FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com SAMPLE TEST - 3 JEE - MAIN 2017 SET - A Time Allotted: 3 Hours Maximum Marks: 360 Do not open this Test Booklet until you are asked to do so. Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. Important Instructions : 1. Immediately fill in the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited. 2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out the Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three sections in the question paper I, II, III consisting of Physics, Chemistry and Mathematics having 30 questions in each section of equal weightage. Each question is allotted 4 (four) marks for correct response. 6. Candidates will be awarded marks as stated above in instruction No.5 for correct response of each question. ¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet. 7. There is only one correct response for each question. Filling up more than one response in any question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above. 8. Use Blue / Black Ball Point Pen only for writing particulars / marking responses on Side-1 and Side-2 of the Answer Sheet. Use of pencil is strictly prohibited. 9. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination hall / room. 10. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the Room / Hall. However, the candidates are allowed to take away this Test Booklet with them. 11. Do not fold or make any stray marks on the Answer Sheet. Name of the Candidate (in Capital Letters) :_____________________________________ Enrolment Number :_________________________________________________________ Batch :________________________ Date of Examination : ________________________ FIITJEE - JEE (Main)
Do not open this Test Booklet until you are asked to do so. Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions:
1. Immediately fill in the particulars on this page of the Test Booklet with Blue / Black Ball Point Pen. Use of pencil is strictly prohibited.
2. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take out
the Answer Sheet and fill in the particulars carefully. 3. The test is of 3 hours duration. 4. The Test Booklet consists of 90 questions. The maximum marks are 360. 5. There are three sections in the question paper I, II, III consisting of Physics, Chemistry and Mathematics
having 30 questions in each section of equal weightage. Each question is allotted 4 (four) marks for correct response.
6. Candidates will be awarded marks as stated above in instruction No.5 for correct response of each question.
¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.
7. There is only one correct response for each question. Filling up more than one response in any question will
be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 6 above.
8. Use Blue / Black Ball Point Pen only for writing particulars / marking responses on Side-1 and Side-2 of the
Answer Sheet. Use of pencil is strictly prohibited. 9. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone,
any electronic device, etc. except the Admit Card inside the examination hall / room. 10. On completion of the test, the candidate must hand over the Answer Sheet to the Invigilator on duty in the
Room / Hall. However, the candidates are allowed to take away this Test Booklet with them. 11. Do not fold or make any stray marks on the Answer Sheet.
Name of the Candidate (in Capital Letters) :_____________________________________
Enrolment Number :_________________________________________________________
Batch :________________________ Date of Examination : ________________________
Section – I (Physics) 1. If L, R, C, V respectively represent inductance, resistance, capacitance and potential difference,
then the dimensions of L
RCV are the same as those of
(A) Current (B) 1Current
(C) Charge (D) 1Charge
1. B
Sol. Dimensions of LR Time
Dimensions of CV = Charge 2 A body starting from rest with acceleration which varies with time t according as At B where A and B are the constants. The velocity of the particle after time t is:
(A) A B t (B) 2
22
At Bt (C) 2
2At Bt (D) 2
2BtAt
2. C
Sol. 2
2Atdt At B Bt
3. A particle starts from origin at t = 0 and moves in the x – y plane with a constant acceleration 6
m/s2 in y-direction. The equation of motion is 2
3xy , then its velocity component along x-
direction at x = 2 is (in m/s)
(A) 32
(B) 3 (C) 6 (D) 2
3. B
Sol. 2
3xy
23
dy x dxdt dt
22 2
2 2
2 23 3
d y dx d xxdt dt dt
2 2
2
26 03
dx d xdt dt
4. From the surface of a certain planet a body is projected with a certain velocity at a certain angle from the horizontal surface. The horizontal and vertical displacements x and y are given by
10 3x t and 210y t t where t is the time in second and x and y are in meter. The magnitude and direction of the
velocity of projection are: (A) 10 ms-1 at 300 from the horizontal (B) 20 ms-1 at 600 from the horizontal (C) 10 ms-1 at 600 from the horizontal (D) 20 ms-1 at 300 from the horizontal 4. D Sol. 3 10 3s t and 210ys t t
5. From the top of a tower of height 40m, a ball is projected upwards with a speed of 20 ms-1 at an angle of 300 to the horizontal. If 210 ,g ms after how long will the ball hit the ground?
