It's a Question of your Career!

Hello student ,

Here you have not mentioned for which have forgot to download the application form . That's why I am unable to guide you to download your application form . So first of all ask your question again mentioning some details about your application form so that I can help you to download your application form which you have forgot to download .

Hope this information will help you !

Good luck !

Hello,

Since you have not mentioned your category, assuming that you are from general category, with a score of 520 it is not possible to a state quota seat in government colleges in Jharkhand. However there are chances for private colleges for state quota seats.

There are also chances for BDS course in government and private colleges both.

To get the list of the colleges where you have chances of admission use our college predictor link -

https://medicine.careers360.com/neet-college-predictor?icn=QnA&ici=qna_answer

Hope this helps!

Thank you.

Hello Aspirant,

According to the marks you have mentioned in the question your percentile would be in the range of 99.04-99.24 and your rank would lies in between 6854-8504. On the basis of this rank and percentile you can easily get into top NITs like NIT Uttarakhand,NIT Jaipur,NIT Jamshedpur and so on as per previous year's cut-offs. You can check out the cut-offs for top NITs from the link given below :-

https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-cutoff-for-top-nits/amp

Note that the cut-offs are bound to change evry year depending upon various factors and you might be qualifying for JEE-Advance so you be prepared for that also.

You can also use JEE-Main College Predictor to predict the college :-

https://engineering.careers360.com/jee-main-college-predictor

I hope this information helps you.

Good Luck!!

Hello student

TS PG CET 2019 cut off marks is at least of total marks 25% that is 30. Total marks for TS PGCET is 120. The candidates have to to qualify in 10 plus 2 level exam and Graduation with 50% marks with science or Technology for MTech environmental scienc e for UG or PG courses.

Cut off rank For M  Tech environmental science of Osmania University is near about 8265 around.  There is 18 seats for MTech environmental science. Bed course fee is 1 lakh per year. The courses of 2 years. The improvement package is off  1;5 lacs of rupees per annum

The National Education Policy 2020 (NEP 2020), which was approved by the Union Cabinet of India on 29 July 2020, outlines the vision of India's new education system.The new policy replaces the previous National Policy on Education, 1986. The policy is applicable  for elementary education to higher education .

But while coming to topic the new educational policy will be implement entire NEP policy  by 2040.The government till then slowly implement the policy key points one by one, while taking care of funds to maintain the NEP proposal. Shortly after the release of the policy, the government clarified that no one will be forced to study any particular language.

*So the new educational policy will not implement completely till 2040.

Hope it would be helpful.

Hello Aspirant,

For Biology notes and questions ,do check the link below to get the notes. you have to just create the account on the website ,register yourself and then you can access the material available there.

https://www.genextstudents.com/studyzone/Board/UP-BOARD/12-%E0%A4%B5%E0%A4%BF%E0%A4%9C%E0%A5%8D%E0%A4%9E%E0%A4%BE%E0%A4%A8/%E0%A4%9C%E0%A5%80%E0%A4%B5-%E0%A4%B5%E0%A4%BF%E0%A4%9C%E0%A5%8D%E0%A4%9E%E0%A4%BE%E0%A4%A8/607

Hope it helps

Hello,

a)Distance from earth's centre = 6400 + 640=7.04 x 10^6m

Now,

mv^2/r= GMm/r^2

V =   √GM/r  = √56.8 x 10^6

= 7.56 x 10^3 ms ^-1

(B) Total energy after 1500 rounds = Initial energy - Energy loss

= - GMm/2r- 1.42 x 10^5x 1500

=(- 6253-0213) x 10^9

= -6.466 x 10^9J

So, r = - GMm/- 6.466 x 10^9

= 6808 km

Its height above the earth's surface = 6808-6400

= 408 km

(C) Average retarding force, |F |=| dU/dr|

F = Energy loss/Distance

= 1.42 x 10^5/ 2 x 3.14 x 6.924 x 10^3 = 3.3 x 10^-3N

Hope it helps

Hello,

You can use the Careers360 Rank Predictor Tool to predict the rank & percentile.

Also you can use,

JEE Main college predictor tool to predict some good colleges for your percentile.

https://engineering.careers360.com/jee-main-college-predictor?icn=QnA&ici=qna_answer

Good luck!