INDIA Sec: Jr IIT_IZ CUT-18 Date: 18-1-17 Time: 07:30 AM to 10:30 AM 013_P MaxMarks:180 KEY SHEET PHYSICS 1 ABCD ACD 3 AC 4 BD 5 AC 6 ABC 7 ACD 8 ABC 9 A 10 A 11 A 1 C 13 B 14 C 15 B 16 C 17 A 18 B 19 D 0 C CHEMISTRY 1 AB BD 3 ABC 4 C 5 ABD 6 ABCD 7 ACD 8 BCD 9 D 30 A 31 D 3 C 33 B 34 C 35 A 36 D 37 B 38 A 39 B 40 A MATHS 41 BD 4 AB 43 AB 44 BC 45 D 46 BC 47 AD 48 AB 49 C 50 B 51 C 5 A 53 A 54 B 55 C 56 B 57 C 58 B 59 B 60 A
SOLUTIONS PHYSICS F L 01 acm Torque about common center of mass F I 4m 4 Angular momentum conservation about point P ' ' L I I Mv 3 Net force acting on the box is equal to Mg 4 MA b A a 5 Relative velocity of object wrt image will be zero for the 3 rd time when particle reaches B in nd rotation 1 1 t t 5 sec 6 Conceptual 7 Conceptual 8 Conceptual 9 & 10 FBD showing impulsive forces along line of impact are shown in figure In horizontal direction, no external impulsive force is acting on system of spheres so we can conserve momentum in horizontal direction In vertical direction momentum can t be conserved as N (impulsive in nature)is acting on M For bigger sphere: N cosθ = Mv 11 Let m 1 and m do not accelerate up or down, then F1 m1 g, F m g But m1 m, so F1 F Hence net horizontal force on M is F1 F So M cannot be in equilibrium If M accelerates horizontally, then m 1 and m also accelerate horizontally Net force F 1 1 F m1 g mg a Net mass M m m M m m 1 1 15 & 16 Torque about left end Tb L Wx Torque about right end Tb L W L x Sec:Jr IIT_IZ Page
CHEMISTRY 1 Mol of Na atom in 53 gm NaCO3 = 53 106 (A) Mol of Na atom in 4 gm NaOH = 4 1 01 40 (B) Mol of Na atom in 585 gm NaCl = 585 01N A 585 (C) Mol of Na atom in 05 Mol NaSO4 = 05 = 05 NA 56 (D) Mol of Na atom in 56 gm of Na3PO4 = 3 N A 010N 164 c1 c, c c3 3 Conceptual 4 Conceptual 5 Conceptual 6 Conceptual 7 Conceptual 8 Conceptual NaOH CH3COOH CH3COONa H O 9 150 100 100mole 0 0 50 50 ka ph p log WA salt =47 30 H CO3 NaOH NaHCO3 H O 31 NaHCO3 NaOH NaCO3 H O N A A 3 Sec:Jr IIT_IZ Page 3
33 34 Conceptual 35 Conceptual 36 Conceptual 37 Y x z IF 7 1 6 SF 6 3 5 PCl 5 4 4 BF 3 1 4 38 In column I there are given 39 Conceptual 40 Conceptual MATHS Sec:Jr IIT_IZ Page 4
41 BD Given 3a1 7a 3a 3 4a 5 0 7 a a a 4a a a 1 3 1 3 5 4 7 1 r r 4 1 r r 7 4 r r 1 4r 4r 1 4 r 1 4 r 1 3 1 r, 4 AB X can not be odd integer for if x is odd, x is odd but px So x px q 0 X can not be even integer for if x is even, x So x px q 0 Also x p p q q is even; px is a multiple of 4 but q is not If x is fraction then x p is also a fraction but p q is an integer So, roots cannot be integer or rational numbers 43 AB PA PB PA PB with equality iff PA = PB But PAPB area PAB with equality iff APB 90 Hence PA PB PC PD (area PAB + area PBC + area PCB + area PDA) With equality iff APB DPC CPD DPA 90 and PA=PB=PC=PD Hence ABCD is square with centre X 44 BC 1 ka XY 1 ka XY 4 9 5 k, 17 17 45 D Sec:Jr IIT_IZ Page 5
