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JIMS Homoeopathic Medical College and Hospital's application window is open. Students can check the steps below to know the step-wise JIMS Homoeopathic Medical College and Hospital application process 2025:

Step 1: Visit the official website.

Step 2: Fill out the application form by providing personal details.

Step 3: After filling out the application form, candidates have to upload the required documents.

Step 4: Pay the application fee through debit, credit card or net banking.

The Neotia University courses are offered to students at the UG, PG, and PhD levels. The university offers a wide range of programmes to students in various disciplines such as Arts, Maritime Studies,  Sciences, Agriculture,  Pharmaceutical Science, Health Sciences, and Commerce. Further, the flagship programmes offered at Neotia University are BTech and MBA. 

JIMS Homoeopathic Medical College and Hospital is one of the prominent institutions for medical education. The college is affiliated with Kaloji Narayana Rao University of Health Sciences, Dr. NTR University of Health Sciences. Ministry of AYUSH, Central Council of Homoeopathy, Central Council for Research in Homoeopathy, and National Commission for Homoeopathy. The college offers a well-structured curriculum, clinical exposure, and experienced faculty. The college prepares students for successful careers in homoeopathic medicine.

JIMS Homoeopathic Medical College and Hospital appoints well-experienced and trained faculty members with years of experience in the related field. The faculty engages the students with the latest activities and innovative teaching, and quality research. The faculty are passionate and committed to helping the students in their career pathways. The faculty focuses on theoretical and practical knowledge for the students.

Yes, it is worth considering Neotia University as a good university, as it has been recognised by the UGC and is one of the renowned universities in Kolkata to offer quality education to students in the fields of Engineering and Management. It is one of the private universities and among the 9 institutions selected in India for 'Mission Innovation' by the Government of India for IC-MAP.

International students are required to submit the official transcripts from all the previously attended institutions. 

They are to be translated into either English or German by a certified agency before an applicant gets to the uni-assist stage of application. Once these requirements are met, students can submit the scanned copies of these transcripts via the portal. The students might be requested to submit more of these documents if requested. 

The Neotia University (TNU) is a private university and was established in 2015. It is recognised by the University Grants Commission  (UGC) and has also been approved by the AICTE, BCI, PCI, Indian Nursing Council, and the Directorate General of Shipping (DGS). Further, the Neotia University Kolkata has more than 600 industry tie-ups. 

Radius of gyration of a solid surface,

${\mathrm{K}}_{\mathrm{S}}=\sqrt[]{\frac{2}{5}\mathrm{R}}$

Radius of gyration of a hollow surface,

${\mathrm{K}}_{\mathrm{H}}=\sqrt[]{\frac{2}{3}}\mathrm{R}$

$\Rightarrow \frac{{\mathrm{K}}_{\mathrm{S}}}{{\mathrm{K}}_{\mathrm{H}}}=\sqrt[]{\frac{3}{5}}=\frac{\sqrt[]{3}}{\sqrt[]{5}}$

Yes, Sai Tirupati University Admissions are open for the academic year 2025. Candidates willing to take admission in the university must fill out the application form online through the official website or offline at the admission office. The university offers a wide range of courses to candidates such as BPT, BBA, MBA, BCA, GNM, etc. Admissions in the university are both merit and entrance-based depending from course to course. 

Sai Tirupati University offers the following courses to candidates:-

• BA
• MA
• MBA
• BBA
• BPT
• GNM
• BPharma, etc. 

The angular acceleration direction is given along angular velocity or opposite to angular velocity depending upon whether angular velocity magnitude is increasing or decreasing and this direction remains along the axis of circular motion.

Potential energy stored in the spring $=\frac{1}{2}{\mathrm{k}\mathrm{x}}^2$

$\begin{array}{rr}& \text{Now}\frac{1}{2}\mathrm{k}(2{)}^2=\mathrm{U}\\ & \&\frac{1}{2}\mathrm{k}(8{)}^2=U^{\mathrm{\text{'}}}\mathrm{}\text{(say)}\\ & \Rightarrow U^{\mathrm{\text{'}}}=\frac{64}{4}U=16U\end{array}$

${\mathrm{h}}_{\text{max}}=\frac{{\mathrm{u}}^2{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta }{2\mathrm{g}}=\frac{280\times 280}{2\times 9.8}\times \frac{1}{4}$

=1000 m

Initial velocity $=-v\hat{j}$

Final velocity $=-v\hat{j}$  

Change in velocity $=v\hat{i}-(-v\hat{j})$

$=v(\hat{i}+\hat{j})$ Momentum gain is along $\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}$

$\Rightarrow$ Force experienced is along $\stackrel{\mathrm{ˆ}}{\mathrm{i}}+\stackrel{\mathrm{ˆ}}{\mathrm{j}}$

$\Rightarrow$ Force experienced is in North-East direction.