how to prepare for iit jee?

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work hard ..............

3 years ago

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is optandon for – physical chemistry morrison and boyd for – organic chemistry ml khanna for- mathematics hc verma for – physics
 
 
dear student http://www.askiitians.com/important-books-iit-jee.aspx , read this article you will find all the good books mentioned , it will be very helpful for you.
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Sher Mohammad one month ago
Sir What is rules for branch change in IIT kharagpur? how many students can change the branch? Is there any reservation criteria ?
 
 
I am attaching you the file regarding the branch change terms and conditions. There are many clauses attached with branch change that can effect the comfortable branch change process in IIT...
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Aziz Alam one month ago
how to study for jee in 8 months as in my case...... i am really confused plz guide me i have dropped a year for jee 2015.shoul i join fiitjee and prepare with them. what are the important...
 
 
For JEE one shouldn’t compromise the quality of study i.e. you have to make sure that the topic which you are studying should be perfect in context to JEE.You should be able to solve...
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Aziz Alam one month ago
a line x=k intersects graph of y=log 5 x and the graph of y=log 5 (x+4) .the distance b/w points of intersection is 0.5 , given k=a+(b) 1/2. .where a & b are integers..the value of a+b...
 
 
Dear rahul given distance b/w points of intersection is 0.5 ie log 5 (k+4) -log 5 (k) =1/2 or log 5 (k+4)/k =1/2 or (k+4)/k =√5 k=4/(√5 -1) =√5 +1 so a=1 and b=5 a+b=6...
 
Badiuddin askIITians.ismu Expert 4 years ago
Sir, I am gettting 86-87% in cbse boards.How much minimum marks is needed in jee mains 2014 to get admission in mnnit with cse
 
 
You have to score above 250 in jee main to get selected in top nit
 
rishikesh mukkagalla 4 months ago
 
To get enter into mnit you have to score atleast 240 marks.
 
Bharat Singh 4 months ago
 
MORE THAN 250 AND YOUR EXPECTED RANK WILL BE AROUND 10K
 
aman kumar jha jha 4 months ago
DEAR SIR , 1. calculate the force of attraction between the sun and the earth . mass of earth is 6 x 10 24 kg and that of sun is 2 x 10 30 kg. the average distance between them is 1.5 x 10...
 
 
F = Gm 1 m 2 /r 2 => F = (6.67*10 -11 * 6*10 24 *2*10 30 )/(1.5*10 11 ) 2 F = 3.5573 * 10 22 N Thanks & Regards askIITians Faculty
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Aziz Alam 7 days ago
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