Solutions to Mock IIT Advanced/Test - 3[Paper-2]/2013

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Λαμία Αλαβάνος
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1 Solutions to Mock IIT Advanced/Test - [Paper-]/0 [CHEMISTRY] VMC/0/Solutions 6 Mock IIT Advanced/Test - /Paper-

2 VMC/0/Solutions 7 Mock IIT Advanced/Test - /Paper-

3 VMC/0/Solutions 8 Mock IIT Advanced/Test - /Paper-

4 VMC/0/Solutions 9 Mock IIT Advanced/Test - /Paper-

5 VMC/0/Solutions 0 Mock IIT Advanced/Test - /Paper-

6 Solutions to Mock IIT Advanced/Test - [Paper-]/0 [PHYSICS].(B) K W g + W f µ θ µ Wf mgcos S mg x x is smaller for 60.(C).(C) πr g πr g πηrv πr a r tan7 8 4 r 6m v 00 v 8 m/s Ans ( ρ ) π F P0 + h g r 5 ( ) π π (D) Imagine loops. B B0 + B0 + B0 B0 towards F 5.(A) kx 0.06 g k k N/m W ext + W g + W string ΔK 0 ( ) W ext 006. g W ext 0.54 J 006. g v v 6 v 6m/s 6.(D) ( ) 7.(A) Reading of C V {i in that branch 0} V Reading of A R R 8.(C) At t 0 inductor is open circuit i V R V V P.d. Va Vb ir + V + V VMC/0/Solutions Mock IIT Advanced/Test - /Paper-

7 9.(ABC) Vidyamandir Classes Moving plane is like a moving observer. Therefore, number of waves encountered by moving plane, v+ v0 c+ v f f f v c v c + v Frequency of reflected wave, f f f v vs c v v c c c v Wavelength of reflected wave λ f f f c+ v 0.(BD) C is decreased Q CV Q is decreased.(b) and U CV is also decreased and charge flows back in battery..(bcd) ωl 90 0 π π π ωc π circuit is inductive VL> VR voltage leads the current ωl ωc 80π 80π tanφ ; R R R At resonance, ω LC () If rod is in middle, i 0 F 0 x ε ε ε ρ( L x) ρ( L+ x) L x εx Eq. emf l + L ρ( L x) ρ( L+ x) L x + Req. ρ( L x) ρ( L+ x) ρ L L x ρ( L x ) Req. L εx L εx i ρ( L x ) ρ( L x ) + RL + R L εxlb εlb ma F il B x ρ( L x ) + RL ρl + RL εlb a x m( L + RL) ρ T π m( ρl + RL) lb ε T sec. 4.(6) By linear momentum conservation impulse (J) mv. l By angular momentum conservation, angular impulse J Iω l mv mvl So mv Iω or ω Il 6v ml l 6 rad/s VMC/0/Solutions Mock IIT Advanced/Test - /Paper-

8 5.() kq r r m(v ) So, kqr mv kq r 6.() 64 σt 4 ( πrl ) m ( ) V m ( V max) V max V ( 0 ) 0 m/s. V 0 m/s r 0 m mm 7.() By Gaus's law of magnetism. φt φ φ Wb Vidyamandir Classes 8.(9) x y t 0 6% N 0 0 t t % N 4% N N n n t T / 5 yrs. 9. [A-p, s] [B-p, s] [C-r, t] [D-q, r, t] P K constant Fv k mav k dv m v ds v k mv v S ks F k constant ma k dv mv ds k v S m dv ds k dv m v ds v k v t mv kt P Fv v t P v F v F andf E / S P S P t 40. [A-r, s] [B-q, s] [C-p, t] [D-p, s, t] VMC/0/Solutions Mock IIT Advanced/Test - /Paper-

