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I would suggest you Go for A text book of Probability by K C Sinha
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momentarily means "at that instant" .... meaning as in your question,,,,at the time when the direction of velocity reverses => at the time velocity reverses its direction it comes to an instantaneous zero in magnitude as well as vectorially...since it does not poses any scalar nor any vector quantity at that moment of time..... if you are still unclear then do reply i will try to xplain again.... As your question is concerned its a simple question on SHM.... use the general equation of SHM { x = A sin (wt + q) } and you can solve it since u hav three unknowns and you hav three instances of time.......
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Assume car A to be at rest....this implies CAr B has decelleration of 4 m/s2 apply: S = ut + at2/2 where u = 1 m/s a = -4 m/s2 S = -10 m => -10 = 1xT - 4T2 /2 => 2T2 - T - 10 = 0 => 2T2 +4T - 5T - 10 = 0 => 2T ( T + 2) - 5 (T+2) = 0 => T = -2 (neglectd) or T = 5/2 sec (ans)
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Relative velocity of A wrt B = V – Vsin30 = V/2 Time = distance / speed Time = 2a/V In reality each particle will follow a curved path and eventually meet at the center of the hexagon. 
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Figure shown is immediately after collision: 
Conservation of Linear momentum in horizontal direction: mu + mu = - mu1 - mu1 +2mV => 2mu = -2mu1 + 2mV => u = -u1 + V .........................(1) Velocity of seperation = e x (Velocity of approach) (takin one particle at a time) u1 + V = e (u) => u1 + V = u .....................(2) similar would be for other particle i.e. u1 + V = u adding 1 and 2: 2V = 2u => V = u
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In figure FG=GI=IA=AE=EC=CH= X Thus X = 4/6 = 2/3 ft BE= Y = 2 ft (a) Displacement of queen from center to front Edge = AB AB = root (X2 + Y2 ) AB = 2 (root 10) / 3 (b) Displacement of queen from front Edge to hole = BD BD = 2 AB BD = 4 (root 10) / 3 (c) Displacement of queen from center to hole = AD AD = root (22 + 22 ) AD = 2 root (2) 
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As i cud understand the problem....the soln is in figure.... requird answer = root [ (pi r) ^ 2 + (2r)^2 ]
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faliure of a structural member when subjected to highly compressive load, it is also due to members elastic instability.
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price of 1st bike = x price of 2nd bike = (80000-x) selling price of one bike = 1.15 x ................................1 selling price of second bike = 0.9 (80000-x) .............................2 difference of eqn 2 and eqn 1 is given as 1800 => 1.15 x - 0.9 (80000-x) = 1800 => 1.15 x - 72000 + 0.9 x = 1800 => x = 36000 therefore othr bike 44000
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Here i hav attached the graph for | log |x| | .... which clearly indicates it is continuos except at X=0
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Distance covered in last t seconds wud be same as the distance covered when it starts descending from its maximum hieght..... S = ut + gt2 / 2 so u = 0 at max hieght S = gt2 / 2 I think i am right conceptually,,,,,bt correct me if i am wrong
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odd digits = 3 3 5 5
even digits = 2 2 8 8 8
O E O E O E O E O
no of even places = 4
no of odd places = 5
ways = (4! / 2!*2!) * (5! / 2!*3!) = 60
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answering second question,,,u can take clue for the 1st one from this one....
In how many ways can the word 'arrange' be arranged such that neither 2 'r' nor 2'a' are allowed to come together ?
total number of letters= 7
total ways of arranging "arrange" = 7! / 2!*2! = 1260
now take 2 "r" and 2 "a" in one packet each...then total ways of arranging these 2 packets and the remaining 3 letters = 5! = 120
now ways in which 2 "r" and 2 "a" do not come together = 1260 - 120 = 1140
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Total number of ways of arranging = 9! = 362880
now we club best and worst paper together in one packet, and the remaining are 7.
so total ways of arraning 7 papers and the packet containing best and worst = 8! x 2 = 80640 ............ this "x 2" is because worst and best can arrange themselves in 2 ways.
now in which they do not occur together = 9! - (8! x 2) = 282240
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according to me the answer is coming out to be 840...so if there is a correction thn do tell me.
i did it like this
Fixing C in 7th position:
1 2 3 4 5 6 7
X X X X X X X
A B C .......... B in 1 way and remainng in 4!
A C ......... B in 2 ways and remaing in 4!
A C ......... .B in 3 ways and remaing in 4!
A C .......... B in 4 ways and remaing in 4!
A C .......... B in 5 ways and remaing in 4!
total 4!(1+2+3+4+5) = 360
Fixing C in 6th position:
1 2 3 4 5 6 7
X X X X X X X
A B C .......... B in 1 way and remainng in 4!
A C ......... B in 2 ways and remaing in 4!
A C ......... .B in 3 ways and remaing in 4!
A C .......... B in 4 ways and remaing in 4!
total 4!(1+2+3+4) = 240
Fixing C in 5th position:
1 2 3 4 5 6 7
X X X X X X X
A B C .......... B in 1 way and remainng in 4!
A C ......... B in 2 ways and remaing in 4!
A C ......... .B in 3 ways and remaing in 4!
total 4!(1+2+3) = 144
Fixing C in 4th position:
1 2 3 4 5 6 7
X X X X X X X
A B C ......... B in 1 way and remainng in 4!
A C ......... B in 2 ways and remaing in 4!
total 4!(1+2) = 72
Fixing C in 3rd position:
1 2 3 4 5 6 7
X X X X X X X
A B C .......... B in 1 way and remainng in 4!
total 4!(1) = 24
Total = 360+240+144+72+24 = 840
But as pannaguma said tht Ajust before B and B just before C then the answer coming out to be 5! = 120....which is one of the options.....so chk out
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i think u should try to solve irodov...coz most of the jee problems are linked with irodov and the concepts used to solve an irodov problem,,,,and moreovr whether the book is hc verma or dc pandey they all hav taken problems from irodov and jst modified thm to look like an original problem....SO IRODOV IS A MUST SOLVE FOR JEE......
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first draw the graphs.....
ne point on line y = x-3 ...... (k,k-3)
ne point on line 2y+x-10=0 corresponding to this above point (16-2k, k-3)
area = A = (16-2k-k)(k-3)
A = -3k2 + 25k - 48
dA/dk = 0 ......... gives k=25/6
puthis this k in A
gives A = 49/12
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solve irodov....got a nice collection of problem....
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yes 100%.....but u hav be among the top 10% students in your branch and you should have a CPI of not less than 8.....and according to rule the branch u opted for a chg can increase its strength by 10%.
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yes u can take gamma effective.....gamma is calculated frm the degree of freedon of gas....so u can also calculate for a mixture
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