MODEL SOLUTIONS TO IIT JEE 2011

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1 MDEL SLUTINS T IIT JEE Paper I A D A B D 8 9 A, B, D A, D A,, D A, B, 4 6 D B B A Section I. 7 Al + 4 e P + n(y) 4 Si + (X) 4 Si + e(z) N Br. Since the mobility of K + is nearly same as that of Ag + which it replaces, the conductance will remain as more or less constant and will increase only after the end point. 4. [Nil 4] is tetrahedral, [Ni(N) 4] is square planar & [Ni( ) 6] + is octahedral... Ba(N ) Ba + N Very pure N is produced by the thermal decomposition of Barium or sodium azide. K N Br NK l. 6. M. M 7. o-hydroxy benzoic acid: pka.99 p hydroxy benzoic acid: 4.8 p toluic acid 4.4 p nitrophenol 7. Section II 8. Adsorption is always exothermic. hemisorption is more exothermic than physisoprtion and it requires activation energy.

2 9. A, D. assiterite contain. % of metal as Sn the rest being impurities of pyrites of Fe, u & wulframite.. In a, b & c all atoms are in the same plane... (P) Section III dil. S 4 gs 4 (i) NaB 4 (ii) dil.acid (Q) ( ) (- ), -shift of - ozonolysis 4. AgN + u u(n ) + Ag The metal dipped is u. Blue colour due to formation of u +. The solution in which the metal is dipped in AgN. 6. The deep blue colour due to formation of [u(n ) 4] + and the white precipitate of Agl dissolve due to formation of [Ag(N ) ] + Section IV 7. Maximum no. of electron with n is 8, of which are having spin 8. Na + Br NaBr + NaBr + 9. Br. hυ alc.k ( isomers) ( isomers) ( isomer) ev E hυ hυ For Li, Na, K and Mg hυ values are less than 4.4 ev. Let the number of glycine units be x Total mass of the hydrolysed products x % by mass of glycine x 6 S 4 6 has sulphur atoms with oxidation states & +.. Vol. of. mole at. atm.4 7 L.

3 PART II D D D A A A B 4 B, D A, B,, D A, B,, D A B D B Section I 4. x component of area vector is a φ E a. T sinθ mω Lsinθ ω L T θ. Z R λ 4 λ λ 6 6 X A mg 6. Position (): Let charge on µf be Q. Then Section II energy Q µj 4 Position (): eq µf. electron B Q Total charge Q. Energy µj Loss % 8% proton 7. n 4 (Θ.4 λ mole) W n V T R. T 4 γ γ T V T V T 4T T T 8. f f v + v c s c 8 ( + ) 9. R λ. T πm, different for them. qb dq is same for A and E and both are maximum. dt λ Thermal resistances are as below. KA R A R B R R D 4 RA, RB, R, RD 8 R E 4. So is also correct. R E 4

4 (Θ Eq.R (R, R B, R D) is 4. D also correct Θ R B parallel R D is equal R Note: It is assumed that there is no radiation loss.. n interconnecting V same Q A Q B (B) correct. RA RB σr V is standard formula ε () correct V same E A (D) correct. V, EB R A V R B 4. Under case (B) eventhough the disc is free to rotate, it will not rotate. ases (A) and (B) are identical in all respects. Section III. [N] L e Force area ε 6. Find ω by the given formula c λ ω π 7. As ball goes up, x is positive and increasing, v is positive and decreasing. Symmetrical for ball coming down. 8. At x, E K.E E varies as p E 4E p m 9. Starting from positive positions (i.e. in air), amplitude in air will be more than amplitude in water. Momentum is negative for downward journey. Also water produces damping. graph is spiralling in. / a q 4 + r 6πε N 4. F d mg sinθ µmg cosθ F u mg sinθ + µmg cosθ F u F d µ 4. N µ B µ Ι L φ coil Bπr µ r g π Ι L dφ E E, i ; M iπr dt R N 6 4. λ MgL AY L L + λ L' Mg + L AY L' L α ( L 4 L' ) Mg Mg AY 4 + AY Solving, we get M kg 44. N 4 mr + md + mr ; ere d r N 9 L x x x x x x x x B 4 m m; m kg Section IV 4. a. q 4. U [ 4 + ] 4πεa + a r du For equilibrium, da α N f f

5 f a α R fr, α mr R f f a. m m f, m f.4, f.8 µ..8 µ.4 M mn 9 kg 9 6 mg mg 46. A λn s N λ 9 9

