Physics

f₁ = 15 cm

$P_1=\frac{1}{15}*100=\frac{100}{15}D$

$P_3=\frac{100}{15}D$

$\frac{1}{f^2}=\left({\mu }_2-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$=\left(1.25-1\right)*\frac{-2}{R}$

$\frac{1}{f_2}=\frac{0.25*-2}{15}$

$\frac{1}{f_2}=\frac{-1}{30}c{m}^{-1}$

$\begin{array}{l}{p}_2=\frac{1}{f_2}=-\frac{100}{30}\\ P_R=P_1+P_2+P_3\\ P_R=\frac{100}{15}-\frac{100}{30}+\frac{100}{15}\\ P_R=10D\end{array}$

$P_R=\frac{1}{f_R}$

$f_R=\frac{1}{10}*100cm$

$f_R=10cm$

$V_{rms}\alpha \sqrt[]{T}|T_i=127°C=400k.$

$V_{rm{s}^1}=2V_{rms},m=0.056kg=56g$

T_f = 4T_i = 4 * 400 = 1600 k

$Q=n{c}_v\Delta T=\left(\frac{m}{M}\right)C_v\Delta T$

= $\left(\frac{56}{28}\right)\frac{5}{2}R\Delta T$

= $2*\frac{5}{2}*2*1200$

Q = 12 * 10³ cal

= 12 k cal

Answer (3) 

Particles having phase difference of $?$ will move with same speed.

Kindly go through the solution

Answer (12.00)

$I_{S_4}=4I_0$

$I_{S_3}=I_0$

$\therefore \frac{I_{max}}{I_{min}}=\frac{{\left(\sqrt[]{I_{S_4}}+\sqrt[]{I_{S_3}}\right)}^2}{{\left(\sqrt[]{I_{S_4}}-\sqrt[]{I_{S_3}}\right)}^2}={\left(\frac{2+1}{2-1}\right)}^2=\frac{9}{1}$

${\lambda }_1=\frac{h}{\sqrt[]{2m{E}_1}}={\lambda }_2=\frac{h}{\sqrt[]{2m{E}_2}}$

$=E_2=\frac{4}{9}{E}_1=4eV$

$E_2=E_1-e{V}_0=V_0=5V$

$dm=\left(\frac{m}{L}\right)dx$

$\therefore T=\frac{m{\omega }^2}{2L}\left(L^2-x^2\right)$

$\therefore ?L={\int }_0^L\frac{m{\omega }^2}{2L\pi r^{2}Y}\left(L^2-x^2\right)dx$

$=?L=\frac{m{\omega }^2{L}^2}{3\pi r^{2}Y}$

$i=\frac{v_0}{R}\left(1-e^{\frac{tR}{L}}\right)$

$U=\frac{1}{2}L{i}^2$

$\frac{dU}{dt}=Li\frac{di}{dt}$

$2L*\frac{R}{L}*\frac{v_0^2}{R^2}\left(1-e^{\frac{tR}{L}}\right)e^{\frac{tR}{L}}=\frac{v_0^2}{R}\left(1-e^{\frac{tR}{L}}\right)$

$t=\frac{L}{R}In2$

$y=A\sin ?\frac{2\pi t}{T}$

$t_2-t_1=\frac{T}{2\pi }\left. [{\sin }^{-1}?\frac{x_1}{A}-{\sin }^{-1}?\frac{x_2}{A}\right]$

Deformation energy per unit volume,

$U=\int dU=\int \frac{P^2}{2B}Ady$

$=\int \frac{{\rho }^2{g}^{2}y}{2B}Ady$

$U=\frac{{\rho }^2{g}^2}{2B}\frac{h^3}{3}A$

$=\frac{{\rho }^2{g}^2{h}^2}{6B}\left(volume\right)$

$=1.5\mu J$