did it works plzz reply am also going thorugh this prb ?

Hey Kratos! MHT CET or Maharashtra Common Entrance Test is a state-level entrance exam conducted for admission into Engineering and Allied Health Science courses offered by colleges of Maharashtra. You can check rank filling according to the percentile secured by you. 

https://cetking.com/mah-mba-mms-cet-result-analysis/

hello aspirant,

the marks conversion criteria should be present on boards website. you can download it from the website and signed it by the principle of your institution. if that not availb;e directly go to the principle of your institution and explain the issues and request him to give you

Hello,

There might be some problem. This can be sorted out if you contact to the board as fast as possible. There have made the website so they know to handle the problems. So put a mail or make a call and share your issue with them and ask them

CBSE gives your result of 10th class in CGPA. However shortlisting is done in terms of a unique and universally accepted method of calculating your marks.

10th document conversion is basically another sheet which has to be given by the school which indicated marks conversion form CGPA to percentage by

Hey Anvay! Definitely yes a Central Board of Secondary Education (CBSE) student can register and is eligible for giving Maharashtra Common Entrance Test (MHTCET) whether his or her parents are born in maharashtra or not. But for home state reservations you need to fulfill the criteria. Hope this may help

Maharashtra Common Entrance Test is conducted by DTE in online (CBT) mode for the candidates to take admission in various colleges in Maharashtra in their B.Tech, B.Pharm, LLB, MBA, etc. Candidates who are eligible as per the criteria determined by the exam conducting authority have to fill the application form

Hi there Aspirant. Hope you're having a great day. 

The important dates for MHT CET is-:

Hey Kanhaiya!

Displacement= a/2

Total energy=1/2 m* (omega)^2* (a)^2 ....(1)

KE when displacement= x is

1/2 m* (omega)^2* [(a)^2-(x)^2]

1/2 m* (omega)^2* [(a)^2-(a/2)^2]

3/4(1/2*m* (omega)^2* (a)^2) .....(2)

By dividing 2 by 1, we get: 3/4

Therefore total fraction is 3/4

Sorry to inform you but it's a career counseling website,

Hello Swati,

The lowe and upper age limit of the MHT-CET Exam is 17 and 25 years respectively. And a candidate can give this Exam only three times during this interval if age 17 years to age 25 years.

Now, if you are under 25 and have given it only