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Chandigarh University is the top private Hotel Management college in India according to The Week for the year 2025.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill bc-a^2\hfill & \hfill ca-b^2\hfill & \hfill ab-c^2\hfill \\ \hfill ca-b^2\hfill & \hfill ab-c^2\hfill & \hfill bc-a^2\hfill \\ \hfill ab-c^2\hfill & \hfill bc-a^2\hfill & \hfill ac-b^2\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ \Rightarrow \left[\begin{array}{ccc}\hfill ab+bc+ac-a^2-b^2-c^2\hfill & \hfill ca-b^2\hfill & \hfill ab-c^2\hfill \\ \hfill ab+bc+ac-a^2-b^2-c^2\hfill & \hfill ab-c^2\hfill & \hfill bc-a^2\hfill \\ \hfill ab+bc+ac-a^2-b^2-c^2\hfill & \hfill bc-a^2\hfill & \hfill ac-b^2\hfill \end{array}\right]\\ \Rightarrow \left(ab+bc+ac-a^2-b^2-c^2\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill ca-b^2\hfill & \hfill ab-c^2\hfill \\ \hfill 1\hfill & \hfill ab-c^2\hfill & \hfill bc-a^2\hfill \\ \hfill 1\hfill & \hfill bc-a^2\hfill & \hfill ac-b^2\hfill \end{array}\right|\left(Takingab+bc+ac-a^2-b^2-c^{2}commonfrom{C}_1\right)\\ R_1\to R_1-R_{2}and{R}_2\to R_2-R_3\\ \Rightarrow \left(ab+bc+ac-a^2-b^2-c^2\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill ca-b^2-ab+c^2\hfill & \hfill ab-c^2-bc+a^2\hfill \\ \hfill 0\hfill & \hfill ab-c^2-bc+a^2\hfill & \hfill bc-a^2-ac+b^2\hfill \\ \hfill 1\hfill & \hfill bc-a^2\hfill & \hfill ac-b^2\hfill \end{array}\right|\\ \Rightarrow \left(ab+bc+ac-a^2-b^2-c^2\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill a\left(c-b\right)+\left(c+b\right)\left(c-b\right)\hfill & \hfill b\left(a-c\right)+\left(a+c\right)\left(a-c\right)\hfill \\ \hfill 0\hfill & \hfill b\left(a-c\right)+\left(a+c\right)\left(a-c\right)\hfill & \hfill c\left(b-a\right)+\left(b+a\right)\left(b-a\right)\hfill \\ \hfill 1\hfill & \hfill bc-a^2\hfill & \hfill ac-b^2\hfill \end{array}\right|\\ \Rightarrow \left(ab+bc+ac-a^2-b^2-c^2\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill \left(c-b\right)\left(a+b+c\right)\hfill & \hfill \left(a-c\right)\left(a+b+c\right)\hfill \\ \hfill 0\hfill & \hfill \left(a-c\right)\left(a+b+c\right)\hfill & \hfill \left(b-a\right)\left(a+b+c\right)\hfill \\ \hfill 1\hfill & \hfill bc-a^2\hfill & \hfill ac-b^2\hfill \end{array}\right|\\ \Rightarrow \left(ab+bc+ac-a^2-b^2-c^2\right)\left(a+b+c\right)\left(a+b+c\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill c-b\hfill & \hfill a-c\hfill \\ \hfill 0\hfill & \hfill a-c\hfill & \hfill b-a\hfill \\ \hfill 1\hfill & \hfill bc-a^2\hfill & \hfill ac-b^2\hfill \end{array}\right|\\ \Rightarrow {\left(a+b+c\right)}^2\left(ab+bc+ac-a^2-b^2-c^2\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill c-b\hfill & \hfill a-c\hfill \\ \hfill 0\hfill & \hfill a-c\hfill & \hfill b-a\hfill \\ \hfill 1\hfill & \hfill bc-a^2\hfill & \hfill ac-b^2\hfill \end{array}\right|\\ Expandingalong{C}_1\\ \Rightarrow {\left(a+b+c\right)}^2\left(ab+bc+ac-a^2-b^2-c^2\right)\left[1\left|\begin{array}{cc}\hfill c-b\hfill & \hfill a-c\hfill \\ \hfill a-c\hfill & \hfill b-a\hfill \end{array}\right|\right]\\ \Rightarrow {\left(a+b+c\right)}^2\left(ab+bc+ac-a^2-b^2-c^2\right)\left[\left(c-b\right)\left(b-a\right)-{\left(a-c\right)}^2\right]\\ \Rightarrow {\left(a+b+c\right)}^2\left(ab+bc+ac-a^2-b^2-c^2\right)\left(bc-ca-b^2+ab-a^2-c^2+2ac\right)\\ \Rightarrow {\left(a+b+c\right)}^2(ab+bc+ac-a^2-b^2-\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- from conservation of momentum

