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This is a multiple choice answer as classified in NCERT Exemplar

(c) According to VIBGYOR, among all given sources of light, the blue light have the smallest wavelength. According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle. So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.

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

Sol:

$\begin{array}{l}Giventhatx=sintandy=sinpt\\ Differentiatingbothsidesw.r.t.t\\ \Rightarrow \frac{dx}{dt}=costand\frac{dy}{dt}=cospt.p=p.cospt\\ \Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{p.cospt}{cost}\\ \Rightarrow \frac{dy}{dx}=\frac{p.cospt}{cost}\\ Againdifferentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=p.\frac{d}{dx}\left(\frac{cospt}{cost}\right)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{cost.\frac{d}{dx}\left(cospt\right)-cospt.\frac{d}{dx}\left(cost\right)}{co{s}^{2}t}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{cost.\left(-sinpt\right).p\frac{dt}{dx}-cospt.\left(-sint\right).\frac{dt}{dx}}{co{s}^{2}t}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{-pcost.sinpt+cospt.sint}{co{s}^{2}t}\right]\frac{dt}{dx}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{-pcost.sinpt+cospt.sint}{co{s}^{2}t}\right]\frac{1}{cost}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\\ Nowwehavetoprovethat\\ \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+ p^{2}y=0\\ L.H.S.=\left(1-x^2\right)\left[p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\right]-x\left(\frac{p.cospt}{cost}\right)+ p^{2}y\\ \Rightarrow =\left(1-si{n}^{2}t\right)\left[p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\right]-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =co{s}^{2}t\left[\frac{-p^{2}cost.sinpt+pcospt.sint}{co{s}^{3}t}\right]-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =\frac{-p^{2}cost.sinpt+pcospt.sint}{cost}-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =\frac{-p^{2}cost.sinpt+pcospt.sint-p.sintcospt+p^2.sinptcost}{cost}\\ \Rightarrow =\frac{0}{cost}=0=R.H.S.\\ Hence,proved.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) A passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.

The infrastructure of the University of Hyderabad is good. The are a total of 21 hostels. Out of these, 13 are for boys and 8 are for girls. The campus also has 7 canteens to facilitate breakfast, lunch and dinner. Apart from this, the university has different libraries for different departments. Along with this, the university has also built 4 auditoriums with a capacity of 700.

Yes, but you need to cancel it before 9 days of your test date. You will get a refund after they cut some admin charges, around INR 4,500. If you cancel after that, no refund is given.

You'll gain a mix of technical, analytical and business skills in you MBA in Business Analytics Course such as:

• Data Visualization (Tableau, Power BI)
• SQL & basic Python or R
• Predictive & Prescriptive Analytics
• Machine Learning (Basics)
• Business Intelligence & KPIs
• Storytelling with Data
• Communication & Strategic Thinking
• Problem-Solving with Data

These skills make you a data translator someone who understands both the tech side and the business side.

Vidyavardhini College of Engineering and Technology accepts MHT CET scores to provide BE admission to eligible applicants. The percentile required to secure BE admission at VCET Mumbai depends on the BE specialisation selected by the candidates. Hence, candidates may or may not get admission to the college's BE programme depending on their choice of specialisation. The latest cut-off details are not yet available. Meanwhile, candidates can refer to the following table presenting the MHT CET cut-off for some BE specialisations offered at Vidyavardhini's College of Engineering and Technology for General category in 2023:

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

Sol:

 $\begin{array}{l}\left(i\right)Giventhat{x}^m.y^n={\left(x+y\right)}^{m+n}\\ Takinglogonbothsides,weget,\\ \Rightarrow log{x}^m.y^n=log{\left(x+y\right)}^{m+n}\left[?logxy=logx+logy\right]\\ \Rightarrow log{x}^m+log{y}^n=\left(m+n\right)log\left(x+y\right)\\ \Rightarrow mlogx+nlogy=\left(m+n\right)log\left(x+y\right)\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow m.\frac{d}{dx}logx+n.\frac{d}{dx}logy=\left(m+n\right)\frac{d}{dx}log\left(x+y\right)\\ \Rightarrow m.\frac{1}{x}+n.\frac{1}{y}.\frac{dy}{dx}=\left(m+n\right)\frac{1}{x+y}\left(1+\frac{dy}{dx}\right)\\ \Rightarrow \frac{m}{x}+\frac{n}{y}.\frac{dy}{dx}=\frac{m+n}{x+y}.\left(1+\frac{dy}{dx}\right)\\ \Rightarrow \frac{m}{x}+\frac{n}{y}.\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}.\frac{dy}{dx}\\ \Rightarrow \frac{n}{y}.\frac{dy}{dx}-\frac{m+n}{x+y}.\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\\ \Rightarrow \left(\frac{n}{y}-\frac{m+n}{x+y}\right).\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\\ \Rightarrow \left(\frac{nx+ny-my-ny}{y\left(x+y\right)}\right).\frac{dy}{dx}=\left(\frac{mx+nx-mx-my}{x\left(x+y\right)}\right)\\ \Rightarrow \left(\frac{nx-my}{y\left(x+y\right)}\right).\frac{dy}{dx}=\left(\frac{nx-my}{x\left(x+y\right)}\right)\\ \Rightarrow \frac{dy}{dx}=\frac{nx-my}{x\left(x+y\right)}\times \frac{y\left(x+y\right)}{nx-my}=\frac{y}{x}\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}Hence,proved.\end{array}$

