Dear Candidate,

We know that, the rate of effusion is inversely proportional to molecular mass, i.e.

r1 / r2  = √ ( M2 / M1 )

And, the rate of effusion is ratio of the volume effused to the tima taken

i.e.  V1 * t2  / V2  * t1  =

Hi Aspirant,

The certain gas which will take three times as long to effuse out as helium its molecular mass will be 36u since volume is same so 3/1 will be equal to root square of Mw2/4.On calculating the mass is 36u.This is based on calculations by basic formulas and

Hello Aspirant,

There might be some error in the application form due to which it is not showing complete details.

See Staff Selection Commission has extended the last date of application to December 19th 2020 and it does not provide correction facility in application form of  SSC CHSL so if

Hello,

Ashoka school of business, Hyderabad a private college. It was established 2014. And fees will be 1lakh. MBA course for 2 years. But due to research there are no placements offer, but college invited some reputed companies they are provided students training in the institute. College faculties are very

Hi there !

Joint Entrance Examination (Main), or JEE Main will be conducted, four times a year from 2021 in 13 languages including English instead of two languages. You can appear in any one or in all the four sessions.

In JEE Main 2021, you can appear for the exam

Hello Aspirant,

See from this year National Testing Agency will conduct JEE-Main four times, first in February, second in March, third in April and fourth in May  2021 and any candidate irrespective of his category can appear in either of the attempt or maximum of all four attempts.

On December

Hello,

Earlier NTA director Vineet Joshi said that "they are thinking to conduct the NEET twice a year" but there are still no official information released regarding NEET 2021 yet. So, there is no confirmation that NEET will be conducted two times a year. Generally, NTA releases the information brochure

As of there is no update on NEET being held twice. And to be eligible for NEET, you should have completed your 12th, passed in all subjects and secure a minimum aggregate of 50% in Physics, Chemistry and Biology. It is 50% for General Category, 45% for PWD and 40%

Dear student,

The equation is given as :-

x^2 + bx + c = 0

Now, it is given that one root of the equation is n times the other root. So, let one of the root be k.

So, the other root will become nk.

Now sum of root

Hello,

According to the official announcement, if one candidate appear for more than one time (maximum 4 times) in the upcoming JEE 2021 then the highest score obtained by the candidate will be considered for evaluation.

So, if you appear for 4 times then the highest marks obtained by you