9 Solutions to Mock IIT Advanced/Test - [Paper-]/0 [MATHEMATICS] (D) 4.(B) x cos + cos y θ cos θ xy x. y Now x y xy cosθ x 4 y xy cosθ 6 sin θ n 6 x y xy cosθ cos θ and ( cos x) ( sin x) ( cos x sin x) > 0 x, [ p,q) π p,q n 6 0, α 0 and ( ) ( ) q p. x > 0 x [ π π ] sin sinx e e x, 4.(C) Number of solution 4 α+ 0 n c gc c t + gct a 0 t + + t 4 x (i) x c t g (ii) g xx x tt tt tt t yy y t (iii) 4 x y g g (iv) G is the point, 4 4, C is the point ( g, g) c t + gct a t + gct+ c 0 g g M is the point,. Thus M and G coincide OG OM x xx xx x OG g a y y y OM 44.(B) Let f ( x) and g + ( x) x x x x ( + ) e π x A A.dx.dx ( ) + x x + x e π π e e x t t.dx +.dt.dt log + t xx xx c tt a a Putx iniindintegral t ( ) + x + t + t e π e e e loge VMC/0/Solutions 4 Mock IIT Advanced/Test - /Paper-

10 45.(B) z z z z z z z z z z z z Vidyamandir Classes 0 ( z z ) + ( z z ) + ( z z ) 0 z, z, z lies on the circle with centre at the origin (0, 0) z z z z z z z z z Let arg θ arg θ θ+ θ arg + arg arg z z z z z z z z z 46.(D) The unit vector along the angle bisector of vectors a and b a b is ± a b a b a b Consider positive sign and the vector l + m + n ˆ ( laˆ + mb+ ncˆ ) a b a b a,b,c ˆ ˆ ˆ are unit vectors along a,b,c The required projection ( laˆ mbˆ nc ˆ ˆ ). ( aˆ b ) n 47.(C) Case I : 0 < x < 0 n l m x as n f ( x) 0 Case II : x> x + as n f ( x) f has a jump (finite) discontinuity at x xsecθ ytanθ 48.(B) Equation of tangent is If it cuts the coordinate axes at A and B, then ( θ 0) ; B ( 0, cotθ) A cos, S cosθ cotθ ds π sinθ cos ec θ 0 sinθ θ dθ + 4 d S 0 dθ < for π θ 4 S is maximum 49.(ABC) Let y mx+ c be a chord of the given curve. Equation of the pair of lines through the origin and the points of intersection of the chord and the curve is y mx x y ( x 4y) 0 c If these are at right angles then coefficient of m c + c x + coefficient of m ( c+ ) y So the equation of the chord is y ( c ) x c ( y x) c( x ) Showing that the chord passes through the point of intersection of the lines y+ x 0 and x 0 through the point (, ) and the equation of the parabola in (C) is ( x ) 4( y+ ) whose vertex is also ( ) Note : The centre of the circle in (D) is (, ).,. VMC/0/Solutions 5 Mock IIT Advanced/Test - /Paper-

11 50.(BC) Let b xi ˆ+ yj. ˆ Since a is perpendicular to b 4 so 4x+ y 0. Thus b x ˆ i ˆj. Let c ui ˆ+ vj ˆ be the required vector. According to the given condition 4 ux c.a vx c.b 4u+ v 5. Also u 4v± 0 a b 6 x + 9 Solving these equations we have u and v or u,v (BCD) (A) False. P( TTT orhhh ) (B) ( ) c ( ) P A B P( A) P( A B) P( B) P( B) P( B) ( ) P( B) P( B) P( A) P( B) P( A B) P A B P A B P( A B) P( A) P( B) Hence, the given statement is true. (C) Let E be the event that white ball is drawn in first draw ; E be the event that black ball is drawn in first draw ; E be the event that white ball is drawn in second draw. E E E P( E) P P( E ) + P P( E ) E d+ w w w b + w+ b+ d w+ b w+ b+ d w+ b w d+ w b + w+ b w+ b+ d w+ b+ d w w+ b Which is independent of d. (D) To prove that A, B, C are pairwise independent only. Now, P( A B) P( I) + P( II) P( A B C) + P( A B C) P( A) P( B) P( C) + P( A) P( B) P( C) (given) ( ) ( ) ( ) ( ) + P A P B P A P B P C P C ( ) ( ) Similarly, for the other two. Hence, this statement is correct. A I II B C 5.(AC) tan4 cot5 cot + 0 cot.cot0 cot cot0 ( + )( ) + ( + + ) VMC/0/Solutions 6 Mock IIT Advanced/Test - /Paper-

