Class 12th

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat:\left[\begin{array}{cc}\hfill 2x+y\hfill & \hfill 4x\hfill \\ \hfill 5x-7\hfill & \hfill 4x\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 7\hfill & \hfill 7y-13\hfill \\ \hfill y\hfill & \hfill x+6\hfill \end{array}\right]\\ Equatingthecorrespondingelements,weget,\\ 2x+y=7\dots \left(i\right)\\ and4x=7y-13\dots \left(ii\right)\\ fromeqn.\left(ii\right)4x-x=6\\ 3x=6\\ \therefore x=2\\ fromeqn.\left(i\right)2*2+y=7\\ 4+y=7\therefore y=7-4=3\\ Hence,thecorrectoptionis(b).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Totalnumberofpossiblematricesoforder3*3witheachentry0or2=2^{3*3}=2^9=512.\\ Hence,thecorrectoptionis(d).\end{array}$

Sol. (a) In extraction of iron lime stone is added on a flux

$\begin{array}{rr}& {\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3?\mathrm{C}\mathrm{a}\mathrm{O}+{\mathrm{C}\mathrm{O}}_2\\ & {\mathrm{C}\mathrm{O}}_2+{\mathrm{S}\mathrm{i}\mathrm{O}}_2?{\mathrm{C}\mathrm{a}\mathrm{S}\mathrm{i}\mathrm{O}}_3\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}GiventhatA=\left[\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 4\hfill \\ \hfill 0\hfill & \hfill 4\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill 0\hfill & \hfill 0\hfill \end{array}\right]\\ Here,numberofcolumnsandthenumberofrowsareequali.e.,3.So,Aisasquarematrix.\\ Hence,thecorrectoptionis(a).\end{array}$

Modulation Index,  $\mu =\frac{A_m}{A_C}$

Variation = $2A_m=8\Rightarrow A_m=4v$

$A_m+A_c=9$

$A_C=9-A_m=5v$

$\therefore \mu =\frac{4}{5}=0.8$

 $\Delta E_1=-\frac{E_0}{4}+\frac{E_0}{1}=\frac{3}{4}{E}_0$

$\Delta E_2=0-\left(-E_0\right)=E_0$

$\frac{\Delta E_1}{\Delta E_2}=\frac{3}{4}$

$\frac{hc}{\lambda }-?=E$ -(i)

$\frac{hc}{\lambda \text{'}}-?=2E$ -(ii)

$hc\left(\frac{1}{\lambda \text{'}}-\frac{1}{\lambda }\right)=E$

$\lambda \text{'}=\frac{hc\lambda }{E\lambda +hc}$

Based an theoretical data.

L = 1H, R = 100 $\Omega$

$As,i=i_0{e}^{-t/\tau }$

$Fori=\frac{i_0}{2}\Rightarrow \frac{i_0}{2}=i_0{e}^{-t/\tau }$

$\Rightarrow -\mathcal{l}n2=-t/\tau$

$i=\frac{6}{100}{e}^{-\frac{15}{1/100}}=.06e^{-1500*10^{-3}}=0.06e^{-1.5}=0.06*0.25=.0.015-A$

So, $U=\frac{1}{2}L{i}^2=\frac{1}{2}*1*{\left(15*10^{-3}\right)}^2=\frac{1}{2}\left(225\right)*10^{-6}=112.5*10^{-6}J=0.1125mJ$