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Fo Maharashtra LAWCET exam preparation, candidates must know the exam pattern first. Read the main topics that are there in the syllabus. The best way to clear the exam is to prepare from books recommended by the experts.  

Those seeking to join University College of Engineering BTech first year need to fulfil the minimum eligibility requirements. Students must have passed Class 12 or equivalent with a minimum aggregate of 45%-60% (depending on the specialisation). Additionally, candidates also need to submit the scores secured in the JEE Main or equivalent examination. For further details, students may connect with the admission desk of the college.

No, BArch is a five-year course at RNS School of Architecture and not a 3-year course. The curriculum planning for a BArch degree programme includes Architectural Design as the core subject, continuing throughout the course. Hence a candidate cannot complete it before that. 

Maharashtra National Law University, Aurangabad offers various courses at the UG and the PG course levels in the field of Law. Candidates are required to follow the entrance based admission process to get the final admission offer. Check out the courses given below that are on offer in the admission session 2025-26: 

• BA-LLB
• BBA-LLB
• LLM
• LLD
• PhD

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

$\begin{array}{l}Wearegiventhat\left\{\left(x,y\right):y^2\le 6axand{x}^2+y^2\le 16a^2\right\}\\ EquationofParabolais{y}^2=6ax\dots \left(i\right)\\ andequationofcircleis{x}^2+y^2=16a^2\dots \left(ii\right)\\ Solvingeqn.\left(i\right)and\left(ii\right)weget\\ x^2+6ax=16a^2\\ \Rightarrow x^2+6ax-16a^2=0\\ \Rightarrow x^2+8ax-2ax-16a^2=0\\ \Rightarrow x\left(x+8a\right)-2a\left(x+8a\right)=0\\ \Rightarrow \left(x+8a\right)\left(x-2a\right)=0\\ \therefore x=2aandx=-8a.\left(Rejectedasitoutofregion\right)\\ Areaoftherequiredshadedregion=2\left[{\displaystyle \underset{0}{\overset{2a}{\int }}\sqrt[]{6ax}dx}+{\displaystyle \underset{2a}{\overset{4a}{\int }}\sqrt[]{16a^2-x^2}}\right]\\ =2\left[\sqrt[]{6a}{\displaystyle \underset{0}{\overset{2a}{\int }}\sqrt[]{x}dx}+{\displaystyle \underset{2a}{\overset{4a}{\int }}\sqrt[]{{\left(4a\right)}^2-x^2}}\right]\\ =2\sqrt[]{6a}.\frac{2}{3}{\left[x^{3/2}\right]}_0^{2a}+2{\left[\frac{x}{2}\sqrt[]{{\left(4a\right)}^2-x^2}+\frac{16a^2}{2}si{n}^{-1}\frac{x}{4a}\right]}_{2a}^{4a}\\ =\frac{4\sqrt[]{6}}{3}\sqrt[]{a}\left[{\left(2a\right)}^{3/2}-0\right]+{\left[x\sqrt[]{{\left(4a\right)}^2-x^2}+16a^{2}si{n}^{-1}\frac{x}{4a}\right]}_{2a}^{4a}\\ =\frac{4\sqrt[]{6}}{3}\sqrt[]{a}.2\sqrt[]{2}.a^{3/2}+\left[0+16a^{2}si{n}^{-1}\left(\frac{4a}{4a}\right)-2a\sqrt[]{16a^2-4a^2}-16a^{2}si{n}^{-1}\frac{2a}{4a}\right]\\ =\frac{8\sqrt[]{12}}{3}{a}^2+\left[16a^{2}si{n}^{-1}\left(1\right)-2a\sqrt[]{12a^2}-16a^{2}si{n}^{-1}\frac{1}{2}\right]\\ =\frac{16\sqrt[]{3}}{3}{a}^2+\left[16a^2.\frac{\pi }{2}-2a.2\sqrt[]{3}a-16a^2.\frac{\pi }{6}\right]\\ =\frac{16\sqrt[]{3}}{3}{a}^2+8\pi a^2-4\sqrt[]{3}{a}^2-\frac{8}{3}\pi a^2\\ =\left(\frac{16\sqrt[]{3}}{3}-4\sqrt[]{3}\right)a^2+\frac{16}{3}\pi a^2=\frac{4\sqrt[]{3}}{3}{a}^2+\frac{16}{3}\pi a^2\\ =\frac{4}{3}\left(\sqrt[]{3}+4\pi \right)a^{2}sq.units\\ Hence,therequiredarea=\frac{4}{3}\left(\sqrt[]{3}+4\pi \right)a^{2}sq.units.\end{array}$

Aspirants seeking admission in Arena Animation Jaipur have a variety of course options available. Aspirants can apply for BVoc or certificate programmes offered at the institute. Candidates have to fulfill the eligibility requirements to apply for a course. Below is the eligibility criteria for Arena Animation Jaipur courses:

