School of Business, UPES Dehradun All Questions

Hey, hope you are doing well!

I went through the website and due to some technical issue the syllabus link is not working. But I am still adding the link here, please check it time to time if it opens.

MCC syllabus (https://www.mcc.edu.in/images/IQAC/NAAC/Criteria1/M.Sc._Plant_Biology__Plant_Biotechnology.pdf)

Plant biology and botany are same. Both of these are studies on plants. Plant biology and botany focuses on the total plant kingdom.

But plant biotechnology consists of introduction of desirable traits into plants through genetic modification. So plant biotechnology is different from plant biology or botany.

And yes its definitely a good option after BSc in Zoology. But only if you are actually interested in the subjects and the study area.

Always pursue a course only if you fully are interested in it.

All the best!

Hello student Bunts community fall under OBC category in Central list. So while filling the Jee application Form you need to select OBC as your category.

Eligibility - Candidates must passed or appearing class 12 in Physics chemistry and Mathematics with minimum 75 percent marks aggregate from a recognized board.

Feel free to comment if you've any doubt

Good luck

Hi Dear,

Following is list of few private colleges offering BAMS course in Maharashtra:

• Jupiter Ayurved Medical College, Nagpur,
• Ayurved Mahavidyalaya & Seth R. V. Ayurved Hospital, Sion
• BHARATI VIDYAPEETH COLLEGE OF AYURVED - [BVCA]
• DR. DY PATIL COLLEGE OF AYURVED, PUNE

For more visit following link:

https://medicine.careers360.com/colleges/list-of-bams-colleges-in-maharashtra

Best of Luck

Hello student. 60610 rank is a bit high. But since there is a reservation in 2A category, you might have chances to get a seat. Since the competetion is going higher, I suggest you to apply for pother colleges too to be on the safer side. If there is any state quota then your chances might improve.

hi,

see it would be best if you directly go to the institute and discuss the reasons with the college authority. however what ever the reason, due to covid 19 pandemic its getting quite delay to process the admission related steps and works. however you may also email your concern to the registered email address of the Institute.

hope you may understand.

Hello Aspirant,

Yes it do comes under NEET.

Perhaps, the Commissioner for Entrance Examinations (CEE) Kerala confirmed the list of medical and allied courses that  come under  the ambit of  NEET  -UG. Candidates who are planning to study courses such as  B.Sc Agriculture  ,  Bachelor of Fisheries  Sciences,  B.Sc  Forestry etc will be required to clear the  NEET  exam.

This may help.

Happy learning!

Hello

Your admission in MBBS is solely dependent on the marks obtained by you in NEET examination there's no role of marks obtained by you in class 12th board examination in getting a seat in a medical college.

The only role of your class 12th board examination marks is to make you eligible for neet examination and that's it .

So, you should prepare well for your neet examination as admission completely based on it, go through NEET mock test , previous year papers , it will let you know your preparation level and and weak areas so that you can work on them and perform well in your examination .

Thankyou

Generally cutoffs changes every year it depends upon various factors like, difficulty level of the entrance exam, number of students applying for the exam, performance of the students in the exam, etc.

But in general if you consider previous year cutoffs then you can easily get CSE in Reva as the closing cutoffs was around 25300, last year.

All the best!!

V=0+αt1

t1=V/α

Similarity, t2=V/β

or t1+t2=V/α+V/β

=αβV/(α+β)

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 .