XtraEdge Test Series # 4
Based on New Pattern
Time : 3 Hours
Syllabus (Revision – 1) : Physics : Essential Mathematics, Vector, Units & Dimension, Motion in One dimension, Projectile motion, Circular motion. Chemistry : Mole Concept, Chemical Bonding, Atomic Structure, Periodic Table. Mathematics: Trigonometric Ratios, Trigonmetrical Equation, Inverse Trigonmetrical Functions, Properties of Triangle, Radii of Circle
Instructions :
Section - I
• Question 1 to 8 are multiple choice questions with only one correct answer. +3 marks will be awarded for correct answer and -1 mark for wrong answer.
Section - II
• Question 9 to 12 are passage based single correct type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer.
Section - III
• Question 13 to 18 are passage based single correct type questions. +4 marks will be awarded for correct answer and -1 mark for wrong answer.
Section - IV
• Question 19 to 20 are Column Matching type questions. +8 marks will be awarded for the complete correctly matched answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer..
8. sin2x dx
d
(A) (sin 2x)–1/2 (B) cos 2x (sin 2x)–1/2 (C) 2 cos 2x (sin 2x)–1/2 (D) cos 2x (sin 2x)1/2 Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.
9. The units of electrical permittivity
= π
ε 2
2 0 4 Fr
q
are-
(A) N–1m–2C2 (B) Nm–2C2 (C) C2/Nm2 (D) N/Cm2
10. If →a and →b are two vectors with |→a|= |→b| and
| b a
|→+→ + |→a–→b| = 2 |→a|, then angle between →a and →b -
(A) 0º (B) 90º (C) 60º (D) 180º 11. Two bodies P and Q are moving along positive x-axis
their position-time graph is shown below if V→PQ is velocity of P.w.r.t Q and V→QP is velocity of Q w.r.t P then –
x
t P
Q
(A) |V→PQ| = |V→QP | constant (B) V→PQ is towards origin (C) V→QP is towards origin
(D) V→PQand V→QP both can be towards origin at same time
12. A particle of mass m is released from height h on a smooth curved surface which ends into a vertical loop of radius r as shown. If h = 2r then,
O r θ h
m u = 0
(A)The particle reaches the top of the loop with zero velocity
(B) The particle cannot reach the top of the loop (C) The particle breaks off at a height h = r from base (D) The particle breaks off at a height r < h < 2r This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.
Passage : I (Ques. 13 to 15)
P O
R
θ
→B
→A
| A
|→ = a |B→| = b
The second law of vector addition is triangle law, which says that if we take →A and →B as two vectors acting at point O as shown in figure, then the resultant of vector is get by taking →A and →B as adjacent sides of a triangle and the 3rd side of the triangle as the resultant, then if θ is angle between →A and →B then.
13. if α is the angle made by resultant vector with →A; then tan α =
(A) + θ
θ cos A B
sin
A (B)
θ +
θ cos B A
sin
B
(C) θ +
θ cos B A
cos
A (D)
θ +
θ sin A B
cos
B
14. If the magnitude of both the vector |A→| & |→B| is A, then the resultant will have magnitude -
(A) A cos θ/2 (B) 2A cos θ/2 (C) 3A cos θ/2 (D) 3A cos θ/3
15. If |A→| = |→B| = a and θ = 120º, then the two vectors and the resultant will form a -
(A) Acute angle triangle (B) Obtuse angle triangle (C) Right angle triangle (D) Equilateral triangle
Passage: II (Ques. 16 to 18)
Height of a tower is 80 m. Two balls A and B are thrown simultaneously. Ball A is thrown upwards with speed u from top of tower while ball B is thrown upwards with speed of 50 m/s from the foot of tower.
16. Ball A will reaches ground in 8 sec if ball do not collide then u is equal to -
(A) 20 m/s (B) 25 m/s (C) 30 ms/s (D) 35 m/s 17. If the balls meets in air then the height from foots of
tower is -
(A) 40 m (B) 80 m (C) 120 m (D) 160 m 18. The time after which balls will meet in air is -
(A) 3 sec (B) 4 sec (C) 5 sec (D) 6 sec This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T;
B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :
A B C D
P
Q R S T
S T P
P P Q R
R R Q Q
S S T
T P Q R S T
Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.
19. Match the column :
Column-I Column-II
(A) (sin θ + cos θ)2 (P) 1 – sin 2θ (B) (sin θ – cos θ)2 (Q) 1 + sin 2θ (C) cos4θ – sin 4θ) (R) cos 2θ (D) cos4θ + sin4θ (S) 1 +
2 2 sin2 θ (T) None of these
20. The displacement - time graph of a body moving on a straight line is given by
x
0 T 2T t Parabola
Column-I Column-II
(A) (P)
T 2T
(B) (Q)
T 2T
(C) (R)
(D) (S)
T 2T
(T) None of these
C HEMISTRY
Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.
1. According to Bohr’s theory, angular momentum of an electron in fourth orbit is -
(A) π 2
h (B)
π 4
h (C)
π h
2 (D)
π h 4
2. Calculate the maximum no. of possible e– for which 4 < n + l ≤ 6 -
(A) 18 (B) 36 (C) 72 (D) 4
3. If the De –Broglie wavelength of an electron in first Bohr's orbit be λ then the minimum radial distance between the electrons in the first and second Bohr's orbit is –
(A) λ (B) 2
λ (C) 2λ (D) π λ 2 Velocity – time
graph
Acceleration-time graph
Distance – time graph
Speed – time graph
4. Hexamethylenediamine, C6H16N2, is one of the starting materials for the production of nylon. It can be prepared from adipic acid C6H10O4, by the following reaction
C6H10O4(l) + 2NH3(g) + 4H2(g) →
C6H16N2(l) + 4H2O(l)
If 385 g of hexamethylenediamine is made from 5.00 × 102 g of adipic acid, the percent yield is (A) 24.2 % (B) 75.0%
(C) 96.9 % (D) 99.9%
5. One mole of a mixutre of CO and CO2 requires exactly 20 g of NaOH to convert all the CO2 into Na2CO3. How many more grams of NaOH would it require for conversion into Na2CO3, if the mixture (one mole) is completely oxidised to CO2 ?
(A) 60 g (B) 80 g (C) 40 g (D) 20 g
6. Rutherford’s experiment, which estabilished the nuclear model of the atom, used a beam of -
(A) β–particles, which impinged on a metal foil and got absorbed
(B) γ–rays, which impinged on a metal foil and ejected electrons
(C) helium atoms, which impinged on a metal foil and got scattered
(D) helium nuclei, which impinged on a metal foil and got scattered
7. Lattice energy of BeCO3 (I) , MgCO3 (II) and CaCO3 (III) are in the order -
(A) I > II > III (B) I < II < III (C) I < III < II (D) II < I < III
8. Specify the coordination geometry around the hybridization of N and B atoms in a 1 : 1 complex of
BF3 and NH3 - [IIT- 2002]
(A) N : tetrahedral, sp3 ;B : tetrahedral, sp3 (B) N : pyramidal, sp3 ; B : pyramidal, sp3 (C) N : pyramidal, sp3 ; B : planar, sp2 (D) N : pyramidal, sp3 ; B : tetrahedral, sp3
Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.
9. Select the correctly presented graph if v = velocity of e in Bohr's orbit r = radius of Bohr's orbit
U = potential energy of e– in Bohr's orbit
(A) n2 r
(B) 1/n2
U
(C)
n2 T
(D)
1/n v
10. Which of the following have the same mass ? (A) 0.1 mole of O2 gas
(B) 0.1 mole of SO2 gas
(C) 6.023 × 1022 molecules of SO2 gas (D) 1.204 × 1023 molecules O2 gas
11. If n and l are principal and azimuthal quantam no.
respectively, then the expression of calculating the total no. of electrons in any energy level is :
(A)
∑
== n +
0
) 1 2 (
l 2
l
l (B)
∑
== 1 +
– n
1
) 1 2 (
l 2
l
l
(C)
∑
= += 1 +
n
0
) 1 2 (
l 2
l
l (D)
∑
== 1 +
– n
0
) 1 2 (
l 2
l
l
12. Which of the following is/are the correct order of mobility ?
(A) Li+ < Na+ < K+ (B) Na+ < Mg2+ < Al3+
(C) Al3+ < Mg2+ < Na+ (D) Li+ > Na+ > K+
This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.
