CHEMISTRY 1. A. Paper-2 HINTS & SOLUTIONS C 3. D 4. C + + Sol: Mg N Mg N Mg ( OH) NH Cu ( NH ) 5. AD Sol: α hydrogen absent

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1 CHEMISTRY. A Sol: hν hν + KE Paper- HINTS & SOLUTIONS. C. D 4. C HO CuSO4 Sol: Mg N Mg N Mg ( OH) NH Cu ( NH ) Blue 5. AD Sol: α hydrogen absent 6. CD 7. CD Sol: Spin multiplicity n +, n number of unpaired electrons. 8. ABD 9. BC Sol: MgCl Mg + Cl cathode + anode. A-R; B-PS; C-PST; D-QR; Sol: Specific conductance decreases with dilution, increase with increase in conc. molar conductance increases with dilution, decreases with increase in concentration α increases with dilution, decreases Λm with increase in concentration. α. Resistance increases with increasing the distance between Λ parallel plates. A-PT; B-R; C-Q; D-S LiAlH4 or Zn+ NaOH+ CHOH Sol: CHNO CHN N CH NaNO,HCl CHNH CH N CH Cl Aniline carbyl amine α m NaNO, HCl N HO, H 5 CH CH NH CH CH N CH CH C H OH +. wt. Sol: wt. of HO. Mol.wt.

2 . Sol: [ ] a X [ Y ] a b [ X] [ Y] b Sol: No. of α XSO.Y SO.4HO Sol: ( ) Sol: Ksp Ag Cl solubility of AgCl in.m AgNO Ksp Ksp S; S..5 S Cl S + Ag 7. Sol: K ( ) n p Kc RT 8. 6 Sol: NH NH4 + NH 4 7 NH K No. of amide ions in cm 9. + Zn 4 Sol: S MATHEMATICS. Ans. : D Sol. Since the roots are real and distinct, dis >, since α <, β <, (α ) (β ) > αβ (α + β) + 9 C c b + 9> K K if g() k + b + c 9a + b + c K > f() > on solving we get a < as the range of a. St is wrong and St- is true.. Ans. : A Sol. Observe that K + if K < + <

3 or K < K,,,, if 67 K 99 observe that 67 K 99 K < < + K + for K,,,.. 66 and for K 67, 68, S ( terms) + ( + +. terms).. Ans. : A Sol. y The ellipse is Here a 5, b 5/4 4 a by Equation to normal at (, y ) is a b y 8 y 8. (6, ) lies on it. (6, ) is nearest to (, ).. Ans. : C 4. Ans. : BD Sol. Equation of circle with A(, ), B(, ) as ends of diameter is ( ) ( ) + (y ) (y ) and equation to line AB is y. Hence any circle through A and B is ( ) ( ) + (y ) (y ) + λ( y). () If this C lies in st quadrant, it will not cut or y ais in real point λ and + y y + λy will not have real roots on solving we get λ radius of () is λ + will be greatest when λ ±. Its centre is (, ) or (, ) 5. Ans. : ABCD dy Sol. Given equation is ytan sin d + tan d I. F is e sec Solutions y. sec sin.sec d + C y. sec π sec + c. If it is passes through, C. y sec sec or y cos cos cos cos.

4 dy sin + sin d dy -sin ( - 4cos ) d d y cos + 4cos sin or cos /4 or π d d y d y > when, or π, < when cos ¼ d d local min,, local maimum Period of the function is π. 6. Ans. : BD Sol. a b c Given A b c a c a b Since A T A -. (A T ) (A) I det (A T ). dt (A) det I. (det A) (a + b + c abc) a + b + c abc ± a + b + c abc ±. But abc a + b + c ± 4 or 7. Ans. : AB Sol. f() for < for < ( ) for < For <, f o f() f[f()] f(). Now let < then f o f() f[f()] f() Now let < 5 then < f o f() f[f()] f( ) ( ) 4 6. and if 5,, so f() is undefined. Hence g() f o f() for < for < 4 6 for < 5. Domain of f o f() is [, 5/). Observe that g() is continuous but not differentiable at. Now g () > for < 4 > for < 5/ g() g() has no ma or min at 8. Ans. : ABC Sol. Given (aω + b + cω) + (aω + b + cω ) and a, b, c R. Let aω + b + cω and y aω + b + cω then LHS + y ( + y) ( + ωy) ( + ω y) + y or + ωy or + ω y + y a(ω + ω) + b + c(ω + ω )

5 a( ) + b + ( ) or a + c b Similarly + ωy aω + b + cω + ω(aω + b + cω) ) aω + b( + ω) + c( + ω) or ω (b + c a) or b + c a similarly a + b c. a, c, b are in A.P. Hence A is correct. a + by + c using C b a a + by + b a or a( ) + b(y + ) It passes through (, ) Let f() a b c, or (, ), f() and f() a (b + c) f () has a root in (, ) 6a b c has a root in (, ). 9. Ans. : A QS, B P, C P, D R. Sol A. C Sin + sin + sin, using the formula 5 65 θ sin - + sin - y sin - y + y simplify to ( ) get C π/ then cot A + cot B q A and cot p. cot B r p A + B π C π A B cot + on simplification, we get p r q. 4 p + qy + r p + qy + p q p( + ) + q(y ). It passes through (, ). B. Using formula find images to get B (, 8) and C (7, ). A is (, 4). Circumcentre is (8, 9). C. Any normal to the hyperbola y is t yt t 4 +. If it passes through (h, k) then t 4 ht + kt. Co-normal points are ti, /ti). Then t + t + t + t 4 h or h and /t + /t + /t + /t 4 k or y + y + y + y 4 k Let the variable line bc a + by + c. i yi h k Given that a + b + c or a + b + c h k It passes through, 4 4 Here (h, k) (, 6). 4 + sin + asin + bsin D. f() when, is cont at. Lt f() should eists and finite. Lt 4 + sin + asin + bcos is finite using L. Hosp value 4 + b 4 + b or b 4 cos + a cos bsin then Lt + a a substitute these values and final limit which is. f() Lt f(). z + zi Im z z ( + zi) Im z is a parabola whose focus is (, )

