It is not possibl to find flu through biggr loop dirctly So w will find cofficint of mutual inductanc btwn two loops and thn find th flu through biggr loop Also rmmbr M = M ( ) ( ) EDT- (JEE) SOLUTIONS Answr : Option () 8 Enrgy will b qually distributd whn Answr : Option () 9 R = 8Ω Out of V only 8V should appar across bulb Answr : Option () Answr : Option () Th magntic fild is constant with tim So it will not produc any lctric fild or inducd currnt through th wir Similarly, th motion of conductor is compltly outsid of th magntic fild SO thr is no qustion of motional EMF ithr Answr : Option () Both th sctions will dvlop EMF Bvl and thy will assist ach othr So nt EMF will b Bvl Answr : Option () W can rmov R as it will not hav any ffct on LR branch ( ) ( ) ( ) ( ) Potntial Drop across L = Answr : Option () 6 Considr a concntric circl passing through P Lt inducd fild b E at vry point on this circl ( ) Answr : Option () 7 Whn capacitor is rmovd, Whn inductor is rmovd, In th givn situation circuit is in rsonanc (X L = X C ) Hnc powr factor (cosφ) = ( ) Answr : Option () Answr : Option () Rfractiv ind of matrial n = Hnc, Answr : Option () Conduction currnt and displacmnt currnt wil b qual Answr : Option () Answr : Option () E Hnc, Amplitud of lctric fild at dist of m from sourc = = 6 V/m Hnc B = = -8 T Answr : Option () a Considr a strip of lngth d at distanc as shown in th figur Flu through this strip is dφ = BA = (B ) (a d) = B a d Intgrating this from = to = a, w gt, d
Projctd lngth EDT- (JEE) SOLUTIONS Φ = B a / Answr : Option () 6 As inducd lctric fild is non-consrvativ, potntial is not dfind Answr : Option () 7 Magntic fild insid th toroid is givn by Hnc flu is Φ = a Not that w will us ara of toroid and not th ring This is bcaus part of ara nclosd by th ring has no flu whil only that part which is insid has uniform magntic fild EMF = = Answr : Option () 8 Us th Projctd lngth concpt or Effctiv lngth concpt V = V + V sin ωt + V V sin ωt V RMS = = ( ) = V RMS = Answr : Option () A V = V A sin (ωt + ϕ) B V = V B sin (ωt) Answr : Option () As inductor is strtchd, its inductanc will dcras and so will th impdanc of circuit Answr : Option () 6 V t = ( ) => V R = V Answr : Option () 7 Assum that rsistanc R = r Whn capacitor is rmovd, cos ϕ = = = ( ) => X L = r v H Similarly whn inductor is rmovd, cos ϕ = = = ( ) => X C = r L That is considr this as rod with a lngth qual to th projction of rod prpndicular to its vlocity In th givn cas it is H So EMF will b ε = vbl = vbh Answr : Option () 9 Projctd lngth is R So EMF will b ε = vrb Answr : Option () Vrtical componnt of th Earth s magntic fild is Bsinδ Forc on th conductor will b F = BLv sinδ Answr : Option () Currnt gnratd in th wir will b I = = Whn both ar prsnt, cos ϕ = = = = ( ) ( ) Answr : Option () 8 Answr : Option () 9 ηv P I P = V S I S whr η is fficincy Hnc, I S = A Answr : Option () Phas diffrnc btwn currnt and voltag is π/ Posr factor = cos ϕ = / P avg = I m V m cos ϕ = W Answr : Option () Wight of th wir will b balancd by th forc du to magntic fild Hnc, mg = BIL = B L v/r v = mgr/b L Answr : Option () ε ma = NABω = NAB(πf) Answr : Option ()
EDT- (JEE) SOLUTIONS
EDT- (JEE) SOLUTIONS
EDT- (JEE) SOLUTIONS
