Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (a, c)

Explanation- when we apply temperature across semiconductor then electron will starts from lower energy to high energy level that is from valence band to conduction band. When electron goes from lower to higher then holes that is left behind they goes to lower energy levels.

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill sin\alpha \hfill & \hfill cos\alpha \hfill & \hfill cos\left(\alpha +\delta \right)\hfill \\ \hfill sin\beta \hfill & \hfill cos\beta \hfill & \hfill cos\left(\beta +\delta \right)\hfill \\ \hfill sin\gamma \hfill & \hfill cos\gamma \hfill & \hfill cos\left(\gamma +\delta \right)\hfill \end{array}\right|\\ =\frac{1}{sin\delta cos\delta }\left|\begin{array}{ccc}\hfill sin\alpha sin\delta \hfill & \hfill cos\alpha cos\delta \hfill & \hfill cos\left(\alpha +\delta \right)\hfill \\ \hfill sin\beta sin\delta \hfill & \hfill cos\beta cos\delta \hfill & \hfill cos\left(\beta +\delta \right)\hfill \\ \hfill sin\gamma sin\delta \hfill & \hfill cos\gamma cos\delta \hfill & \hfill cos\left(\gamma +\delta \right)\hfill \end{array}\right|\end{array}$

Applying $cos\left(A+B\right)=cosAcosB-sinAsinB$ in $c_3$

$\begin{array}{l}{c}_1\to c_1+c_3\\ ?=\frac{1}{sin\delta cos\delta }\left|\begin{array}{ccc}\hfill cos\alpha cos\delta \hfill & \hfill cos\alpha cos\delta \hfill & \hfill cos\alpha cos\delta -sin\alpha sin\delta \hfill \\ \hfill cos\beta cos\delta \hfill & \hfill cos\beta cos\delta \hfill & \hfill cos\beta cos\delta -sin\beta sin\delta \hfill \\ \hfill cos\gamma cos\delta \hfill & \hfill cos\gamma cos\delta \hfill & \hfill cos\gamma cos\delta -sin\gamma sin\delta \hfill \end{array}\right|\\ c_1=c_2\\ \therefore ?=0=R.H.S.\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation- C=A.B and D=A'B

E= C+D = (A.B)+ (A'.B)

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 2\hfill & \hfill 3+2p\hfill & \hfill 4+3p+2q\hfill \\ \hfill 3\hfill & \hfill 6+3p\hfill & \hfill 10+6p+3q\hfill \end{array}\right|\\ R_2\to R_2-2R_1\\ R_3\to R_3-3R_1\\ =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 7+3p\hfill \end{array}\right|\\ R_3\to R_3-3R_2\\ =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1+p\hfill & \hfill 1+p+q\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|\end{array}$

Expanding along $c_1$

$?=\left|\begin{array}{cc}\hfill 1\hfill & \hfill 2+p\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=1=R.H.S.$

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 $c_3$

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