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Also , remember one thing - ' In Physics any question can be solved , if your concepts are perfect '. So 1st try to perfect your concepts . Try to solve the question more than asked , that is the question may be simple but their may be many things in it to think upon , using which good questions can be framed . :)

I believe in cengage as 1st priority to gain confidence in basic , medium and some hard problems . Then I complete all q from it ( approx 200 to 300 q per chap) . Then I move to solve harder problems , I solve Irodov , Krotov , Brilliant , and I also try to make tougher questions on a topic which I didn't face in a book so as to check my knowledge and also to gain perfection! . :)

If you see geometry carefully , A0A1:A0A2:A0A3 = 3 : 5 : 6 .

How ? , see this - mark pts. Between A and A1 as B1 (above ) and B2 ( below) , similarly between A1 and A2 as B3 and B4 and between A2 and A3 as B5 and B6.

By geometry , angle A1A0B1= A2A1B3 = A3A2B5 = let x .

A0A1 = 2 × 3lcosx, A1A2 = 2 × 2lcosx, A2A3 = 2 × lcosx. And hence the ratio , A0A1:A0A2:A0A3 = 3 : 5 : 6 .

Differentiating both sides w.r.t t , we get v1 : v2 : v3 = 3 : 5 : 6

In order to find lengths AOA1 , AOA2 I Used slightly different approach . I firstly used the fact that diagonals of rhombus bisect each other and then the property of parallogram sum of squares of sides = sum of squares of diagonals .

@Harsh Shrivastava
–
For physics i suggest you to do HCV First that will help you to grasp concepts. Then move to fiitjee study material and Then Finally Archive . Doing this wont leave any concept behind . Then you can keep one reference book for harder problems of JEE Level (like I Prefer DC Pandey) . This will be more than enough and i guess it will be difficult to do all these simultaneously hence i did DC pandey after 2-3 chapters after i got sufficient time. i give more emphasis on packages and archives.
And finally solve some problems of brilliant to challenge yourself.

@Harsh Shrivastava
BTW I Did not got 23 in AIITS :p (as you told to look at questions of aniket)

@Prakhar Bindal
–
Hello Prakhar. I wanted just one book for JEE Advanced Physics for difficult problems as we are almost through with the syllabus and need practice. Which according to you is the best and contains hard problems of every topic like here on brilliant.

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestBtw , which book ? @Harsh Shrivastava

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Any tips for problem solving in physics?

I have seen your name in recent solvers of many hard physics problems.

Please share some tips, thanks!

Also congrats for an amazing air 23 in aiits.

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It is from a magazine named "Physics for you".

Btw can you please tell what sources you use for physics problems? Which books?(not for theory but

goodproblems.)Thanks.

@Aniket Sanghi

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Also , remember one thing - ' In Physics any question can be solved , if your concepts are perfect '. So 1st try to perfect your concepts . Try to solve the question more than asked , that is the question may be simple but their may be many things in it to think upon , using which good questions can be framed . :)

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I believe in cengage as 1st priority to gain confidence in basic , medium and some hard problems . Then I complete all q from it ( approx 200 to 300 q per chap) . Then I move to solve harder problems , I solve Irodov , Krotov , Brilliant , and I also try to make tougher questions on a topic which I didn't face in a book so as to check my knowledge and also to gain perfection! . :)

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If you see geometry carefully , A0A1:A0A2:A0A3 = 3 : 5 : 6 .

How ? , see this - mark pts. Between A and A1 as B1 (above ) and B2 ( below) , similarly between A1 and A2 as B3 and B4 and between A2 and A3 as B5 and B6.

By geometry , angle A1A0B1= A2A1B3 = A3A2B5 = let x .

A0A1 = 2 × 3lcosx, A1A2 = 2 × 2lcosx, A2A3 = 2 × lcosx. And hence the ratio , A0A1:A0A2:A0A3 = 3 : 5 : 6 .

Differentiating both sides w.r.t t , we get v1 : v2 : v3 = 3 : 5 : 6

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Thank you very much bro!

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@Aniket Sanghi @Prakhar Bindal

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Aniket has done absolutely correct!.

In order to find lengths AOA1 , AOA2 I Used slightly different approach . I firstly used the fact that diagonals of rhombus bisect each other and then the property of parallogram sum of squares of sides = sum of squares of diagonals .

That will yield you same answer @Harsh Shrivastava

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Same questions to you that I have asked aniket, please reply thanks @Prakhar Bindal

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@Harsh Shrivastava BTW I Did not got 23 in AIITS :p (as you told to look at questions of aniket)

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