(a) Lt (c) / cos I d cos ( sin ) / cos d ( sin )( sin / t EDT- (JEE) SOLUTIONS t By partial fractions, whr ( log ( t log t ) ) tan t tan / ) dt t sin log( ) log( ) log ( u du ) / 6 as u t t dt t 6 (tan t ) tan 6 tan sin cos d cos sin / (d) I (i) (b) (d) / cos sin sin cos (ii) / cos By adding (i) and (ii), w gt I cos / tan I Now, Put tan sc tan t d, w gt dt I [tan t] 8 t 8 6 d d d d = ) d ( ) d = tan ( ( ) d = d tan tan d [tan d tan log( ( ) d )] log sin sin d 6
6(b) 7(b) Lt n! P lim n n / n lim n n n EDT- (JEE) SOLUTIONS n n n log P limlog log log n n n n n n log P lim n r r log n n log P log d ( log ) ( ) n L lim n r / n n r ( r / n) / / / 8(c) ( sin cos) d sin Applying gamma function, / sin / cos d n n P = d cos ( / ) ( / ) 9 9(c) Put a( cos ) d a sin d Thrfor, / a ( cos Now again, put a / a sin d ( a ) / )sin d / d d / n sin cos d a 6 (d) For, w hav, so that Hnc Also cos, cos d cos d Thrfor d / d d 8 cos I Hnc I is th gratst intgral 7
(b) Solving th quations y EDT- (JEE) SOLUTIONS of th parabola and th lin ar A(,) and, =y Y and y simultanously Th points of intrsction B, B O =y A(,) X Th rquird ara = shadd ara y d( for y ) yd (for ( ) d d (a) y (i) y (ii) y) By quation (i) and (ii), w gt, y Y (, ) y = 9 8 sq unit (, ) (, ) y = X Rquird ara ( ) d d 8 8
EDT- (JEE) SOLUTIONS (a) Solving y and y y ( ), 6 9 9 ( )( 9), 9 Y O A B (,) = 9 X = = [ log ] ( ) sq unit, Rquird ara = A+B 9 d [ / ] [ / 9 ] 9 8 9 [9 ] 7 9 8 [6 8] 8 9 9 sq unit (c) Rquird ara log ( ) d d ( lim log ) (, ) (, ) log t dt [ t logt t] sq unit, (Put t ) 9
(d) Rquird ara ( y / EDT- (JEE) SOLUTIONS ) dy Y y =(+) y=( ),(From th symmtry) (,) y = X On solving, w gt rquird ara 6(d) W hav f' Also, 9 log f' ( ) P (log) 8P Q R f() P Q Q log R [ f( ) R] d ( P Q ) d P 9 Q sq unit Now on solving th abov quations, w gt P, Q 6 and c 7(c) y ( c c )cos( c c 8(c) ) y y ( c c)sin( c) c c C ( c c)cos( c) c y c c R c y y c y y y Hnc th diffrntial quation is y y y y, which is of ordr dy y dy y y y tan or tan d d It is homognous quation, hnc put w gt, dv v v tan v d cot vdv d log( sinv) logc y sin c y v
9(a) a) f( ) d f( ) d a f( ) d BONUS EDT- (JEE) SOLUTIONS ( af( ) d = a a a a (b) IF d log( ) = log( ) ( ) t (a) f ( ) d t, t Diffrntiat both sids wrt t, w gt t f( t )t t f t ( ) Put t, w gt t f (b) Th quation of curv is y y y This is a parabola whos vrt is, Y, y = m O m X Hnc point of intrsction of th curv and th lin m ( m) i, or m 9 m ( m) d m m ( m) ( m) ( m) ( m) 6 6 9 / ( m ) 7 m 7 m Also, ( m ) ( m )[( m) 9 ( m)] ( m ) ( m) 9
(d) EDT- (JEE) SOLUTIONS m m m 9 m m m Hnc, m i, m is imaginary sin d sin d sin d sin d sin d sin d sin d sin d 9 sin d [ sin 9[ cos ] 9 ( cos cos ) 9( ) 8 (a) Th quation is IF dy d d log y log log Hnc solution is y y log c y c ( log) is priodic with priod and in [, ],sin ] d 6(a) Lt F ( ) y (t ) dt and F ( ) y t dt Now point of intrsction mans thos point at which y y y y 6 and y 6 6 On solving, w gt 6 and y Thus point of 6 6 intrsction is, 7(b) f( ) f (i) Rplacing by Eliminating f in (i), w gt f f( ) (ii) from (i) and (ii), w gt f ( ) f( ) f( ) d d [ log ]
8(d) b a d b [ log ] ln, (Givn) EDT- (JEE) SOLUTIONS ( b a ) b a b a a d b a b a b a ( b a )( b a ) or ( b a)( b a) b a but b a dos not satisfy th quation, b a Now, b a ( a ) a a or a a b a Hnc b ( ) a, 9(c) Applying R R, w gt f( ) sin cos sc R / / Thus f ( ) d (sin 8 tan y dy (b) ( y ) ( ) ( y d dy IF ) d dy ( y dy y tan y d tan y tan y ) ( y tan y ) tan y y = tan y tan y y tan c, tan y tan y k b ( a, b) (, ) cos 8 dy ) d