A 1 s B 2 s C 3 s D 4 s 5. D Sol. The horizontal range R = AC is
2 0sin 2 20 20 sin 60 20 3
10u m
g
At C, the velocity of the ball is again 20 ms-1 directed down as shown in Fig. The downward vertical component of this velocity is 20sin300 = 10 ms-1. The ball will hit the
ground at D after travelling a vertical distance h = 40 m. If t1 is the time taken for this, then
1 1
12
h ut gt
2
1 140 10 5t t
2
1 12 8 0t t
The positive root of this quadratic equation is 1
2t s . Now, the time of flight from A to C via B is
6. A particle is moving in a circle of radius R. At t = 0 its speed is zero and during its motion speed varies as 22v s where s is the distance travelled. The angle made by acceleration vector with radial direction after one revolution.
(A) 2
(B) 4
(C) 1 1tan
(D)
6. C Sol. 38t
dv vdva sdt ds
2 44c
v saR R
3
4
8 2tan4 /
t
c
a s Ra s R s
After one revolution 2s R 1tan
ca
a
v
ta
7. The displacement x of a particle at any instant is related to its velocity as 2 9.v x Its
acceleration and initial velocity are: (A) 1 unit and 3 unit (B) 3 unit and 9 unit (C) 9 unit and 3 unit (D) 2 unit and 9 unit 7. A Sol. at x = 0 2 9v initial velocity v = 3 units
1/21 2 9 .22
dv dxa xdt dt
a = 1 unit 8. A body is moving down a long inclined plane of angle of inclination . The coefficient of friction
between the body and the plane varies as 0.5x , where x is the distance moved down the plane. The body will have the maximum velocity when it has travelled a distance x given by
A 2 tanx B 2
tanx
C 2 cot D
2cot
x
8. A Sol. The acceleration of the body down the plane is
sin cos sin cos sin 0.5 cos .g g g g x Therefore, the body will first accelerate up to x < 2 tan . The velocity will be maximum at x = 2 tan , because for x > 2 tan , the body starts decelerating. Hence, the correct choice is A
9. At the instant t = 0 a force is a constantF kt k acts on a small body of
mass m resting on a smooth horizontal surface. The time, when body leaves the surface is: (A) sinmgk (B) sin /k mg (C) sin /mg k (D)
M and N being positive constants, then the potential energy at equilibrium must be: (A) zero (B) 2 / 4M N (C) 2 / 4N M (D) 2 / 4MN 10. B
Sol. 0
7 136 12 0
r r
du M NFdr r r
60
2NrM
So, 2
0 4MU
N
11. A body is moving up an inclined plane of angle with an initial kinetic energy E. The coefficient of friction between the plane and the body is . The work done against friction before the body comes to rest is
A cos
cos sinE
B cosE C cos
cos sinE
D
coscos sin
E
’
11. D Sol. The retardation is given by (see fig) cos sin ...a g i Let u be the initial velocity of the body. If it is stopped after moving a distance s up the plane, then
12. A body P strikes another body Q of mass that is p times that of body P and moving with a velocity
that is 1q
of the velocity of body P. If body P comes to rest, the coefficient of restitution is
A p qp q
B 1p q
q p
C 1p q
p q
D 1p q
p q
12. D Sol. Given / .
Q P Q Pm pm and v v q From the principle of conversion of momentum, we have (since
body P comes to rest after collision)
P P Q Q Qm v m v m v
Where v is the velocity of body Q after collision. Thus
PP P P P
vm v Pm pm v
q
Which gives ...P
v p q iv pq
Now, the coefficient of restitution is given by
PP Q
P
v ve vv v vq
Which gives 1 ...P
v e q iiv q
Equating (i) and (ii), we get 1p qe
p q
which is choice D.