y O P a, a n x n Slope of normal at, P a a is 1 n n a 1 n 1 x 0 to get y-intercept as b a n na 1 0, if n 1 lim b, if n a0, if n 46 BC 47 AD As n,n in radius sin n 1,1 and varying 48 AB 0 0 0 0 sin 47 sin 61 sin11 sin 5 sin54 cos7 sin18 cos7 0 0 0 0 0 0 0 cos 7 sin 54 sin18 0 5 1 5 1 cos 7 4 4 0 = cos 7 0 cos7 0 0 Clearly, cos7 cos15 0 Also, cos36 and cos1 (As 1 is in radian) 49 C Projection of x and y xy 1 tan / is x cos x y 1 tan / 50 B 1 Put n 1 na n Equation of normal is Y a X a tan / Area of parallelogram is given by x y sin x y 1 tan / 51 C k k k1 k Sol: If n n 1 Sec:Jr IIT_IZ Page 6
So, 5 A 1 0 k1 k1 k k n n 1 1 1 3 3 We have x x x x x If x k f, 0 f then 3 1 3 So, x x 3k 1 3k If 53 A x k f 1 x k 3 Sec:Jr IIT_IZ Page 7 and x k 1 1 1, then x k 1 3 3 and x k 1 1 3 So, x x 3k 3x If x k, then 1 3 1 x k 3 So, x 3k 3x f (1) 0 f (0) f (1) f (0) f (1) f (1) f () and x k Similarly we can prove f (n) f (n 1) Hence increasing 54 B We know f (1) f (1) f (1) f () Similarly, f (n) lim f (x) x 55 C 56 B Sol: Let A and B be the feet of perpendicular from the points A and B to the line of intersection of the two planes x 3 y 1 x 1 equation of AA is r 1 1
For A r r r ', 3 1 1 5 1 1 1 point A is (,, ) In the same way point B can be found as,0, 1 Let A be the point corresponding A when the plan e x y z 1 is rotated through an angle / about its line of intersection with the plane x y z 5 The perimeter of ABC would be minimum when A, C, B are collinear From the similar triangles AA C and BB C, we can observe that C divides AB in the ratio P1 : P where 1, P P are the perpendicular distance from A, B to the planes x y z 5 and x y z 1 respectively 6 11 5 3 3 P1 6 6 and P 3 1 0 1 3 3 3 3 57 C (A) p 3q 5r 3q5r p5r p3q p 3q 3q 5r 5r p 0 p 3q 5r x(say) x x p x,q and r 3 5 (B) log r p log p r log p q r r p p r p q r 1 1 q p r (C) f ' q f '(q) f '(r) pq q p q pr q q p r (D) P 5,q 8and r 1 58 B (C) The point (6, 8 ) divides the line segment joining (, 4) and (8, 10) in the ratio : 1 internaly The harmonic conjugate of (6, 8) divides the line segment joining (, 4) and (8, 10) in the ratio : 1 externally the point is (14, 16) (D) let be the angle of which line segment AB taken in the direction from A to B makes with the positive direction of x-axis then Sec:Jr IIT_IZ Page 8
1 1 cos and sin Let the coordinates of the new positions of B be (x, y) then x cos 90 sin 1 4 y 4 1 x and sin 90 cos 4 y 0 59 Consider the multiplication of two quadratic expression P(x) x ax c x bx d 4 3 P(x) x 8x 3x 8x 1 Or P(x) x 1x x 3 P(x) x 4x 3 x 4x 4 or P(x) x 3x x 5x 6 So the ordered solutions for (a, b, c, d) can be (4, 4, 3, 4); (4, 4, 4, 3); (3, 5,, 6); (5, 3, 6, ) 60 A sin1 sin 5 (A) sin sin 6 sin x Take f (x) sin x 1 sin(x 1) cos x cos(x 1) sin x sin1 f '(x) 0 f (x) is increasing f (1) f (5) (B) take f (x) tan x x f '(x) sec x x f ''(x) sec x tan x sin x 1 sin x 1 f '(1) 0 and f ''(x) 0 for x > 1 f (x) is increasing in 1, 3 9 tan 4 x e 1 (C) lim x0 x does not exist as left hand and right hand limits are not equal (D) f '( ) 0 and f '(x) 0 x R n f '(x) x g(x) f ''( ) 0 Sec:Jr IIT_IZ Page 9