Class 12th

Kindly go through the solution

Kindly go through the solution

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, c

Explanation- I_C= 10mA

I_c= 95/100  I_e

I_e= $\frac{10*100}{95}$ = 10.53mA

I_b=I_e-I_C= 10.53-10= .53mA

A $\left|\begin{array}{ccc}\hfill x+2\hfill & \hfill x+3\hfill & \hfill x+2a\hfill \\ \hfill x+3\hfill & \hfill x+4\hfill & \hfill x+2b\hfill \\ \hfill x+4\hfill & \hfill x+5\hfill & \hfill x+2c\hfill \end{array}\right|=0$

This is a Multiple Choice Questions as classified in NCERT Exemplar 

Answer- a, c

Explanation- Here emitter-base junction is forward biased i.e., the positive pole of emitter base battery is connected to base and its negative pole to emitter. Also, the collector base junction is reverse biased, i.e., the positive pole of the collector base battery is connected to collector and negative pole to base.

Thus, electron move from emmiter to base and cross over from emitter to collector.

Given equations are :-

$\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$

$\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$

$\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$

This system of equation can be written, in matrix form, as AX= B, Where

⇒ x = 2, y = 3, z = 5.

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 $c_1$

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