Physics

$\gamma =1+\frac{2}{f}\Rightarrow \frac{2}{f}=\gamma -1\Rightarrow f=\frac{2}{\gamma -1}$

Let the distance travelled is x, so

$$ $\left|{\overrightarrow{v}}_{Eg}\right|=\frac{x}{t_2}\Rightarrow$ speed of escalator for ground

${\stackrel{}{}}_{}$ $\left|{\overrightarrow{v}}_{BE}\right|=\frac{x}{t_1}\Rightarrow$ speed of boy concerning the escalator

${\stackrel{}{}}_{}$ $\left|{\overrightarrow{v}}_{Bg}\right|=\frac{x}{t_1}+\frac{x}{t_2}\Rightarrow$ speed of boy concerning ground

The time taken by him to walk up the moving escalator = t =  $\frac{x}{\left|{\overrightarrow{v}}_{Bg}\right|}$ $\frac{}{{\stackrel{}{}}_{}}$

$\Rightarrow t=\frac{x}{\frac{x}{t_1}+\frac{x}{t_2}}=\frac{t_1{t}_2}{t_1+t_2}$

$\left|\overrightarrow{P}\right|=\left|\overrightarrow{Q}\right|=x$

let the angle between $\overrightarrow{P}and\overrightarrow{Q}$ $\stackrel{}{}\stackrel{}{}$ is , so according to question, we can write $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\text{exception 25:}\text{exception 25:}$

^{$\left|\overrightarrow{P}+\overrightarrow{Q}\right|=n*\left|\overrightarrow{P}-\overrightarrow{Q}\right|\Rightarrow \sqrt[]{P^2+Q^2+2PQcos\theta }=n*\sqrt[]{P^2+Q^2-2PQcos\theta }$}

$\Rightarrow \sqrt[]{x^2+x^2+2x^{2}cos\theta }=n*\sqrt[]{x^2+x^2-2x^{2}cos\theta }$

$\Rightarrow \theta =co{s}^{-1}\left(\frac{n^2-1}{n^2+1}\right)$

Total surface energy before coalesce = $U_i=2*4\pi R^2*\sigma =8\pi R^2\sigma ..........\left(i\right)$ ${}_{}{}^{}{}^{}$

Let new radius becomes r, so according to conservation energy we can write

$2*\frac{4\pi R^3}{3}=\frac{4\pi r^3}{3}\Rightarrow r=2^{\frac{1}{3}}R$

Total surface energy after coalesce = $U_f=4\pi r^2*\sigma =2^{\frac{2}{3}}*4\pi R^2\sigma ..........\left(ii\right)$ ${}_{}{}^{}{}^{\frac{}{}}{}^{}$

$\Rightarrow \frac{U_i}{U_f}=\frac{8\pi R^2\sigma }{2^{\frac{2}{3}}*4\pi R^2\sigma }=2^{\frac{1}{3}}:1$

According to de-Broglie's hypothesis

$\lambda =\frac{h}{mv}\Rightarrow p=mv=\frac{h}{\lambda }\Rightarrow K=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}$

Cut-off wavelength of emitted X-ray = $\frac{hc}{K}=hc*\frac{2m{\lambda }^2}{h^2}=\frac{2mc{\lambda }^2}{h}$

Let Mass α  $\left[t^a{\upsilon }^b{\mathcal{l}}^c\right]\Rightarrow \left[M^1{L}^0{T}^0\right]\equiv T^a*{\left[L{T}^{-1}\right]}^b*{\left[M{L}^2{T}^{-1}\right]}^c$

=> a – b – c = 0    c = 1, and b + 2c = 0 Þ b = -2c = -2

=> a = b + c = 1 – 2 = -1

$Y=\frac{Mg{L}^3}{4b{d}^3\delta }$

For significant error in Y

$=\frac{\Delta M}{M}+\frac{3\Delta L}{L}+\frac{\Delta b}{b}+3\frac{\Delta d}{d}+\frac{\Delta \delta }{\delta }$              

$=\frac{1*10^{-3}}{2}+\frac{3*10^{-3}}{1}+\frac{10^{-2}}{4}+3*\frac{0.01*10^{-1}}{0.4}+\frac{10^{-2}}{5}$             

= 0.0155

$p=\sqrt[]{2mK}\Rightarrow \frac{p_2}{p_1}=\sqrt[]{\frac{K_2}{K_1}}\Rightarrow \frac{p_2}{p}=\sqrt[]{\frac{4K_1}{K_1}}=2\Rightarrow p_2=2p$

The percentage change in its momentum = $\frac{\Delta p}{p}*100=\frac{p}{p}*100=100\mathrm{\%}$

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m      So

$E=150*{\left(0.5\right)}^2=\frac{150}{4}$

$fluxflowing?=EA=\frac{150}{4}*{\left(0.5\right)}^2$              

$=\frac{150}{16}$             

Gausses law $?=\frac{q}{{\epsilon }_0}$  

  $\frac{150}{16}=\frac{q}{{\epsilon }_0}$            

$=8.3*10^{-11}C$              

  $\omega =100\pi ,R=100\Omega ,X_L=L\omega =0.5*10^{-3}*100\pi =5\pi *10^{-2}\Omega ,$ and

  $X_C=\frac{1}{C\omega }=\frac{1}{0.1*10^{-12}*100\pi }=\frac{10^{11}}{\pi }\Omega$            

Since X_C is approximately infinite, so the phase angle between current and supplied voltage and the nature of the circuit is $\approx 90°,$ predominantly capacitive circuit.