Sec: Sr. IZ Date : Time : 02:00 PM to 05:00 PM 2012-P2 Max Marks : 198 KEY SHEET PHYSICS 1 B 2 C 3 B 4 A 5 B 6 B 7 B 8 C 9 A 10 B
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1 Sec: Sr. IZ Date : --8 Time : : PM to 5: PM -P Ma Marks : 98 KEY SEET PYSIS B B 4 A 5 B 6 B 7 B 8 9 A B 4 B 5 A 6 AD 7 AD 8 A 9 AB BD EMISTY D B 4 5 A 6 A 7 8 B 9 B A 4 B 5 ABD 6 AB 7 B 8 AB 9 AB 4 ABD MATS 4 B 4 A 4 D 44 D 45 D 46 B B 49 A 5 A 5 A 5 D 5 A 54 D 55 ABD 56 ABD 57 A 58 AB 59 B 6 AD
2 SLUTIS PYSIS. There should be continuity of path. --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s. The intensity of light emerging from a slit is proportional to its width. Since the I I I ma n Imin I n I cosi= sini= = = = 5 5. ( ) ( ) nl ( + t) ( nl ) l + t l + l nl = r 5. B only relative motion generated frequency difference and normal relative velocity for movementum. 6. B as A,, D are be eliminated D is for adiabatic for all is for isothermal for all. 7. nly D is wrong all others are correct. 8. Let the side of the cube be lc. The volume of the cube above the surface of water 9. =volume of water displaced due to mass of g. therefore mass of displaced water is g, its volume is cm. ence l l or l cm. ence the correct choice is (c) mv KP cos =. ormal reaction will have a factor of. Initial position is etreme position. dv dt Mv also. mg sin IBl = m ( i) di L = Bl ii dt ( ) Differentiating (i) w.r.t time Sec: Sr. IZ Page
3 di d v Bl = m dt dt --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s B l d v = m L dt B l d v = ml dt Bl = ml So V = VS int v = cost When V is maimum t = / V = mg sin = ibl = v cost at t =, l = ( ) mg sin = v mg sin = v mg sin =.. Directly from Graph. 4. Directly using theory rules. GM r 5. Inside sphere ( c ) 6. GM M = 7 5 uniform Sec: Sr. IZ Page
4 --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s T T r T d In steady state heat current dt dt I = KA = K 4 d d I I dt d = 4 r T d = 4 dt T ( T T) ( r) 4 I = ln / b) d) dt I = d 4 ( T T ) ( T T) ln ( / ) 4 4 I = = ln r Sec: Sr. IZ Page 4 T + T = = r at T 7. Suppose field in air gap is E then in dielectric it is For dielectric medium E E dv = d V = K K For air gap medium V = E 8. harge on capacitor increases with time q potential V = increases too urve I is for V and curve II is for V as current will decrease with time. Also V + V = V + V = V + V = V (when both were equal) ( ) V + V + V = V E K
5 V + V = V --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s 9. + a = d dt Integrate a t =. The value of n is proportional to the intensity of incident light. If the distance of the. D lamp is halved, intensity becomes four times, but K ma is independent of the intensity of light. ence the correct choices are (b) and (d) (i) ab4 (ii)mes in Et EMISTY Pl 5 Et l l a. B P =k= =costant V V V A B V V V w = P.dV = dv = = 8atm Litre = 8.KJ V V 4. Migration of group in offman Bromamide reaction is with retention of configuration A K sp,agl + 9 Ag = = = mol / L l.5 6 K sp,agi Ag = = = 8 mol / L l.5 l Pd Δ l D (A) Sec: Sr. IZ Page 5
6 --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s + l ab 4 Et (B) + () In will be the product + + Θ I (D) 6. A Pinitia = 76cm + 8cm = 4cm g will be obtained 7. Pequilibrium = 76cm + 76cm = 5cm g A B A B + B = 5 = = 9mm g 9 % dissociation = = 6.67% 4 MX must be formula since X is forming hcp. + M + e M(s). moles per unit cell o.59 E = E log M +.8 =.5.59 log. Sec: Sr. IZ Page 6
7 8. B =.44 V --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s = e a r a At nucleus r = and in st orbit r = a = e a a at a= a 9.- n = e ; n a = e Δ yloy - 6is the Polymer (A) (B) () (D) aprolactum (E).. A -4 Me I Et S Me (P) + Et - I E (i)moist Ag (ii)δ Me (B) Et Et + = (A) + I 5. ABD 6. AB Factual 7. B 8. AB - - Ag S Me 4Mn + Mn + 5Mn + n Δ E i Me () + (D) mol Mn 5Mn + 4 ypo = 5. 5mmol 5mmolMn + 9. AB mmolmn V = 5ml 4 Sec: Sr. IZ Page 7
8 X is --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s i / - S S S I S S gl +d, - 4. ABD 4. Put t in I and S t in I D MATS D S S gl +d, D We get d 4. Let u ln k k So, k f k k k du then u ln f f e e 4. image of (,) in -y+= is y y Sec: Sr. IZ Page 8
9 y Slope of directri = Equation of ais, y y _Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s 44. ˆ aˆ bˆ. ac ˆˆ. b aˆ. bˆ cˆ aˆ. b ˆ aˆ bˆ cˆ ˆ aˆ b cˆ a b c a b c a b c where a, b, c are elements of 7, ase I: All seven entries are and rest of all zero ase II: nly one entry is &three entries are and rest of all s Total = 54 Put t tdt dt f t t t u th i row. Sec: Sr. IZ Page 9
10 u udu u u ln ln t 6 u du ln 9 u 6 lnu _Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s ln ln ln ln Equation of B is +y=7 so Let slope of AB is m so, m m m 7 B 4 7, So equation of AB is +y-= 7 So A= A, midpoint of AB AB 4 4 Equation of median through is 7+y=5 48. equired area is times the area of loop 67, 4 4 d Sec: Sr. IZ Page
11 --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s 49 & cos y cos y 7 So, g 7 7 ; ; So, 5 & 5 g min 7 So, f cos / A cos d / / sin / 6 intersect at points ABP and DP are similar if AP=, BP=y then P=5, DP=5y Area of trapezium ABD= 49 y tan y,tan 5y 5 also 45 Sec: Sr. IZ Page
12 5&54 Also Ar y y 4 AB AP BP y 4 y PD 5y Equation of line L is y 4 z () equation of plane containing L and L --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s y z 6 y z 5 () volume of tetrahedron = Shown in fig 56. A) 4 4 / So, Sec: Sr. IZ Page
13 So, f cos 4 --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s sin cos sin many one onto 57. A) let n sin t cost dt A t n dt A sin t cos t n sin t cos t dt n... n ) sin 8 sin sin 8 sin D) tan and 4 So, tan /8 I f f d... /8 7/8 7/8 I f f d f f d... So, /8 /8 I 59. For, A, For, B, 4 z ez z y z y For, Sec: Sr. IZ Page
14 z z z z z z z z --8_Sr.IZ_Jee-Adv_-P_TA-6_Key Sol s B. equired area d 6.. z iis comple number in region A B having ma amplitude D. (-,), (-,),(-,-),(,)(,)(,-) are point but z=- is not in domain of So total points 5 y 6 a 4 Slope of tangents at t is tan, tan ;tan t t t t t, t, t are in A.P with d= t t ; t t att a t t Area of PQ = at t a t t at t a t t a d Area of AB is =. PQ =864 Sec: Sr. IZ Page 4
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