Physics NCERT Exemplar Solutions Class 12th Chapter Fifteen

$\frac{1}{f}=(\mu -1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

${\mathrm{R}}_1=\propto$

${\mathrm{R}}_2=-30\mathrm{c}\mathrm{m}$

$\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{\infty }}-\frac{1}{-30}\right)$

$\frac{1}{f}=\frac{0.5}{30}f=60\mathrm{c}\mathrm{m}$

$\mathrm{}V=\sqrt[]{\frac{B}{?}}$

$\frac{V_{\text{pipe}}}{V_{\text{air}}}=\frac{\sqrt[]{\frac{B}{2?}}}{\sqrt[]{\frac{B}{?}}}=\frac{1}{\sqrt[]{2}}$

${\mathrm{V}}_{\text{pipe}}=\frac{{\mathrm{V}}_{\text{air}}}{\sqrt[]{2}}$

${\mathrm{f}}_{\mathrm{n}}=\frac{(\mathrm{n}+1)V_{\text{pipe}}}{2\mathcal{l}}\mathrm{};\mathrm{}{\mathrm{f}}_1-{\mathrm{f}}_0=\frac{V_{\text{pipe}}}{2\mathcal{l}}=\frac{300}{2\sqrt[]{2}}$

$=105.75\mathrm{H}\mathrm{z}$  (If $\sqrt[]{2}=1.41$ $\text{exception 25:}$ )

$=106.05\mathrm{H}\mathrm{z}$  (If $\sqrt[]{2}=1.414$ $\text{exception 25:}$ )

For elastic collision   ${\mathrm{K}\mathrm{E}}_{\mathrm{i}}={\mathrm{K}\mathrm{E}}_{\mathrm{r}}$

$\frac{1}{2}\mathrm{m}*25+\frac{1}{2}*\mathrm{m}*9=\frac{1}{2}\mathrm{m}*32+\frac{1}{2}m{v}^2$

$34=32+v^2$

$\mathrm{K}\mathrm{E}=\frac{1}{2}*0.1*2=0.1\mathrm{J}=\frac{1}{10}$

$x=1$

$\mathrm{}580$

$\mathrm{}{X}_1=-3t^2+8t+10$

${\stackrel{?}{v}}_1=(-6t+8)\hat{i}=2\hat{i}$

$Y_2=5-8t^3$

${\stackrel{?}{\mathrm{v}}}_2=-24{\mathrm{t}}^2\stackrel{\mathrm{ˆ}}{\mathrm{j}}$

$\sqrt[]{v}=\left|{\stackrel{?}{v}}_2-{\stackrel{?}{v}}_1\right|=|-24\hat{j}-2\hat{i}|$

$\sqrt[]{v}=\sqrt[]{{24}^2+2^2}$

$v=580$

$dM=\sqrt[]{2R{h}_T}+\sqrt[]{2R{h}_R}$

$?h_T+h_R=160$

[Given]

So, hR = 160 - hT

$d_M=\sqrt[]{2R{h}_T}+\sqrt[]{2R\left(160-h_T\right)}$

$\therefore d_M=\sqrt[]{2*6400000*80}+\sqrt[]{2*6400000*80}$

$=2*4*8*10^3$

dM = 64 km

E = (50 N/s) sin (t – x/c)

Energy 5.5 * 10^-12 J., vol v =?

$u=\frac{1}{2}{\epsilon }_0{E}^2\to$ energy density

$?u=\frac{U}{v}\Rightarrow U=uv$

= $\frac{1}{2}{\epsilon }_0{E}^2.v$

$v=\frac{5.5*10^{-12}*2}{8.55*2500*10^{-12}}*10^{6}c{m}^3$

= 497

$\cong 500c{m}^3$

$?\frac{P_i=P_f}{m*40}=\frac{m}{2}v\text{'}+\frac{m}{2}*60$

v’ = 20 m/s

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$\frac{\Delta KE}{KE}=\frac{\frac{1}{2}m\left(2000\right)-\frac{1}{2}m\left(1600\right)}{\frac{1}{2}m\left(1600\right)}$

$\therefore x=1$

$x=Asin\omega t$   (Eq. of a particle executing SHM)

When KE = PE

$\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2$

$\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)=\frac{1}{2}k{x}^2\left[?k=m{\omega }^2\right]$

A2 – x2 = x2

 $So,\frac{A}{\sqrt[]{2}}=Asin\omega t$

$\frac{1}{\sqrt[]{2}}=sin\omega t$

$\therefore t=\frac{T}{8}$

$P=-B\left(\frac{\Delta v}{v}\right)$

or  $B=\frac{-P}{\left(\frac{\Delta v}{v}\right)}$

$B=\frac{-\rho gh}{\left(\frac{-0.005V_0}{V_0}\right)}\Rightarrow h=\frac{908*10^8*0.005}{10^3*9.8}$

$\Rightarrow h=500m$