It's a Question of your Career!

You get seats based on your KCET merit score so if that score is less, then you wont get a decent college. If you are talking about failing in boards, then if this happens then too you wont be able to  get into colleges through the KCET score. Because as per the KCET eligibility criteria, you need a minimum of aggregate 45% and above in your 12th boards to fulfill the criteria (for general category) and 40% and above for reserved category.

Refer to the site below for more information about the same...

https://engineering.careers360.com/articles/kcet-eligibility-criteria

Best wishes. Thank you.

Hello Aspirant,

Some top government colleges for science which takes direct admission based on your 10+2 score are enlisted below :-

• Lady Shri Ram College (Delhi University)
• Hansraj College (Delhi University)
• Nizam College (Osmania University)
• Stella Maris College (University of Madras)
• Fergusson College (Mumbai University)

And many more.

I hope this information helps you.

Good Luck!!

All subjects are good if you study with happiness. I am also a B.Com (//B.Com) Student I am having proud for studying this, so all of my subjects are great.

hello Maria,

ICAI conducts the CA exams list in India consists of 3 names – CA Foundation, IPCC and CA Final. Candidates need to clear the entire CA exam list in order to become a professional in this field.

where as, BCom is a 3-year Undergraduate program that focuses on subjects related to commerce, Economics, Business Law, Accountancy, Taxation and finance.

If you are a disciplined and focused kind of student, I suggest you to pursue CA and bcom simultaneously. CA foundation and 1st semester bcom are almost the same but the only thing is CA is a bit difficult compared to bcom as far as CA Foundation is concerned. BAF will absolutely right choice if we want to pursue CA with b.com

BCom with CA is divided into six semesters. Various topics such as financial accounting, economics, company law, corporate tax, auditing, business management, etc., and many more are present in BCom Syllabus to increase the skill threshold of the student.

for more information please visit the following link

https://dqxeclau.top/

Hope this will help you a lot,

Thank you

Hello,

Oil and Natural Gas Corporation (ONGC) has released a notification for ONGC recruitment through GATE 2020. To be eligible Candidates need to score minimum 60% marks in the undergraduate engineering programmes, to be eligible to apply for the recruitment. If you get a good score in GATE and have 60% marks in your UG course then you will be eligible for ONGC. Please click the below link to know more.

https://engineering.careers360.com/articles/ongc-recruitment-through-gate

Hello,

Yes it is possible to give fresh registration for every round in Manipal Institute of technology counselling. Manipal Academy of Higher Education, MAHE has started MET 2020 counselling registration in online mode on August 20 which is available till August 24. Students who qualified the entrance exam can participate in MET counselling and seat allotment 2020. After every round the fresh registration opens up for two days. Please click the below link to know more.

https://engineering.careers360.com/articles/met-counselling-seat-allotment

Hindu College in delhi

DAV College in Abohar

Jaipur National University

Hello!

Let us consider the first statement. The train starts from rest and acquires a speed v. So, initial velocity is 0 and final velocity is v.

We know that for uniform motion,

v = u + at. Here a = α .

Therefore,

v = u + α t

Since u = 0,

v = α t

α = v/t

Let t = t1

Thus

α  = v/t1 ..........(1)

Also, v^2 - u^2 = 2as

Since a = α, u = 0 and s = s1 is the distance travelled

v^2 = 2 α s1 ...........(2)

From (1) and (2), equating the values of  α, we get

s1 = v t1/2 ............(I)

Before it comes to stop,

v = u + at

Since it comes to stop, v =0; also here, a = β

Thus

0 = u + βt

Putting t = t2

0 = u + β t2

Now, the final velocity when the train starts from rest and the initial velocity before the train comes to rest are the same.

That is, u = v

Therefore,

0 = v + β t2

β = -v/t2 .............(3)

Also, v^2 - u^2 = 2as

Since a = β, final velocity v = 0 (as the train comes to a stop) and s = s2 is the distance travelled,

0 - u^2 = 2 β s2

We know that u = v. Thus,

-v^2 = 2 β s2 ........(4)

From (3) and (4), equating the values of β , we get

s2 = v t2/2 ............(II)

The average velocity can be calculated by

Avg. Velo. =  Total Distance Travelled/Total Time Taken

=  s1+s2/t1+t2

Substituting the values of s1 and s2 from (I) and (II),

Avg. Velo. =  (vt1/2 + vt2/2) / t1+t2

=   v/2 (t1+t2) /(t1+t2)

= v/2

Thus, the average velocity is v/2 .

Hello Riza,

There are many exams you can apply for after 12.Below some of the exams which you can apply as an medical student.

1.Joint Entrance Examination (JEE) Main

2.JEE Advanced

3.State Level Engineering Entrance Exams

4.National Eligibility Cum Entrance Test (NEET)

5.Indian Maritime University Common Entrance Test

6. Defence

7. All India Entrance Examination for Design (AIEED)

8.Indian Statistical Institute Admission