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@Harsh Shrivastava
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Irodov(that's too good a book as I felt for rotational mech),and hc verma is good as well.For more tough prob you can refer to 300 creative physics problems by Laszlo Holics.

@A Former Brilliant Member
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Then try this "A guide to physics problems"... May be the second part be a bit tough as its Mostly Quantum Electrodynamics,Quantum Mechanics and all those stuffs but part 1 is good...... You will surely like it..And also try the integral I have posted

Use the fact that since we assume strings to be inextensible, velocity of M at bottommost point will be only in horizontal direction, and use constraint relation between the velocity of M and m by equating their velocities along the thread connecting them.Finally use energy conservation.

Which part...I guess straight lines and circles you must be well acquainted with......Coordinate needs some heavy practice....Formulas and all those stuffs should be known all the time..?As such I don't follow any special book only fiitjee package.

@Harsh Shrivastava
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OK ok.That will lead to answer of $I=1/ab$.If a and b are of opposite sign then the integrals value is Negative $(ab<0)$.But the integrand be a square the area can never be negative.How do you resolve this...Here you integrating through Singularities.Pretty much the same thing as $dx/x^2$ integral from (-1 to 1) isn't -2.The function blows up when x-->0...That's why its a Feynman's Puzzle...Try using limits in integrand.

Hey in your q you talked abt Max area....I think it should be min.I did it as the req point of intersection is $apq,a(p+q)$ where the points P and Q are $(ap^2,2ap)$and $(aq^2,2aq)$.Then the cond yields $apq=-1$.Vertex being $(0,0)$The Area of $PCQ$ is $a^2|pq(p-q)|$.Puting $pq=-1/a$ And later applying AM-GM yields the ans as $Min=1$.$a=1/4$ from $y^2=4ax$

@Harsh Shrivastava
–
Ya I see the problem has been changed.Thanks....I was surprised because of the title "its right"....I accepted my ans to be wrong....

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## Comments

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TopNewest@Spandan Senapati

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Ya I know this q.Follow Harsh s solution.And you will get it.And if any problem then tag me and I will post it.Ok

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Hey Spandan ,AE can you guyz tell me good source for practicing mechanics questions, especially rotational mechanics?

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I got it. Will post a solution by night.

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Use the fact that since we assume strings to be inextensible, velocity of M at bottommost point will be only in horizontal direction, and use constraint relation between the velocity of M and m by equating their velocities along the thread connecting them.Finally use energy conservation.

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Thanks Harsh & Spandan.

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Any suggestion on coordinate geometry, I am very weak in it?

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TMH and archive will suffice.

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Which part...I guess straight lines and circles you must be well acquainted with......Coordinate needs some heavy practice....Formulas and all those stuffs should be known all the time..?As such I don't follow any special book only fiitjee package.

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Hey you like integrals.I have a good one1)$int1/(a+b(1-x))^2dx$with x varying from $0-1$.This is a Feynman's Puzzle.Looks easy but needs some more....

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use the transformation x-> 1-x and then differentiate the function 1/(a+bx) wrt a and then done.

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$I=1/ab$.If a and b are of opposite sign then the integrals value is Negative $(ab<0)$.But the integrand be a square the area can never be negative.How do you resolve this...Here you integrating through Singularities.Pretty much the same thing as $dx/x^2$ integral from (-1 to 1) isn't -2.The function blows up when x-->0...That's why its a Feynman's Puzzle...Try using limits in integrand.

OK ok.That will lead to answer ofLog in to reply

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Hey in your q you talked abt Max area....I think it should be min.I did it as the req point of intersection is $apq,a(p+q)$ where the points P and Q are $(ap^2,2ap)$and $(aq^2,2aq)$.Then the cond yields $apq=-1$.Vertex being $(0,0)$The Area of $PCQ$ is $a^2|pq(p-q)|$.Puting $pq=-1/a$ And later applying AM-GM yields the ans as $Min=1$.$a=1/4$ from $y^2=4ax$

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Yeah you are right, my mistake, edited!

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