\frac{I^3{T}^3}{M^2{L}^4{T}^{-4}}\Rightarrow \left[\lambda \right]=\left[M^{-2}{L}^{-4}{T}^7{l}^3\right]$                                                                             

$\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x^2}{\int }}\left(sin\sqrt[]{t}\right)dt}}{x^3}$     $\left(\frac{0}{0}\right)$  applying L' Hospital rule

$=\underset{x\to 0}{lim}\frac{sin\left(\sqrt[]{x^2}\right).2x}{3x^2}$

$=\frac{2}{3},forx>0$

Modulation Index $\left(\mu \right)=\frac{A_{max}-A_{min}}{A_{max}+A_{min}}=\frac{16-8}{16+8}=\frac{1}{3}=0.33=33*10^{-2}$

$W=q\Delta V=20*15=300J$

D₁ is forward biased & D₂ is reverse biased.

$l=\frac{6}{120+130+50}=\frac{6}{300}=2*10^{-2}A=20*10^{-3}A=20mA$

$\widehat{r}-\widehat{i}=2cos\theta \widehat{n}......\left(1\right)$

$\widehat{i}.\widehat{n}=-cos\theta ......\left(2\right)$

$\Rightarrow \widehat{r}=\widehat{i}-2\left(\widehat{i}.\widehat{n}\right)\widehat{n}$

$\Rightarrow \overrightarrow{b}=\overrightarrow{a}-2\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{8-sin2x}}=asi{n}^{-1}\left(\frac{sinx+cosx}{b}\right)+c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{9-{\left(sinx+cosx\right)}^2}}=}$  

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$={\displaystyle \int \frac{dt}{\sqrt[]{3^2-t^2}}=si{n}^{-1}\frac{t}{3}+c=si{n}^{-1}\left(\frac{sinx+cosx}{3}\right)+c}$           

= a = 1, b = 3

$\therefore \left(a,b\right)\equiv \left(1,3\right)$       

$\beta =\frac{\lambda D}{d}=\frac{500*10^{-9}*1}{2*10^{-3}}=250*10^{-6}m=250*10^{-3}mm=0.25mm$

$N=mgcos30°+qEsin30°$

$a=\frac{mgsin\theta -qEcos\theta -\mu N}{m}=2.30m/s^2$

$S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow t=\sqrt[]{\frac{2\mathcal{l}}{a}}=1.31sec$

$Z=\sqrt[]{R^2+{\left(X_L-X_C\right)}^2}$

$Z=\sqrt[]{{\left(120\right)}^2+{\left(10-100\right)}^2}=150\Omega$

$\omega =\frac{1}{\sqrt[]{LC}}=\frac{1}{\sqrt[]{10^{-1}*10^{-4}}}=\sqrt[]{10^5}$

$?\omega =2\pi f$

$\Rightarrow f=\frac{10^3}{2\pi \sqrt[]{10}}=50Hz$