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

$\begin{array}{l}Thecoordinatesoftheverticesof\Delta ABCaregivenbyA\left(-1,1\right),B\left(0,5\right)C\left(3,2\right).\\ EquationofABisy-1=\frac{5-1}{0+1}\left(x+1\right)\\ \Rightarrow y-1=4x+4\Rightarrow y=4x+4+1\\ \Rightarrow y=4x+5\dots \left(i\right)\\ EquationofBCisy-5=\frac{2-5}{3-0}\left(x-0\right)\\ \Rightarrow y-5=-x\\ \Rightarrow y=5-x\dots \left(ii\right)\\ EquationofCAisy-1=\frac{2-1}{3+1}\left(x+1\right)\\ \Rightarrow y-1=\frac{1}{4}x+\frac{1}{4}\Rightarrow y=\frac{1}{4}x+\frac{1}{4}+1\\ \Rightarrow y=\frac{1}{4}x+\frac{5}{4}=\frac{1}{4}\left(5+x\right)\dots \left(iii\right)\\ Areaof\Delta ABC={\displaystyle \underset{-1}{\overset{0}{\int }}\left(4x+5\right)dx}+{\displaystyle \underset{0}{\overset{3}{\int }}\left(5-x\right)dx-}{\displaystyle \underset{-1}{\overset{3}{\int }}\frac{1}{4}\left(5+x\right)dx}\\ =\frac{4}{2}{\left[x^2\right]}_{-1}^0+5{\left[x\right]}_{-1}^0+5{\left[x\right]}_0^3-\frac{1}{2}{\left[x^2\right]}_0^3-\frac{1}{4}{\left[5x+\frac{x^2}{2}\right]}_{-1}^3\\ =2\left(0-1\right)+5\left(0+1\right)+5\left(3-0\right)-\frac{1}{2}\left(9-0\right)-\frac{1}{4}\left[\left(15+\frac{9}{2}\right)-\left(-5+\frac{1}{2}\right)\right]\\ =-2+5+15-\frac{9}{2}-\frac{1}{4}\left(\frac{39}{2}+\frac{9}{2}\right)\\ =18-\frac{9}{2}-\frac{1}{4}\times \frac{48}{2}=18-\frac{9}{2}-6=12-\frac{9}{2}=\frac{15}{2}sq.units\\ Hence,therequiredarea=\frac{15}{2}sq.units.\end{array}$

Arena Animation Jaipur offers short-term courses of one to six months. The institute offers short-term courses in Editing/ VFX, Web Design, Animation fields. Aspirants meeting the eligibility requirements can apply for any of the short-term programme offered at Arena Animation Jaipur. 

Yes, candidates can enrol for the BTech course at Rungta International Skills University through the JEE Main entrance exam score. Candidates from states other than Chhattisgarh who wish to pursue a BTech course at this university must have valid JEE Main entrance exam scores. Admission is carried out through Online Counselling conducted by the Directorate of Technical Education (DTE), Govt of Chhattisgarh. The Counselling is based on the marks obtained in Class 12 and the score obtained in the Joint Entrance Examination - Main (JEE Main) at the National Level.

To get admission into the MBA course at IMS Noida, candidates are required to register themselves by completing an application process. Moreover, to get enorl, candidates must give an admission exam, followed by an in-person interview. Conclusively, the whole admission procedure includes exam scores, interview ratings, and academic achievement. Additionally, the official website of the institute does not mention that the CUET PG exam is compulsory to get admission into the MBA. 

Admission to Maharashtra National Law University, Aurangabad is entrance-based. Candidates are shortlisted based on their performance in various entrance exams such as the CLAT or the MHCET Law. Candidates who clear the shortlist for he entrance exams are called for the next admission rounds such as the GD and the PI. 
The final merit list is released once all the admission rounds have concluded. 

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

$\begin{array}{l}Requiredarea={\displaystyle \underset{0}{\overset{\pi }{\int }}sinxdx}+{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|sinx\right|dx}\\ ={\left[cosx\right]}_0^{\pi }+{\left|\left(-cosx\right)\right|}_0^{2\pi }\\ =-\left[cos\pi -cos0\right]+\left[cos2\pi -cos\pi \right]\\ =-\left[-1-1\right]+\left[1+1\right]=2+2=4sq.units\\ Hence,therequiredarea=4sq.units.\end{array}$

Arena Animation Jaipur offers one degree and certificate courses. The institute offers BVoc at the UG level for three-years. The institute also offers certificate in Motion Graphic Design, Graphics & Web Design, Web Design & Development, VFX, etc. Besides, Arena Animation Jaipur also offers short-term programmes.

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

$\begin{array}{l}Equationsofthecurvearegivenby\\ x^2+y^2=4x\dots \left(i\right)\\ and{y}^2=2x\dots \left(ii\right)\\ \Rightarrow x^2-4x+y^2=0\\ \Rightarrow x^2-4x+4-4+y^2=0\\ {\left(x-2\right)}^2+y^2=4\\ Clearlyitistheequationofacirclehavingitscentre\left(2,0\right)andradius2.\\ Solving{x}^2+y^2=4xand{y}^2=2x\\ x^2+2x=4x\\ \Rightarrow x^2+2x-4x=0\\ \Rightarrow x^2-2x=0\\ \Rightarrow x\left(x-2\right)=0\\ \therefore x=0,2\\ Areaoftherequiredregion=2\left[{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{4-{\left(x-2\right)}^2}dx-}{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x}dx}\right]\\ \left[?Parabolaandcirclebotharesymmetricalaboutx-axis\right]\\ =2{\left[\frac{x-2}{2}\sqrt[]{4-{\left(x-2\right)}^2}+\frac{4}{2}si{n}^{-1}\frac{x-2}{2}\right]}_0^2-2.\sqrt[]{2}.\frac{2}{3}{\left[x^{3/2}\right]}_0^2\\ =2\left[\left(0+0\right)-\left(0+2si{n}^{-1}\left(-1\right)\right)\right]-\end{array}$