ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST VII (Paper-2)

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1 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/ FIITJEE Students From All Programs have bagged in Top 00, 66 in Top 00 and 7 in Top 00 All India Ranks. FIITJEE Performance in JEE (Advanced), 0: FIITJEE Students from Classroom / Integrated School Programs & 79 FIITJEE Students from All Programs have qualified in JEE (Advanced), 0. FIITJEE JEE(Advanced), 0 ALL INDIA INTEGRATED TEST SERIES ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST VII (Paper-) Q. No. PHYSICS CHEMISTRY MATHEMATICS. C A D. A A C. D A B. D D A. C B D 6. C A A 7. B C A 8. A C A 9. A C B 0. D A A. A A C. B A A. A C B. A C A. A A D 6. C D B 7. C C C 8. C D B 9. D B A... (A) (q) (B) (q) (C) (p) (D) (q) (A) (q) (B) (r) (C) (p) (D) (s) (A) (q, r) (B) (q, r) (C) (q, r) (D) (p, q) (A) (s) (B) (q) (C) (p) (D) (r) (A) (r) (B) (q) (C) (p) (D) (q) (A) (q) (B) (p) (C) (r) (D) (s) (A) (r) (B) (s) (C) (p) (D) (r) (A) (q) (B) (r) (C) (s) (D) (p) (A) (p, r, s, t) (B) (q) (C) (p, q, r, t) (D) (p) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

2 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/ Phsics PART I. In equilibrium, there will not be an friction between the clinder and the wedge. If be the required angle Clinder A: mg sin 60º = kx cos (60º ) ( = mg cos 0º) Clinder B: mg sin 0º = kx cos (0º + ) = kx sin (60º ) cot 0º = cot (60º ) = 0º.. N mv R f max = mg N = mg mv R = mg v = Rg 0 0. = 0 m/s. The impulse due to friction = k impulse delivered b collision. The z-component of the force and the x-component of displacement are ineffective here. dw F d x.d z 0 = 6x dx ( = x ) Integrating between x = 0 and x = gives the result.. km/hr = = 0 m/s 8 V V 0 sin 7 sin(90 7 ) 0 sin7 cos 7 V V = 8 m/s cos7 0 m/s 7 N V V 7 0 m/s E Change in momentum = 8 = 6 6 f avg = 800 N Solve the problem using as the coefficient of kinetic friction. 8. If if u cos g u sin g tan u cos g u sin g tan. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

3 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/. Time after which velocit vector becomes perpendicular to initial velocit vector is u 0 t seconds gsin 0 sin 60 Let v be the vertical component of velocit at that instant then v = u + at 0 v = v v v gcos R v R gcos 0 R m. m/s 0 m/s 60 m/s 0 v m/s v x = v 0 v x v = v = 0 v 6. a t = g sin 0 = m/s T 9. g cos and T FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

4 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/ Chemistr PART II SECTION A. [C][D] Q 0 [A][B] Kc 0 Q K c Therefore equilibrium will shift from left to right.. Buffer capacit = Buffer capacit = moles of acid or base added per litre change in ph HBr CH C HONa H C Peroxide anti M.K. addition H C Br SN H C OC H CH 6. will be most reactive in electrophilic substitution reaction due to hperconjugation. It is having three hdrogen for hperconjugation. 7. Me The product is H H Br Br 8. A(g) B(g) C(g) Me (Meso compound) P at initial P x x x at 0 minutes 0 P x x x 00 0 P 00 x 0 0 It shows 0 minutes is the half life time of the reaction. 9. u RMS = RT M P M / V m P FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

5 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/. Hdrocarbon (C 0 H 6 ) CH H C CH H C cis trans SECTION B. Reaction (a) is ECB, hence RI is a carbanion. Reaction (b) is E, hence RI is a carbocation. Reaction (c) is E, hence RI is a TS. Reaction (d) is E, hence RI is a carbocation. FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

