Class 11th

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

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

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

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

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

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 $?=\frac{q}{{\epsilon }_0}$  

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

$=8.3*10^{-11}C$              

Maximum energy = 10 J

$\frac{1}{2}K{x}^2=10$              

K = 5

Given T_pendulum = T_spring

$2\pi \sqrt[]{\frac{\mathcal{l}}{g}}=2\pi \sqrt[]{\frac{m}{K}}$             

$\sqrt[]{\frac{4}{g}}=\sqrt[]{\frac{5}{5}}$           

g = 4m/s²

Range R = $\frac{u^{2}sin2\theta }{g}$

For 42° and 48° Range will be same

$H_{max}\alpha$  maximum q

So maximum height will be for 48°

Comparing E = 20 cos (2 * 10¹⁰ t – 200x) V/m to

$E=E_{0}cos\left(\omega t-kx\right)v/m$              

$\omega =2*10^{10},K=200$               

Speed = $\frac{2*10^{10}}{200}=10^{8}m/s$  

R.I. $=\frac{C}{speed}=\frac{3*10^8}{10^8}=3$  

$NowR.I.=\sqrt[]{{\epsilon }_r{\mu }_r}$

$3=\sqrt[]{{\epsilon }_r*1}$

${\epsilon }_r=9$              

Energy required to melt

Q = $MS\Delta T+ML$  

$\Rightarrow 10^{-1}*2*10^3*10+10^{-1}*3.33*10^5$      

->3.53 * 10⁴ J

Heat produce in wire

H = l²RT

  $Q=3.53*10^4={\left(\frac{1}{2}\right)}^2*\left(4*10^3\right)*t$

$t=\frac{3.53*10^4*4}{4*10^3}=35.3sec$             

De Broglie wavelength

$\lambda =\frac{h}{mV}=\frac{h}{\sqrt[]{2mE}}E\to K.E.$              

$E=\frac{3}{2}$  KT for gas

$So\lambda =\frac{h}{\sqrt[]{3mKT}}=\frac{6.6*10^{-34}}{\sqrt[]{3*9*10^{-31}*1.38*10^{-23}*300}}$

$\begin{array}{l}\lambda =6.26*10^{-9}m\\ \lambda =6.26nm\end{array}$