$\begin{array}{l}\left(ii\right)Giventhat\frac{dy}{dx}=\frac{y}{x}\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{y}{x}\right)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{x.\frac{dy}{dx}-y.1}{x^2}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{x.\frac{y}{x}-y}{x^2}\left[?\frac{dy}{dx}=\frac{y}{x}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{y-y}{x^2}=\frac{0}{x^2}=0\\ Hence,\frac{d^{2}y}{d{x}^2}=0Hence,proved.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(c) When an object approaches a convergent lens from the left of the lens with a uniform speed of 5 m/s, the image away from the lens with a non-uniform acceleration.

As per the stream-wise NIRF reports 2024 released by UoH, the key highlights of Hyderabad University placements 2023 as compared with 2022 are presented below:

For MSc course details, please visit the official website of the university. 

This is a multiple choice answer as classified in NCERT Exemplar

(d) Since v ∝ λ, the light of red colour is of the highest wavelength and therefore of the highest speed. Therefore, after travelling through the slab, the red colour emerge first.

The University of Hyderabad popular programmes are MSc, MBA, and MTech. Central University Hyderabad places special emphasis on critical and analytical thinking, knowledge creation, and discovery. There are focused research programmes conducted in the various fields of modern Biology, which makes the department a hub of basic fundamental research. MBA students are taught presentation and technical skills. The University of Hyd M.Tech students are placed in top companies such as GE Digital, One Convergence, Marvell, TechSophy,  and others.

You can visit the IDP IELTS India website, select your city, and it will show you all the upcoming available dates and test types (Academic or General Test).

This is a Long Answer Type Questions as classified in NCERT Exemplar $\begin{array}{l}Giventhatf\left(x\right)=\{\begin{array}{cc}\hfill x^2+3x+p,\hfill & \hfill x\le 1\hfill \\ \hfill qx+2,\hfill & \hfill x>1\hfill \end{array}atx=1.\\ L.H.L.f\text{'}\left(c\right)=\underset{x\to 1^{-}}{lim}\frac{f\left(x\right)-f\left(c\right)}{x-c}\\ f\text{'}\left(1\right)=\underset{x\to 1^{-}}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1^{-}}{lim}\frac{\left(x^2+3x+p\right)-\left(1+3+p\right)}{x-1}\\ =\underset{h\to 0}{lim}\frac{\left[{\left(1-h\right)}^2+3\left(1-h\right)+p\right]-\left(1+3+p\right)}{1-h-1}\\ =\underset{h\to 0}{lim}\frac{\left[1+h^2-2h+3-3h+p\right]-\left(4+p\right)}{-h}\\ =\underset{h\to 0}{lim}\frac{\left[h^2-5h+4+p\right]-\left[4+p\right]}{-h}\\ =\underset{h\to 0}{lim}\frac{h^2-5h+4+p-4-p}{-h}\\ =\underset{h\to 0}{lim}\frac{h^2-5h}{-h}=\underset{h\to 0}{lim}\frac{h\left[h-5\right]}{-h}=5\\ R.H.L.f\text{'}\left(1\right)=\underset{x\to 1^{+}}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1^{+}}{lim}\frac{\left(qx+2\right)-\left(1+3+p\right)}{x-1}\\ =\underset{h\to 0}{lim}\frac{\left[q\left(1+h\right)+2\right]-\left[4+p\right]}{1+h-1}\\ =\underset{h\to 0}{lim}\frac{q+qh+2-4-p}{h}=\underset{h\to 0}{lim}\frac{q+qh-2-p}{h}\\ Forexistingthelimit\\ q-2-p=0\Rightarrow q-p=0\dots \left(1\right)\\ \underset{h\to 0}{\Rightarrow lim}\frac{qh-0}{h}=q\\ IfL.H.L.f\text{'}\left(1\right)=R.H.L.f\text{'}\left(1\right)thenq=5.\\ Nowputt\end{array}$