6 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/ 6 Mathematics PART III SECTION A. All the co ordinates of vertices are at a distance of units from origin. Hence circumcentre of the triangle is (0, 0) cos sin sin cos Centroid =, Centroid divides the line joining circumcentre and orthocentre in the ratio : Suppose co-ordinate of orthocentre is (x, ) x = + cos + sin... () = + sin cos... () x 7 x Solving () and () sin cos 0 0 (x + 7) + (x + + ) = 00 Replace x b x and b Locus of orthocentre in (x + 7) + (x + ) = 00. P(A) P(B) P(A) P(B) 6 P(A) P(B) a b a b. For arg 0 0 a c a c Select numbers from the set and call the greatest as a and remaining two as b and c. This can 0 be done in 0 C C was. Therefore the probabilit shall be 0 C!. Put x and solve for C. f(x) = (x a)(x c) + (x b) a So f a b c f c b a c Now f.f (a b)(c b) 0 Hence exactl one root lies between c and a 6. Let P(x, ) be an point on the parabola. Then b definition (x 0) ( ) x FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

7 7 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/ 7. z z z z zzzz 60 z z z z Now, zz,zz,z z and zz z z z z z z z z k k 8. PF PF,PF.PF Let PF x,pf x + =, x = (FF ) 0 (FF ) FF 9. Let us draw tangent at P and Q which meet in T. Let Pat,at,Qat,at and Rat,at So T att, at t An circle which passes through points P, Q and R must also pass through the points T as the four points are concclic Circumcentre is mid-point of TR that is M(h, k) As normal chords at P and Q meet in R t t t tt t t a at a a(t t ) h bt 0 k t t a From equation () and () we get k (h a) a a(x a) a 0. Let P(h, 0) be a point on axis of parabola the straight line passing through P cuts the parabola at a distance r (r sin ) = a(h + r cos ) r sin (a cos ) r ah = 0.(i) acos Where, r r sin ah and r r sin r r cos sin PQ PR r r r r h ah Which is constant onl, if h = ah i.e. h = a cos sin a a a Thus, constant for all chords QPR, if h = a PQ PR Hence, (a, 0) is the required point on the axis of parabola Statement- and statement- are true and statement- is correct explanation of statement- FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

8 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/ 8. f(x) = (x ) (x + ) is a multiple root of order f(x) = x 0x + f(x) = 6x 0 f() = 0, f() 0 Statement- is true, statement- is false. Statement- is correct as through x x + is non-differentiable at x, but cos x = 0 these points. So, f and f exists. If each of x,, z is less than / then, statement- is obviousl true Also, x + + z = = (x + + z) = The sum of the three given number is positive also at must one of x,, z can be more than If one of x,, z is more than or equal to, then their product is less than equal to zero hence still remains true Statement- is alwas true but it does not explain statement- SECTION B. (A) Write the equation of normal and use the given condition b Normal at ae, is a (C) b x ae a b ax ae ea ae b a a a b b x e ea a a b a e It passes through (0, -b), b ae a a b a e a e a e x e e e e (D) Find the line in terms of m and satisf the condition to solve for m An line through (, ) is = m(x ) According to given condition m 6 8 s m 0 m = m Y B O L ae, b / a X S A B L (0, b) FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

9 9 AIITS-HCT-VII-(Paper-)-PCM(S)-JEE(Advanced)/. We have, P(A B) P(A) P(B) (A) P(A B) P(A) P(B) P(A B) P(A B ) P(B) P(A B) = /6 (B) (C) P(BA) = / (D) P(A B) P(A).P(B) /. (A) x d = dx + d xd dx d x c put x =, = c = x x (B) d t dt t t t dt t tn(t) t I.f e e (t )e x d d Solution is (t + )e t = e t + c put t = 0 and = c = 0 e = e (C) (x + )d = x dx d x put = vx, d v x dv dx x dx dx nv nx c v x c n nx put = e x x e (D) d 0 dx x x = C put x =, = and we get C = put x = FIITJEE Ltd., FIITJEE House, 9-A, Kalu Sarai, Sarvapria Vihar, New Delhi -006, Ph , 6699, Fax 69

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