SOLUTIONS TO MOCK CET -3

Transcription

1 SOLUTIOS TO MOCK CET -3 Hints and expanations: sin. i sin 3 sin, r µ, r µ µ sin r µ sin r µ sini µ 3 sini sin r Or sin r µ µ 3 µ µ µ µ sin r sin r sin ri 3 sini µ Or µ sini Or µ µ or option () µ µ µ 3 C D µ. (a) For the fish, the bird appears at a height of m from water surface 3 3 (b) For the bird, the fish appears at a depth of m beow water surface 3 Hence the bird sees the fish 6.60 m away whie the fish sees the bird 8.8 m away or option () 3. s the ens can form both rea and virtua images, it is a convex ens and u ve 6f + v 6 f 6 f we get v nd v 6 f + and v 6 f 6 f s the images are of the same size v v u u Or 6 f 6 f 6 f 6 f 6 6 Or ( f ) 6 6 f Or f cm or option (). r + r for emergent ray, For no emergent ray, r > C, Max vaue of r is when i 90 or sin r sinc or µ i r r Or r C r+ r > C+ C OR > C is the condition for no emergent ray as per option (3 )

2 5. Given I I ; the required ratio is given by ( I I ) ( I I ) ( ) I + + max 3+ I min 3 I Imin or by max r option() th 6. Since the n red band coincides with the ( n + ) th bue band, we get n 7800 D 500 D d d ( n+ ) or 78n ( n+ ) 5 or n. Or option (3) 7. Poarisation means restriction of transverse vibrations to one pane perpendicuar to the direction of propagation. This pane is caed the pane of vibration. The pane perpendicuar to both the pane of vibration and the direction of propagation is caed the pane of poarization or option () 8. Option () 9. λ dx n sinθ dx λ dx λ dx λ dx m or option (). In grating, the diffraction bands are cooured, whie zero order principa maxima is white or option () r r r r r mg. F Eq mg F ee mg or E or option () e. Potentia at Or option () + kq kq 0 r + r + +q q r 3. Tota voume of dropets voume of coaesced drop 3 3 or n π r π R 3 3 or R 3 nr Capacity of drop πε nr n πε r n capacity of one dropet or option (3) Ceff C3 C+ C + k3ε 0 kε 0 k ε0 d + d d d C k k k 3 C C 3

3 or d d + keff ε0 ε0 k3 k+ k + keff k3 k+ k or option () ρ 5. ρ R ρ R... () ρ R ρ ρ R... C ρ RC ρ ρ R... 3 C o RO ρ R... ( ) ρ ρ o Hence max resistance R or option () 6. Since the sequence is infinite; it can be repaced by Reff Reff Ω+ + + Reff Reff + Reff + Reff + Reff + Ω Ω Ω R eff Or R + R + R eff eff eff Reff Reff 0 Or R eff + ± + 6 ± 5 taking positive vaue R + 5 ( +.36) 3.3Ω or option () eff 7. In the absence of any resistance in the eft gap, there wi aways be a current fow in the bridge wire. Hence no baance point wi be found. nswer option () 5Ω 8. I R IR R I > R I Or Or R > R or option () I I I V

4 9. Initiay the partice wi move aong the y-axis due to the force exerted by the eectric fied; then it gets bent in the y-z pane as the magnetic fied imparts a radia force. However the effect of the magnetic fied is to ony change the direction not the magnitude of the partice s veocity-which is affected ony by the y-component of the eectrica force. This component increases the vaue of the y co-ordinate of the partice. The onger the eectric fied operates, the higher the vaue of its veocity i.e. veocity depends ony on y or option () 0. The fied due to segment O is opposite to the fied due to the segment MP as the current fows in opposite directions through them. The portions M and OP do not create any fied at C, which is in ine with them. Hence nett fied at C 0I 0I 0I O mp µ µ µ r r r r or option (). The shape of the Hysteresis oop depends on the retentivity and coercivity of the materia. These depend on the composition, temperature and the degree of saturation of the materia. The size of the oop depends on the dimensions of the materia in addition to the above factors or option (). µ 0 I (due to wire ) π r 7 5 T F I sinθ (force experienced by wire ) 5 5 sin90 8 or option () 3. Required R ( n ) R () n 5 V 0 sub in () R Ωor option () φ as is given constant.. d φ d d dt dt dt dφ ε 0 0.V dt + 8 V 8 mv or option ( 3)

