Colleges in Hyderabad

No, you won't. I'm afraid the free seat you were expecting lies on the rank of 1-250 rank in TS-ICET.

Check this link to know all about placements at Hyderabad campus:
http://www.bits-pilani.ac.in/hyderabad/placementsOverview
Check all the tabs for complete information.

Hi Shiva,
The total fees charged for B.Arch in Sri Venkateswara College of Architecture is INR 2.25 LAKHS. Per year fees would be approximately INR 45,000/-
Regards

You can check out the link below to confirm whether you can get the desired college or not:
TS EAMCET: https://www.shiksha.com/b-tech/resources/ts-eamcet-college-predictor
For more details, check link below :
https://www.shiksha.com/university/jawaharlal-nehru-technological-university-hyderabad/exams/ts-eamcet/counselling

Hi Santosh,
The allotment list for CLAT is available on the CLAT website, please check.
Last year for SC, the cut-off started at Rank 1704 in round 1 and closed at 2670 in round 4. So, be hopeful.

Yes, admissions are open you can apply.

As Admission is through web counseling based on the rank obtained in Telangana State Engineering, Agriculture & Medical Entrance Test TS EAMCET its difficult to get admission without TS EAMCET. You can try in management quota
but Management quota admission and fee cost is not advertised openly by any of the colleges/universities across India as being a paid seat it totally depends on the discretion of the college authorities. Please contact the college authorities for all the details.

Hi,
On a personal note, I'd advise that stay updated with the current affairs because it's an online paper so current affairs of previous day can also come as a surprise. Another stress more on the legal section and try to solve as many principles- facts- problems. For maths and logical reasoning solve as many questions as you can and try and finish all the variations. For English please develop a habit of reading Newspaper daily, specifically The Hindu and in that the Editorial. Work on basic grammar and be confident. I think that will be enough.