6 PART III B B A D QUESTIN INRRET B,D A,D B, B D D A B Let v ρ A i + Bj + k 48. a b. v A A B B Section I ( B) + (B + A) A For the vector in (), A Projection of ( i j + k) on + b R ( x ) y b dx x (x ) b (b ) + (x ) R (x ) dx b b (b ) R R gives 4 (b ) + 4 b satisfies the above equation. 49. y + m (x ) and m y + (x ) y - x + + ln x sin x. Let I dx sin x + sin (ln 6 x ) ln Set x dx t dt t x ln x ln, t ln, t ln ln t sin t I dt sin t + sin(ln 6 t) t ln ln sin t dt sin t + sin ( ln 6 t) ln

7 . ln sin(ln 6 t) dt sin (ln 6 t) + sin t I ln dt ln log I log 4 log x log log y log log x k log x k log y k log y k k + k+ ( ) log ( ) log k log. P. + k log + ( k +) (log) (k+) (log ) k x P π θ : sin θ cos θ 4 π θ : cos θ cos θ 4 π P θ : θ nπ ± θ 4 π θ : θ nπ + 8 π π Q θ : θ nπ ± θ 4 π θ : θ nπ + 8 P Q a a8 a9 8 8 α β ( α β ) ( α β ) 8 8 α ( α ) β ( β ) ( α β ) 8 8 α,6α β 6β ( α β ) 6( α β ) ( α β ) But if The question is reframed in the following manner; Let M and N be two nonsingular skewsymmetric matrices The solution is M N (M - N) - (MN - ) T M N (-MN) - (N - ) T M T M N (-N - M - ) - (N T ) - (-M) M N N - M - (-N) - M - M N N - M - N - M - M N M - N - M - M N (N M) - M - M N(M N) - M - M N N - M - M - M hoice (c) x y. For + 4 e for b a and focus (±,) x y,e a b Substitution (,), a, b Focus (, ) required hyperbola is x y 6. j k i + j + k (i + j + k) & k j i + j +k (i + j +k) j k, k j coplanar with the given vectors and ± ± (A) and (D) true. 7. f (x +y) f (x) +f(y) f(x) kx f(x) is continuous x R and f (x) k Section II 4. Skew symmetric matrix of order is singular. Its inverse does not exist. Therefore there is N SLUTIN T TIS QUESTIN.

8 Section III For problems (8) and (9) W R a c a c Solving a and c - 7. The equation is x + 6x 7 α + β 6 + < α β αβ 7 n + (is an infinite geometric progression) α β n a 7 r 6 7 T W R+W R 8. Probability Probability 6. 7(a + b+ c), a + b + c also a + b + c a, b + c 7a + b + c x and Im(ω) > ω + i a + 8b + 7c 4 + 7b + 7c (taking first and third columns) Solving we get b and c 4 The equation in ω ω ω ω + + ω (ω + ω ) + orrect choice (a) 6. b 6. Taking st and rd columns we get 6. θ n π Section IV + sin θ sin θ sin θ sin θ sin θ sin θ sin θ sin θ sin θ sin θ sin θ cos θ sin θ sin θ sinθ cosθ sinθ sin4θ sinθ 7θ θ cos sin 7π π cos sin n n 7π π ( k + ) n 7 n k + For positive integral values of n k or put k, n m n; n [ 6 + (n )(a ) ] s m sn n [6 + (n )(a ) 9 a + n (a ) 9 a + n (a ) 9 a + (a ) (9 a) + (a ), 9 a + a 9 a + (a ) is independent of n 4a 6 9a 6 a + (a ) (a + ) (a 7) a 4a + 4 a a + 7 a, a 9

9 4 8 a + a + a + a + a + + a + a ( a a ( a ) a a )8 minimum value of sum 8. x f(t)dt x f(x) x 6f(x) xf (x) + f(x) x f(x) xf (x) x dy y x x dx dy dy y x y x x dx dx x I. F dx x e y. x dx x + c x x y x(x + c) f(x) f() + c c f(x) x(x + ) f() Extremum point of latus rectum are (,4) and (, 4) Area of triangle so formed with, (y y ) (y y ) (y y ) 8a (4+4) ( 4 ) ( 4) 6 8. Eqn. of tangent at (,4) is y x + () Eqn. of tangent at (, 4) is y x + () Eqn. of tangent at, is y x + () Tangent at the extremities of latus rectum intersect directrix at (, ). Point of intersection of () and () is (,) and of () and () is (, ) f(θ) sin tan - sin θ cos θ sin θ tan tan cos θ sin θ + tan (tan cos θ sin θ cos θ cos θ + sin θ cos θ tan θ d( f( θ) ) d(tan θ) sin θ cos θ sin θ + sin θ 69.,, minimum of z + i +