5 5. ue scatters more compared to any other coour or option (b) q p m q m q m α 6. Required ratio q qα mp qp m α p mα : p p p p : or option ( ) 7. hc λ W + V S e e (Einstein s equations) hc e () hc.7 + V S 00 e from () and () () V 6..7 S V.5eV S V.5Vor option () S V S 8. R H λ n n RH n, n 3 3 5R H λ or option () R 5 H 9. Q ( 7 ) (.) 3.6 MeV or option () Using n n n n.. () n From t nt T nt, n T n( T) n sub in () :or option () 3. Emitter base junction must forward biased (i.e. base shoud be at positive potentia compared to emitter) and coector base junction must be reverse biased (i.e. coector shoud be at greater positive potentia compared to base) or option ()

6 3. For C0, output of G is 0 and that of G is For & C0, output of G is and that of G is or option () 33. Option () 3. Option () 35. Option (3) 36. Option (3) 37. The gain in K.E. of a charged partice after moving through a potentia difference of V is given by ev, that is aso equa to mv where v is the veocity of the charged partice. Disregarding the reativistic effect, qv v mv mqv m mv qv de rogie waveength h λ mv h mqv λ p λ α mqv mqv α α α p p p λ p ()() Putting Vα V p, λ ()() α or option (3) a P a PV ML T L V [ ] [ ] or option () 39. v mqv vα h and as the ba is moving down, its veocity is negative and it is positive whie ba is rising up. Option () tota distance 0. verage speed tota time segmentwise distances segmentwise times average speed 3 km / hr or option (3) 5 5. Since the ba just cears the wa horizontay, the wa s height is equa to the maximum eevation of the ba and the wa is situated at a distance equa to haf the range of the ba. given o θ 37 ; u sin θ u sin θ H sin R and H. θ tanθ g g R sin θ required ratio H H 3 tanθ tanθ tan R / R 8 o or option (3)

7 . s the body has the tendency to move up the pane, friction acts down the pane. F mg sinθ + µ mg sinθ + µ mg cosθ or option () 3. Let be initia veocity of mass ' m ' be the coision v, v find veocities of after coision. 3 Then mu mv + ( m) / and mn mv + ( 3m) v Or mu mv+ ( 3m) v or u v + 3v and u v+ 3v soying we get v v u v + 3v v or v, v initia of mass m mu mu Fina KE of mass m m u 75 of its K.E eas been transferred to mass or option () oth L and E are conserved. Iω Iω vr vr s R (distance of P from S) decreases, its speed increases. K is a maximum at P GMm U K max a U is the east a P GMm E a So, E is ve at any position. Option (3) 5. Option () 6. Force due to surface tension aong the circumference of the is (i) capiary is F πrtand (ii) the wire is F π rt oth are upward. The resutant must support the weight of the iquid coumns. ( ) ( ) π r r T π r r hρg h T r r ρg ( ) Option ()

8 7. W, T O & T O W T T W T T O O 0 6 W 3W T & T 5 5 6W 9W T + T + or option () 5 5 0cm T T cm o 90 6cm O W ( ) K T T t Qsemicircuar rod 8. Q or Option () Q straight road π π 9. ms b b( 30 0) + mv micel ( 30 0)( ca) ( J ) mice80( ca ) ( ca) 30( ca) + mice80( ca). m 5.875g or option () ice 50. Option (3) 5. Work done is positive if expansion curve ies above compression curvre (cockwise) and is represented by the area encosed by the curve. s area of is more than that of, net work done is positive or option () 5. y sin( ωt) In the first case sin( ω ) π T t ωt t 6 In the second case, T 6 t T t T & t or option () 6 t 53. Option (3) 5. V V + f f f. f V or option () V Option () 56. Option (3) xd m x k q q kq x d << d 57. ω ( x d ) ( x+ d ) 3 8kq 8kq md mω x x ω T π 3 3 d md 8kq m Shortcut, T π hence mass must be in the numerator. Option () K ( restoring factor)

9 58. Dieectric strength of air 3x 6 V/m. Then pd required to conduct eectricity through 0.mm air is 300V. If 300V is peak vaue, then rms vaue is 0 V V or option () 59. From the given kinetic energy of the neutrons, we first cacuate their veocity. Thus mu - rms 9 V \ u or u 500 m / s with this speed, the time taken by the neutrons to trave a distance of m is, D t The fraction of neutrons decayed in time s D t second is, D D t & aso, T / \ D D 3 6 t - - ( ) Option (3) T / Measured vaue must be accurate upto the east count of the instrument. Option ()