Class 12th

R_w = 2R_y                                          

  $\Rightarrow \rho \left(\frac{2x}{\raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)=\frac{2\rho \left(1-x\right)}{A}$              

  $\frac{4\rho x}{A}=\frac{2\rho \left(1-x\right)}{A}$  

->4x = 2 – 2x Þ 6x = 2

x =   $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$6$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ $\therefore \frac{L_x}{L_y}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{1-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{3}*\frac{3}{2}$    

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$            

Based on theoretical data.

Efficiency $\eta =50\mathrm{\%}$ $$

$\Rightarrow \eta =\left(1-\frac{T_2}{T_1}\right)*100=50$

$\frac{{}_{}}{}$ $\Rightarrow 1-\frac{T_2}{T_1}=\frac{50}{100}\Rightarrow 1-\frac{T_2}{T_1}=\frac{1}{2}$  -(i)

After the change :- $1-\frac{T_2^1}{T_1}=\frac{50*\left(1.3\right)}{100}=\frac{65}{100}=.65$ $\frac{{}_{}^{}}{{}_{}}\frac{}{}\frac{}{}$ -(ii)

$\Rightarrow 1-\left(\frac{T_2-40}{T_1}\right)=.65$ --(ii)

$\Rightarrow \frac{40}{T_1}=.65-.50=.15\Rightarrow T_1=\frac{4000}{15}=266.7k$

V =   $\frac{q_{\left(capacitance\right)}}{C}=\frac{q_1-q_2}{2c}$

$Dimensionof\frac{B^2}{{\mu }_0}:-$ Magnetic energy density

$u_B=\frac{B^2}{2{\mu }_0}$

Dimension $\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$m^3$}\right.\to \left[M{L}^2{T}^{-2}{L}^{-3}\right]$ $\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{${}^{}$}\right.{}^{}{}^{}{}^{}$

$=\left[M{L}^{-1}{T}^{-2}\right]$

You can obtain Class 12 Physics Set 2 question paper from the school, teachers, or by checking your previous year's board exam paper booklet. Coaching institutes may have printed versions available, or Libraries might hold copies. There are several educational apps and booklet such as CBSE sample paper which contains every set for practice, there is also generally included Set 2 with marking schemes. 

Energy of photon, $E=\frac{1240}{310}=4\mathrm{e}\mathrm{V}>2\mathrm{e}\mathrm{V}\mathrm{}$ ${}_{}{}_{}{}_{}{}_{}$  (so photoelectric effect will take place)

$=4*1.6*{10}^{-19}=6.4*{10}^{-19}\text{Joule}$

Number of photons falling per second

$=\frac{6.4*{10}^{-5}*1}{6.4*{10}^{-19}}={10}^{14}$

Number of photoelectron emitted per second

$=\frac{{10}^{14}}{{10}^3}={10}^{11}$

11

Loss of energy $=\frac{1}{2}{\mathrm{C}\mathrm{V}}^2-\frac{1}{2}\left(2\mathrm{C}\right)*{\mathrm{V}}_{\text{Common}}^2$ $\frac{}{}{}^{}\frac{}{}{}_{}^{}$

$\begin{array}{rr}& =\frac{1}{2}*60*{10}^{-12}*(20{)}^2-60*{10}^{-12}*(10{)}^2\\ & =60*{10}^{-12}(200-100)\\ & =6000*{10}^{-12}\\ & =6\mathrm{n}\mathrm{J}\end{array}$

n=12

Common potential after connection.

${\mathrm{V}}_{\text{common}}=\frac{{\mathrm{C}}_1{\mathrm{V}}_1+{\mathrm{C}}_2{\mathrm{V}}_2}{{\mathrm{C}}_1+{\mathrm{C}}_2}=\frac{60*20+0}{120}=10\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{t}$

Loss of energy $=\frac{1}{2}{\mathrm{C}\mathrm{V}}^2-\frac{1}{2}\left(2\mathrm{C}\right)*{\mathrm{V}}_{\text{Common}}^2$

$\begin{array}{rr}& =\frac{1}{2}*60*{10}^{-12}*(20{)}^2-60*{10}^{-12}*(10{)}^2\\ & =60*{10}^{-12}(200-100)\\ & =6000*{10}^{-12}\\ & =6\mathrm{n}\mathrm{J}\end{array}$