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Amity University Noida BBA admissions are merit-based. Students who apply for the course are evaluated based on merit obtained in Class 12 examinations. Students must also fulfil the eligibility requirements stated by the university. Students who wish to apply for the programme can visit the official website of the university.

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Amity BBA admissions are merit-based. Candidates are shortlisted based on their Class 12 marks. Shortlisted applicants must appear for a Personal Interview (PI) and also submit a video response as part of the final evaluation process.

To apply for the BBA programme, visit the official website of Amity University Noida and fill out the online application form. Register with your basic details, log in, complete the form, upload documents, pay the INR 1,500 application fee, and submit. A printout of the form should be kept for future reference.

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

(a) For process AB

Volume is constant , hence work done dW=0

dQ=dU+dW=dU+0=dU

 = nC_vdT= nC_v(T_B-T_A)

 = $\frac{3}{2}R\left(T_B-T_A\right)$

= $\frac{3}{2}\left(R{T}_B-R{T}_A\right)=\frac{3}{2}\left(P_B{V}_B-P_A{V}_A\right)$

Heat exchanged = $\frac{3}{2}\left(P_B{V}_B-P_A{V}_A\right)$

(b) For process BC , p =constant

dQ= dU+dW  = $\frac{3}{2}R\left(T_C-T_B\right)+P_B(V_C-V_B)$

heat exchanged = $\frac{5}{2}{P}_B\left(V_C{-V}_A\right)$

(c) For process CD , because CD is adiabatic , dQ= heat exchanged =0

(d) DA involves compression of gas from V_D to V_A at constant pressure P_A

heat transferred as similar way as BC₁

hence dQ = $\frac{5}{2}$ P_A(V_A-V_D)

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Kindly go through the solution

$\begin{array}{l}7–6x–x^2=7–\left(x^2+6x+9–9\right)\end{array}$

$\begin{array}{l}=7–\left(x^2+6x+9\right)+9\end{array}$

$\begin{array}{l}=16–\left(x^2+6x+9\right) \end{array}$

$\begin{array}{l}=16–{\left(x+3\right)}^2\end{array}$

$\begin{array}{l}=\left(4^2\right)–{\left(x+3\right)}^2\end{array}$

Sant Baba Bhag Singh University BBA graduates can find job opportunities in the government sector. The selection criteria for government jobs is based on the nature of the job and the qualifications required by the recruiting agency. Some avenues for government jobs that BBA graduates can explore:

• Banking Sector
• Staff Selection Commission (SSC)
• Union Public Service Commission (UPSC)
• State Public Service Commissions (PSC)
• Insurance Sector 

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

pV^1/2= constant

P=k/ $\sqrt[]{V}$

Work done from 1 to 2

W= ${\int }_{V1}^{V2}pdV=K{\int }_{V1}^{V2}\frac{dV}{\sqrt[]{V}}=k{\left[\frac{\sqrt[]{V}}{\frac{1}{2}}\right]}_{V1}^{V2}=2K\left(\sqrt[]{V_2}-\sqrt[]{V_1}\right)$

from ideal equation = pV=nRT

T= pV/nR= $\frac{p\sqrt[]{V}\sqrt[]{V}}{nR}$

T= $\frac{K\sqrt[]{V}}{nR}$

T₁= $\frac{K\sqrt[]{V1}}{nR}$ , T₁= $\frac{K\sqrt[]{V2}}{nR}$

$\frac{T_1}{T_2}=\frac{\frac{k\sqrt[]{V_1}}{nR}}{\frac{k\sqrt[]{V_2}}{nR}}={\sqrt[]{\frac{V_1}{V_2}}}^{}$ = $\sqrt[]{\frac{V_1}{2V_1}}=\frac{1}{\sqrt[]{2}}$

U= $\frac{3}{2}RT$

$?U=U_2-U_1=\frac{3}{2}R\left(T_1-T_2\right)$

= $\frac{3}{2}$ RT₁( $\sqrt[]{V}-1$ )

$?W$ =2p₁V₁^1/2( $\sqrt[]{V_2}-\sqrt[]{V_2}$ )

 = 2p₁V₁^1/2(2 $\sqrt[]{V_1}-\sqrt[]{V_1}$ )

 = 2p₁V₁( $\sqrt[]{2}-1$ )= 2RT₁( $\sqrt[]{2}-1$ )

$?Q$ = $?U+?W$

= $\frac{3}{2}$ RT₁( $\sqrt[]{2}-1$ )+ 2RT₁( $\sqrt[]{2}-1$ )

= $\frac{7}{2}R{T}_1(\sqrt[]{2}-1)$

Yes, fees is one of the significant factors that students can take into account when comparing the Sewa Community College Andheri BBA course with other similar universities. Drawing a comparison based on fees helps students calculate the affordability of a programme. As per official sources, the total fees to pursue an BBA at the university is INR 30,000.

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

$\begin{array}{l}I=\int \frac{1}{9x^2+6x+5}dx=\int \frac{1}{{(3x+1)}^2+2^2}\\ \begin{array}{l}Let \left(3x+1\right) =t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow 3dx=dt\hfill \end{array}\\ I=\int \frac{1}{{(3x+1)}^2+2^2}dx\\ =\frac{1}{3}\int \frac{1}{t^2+2^2}dt\\ =\frac{1}{3}\left[\frac{1}{2}ta{n}^{-1}\left(\frac{t}{2}\right)\right]+C\\ =\frac{1}{6}ta{n}^{-1}\left(\frac{t}{2}\right)+C=\frac{1}{6}ta{n}^{-1}\left(\frac{3x+1)}{2}\right)+C\end{array}$

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Kindly go through the solution

Yes, candidates looking for course admisison at IIP can apply for course admission with 50% aggregte in Class 12. Ishan Institute of Pharmacy offers two courses in the field of Pharmacy. The courses are offered in regular, full-time mode at undergraduate and diploma.Candidates willing to secure admission to Ishan Institute of Pharmacy courses are required to meet the eligibility criteria and appear for the relevant entrance exam. The Ishan Institute of Pharmacy offers BPharm and DPharm courses.