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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 $c_3$

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

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

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- c

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

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

The accepted entrance exams for Asian Business School are MAT, ATMA, XAT, CAT, CMAT, and others. Candidates must visit the official website and apply for admissions with valid scores. 

To crack BNMIT placement interview students can follow the steps given below:

·       Review your resume

·       Practice common interview questions

·       Showcase your strengths

·       Be confident and articulate

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

St. Kabir Institute of Professional Studies MBA/PGDM admission dates are not out yet. Candidates with a valid qualification can apply on the official website of SKIPS. Since the admission portal is open for the academic year 2025, candidates can expect the official dates to be out soon. Insterested candidates can also refer to accepted entrance exam dates. 

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

$R_2\to R_2-R_1$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$

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

The MIT Moradabad has had a strong placement record since its inception. Check out the tabulated data given below to know more about MIT Moradabad placement in 2023-2025:

The highest package recorded during BNMIT placements over the years are given in the table below:

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.

A total of 79 students got placed during the MIT Moradabad placements in 2025. Refer to the table for total students placed details during placement season 2023-2025:

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

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)

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

No, CMAT is not compulsory for admissions to the MBA course at  Priyadarshini Lokmanya Tilak Institute of Management Studies and Research. Because CMAT is one of the accepted entrance exams for admission to MBA courses. As the institute accepts valid scores from several entrance exams, such as, MBA/ MMS CET/ AMMI-MBA-CET/ CAT/ XAT/ ATMA/ MAT. Thus, candidates with valid scores in CMAT can apply for the course. However, it is not compulsory. 

Yes, the placement report at BNMIT has been released. Check out the placement statistics given below:

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

$R_2\to R_2-R_1$ 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 $c_3$

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

The Shri Vishwakarma Skill University has released the application deadline for the MBA programme. The last date to apply for the MBA for the academic year 2025 is 21 July '25. Interested candidates can visit the official website to apply online. However, it is essential for candidates to adhere to this prescribed application deadline. The university will not accept applications after the deadline.

To get a seat at Asian Business School, students have to go through the steps below:

1. Apply online at the official site of the institute.
2. Ensure eligibility and entrance-based selection for the PGDM programme.
3. The selection criteria is entrance-based.
4. Upon selected, pay the specified fee to confirm and secure your seat.

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.