P_c=P_A+P_B

= $\frac{h}{{\lambda }_c}=\frac{h}{{\lambda }_A}+\frac{h}{{\lambda }_B}$

$\frac{h}{{\lambda }_c}=\frac{h{\lambda }_A+h{\lambda }_B}{{\lambda }_A{\lambda }_B}$

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 1 when both momentum are positive

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 2 when both momentum are negative

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A+{\lambda }_B}$

Case 3 when 1 is negative and 2 is positive

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_B-{\lambda }_A}$

Case4 when 1 is positive and 2 is negative

${\lambda }_c$ = $\frac{{\lambda }_A{\lambda }_B}{{\lambda }_A-{\lambda }_B}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l}Giventhat:a+b+c\ne 0and\left[\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill a\hfill \\ \hfill c\hfill & \hfill a\hfill & \hfill b\hfill \end{array}\right]=0\\ C_1\to C_1+C_2+C_3\\ \Rightarrow \left[\begin{array}{ccc}\hfill a+b+c\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill a+b+c\hfill & \hfill c\hfill & \hfill a\hfill \\ \hfill a+b+c\hfill & \hfill a\hfill & \hfill b\hfill \end{array}\right]=0\\ \Rightarrow \left(a+b+c\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill 1\hfill & \hfill c\hfill & \hfill a\hfill \\ \hfill 1\hfill & \hfill a\hfill & \hfill b\hfill \end{array}\right|=0\left(Takinga+b+ccommonfrom{C}_1\right)\\ \Rightarrow a+b+c\ne 0\therefore \left|\begin{array}{ccc}\hfill 1\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill 1\hfill & \hfill c\hfill & \hfill a\hfill \\ \hfill 1\hfill & \hfill a\hfill & \hfill b\hfill \end{array}\right|=0\\ R_1\to R_1-R_{2}and{R}_2\to R_2-R_3\\ \Rightarrow \left|\begin{array}{ccc}\hfill 0\hfill & \hfill b-c\hfill & \hfill c-a\hfill \\ \hfill 0\hfill & \hfill c-a\hfill & \hfill a-b\hfill \\ \hfill 1\hfill & \hfill a\hfill & \hfill b\hfill \end{array}\right|=0\\ Expandingalong{C}_1\\ \Rightarrow 1\left|\begin{array}{cc}\hfill b-c\hfill & \hfill c-a\hfill \\ \hfill c-a\hfill & \hfill a-b\hfill \end{array}\right|=0\\ \Rightarrow \left(b-c\right)\left(c-a\right)-{\left(c-a\right)}^2=0\\ \Rightarrow ab-b^2-ac+bc-c^2-a^2+2ac=0\\ \Rightarrow -a^2-b^2-c^2+ab+bc+ac=0\\ \Rightarrow a^2+b^2+c^2-ab-bc-ac=0\\ \Rightarrow \text{\&thin}\end{array}$

You can take IELTS test in paper mode up to 4 times a month, which makes around 48 fixed test dates all year.

This is a long answer type question as classified in NCERT Exemplar

Time from S to P₁ is

t₁ = $\frac{S{P}_1}{c}=\frac{\sqrt[]{u^2+b^2}}{c}$

$\frac{u}{c}(1+\frac{1}{2}\frac{b^2}{u^2})$

Time from P₁ to O is

t₂ = $\frac{P_{1}O}{c}=\frac{\sqrt[]{v^2+b^2}}{c}$ ; $\frac{v}{c}\left(1+\frac{1}{2}\frac{b^2}{v^2}\right)$

the time required travel through lens is t₁ = $\frac{(n-1)wb}{c}$

so total time is t=1/c(u+v+b²/2D+(n-1)(wo+b²/ $\propto$ ))

after solving we get $\propto =2\left(n-1\right)D$

differentiating with respect to time

t=1/c(u+v+b²/2D+(n-1)K1In(K₂b))

dt/db=0=b/D-(n-1)K₁/b

b²= (n-1)K1D

b= $\sqrt[]{(\mathrm{n}-1)K_1\mathrm{D}}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

Hi, there are about 25+ top private Hotel Management colleges in India. And all of these HM colleges are owned privately.  Admissions to top private Hotel Management colleges in India are primarily based on merit. Army Institute of Hotel Management and Catering, IIHM Delhi, Welcomgroup Graduate School of Hotel Administration, Manipal Academy of Higher Education, AIMS Institutes, and School of Hospitality and Tourism Studies, Padmashree Dr. D.Y. Patil Vidyapeeth are some of the top ranked Hotel Management institutes in India.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- Suppose nA is the number of photons falling per second of beam A and nB is the number of photons falling per second of beam B.

N_A=2n_B

energy of falling photon A=hv_A

B=hv_B

as we know intensities are same

n_Ahv_A=n_Bhv_B

v_a/v_b=n_B/n_A=1/2

v_B=2v_A

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol: