Solution - Career Point

32. Which one of the following is not an amphotericsubstance -−(A) HNO 3 (B) HCO 3 (C) H 2 O (D) NH 333. Which reaction cannot be used for the production ofhalogen acid –(A) 2KBr + H 2 SO 4 → K 2 SO 4 +2HBr(B) NaHSO 4 + NaCl → Na 2 SO 4 + HCl(C) NaCl + H 2 SO 4 → NaHSO 4 + HCl(D) CaF 2 + H 2 SO 4 → CaSO 4 + 2HF34. B(OH) 3 + NaOH NaBO 2 + Na[B (OH) 4 ] + H 2 OHow can this reaction is made to proceed in forwarddirection -(A) Addition of cis 1, 2-diol(B) Addition of borax(C) Addition of trans 1, 2-diol(D) Addition of Na 2 HPO 435. Sodium thiosulphate is prepared by -(A) Reducing Na 2 SO 4 solution with H 2 S(B) Boiling Na 2 SO 3 solution with S in alkalinemedium(C) Neutralising H 2 S 2 O 3 solution with NaOH(D) Boiling Na 2 SO 3 solution with S in acidic medium36. The critical temperature of water is higher than thatof O 2 because H 2 O molecule has -(A) Fewer electrons than oxygen(B) Two covalent bond(C) V-shape(D) Dipole moment37. Zone refining is a technique used primarily for whichone of the following process -(A) Alloying (B) Tempring(C) Sintering (D) Purification38. Which one of the following elements has the highestionization energy –(A) [Ne] 3s 2 3p 1 (B) [Ne] 3s 2 3p 2(C) [Ne] 3s 2 3p 3 (D) [Ar] 3d 10 4s 2 4p 239. The correct order of dipole moment is -(A) CH 4 < NF 3 < NH 3 < H 2 O(B) NF 3 < CH 4 < NH 3 < H 2 O(C) NH 3 < NF 3 < CH 4 < H 2 O(D) H 2 O < NH 3 < NF 3 < CH 440. If N x is the number of bonding orbitals of an atomand N y is the no. of the antibonding orbitals, then themolecule/atom will be stable if -(A) N x > N y (B) N x = N y(C) N x < N y (D) N x ≤ N yMATHEMATICS1. Consider the sequence (angles are measured inradians) sin log 10 2 , sin log 10 3 , sin log 10 4 ….then -(A) all the terms of this sequence are positive(B) all the terms of this sequence are negative(C) 1001 th term is negative(D) 10001 th term is negative2. The order relation between x, sin –1 x & tan –1 xx ∈(0 ,1) is -(A) tan –1 x < x < sin –1 x (B) sin –1 x < tan –1 x < x(C) x < sin –1 x < tan –1 x (D) None3. The smallest positive valve of x satisfying theequation log 2 cos x + log 2 (1 – tan x ) + log 2 (1 + tan x)– log 2 sin x = 1 is -(A) π/8 (B) π/6 (C) π/4 (D) π/64. A pole stands at a point A on the boundary of acircular park of radius r and subtends an angle α atanother point B on boundary. If arc AB subtends anangle α at the centre of the path, the height of thepole is -(A) r sin α/2 tan α (B) 2r sin α/2 tan α(C) 2r sin α/2 cot α (D) None of these5. The base of a triangle lies along the line x = a and isof length 2a. The area of the triangle is a 2 . If the thirdvertex lies on the line -(A) x = 0(B) x = – a(C) x = 2a, or x = 0 (D) x = 0 or x = – 2a6. If y = mx bisects an angle between the linesax 2 – 2hxy + by 2 m 2 –1= 0 then =mb – a b – b a + b(A) (B) (C) (D) Noneh h h7. If the circle x 2 + y 2 + 2gx + 2fy + c = 0 passesthrough all the four quadrant then -(A) g = – b (B) C > 0 (C) C < 0 (D) None8. The equation of the circle which has two normals(x–1) (y – 2) = 0 and a tangent 3x + 4y = 6 is(A) x 2 + y 2 – 2x – 4y + 4 = 0(B) x 2 + y 2 + 2x – 4y + 5 = 0(C) x 2 + y 2 = 5(D) (x –3) 2 + (y – 4) 2 = 59. Circles drawn on the diameter as focal distance ofany point lying on the parabola x 2 – 4x + 6y + 10 = 0will touch a fixed line whose equation is(A) y = 2 (B) y = –1(C) x + y = 2 (D) x – y = 2XtraEdge for IIT-JEE 82APRIL 2010