12 5.(4) f ( x) Vidyamandir Classes ( 6+ )( ) ( 6+ )( + ) ( ) Hence, µ, λ 6, µ + λ x if x + x x if < x< + x x if x x x +, x> ( x + ) x x f ( x ), < x< ( x + ) x, x< f ( x) 0 gives x + or The function has a continuity at x As x,f ( x) 0 The graph is as shown below : dy dy Where, 0 i.e. it exists or is non-existent. dx dx The points x,, and + are the four critical points on the graph of this function. 54.(6) d xa + yb + zc Take dot product with b c, c a,a b one by one d b c a + d c a b + d a b c d a b c + + d b c a d c a b d a b c 6 d d b c a + d c a b + d a b c a b c VMC/0/Solutions 7 Mock IIT Advanced/Test - /Paper-

13 55.(0) (i) x x 0 56.() Vidyamandir Classes (ii) 6 x 0 x [ 4, 4] ( ) > 0 ( 0) ( ) x x x,, πx πx domain is {4} and Range is {5} (iv) sin 0 n π, ( n+ ) q q 5 and p 4 p I dx and x e ( + x) 0 π 4 tan θ (iii) π e sinθ I. dθ θ 0 tan θ cos x x e e e dx I dx dx x x e + x x ( ) ( ) I ei I e I 57.(9) z,pa PB Greatest value of w OA OP+ PA G 5+ And least value of w L 0 G+ L 5 9 ( ) 58.(6) F ( x) f ( x) g( x) h( x) + f ( x) g ( x) h( x) + f ( x) g( x) h ( x) q q q 0 p 4 p p 4 4 and k [ k] So F( x0) F ( x0) f ( x0) g( x0) h( x0) + f ( x0) g ( x0) h( x0) + f ( x0) g( x0) h ( x0) 4f ( x0) g( x0) h( x0) 7f ( x0) g( x0) h( x0) + kf ( x0) g( x0) h( x0) ( + k) f ( x ) g( x ) h( x ) ( + ) ( ) Hence + k k [A-q] [B-s] [C-p] [D-q] (A) Equation of common chord: x + y λ 0 Replace y ( λ x ) in equation of circle I : k F x 0 x + ( λ x) x ( λ x ) 0 ( ) x λ x + λ λ 0 (i) As exactly two common tangent possible roots of equation (i) must be real and unequal. ( ) ( ) (B) ( + ) (C) λ 4 λ λ > 0 0 < λ < 4 π π π/ sinx cosx dx 4 sinxdx + 8 cos xdx 6 π 0 0 dy x + y dy y + [Linear DE] dx x dx x dx I.F. e x x As curve passes through ( ) As it also passes through ( λ, ) x xy + C x, 6 / C 4 xy + 4 λ λ + 4 λ 4, VMC/0/Solutions 8 Mock IIT Advanced/Test - /Paper-

14 (D) DRs of line of intersection of planes x y 5z 6and x + 5y + z 4are(, 9, 4) DRs of normal to required plane: ( 5,, 7) A point in this plane: (,, ) equation of required plane: ( x ) ( y ) ( z ) 5x + y 7z 9 0 (i) As, ( λ, 6, ) lies on (i): 5λ λ 60. [A-r] [B-s] [C-p] [D-q] i z (A) is a circle z+ (B) z i ( i) z ( ) + + i z + SP e.pm, where S is the focus and ( i) z ( ) Note : Distance of point z 0 from line a az + az+ b 0 is e < ellipse + + i z + is equation of directrix az0 + az0+ b a (C) According to the question ( k)( k ) ( k )( k 7) < 0 k (, 7) < 0 k {, 4, 5, 6} (D) According to the question x + y + z x + y + z (i) and x y z x y...(ii) and y z...(iii) Solving (i), (ii) and (iii) we get : x,y,z VMC/0/Solutions 9 Mock IIT Advanced/Test - /Paper-

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