Passage : I (Ques. 13 to 15)
A black coloured compound (A) on reaction with dil H2SO4 form a gas 'B' and a solution of compound (C). When gas B is passed through solution of compound (C), a black coloured compound 'A' is obtained which is soluble in 50% HNO3 and forms blue coloured complex 'D' with excess of NH4OH and chocolate brown ppt. 'E' with K4[Fe(CN)6] 13. 'A' is
(A) CuS (B) FeS
(C) PbS (D) HgS
14. 'D' is
(A) Cu(OH)2 (B) [Cu(NH3)2]SO4
15. 'E' is
(A) Cu2[Fe(CN)6] (B) [Cu4[Fe(CN)6] (C) Cu3[Fe(CN)6]2 (D) None of these Passage: II (Ques. 16 to 18)
Spin angular momentum of an electron has no analog in classical mechanics. However, it turns out that the treatment of spin angular momentum is closely analogous to the treatment of orbital angular momentum.
Spin angular momentum = s(s+1)h Orbital angular momentum = l(l+1)h Total spin of an atom or ion is a multiple of
2 1. Spin multiplicity is a factor to confirm the electronic configuration of an atom or ion.
Spin multiplicity = (2Σs + 1).
Answer the following questions :
16. Which of the following electronic configurations have four spin multiplicity ?
(A) (B)
(C) (D)
17. In any subshell, the maximum number of electrons having same value of spin quantum number is : (A) l(l+1) (B) l + 2
(C) 2l + 1 (D) 4l + 2
18. The orbital angular momentum for a 2p-electron is : (A) 3h (B) 6h
(C) zero (D)
2π 6 h
This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T;
B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :
A B C D
P
Q R S T
S T P
P P Q R
R R Q Q
S S T
T P Q R S T
Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.
19. Column-I Column-II (Ionic species) (Shapes)
(A) XeF5+ (P) Tetrahedral (B) SiF5– (Q) Square planar (C) AsF4+ (R) Trigonal bipyramidal
(D) ICl4– (S) Square pyramidal
(T) Octahedral
20. Column –I Column II
(A) If P.E. = –13.6 eV (P) 21 (B) Ionization energy of (Q) 10 electron from 2nd shell
of Na10+
(C) Number of spectral line (R)T. energy when electron jumps form = – 6.8 eV 7th to 3rd shell
(D) Number of spectral lines (S) 411.4 eV when electron comes form
7th shell to 1st shell
(T) zero
MATHEMATICS
Questions 1 to 8 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.
1. If x ∈
π π 2 2 ,
3 then value of the expression sin–1 (cos(cos–1(cosx) + sin–1(sinx))) equals
(A) – 2
π (B)
2 π
(C) 0 (D) None of these
2. If median AD of a ∆ABC divides the angle BAC in ratio 1 : 2 then
C sin
B
sin is equal to
(A) 3
secA 2
1 (B)
3 cosA 2
1
(C) 3 ecA 2cos
1 (D) None of these
3. If distance between incentre & one of the excentre of equilateral triangle is 4 unit. Then inradius of triangle is -
6. A circle is inscribed in a triangle ABC touching the side AB at D such that AD = 5, BD = 3. If ∠A = 60º circumference in anticlockwise direction & (r, s) is at a distance of 2θ from (p, q) along circumference in anticlockwise direction, then expression sp3 + q3r equals -
Questions 9 to 12 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which MULTIPLE (ONE OR MORE) is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong
9. In a ∆ABC a semi-circle is inscribed, whose diameter lies on side c. If x is length of angle bisector through angle C, then radius of semi-circle is - (A) 4R (sinA sinB)
This section contains 2 paragraphs; each has 3 multiple choice questions. (Questions 13 to 18) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 4 marks will be given for each correct answer and – 1 mark for each wrong answer.
Passage : I (Ques. 13 to 15)
15. Value of tan A, tan B and tan C are -
This section contains 2 questions (Questions 19, 20).
Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T;
Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and No Negative marks for wrong answer.