ANSWERS, HINTS & SOLUTIONS HALF COURSE TEST VII PAPER-2

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1 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7 In JEE dvanced 06, IITJEE Students bag 6 in Top 00 IR, 75 in Top 00 IR, 8 in Top 500 IR. 5 Students from Long Term lassroom/ Integrated School Program & Students from ll Programs have qualified in JEE dvanced, 06 IITJEE LL INDI INTEGRTED TEST SERIES NSWERS, HINTS & SLUTINS HL URSE TEST VII PPER- NSWERS KEY Q. No. PHYSIS HEMISTRY MTHEMTIS. D... D 5. D 6. D 7., D, D, 8.,, D,,, D 9.,,, 0. D. D D.. D... () (q) () (r) () (p) (D) (s) () (r) () (p) () (q) (D) (s) JEE(dvanced)-07 () (r, s, t) () (p, r) () (q, s) (D) (r, s) () (s) () (r) () (p) (D) (q) () (s, t) () (r) () (t, p) (D) (p, q, r, s, t) () (q, r, s) () (q) () (r, s) (D) (p) IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659

2 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7 Physics PRT I SETIN. X = H H = H. uoyant force = Weight of liquid displaced = Viscous force = Stoke s drag force = 6 rv 6rv r g r g. Total mechanical energy = gravitational P.E. + K.E., total mechanical energy is conserved.. alculate the field due to a semi-circular ring and then use superposition principle. 6. orce dm V dt 8. f max > 0 N 9. or : W = nrt 0 ln RT0 ln Q W RT0 ln or, W = nrt 0 = RT Q = n R T0 RT0 or (isochoric process), W = 0 Q = RT 0 T 0 T 0 P 0 P 0 P 0. U T V P P. : U is increasing T is increasing, V is constant : Isobaric with T decreasing V/T is constant Straight line passing through the origin. : Isothermal with P decreasing V increasing. V T. Time after which velocity vector becomes perpendicular to initial velocity vector is u 0 t seconds gsin 0 sin 60 Let v y be the vertical component of velocity at that instant then 5 m/s 0 m/s 60 5 m/s v y 0 v x v IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659

3 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7 v y = u + at 0 v y = 5 5 v y v y = 5 5 v 5 v gcos R v R gcos 0 R m. 0 v m/s v x = 5 v = 0 SETIN. Gm Gm R R mv GM m GM m GM m e e e. E Re Re R e R e GMe GMem E mv0 m.. Re R R e. D L D L T T D T. U = = 50 J ( ) 50(5 / ) 00 T mc nr nr nr V molar specific heat Q 600 R R nt Stress = Y strain = Y strain IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659

4 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7 hemistry PRT II SETIN. n,, m h 0 angular nodes angular nmomentum spherical nodes. Li has highest hydration enthalpy which compensates its high I.E. there by accounting high reducing power.. Moles = Moles = mass molar mass 0 Mass of N + mass of + mass of H = Mn Mn ; nf 5 r r ; nf 0 0. ree expansion work done is 0.. Partial pressure = mole fraction P Kr SETIN. x = 0, y =, z = H H l l H H H H IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659

5 5 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7. l l l l l l l l 6 = 6. ph pk pk pk w a b. LiN Li N 5. Si Si Si IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659

6 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7 6 Mathematics PRT III SETIN. Perimeter of regular polygon of n sides = ansin n (a = distance of centre from any vertex) If radius of circle is r, then its perimeter = circumference = r a ansin = r n r Now required ratio = nsin n r n a sin n = tan n. / n. (x) + (y) + (z) (x)(y) (y)(z) (x)(z) = 0 [(x y) + (y z) + (z x) ] = 0 x = y, y = z, z = x y = x, z = x y x x x x z x, y, z are in H.P.. (x + )(y + ) = 5 (x + )(y + ) = 6 (y + )(z + ) = 8 (y + )(z + ) = 9 (z + )(x + ) = (z + )(x + ) = ((x + )(y + )(z + )) = 6 9 (x + )(y + )(z + ) = z + = 6 z + = z = 0, z = z z =.. P = R(cos + cos + cos) since < cos + cos + cos R < P R. 5. irst boys in! Ways Let x, x, x be no. of places for girls in gaps x + x + x = such that x ; x ;. x X + X + X = (X = x,.) No. of ways 5 Total no. of ways = 5!! x x x IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659

7 7 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7 6. (r r) r r Tr r (r ) r(r ) r(r ) r r 0 r 0 r r 0 7. sin x(sin x )( sin x ) 0 8. learly r = 9, 9 ab 0 5 a 5 b 5 5 Now ab a b a b a b ab a b ab a b ab 8 a b 0 a = a = b 8 b a 5,b a 6,b 8 a b 8 rdered pair (a, b) = (5, ) or (6, 8) Two ordered pair. or maximum perimeter (a, b) = (5, ) Maximum perimeter 5++ = 0 5. ap, bp, cp p, p, p a b c a b c p p p c D a E b D = p E = p = p. a b c a b c s s p p p r a b c. LHS = cos cos cos = (R sin cos + R sin cos + R sin cos ) IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659

8 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7 8 = R (sin sin sin) = R (sin sinsin) = R a b sin = ab sin R R R = R R. () x x (8 + a) + 0a + 8 = 0 Discriminant = a a a = 0, () If =, = z arg z z z arg z z SETIN (z ) (z ) () (z ) r 00 r 00 r r. r 00. r r. r r 0 r0 r 0 = = 0 (D) If ax + (a ) x a = 0, a a x ; i.e. x =, a a a If a > 0, -a < x < a and - x gives a R+ If a < 0, < x < -a and - x gives a R a or a = 0, -x < 0 and - x gives > x > 0 So a can be any real numbers.. () () () (D) (sin x) (sin y) (sin x) (sin y) sin x, sin y x + y =, 0, (cos x) (cos y) (cos x) (cos y) x = y = x 5 + y 5 = (sin x) (cos y) (sin x) and (cos y) (sin x) and x = and y = x y = 0, sin x sin y (cos y) IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659

9 9 IITS-HT-VII (Paper-)-PM(Sol)-JEE(dvanced)/7 sin x and sin y or y () x ( ) or ( ) = or sin x and sin y SETIN. r = 7 entres are (8, 0) and (, 0) r = d = = PQ Q r + r < d R = 90 PQ d (r r ) R Q r r (7 ) 0 9 r P. 8 8 (7 ) (7 ) 9k (8 7). Equation has roots, but when x + 8x + 0 is ve solution rejected), k = 0. D E. r r 8 =, = R R 0 6 R R =6, R 6 E r D learly a n is an.p. with. IITJEE Ltd., IITJEE House, 9-, Kalu Sarai, Sarvapriya Vihar, New Delhi -006, Ph , 65699, ax 659