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New answer posted

2 months ago

0 Follower 11 Views

R
Rahul Verma

Contributor-Level 9

Suppose it is 187 and 61 percentile in JEE Mains 2025 January Exam and it seems that your rank would be between 173239 to 326517. With such strength of merit, it would be extremely difficult to secure an admission in the highly ranked government colleges such as NITS or ILCA IIITs. But you may get an opportunity with some GFTIs or colleges (depending on the branch preferences and quota) in the state government college, if you belong to a particular category and depending upon your home state quota.

New answer posted

2 months ago

0 Follower 4 Views

R
Rahul Verma

Contributor-Level 9

Of course, it is extremely possible to learn AI without JEE or NEET. You may also consider B.Sc. or BCA courses with Artificial Intelligence for specialization, which are usually taken on the basis of 12th marks or your result in the exam, to be conducted by some universities. Moreover, very many online sites provide diplomas, certificates, and post-graduate programs in AI and Machine Learning, where there is no specific eligibility and that one does not need such competitive examination like the ones we have in universities. Emphasize on developing a solid background on mathematics and programming.

New answer posted

2 months ago

0 Follower 8 Views

P
Prayas Pachauri

Contributor-Level 6

With a 94 percentile in JEE B.Arch (Paper 2), you are in a solid position for admission to several architecture colleges.General catagory cutoff for JEE Advance is around 93.2 percentile or higher. If eligible, you could aim for IIT Roorke or IIT Kharagpur, which offer B.Arch via AAT. Your expected rank at 94 percentile is around 90,000, which aligns well with many NITs and SPAs, especially if you have a home state quota or belong to OBC/SC/ST/EWS categories, where cutoffs are significantly lower.

New answer posted

2 months ago

0 Follower 6 Views

D
Dua naqvi

Contributor-Level 9

No, it is unlikely to get top branches of VNIT  like computer Science or electrical branches you may get metallurgy branch maybe from the score of 95.90 in JEE Mains for general unreserved or EWS female student. It doesn't depend on your home state.

Thank you

New answer posted

2 months ago

0 Follower 17 Views

V
Vaibhav Mishra

Contributor-Level 8

Yes a student can get Nit Suruthkal's EEE branch with a score of 80 percentile in JEE Mains 2025 exam because the students are from home state and their category is St the excepted rank for admission in EEE branch in this college is 30000 to 50000 and if you are St category caste and you are from that state then it is a golden chance for you to getting admission in Nit Suruthkal's 

New answer posted

2 months ago

0 Follower 7 Views

R
Rahul Verma

Contributor-Level 9

To avail affordable B.Tech CSE in Andhra Pradesh and Telangana, some of the popular government colleges are Andhra University College of Engineering (Visakhapatnam), JNTUK University College of Engineering (Kakinada), Sri Venkateswara University College of Engineering (Tirupati) and University College of Engineering, Osmania University (Hyderabad). These are characterized by cheaper fees (usually say INR 1.78 Lakhs - 2 Lakhs in total in case of B.Tech) and good academic profile. In case of private ones that charge average rates and offer alright placements, consider such colleges as Velagapudi Ramakrishna Siddhartha Engineering College

...more

New answer posted

2 months ago

0 Follower 6 Views

R
Rahul Verma

Contributor-Level 9

A 96 percentile in JEE Mains will put you in the general percentage of score between 33000 and 48000. The score is not enough to get an admission in best IITs or most popular faculty members (such as CSE ) in best NITs. But you can easily get into mid-tier NITs (e.g. NIT Agartala, NIT Raipur, NIT SIlchar, NIT Srinagar) in core subjects such as Civil, Mechanical, Chemical, or even Metallurgy. You may even secure a place in newer (e.g. IIIT Manipur, IIIT Bhagalpur, IIIT Dharwad) institutes to other branches such as ECE. This is also dependent on your category and quota of home state.

New answer posted

2 months ago

0 Follower 8 Views

N
nehanth Reddy

Contributor-Level 7

  1. Amity University
  2. Bharati Vidyapeeth's College of Engineering
  3. Maharaja Agrasen Institute of Technology
  4. Maharaja Surajmal Institute of Technology

New answer posted

2 months ago

0 Follower 14 Views

S
Sambit Gumansingh

Contributor-Level 9

sin (A + B) = sinA cosB + cosA sinB

sin (A - B) = sinA cosB - cosA sinB

cos (A + B) = cosA cosB - sinA sinB

cos (A - B) = cosA cosB + sinA sinB

tan (A + B) = (tanA + tanB) / (1 - tanA tanB)

tan (A - B) = (tanA - tanB) / (1 + tanA tanB) 

Double Angle Formulas:

sin 2? = 2 sin? cos?

cos 2? = cos²? - sin²? = 2cos²? - 1 = 1 - 2sin²?

tan 2? = 2 tan? / (1 - tan²? ) 

Triple Angle Formulas:

sin 3? = 3 sin? - 4 sin³?

cos 3? = 4 cos³? - 3 cos?

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