Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b

Explanation- r1=5kohm and r2= 5kohm and both are in series

Then V-0.3 = [ (r1+r2)10³] $*$ [0.2 $*$ 10^-3]

 V-0.3= 2

V= 2.3V

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill 3a\hfill & \hfill -a+b\hfill & \hfill -a+c\hfill \\ \hfill -b+a\hfill & \hfill 3b\hfill & \hfill -b+c\hfill \\ \hfill -c+a\hfill & \hfill -c+b\hfill & \hfill 3c\hfill \end{array}\right|\\ c_1\to c_1+c_2+c_3\\ =\left|\begin{array}{ccc}\hfill a+b+c\hfill & \hfill -a+b\hfill & \hfill -a+c\hfill \\ \hfill a+b+c\hfill & \hfill 3b\hfill & \hfill -b+c\hfill \\ \hfill a+b+c\hfill & \hfill -c+b\hfill & \hfill 3c\hfill \end{array}\right|\\ \left(a+b+c\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -a+b\hfill & \hfill -a+c\hfill \\ \hfill 1\hfill & \hfill 3b\hfill & \hfill -b+c\hfill \\ \hfill 1\hfill & \hfill -c+b\hfill & \hfill 3c\hfill \end{array}\right|\end{array}$

$R_2\to R_2-R_1$and $R_3\to R_3-R_1$

$=\left(a+b+c\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -a+b\hfill & \hfill -a+c\hfill \\ \hfill 0\hfill & \hfill 3b+a-b\hfill & \hfill -b+a\hfill \\ \hfill 0\hfill & \hfill -a+b+a-b\hfill & \hfill 3c+a-c\hfill \end{array}\right|$

Expanding along ${}_{}$$c_1$

$\begin{array}{l}=\left(a+b+c\right)\left|\begin{array}{cc}\hfill 2b+a\hfill & \hfill a-b\hfill \\ \hfill a-c\hfill & \hfill 2c+a\hfill \end{array}\right|\\ =\left(a+b+c\right)\left[4bc+2ab+2ac+a^2+ac+ab-bc\right]\\ =\left(a+b+c\right)\left(3ab+3bc+3ac\right)\\ =3\left(a+b+c\right)\left(ab+bc+ac\right)\\ =R.H.S.\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation- due to forward biasing the diffusion current in the circuit is very high and resistance in the circuit is very low. Thus voltage across pn junction is very low but in case of reverse biasing diffusion current very small so resistance becomes very high . and that end making high voltage across pn junction.

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1+p{x}^3\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill 1+p{y}^3\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill 1+p{z}^3\hfill \end{array}\right|\\ =\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill 1\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill 1\hfill \end{array}\right|+\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill p{x}^3\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill p{y}^3\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill p{z}^3\hfill \end{array}\right|\\ ={?}_1+{?}_2-----(1)\\ Now{?}_2=\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill p{x}^3\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill p{y}^3\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill p{z}^3\hfill \end{array}\right|\\ =pxyz\left|\begin{array}{ccc}\hfill 1\hfill & \hfill x\hfill & \hfill x^2\hfill \\ \hfill 1\hfill & \hfill y\hfill & \hfill y^2\hfill \\ \hfill 1\hfill & \hfill z\hfill & \hfill z^2\hfill \end{array}\right|\\ c_1\leftrightarrow c_3\\ =-pxyz\left|\begin{array}{ccc}\hfill x^2\hfill & \hfill x\hfill & \hfill 1\hfill \\ \hfill y^2\hfill & \hfill y\hfill & \hfill 1\hfill \\ \hfill z^2\hfill & \hfill z\hfill & \hfill 1\hfill \end{array}\right|\\ c_1\leftrightarrow c_2\\ =pxyz\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill 1\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill 1\hfill \end{array}\right|=pxyz{?}_1\\ \end{array}$

Putting value in (1)

$\begin{array}{l}\Rightarrow {?}_1+pxyz{?}_1\\ =\left(1+pxyz\right){?}_1-----(2)\\ Now{?}_1=\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \\ \hfill y\hfill & \hfill y^2\hfill & \hfill 1\hfill \\ \hfill z\hfill & \hfill z^2\hfill & \hfill 1\hfill \end{array}\right|\end{array}$

$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$=\left|\begin{array}{ccc}\hfill x\hfill & \hfill x^2\hfill & \hfill 1\hfill \\ \hfill y-x\hfill & \hfill y^2-x^2\hfill & \hfill 0\hfill \\ \hfill z-x\hfill & \hfill z^2-x^2\hfill & \hfill 0\hfill \end{array}\right|$

