106yeah that's the answer
but look at the sol. theyve given
φ = BNA = μnI.nπR2 = μn2IπR2
wrong naa?
circled questions are doubts
ticked answers are my answers
circles options are fiitjee answers
-
UP 0 DOWN 0 1 17
17 Answers
106@eure: agreed there is only one resonant freq..
but the question does not state reso. freq..
for any other value of freq.. max current is same for two values of freq..
301.. shouldnt there be a mod sign over the whole expression?
13300) A will be correct.
308) Assumin' frm ur diag. tht green n blue are getting reflected.
One thing is tht even though TIR ho rahaa hai, Green will come out shortly b4 Blue, but eventually they'll mix 'coz of the same path....so are they regarding this as a separation of green n blue lights...? [12]
13Fr #14....Ya, they are finding the net "f" fr 2 lenses in commbination...but, they can't apply the two-medium focal length formula... one side there is water & another medium (n=2) the other side.
n Yes, tht'll be wrong fr 305, v can't take tht way.
106106I dunno answer given is NOT
but in the solution they have done this:
1f = [1.54/3 - 1][130 - 130] + [24/3 - 1][130 - 130] = 0
=> v =∞
1Q 299)
Answer is b)
P.D.= ∫(V*B).dl
substitute v=lω
Q 300)
Can you give any support to ur answer? I feel b) is correct
Q 303)
We have to assume the length to be same.
Q 304)
Definatly 2.50
I'll give the solution if u need
Rest will do after dinner.
24300)there is only one resonance frequincy ..so b is correct
301 ..ur ans is matching naa ?
302 ur r rite
303 since it is mentioned that both solenoids are loong ...so u cna assume lengths to be same....btw whats the ans given ?
1308)
UR ANS IS CORRECT,
CRITICAL ANGLE FOR RED LIGHT IS APPROX. 46° AND FOR GREEN IS 44° AND THAT OF BLUE IS 42.86°
SO ANS IS (A)
106but another option is there also
∞ l3/2 ... then which one will you select... (i had completely missed that one)
106@manmay: look at my first post..
circled options indicate FIITJEE answer
ticked options indicate MY answer
so ur with my answr?
1FROM CONS. OF MECHANICAL ENERGY,
mgl2sinθ= 12 Iω2 = 12(ml23)ω2
ω2 = 3gsinθl
ω = √((3gsinθ)/l)
potential difference = 12Bωl2
= 12B√((3gsinθ)/l) l2
= 12B√(3gsinθ) ( l 3/2 )
Hence p.d is proportional to ( l 3/2 )
and √sinθ .
106@d'artagnan: 304 is not my doubt (the question is not circled)
Q300.. look at the diag ive given
Q303. why?
Q299.. pls post full solution