6 and directri. y. verte is (, ). Ans. : A S, B P, C Q, D RT. Sol. A. Let tan - π π then θ as 6 6 cos tan cos - tanθ + tan θ + tanθ tan- tan θ cos - (sin θ) + tan - (tan θ) π/ θ + θ π/. 5 as π π, π π π θ θ. 6 6 / π d Given integral, using formula, e + / / / π d d + - e + e + π. B. PA + PB < P lies inside the ellipse ( ) y + /4 and PB + PC < P lies inside the ellipse ( ) y +. /4 Both should hold according to given condition. Required to find should area DQ ERD D (, ) B (/, ). A(/, ) From the second ellipse y O ( ) Required area / ( ) 4 y d / d ( ) ( ) / + sin π 4. C. Given f ( )sec d f() sec + c f() sec f() sec + tan + sin f (). f ( ) f() sec f() tan + f () or sec tan. f ( ) f ( ) sin cos sin cos sin. f ( ) cos cos sin cos + sin cos + sin y D. Hyperbola is latus rectum cot θ cos θ cos θ cotθ cos θ sinθ cosθ 4 Q E(, ) F(, ) D B C R

7 sin θ cos θ sin θ π 5π θ or 6 6 π 5π θ or. Ans. : Sol. AM is + k ; GM k k λ where λ is an integer and also given k 999. Since + k is in the given series, k should be odd. λ λ λ, 4, 5, 6. λ or 5. Hence values.. Ans. : Sol. Let I d ( sec ) + tan ( ) sec + + d( sec + ) / ( + sec ) + C + C + sec + C. + cos + tan + C + + C + cos + cos + C k. + cos. Ans. : Sol. Cosec θ θ sin sinθ θ sinθ sin ( θ θ ) sin / θ sinθ sin θ π 4π 8π cosec θ cot - cot θ. Put θ, and odd Ans. : 6 Sol. Given series can be written as ( C ) ( C 5 ) + ( C ) ( C 49 ) ( C 5 ) ( C ) n C r LHS in the coefficient of 5 in ( + ) ( + ) n C r C 5. n r Ans. : 4 Sol. Given planes are + y, y + z, z + ; + y + z 6

8 Let OA be + y or y z. () y Its opposite edge BC is z 6. l m n l + m + n Let l, m, n be the dc s of line along shortest distance. Then 6 6 required distance 6 ( ) + 6 ( ) + ( 6 ) Ans. : 8 Sol. P 6 C. D + 6 C 4. D + 6 C 5. D + 56 where D n denotes no. of derangenmtamts of n diff. things. Q (6 + 6 ) Q 46 C P 56 4 Q P Let the DRs of the line of intersection be l,m,n. Normal to the palne parallel to i+j+k and i will be in the direction of to (i+j+k) i -j+k and line of intersection of given planes will be perpendicular to this hence m +n similarly l-m+n. On solving l 8. Ans. : 8. Sol P 8. 7 m n. Hence cosθ 6 5 and tanθ

9 PHYSICS 9. Ans. : B Conceptual 4. Ans. : D y y sin (ωt k + π/) y y sin (ωt k + 4π/) 4. Ans. : D PQ R a PQ (PQ)ω Rω. 4. Ans. : B P decay e λt 4. Ans. : AB P AσT 4 A T 4 4 a B T B 44. Ans. : ABCD v iˆ R ˆ ωj iˆ ˆj ( ln ) t t e ¾. t / t φ 5π/6. Wein s law λ A T A λ B T B. v, ω V Q 45. Ans. : BC Q φ/r. 46. Ans. : ABD R ωl R/ ω ωc LC I(R/) V V R IR V 5 Z R V S IZ 5V Power I R 4V /R. 47. Ans. : BCD v a + t, a t 48. Ans. : A P, B P, C P, D S Image by slab m + shift 6 cm towards the slab mirror + v v m u 5 m m m m +. 5 v cm + m + Rω ˆj + v iˆ. 49. Ans. : A P, B R, C T, D S i A V V P i R.6 W Q C 4; R C V R 5. Ans. : 7 W nrt ln W nrt Q C 84.

10 Q n(7r/) (T ) nrt ( ln ) ln 4 η. 7 nrt 7 5. Ans. : 6 Z Mosely law L Z Ans. : 4 Energy of incident photon eV For H-atom E.6eV, E.4eV E.5eV E 4.85eV. Note that.85 (.4).55eV n Ans. : nλ + 9λ n,, 5, 7 9 n λ > n 4 n < 6 Three minimum 54. Ans. : 5 E V p pcosθ + cos θ cos θ r cosθ + cos θ 5 cos θ 4π r 4π r r 5m 55. Ans. : 8 V VR V i V A R + R R + R R + R V V V R R VR. RR R + R + R R + R R + R R + R R + R Solving R 8R R 8R/. 56. Ans. : v + v f + m for m. v y v y y y y m distance moved m. 57. Ans. : ( ) R a + R d R a + R + d Rd a + d R d mv a + d qb d + ( ) R d d a R

11 a + d + 4 mv qb. d 4