Expanding along $c_3$

$\begin{array}{l}{?}_1=\left|\begin{array}{cc}\hfill \left(y-x\right)\hfill & \hfill \left(y-x\right)\left(y+x\right)\hfill \\ \hfill \left(z-x\right)\hfill & \hfill \left(z-x\right)\left(z+x\right)\hfill \end{array}\right|\\ =\left(y-x\right)\left(z-x\right)\left|\begin{array}{cc}\hfill 1\hfill & \hfill y+x\hfill \\ \hfill 1\hfill & \hfill z+x\hfill \end{array}\right|\\ =\left(y-x\right)\left(z-x\right)\left(z+x-y-x\right)\\ =\left(x-y\right)\left(y-z\right)\left(z-x\right)\end{array}$

Putting ${?}_1$ in (2)

$L.H.S.=\left(1+pxyz\right)\left(x-y\right)\left(y-z\right)\left(z-x\right)=R.H.S$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b

Explanation -when a neutral entity is there it contains both equal number of elctron and holes. Also when electron emit from neutral entity then it leave behind a vacancy that is hole.

  $\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill {\alpha }^2\hfill & \hfill \beta +\gamma \hfill \\ \hfill \beta \hfill & \hfill {\beta }^2\hfill & \hfill \gamma +\alpha \hfill \\ \hfill \gamma \hfill & \hfill {\gamma }^2\hfill & \hfill \alpha +\beta \hfill \end{array}\right|\\ c_3\to c_3+c_1\\ =\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill {\alpha }^2\hfill & \hfill \alpha +\beta +\gamma \hfill \\ \hfill \beta \hfill & \hfill {\beta }^2\hfill & \hfill \alpha +\beta +\gamma \hfill \\ \hfill \gamma \hfill & \hfill {\gamma }^2\hfill & \hfill \alpha +\beta +\gamma \hfill \end{array}\right|\\ =\left(\alpha +\beta +\gamma \right)\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill {\alpha }^2\hfill & \hfill 1\hfill \\ \hfill \beta \hfill & \hfill {\beta }^2\hfill & \hfill 1\hfill \\ \hfill \gamma \hfill & \hfill {\gamma }^2\hfill & \hfill 1\hfill \end{array}\right|\end{array}$

$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$

$=\left(\alpha +\beta +\gamma \right)\left|\begin{array}{ccc}\hfill \alpha \hfill & \hfill {\alpha }^2\hfill & \hfill 1\hfill \\ \hfill \beta -\alpha \hfill & \hfill {\beta }^2-{\alpha }^2\hfill & \hfill 0\hfill \\ \hfill \gamma -\alpha \hfill & \hfill {\gamma }^2-{\alpha }^2\hfill & \hfill 0\hfill \end{array}\right|$

Expanding along $c_3$

$\begin{array}{l}=\left(\alpha +\beta +\gamma \right)\left|\begin{array}{cc}\hfill \beta -\alpha \hfill & \hfill {\beta }^2-{\alpha }^2\hfill \\ \hfill \gamma -\alpha \hfill & \hfill {\gamma }^2-{\alpha }^2\hfill \end{array}\right|\\ =\left(\alpha +\beta +\gamma \right)\left|\begin{array}{cc}\hfill \beta -\alpha \hfill & \hfill \left(\beta -\alpha \right)\left(\beta +\alpha \right)\hfill \\ \hfill \gamma -\alpha \hfill & \hfill \left(\gamma -\alpha \right)\left(\gamma +\alpha \right)\hfill \end{array}\right|\\ \Rightarrow \left(\alpha +\beta +\gamma \right)\left(\beta -\alpha \right)\left(\gamma -\alpha \right)\left|\begin{array}{cc}\hfill 1\hfill & \hfill \beta +\gamma \hfill \\ \hfill 1\hfill & \hfill \gamma +\alpha \hfill \end{array}\right|\\ \Rightarrow \left(\alpha +\beta +\gamma \right)\left(\beta -\alpha \right)\left(\gamma -\alpha \right)\left\{\gamma +2-\left(\beta -\gamma \right)\right\}\\ \Rightarrow \left(\alpha +\beta +\gamma \right)\left(\beta -\alpha \right)\left(\gamma -\alpha \right)\left(\gamma -\beta \right)\\ \Rightarrow \left(\alpha +\beta +\gamma \right)\left(\alpha -\beta \right)\left(\beta -\gamma \right)\left(\gamma -\alpha \right)=R.H.S.\end{array}$

Kindly go through the solution

Expanding along C₁

$=2\left(x+y\right)\left\{\left(-x\right)\left(x-y\right)-y.y\right\}$

$=2\left(x+y\right)\left(-x^2+xy-y^2\right)$

$=-2\left(x+y\right)\left(x^2-xy+y^2\right)$

$=-2\left\{x^3-x^{2}y+x{y}^2+x^{2}y-x{y}^2+y^3\right\}$

= $-2\left(x^3+y^3\right)$

Kindly go through the solution

Kindly go through the solution