13. A solid cylinder is rolling without slipping down an incline of inclination . Minimum coefficient of friction so that the cylinder does not slip on the incline is
14. A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in Fig. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is
A 215
32MR B
2716
MR
C 213
32MR D
238
MR
14. C
Sol. Moment of inertial of complete disc about O is 21 .
2I MR Mass of the cut-out part is
4Mm
.
The moment of inertial of the cut-out portion about its own centre 2
2 2
0
1 1 12 2 4 2 32
M RI mr MR
because r = R/2. From the parallel axes theorem, the
moment of inertial of the cut out portion about O is
2
2 2 2
0
1 332 4 2 32c
M RI I mr MR MR
Moment of inertial of the shaded portion about O is 2 2 21 3 13 ,
2 32 32s cI I I MR MR MR which
is choice C. 15. The centres of a ring of mass m and a sphere of mass M of
equal radius R, are at a distance 8R apart as shown in Fig The force of attraction between the ring and the sphere is
A 2
2 227
GmMR
B 8 2GmM
Rh
C 29GmM
R D
2
29 9
GmMR
15. A Sol. Let be the mass per unit length of the ring. 2L R is
the length of the ring. Consider a small element of length dx of the ring located at C. Then, Force along BC is
16. The magnitude of gravitational force on a particle of mass m placed at a distance x from the rod of mass M and length l as shown in the figure is:
(A) 2GMml x
(B) GMm
l l x
(C) 2GMml x
(D) GMm
x l x
x l
16. D Sol.
r l
m x dr
Mdm drl
2 21Gmdm GmMdF dr
lr r
1
21x
x
GmM GmMF drl x l xr
17. Water from a tap emerges vertically downwards with an initial speed of 1 ms-1. The cross-sectional area of top is 10-4m2. Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 0.15 m below the top is (take g = 10ms-2)
(A) 4 25 10 m (B) 5 21 10 m (C) 5 25 10 m (D) 5 22 10 m 17. C Sol. From equation of continuity 1 1 2 2a v a v
1 4 21 1 21 , 10 ,v ms a m v velocity of stream at h = 0.15 m below the tap.
2 22 1 2v v gh
22 4v
5 22 5 10a m
18. One end of a thermally insulated rod is kept at a temperature T1 and
the other at T2. The rod is composed of two sections of lengths l1 and l2 and thermal conductivities k1 and k2 respectively. The temperature at the interface of the section is A 1 2 1 2 1 2 1 1 2 2
/k l T k l T k l k l B 2 1 1 1 2 2 2 1 1 2/k l T k l T k l k l
C 1 1 1 2 1 2 1 2 2 1/k l T k l T k l k l D 1 1 1 2 2 2 1 1 2 2
/k l T k l T k l k l
18. A Sol. Let
0T be the temperature at the interface and A be the cross-sectional area of each rod.
In the steady state, the rate of flow of heat through rod A = rate of flow of heat through rod B, i.e.
19. A body cools from 50.0 0C to 49.9 0C in 5s. How long will it take to cool from 40.00C to 39.90C? Assume the temperature of the surroundings to be 30.00C and Newton’s law of cooling to be valid.
(A) 2.5s (B) 10 s (C) 20 s (D) 5 s 19. B Sol. Rate of cooling
1 2 1 2
2 sT T T T T
t
Case I: 0.1 49.95 305
0.1
5 19.95
Case II. 0.1 39.95 30t
00.1 9.95t
0.1 0.1 9.95 105 5 19.95
t s
20. In the following figures, the block of mass m is slightly
displaced from its mean position. The ratio of time periods of oscillations in Fig(i) and Fig(ii)
is: (A) 1 : 2 (B) 2 : 3 (C) 3 : 2 (D) 1 : 1
2k k
(i)
m
2k m k
(ii) 20. C Sol. In first case
Keff = .2 22 3
K K KK K
In second case 2 2 3Keff K K K K
2 mTK
1
2
32
TT
21. Third overtone of a closed organ pipe is in unison with fourth harmonic of an open organ pipe. Find the ratio of lengths of the pipes:
22. Two point charges q1 and q2 (q1/2) are placed at points A(0, 1) and B(1,0) as shown in the figure. The electric field vector at point P(1,1) makes an angle with the x-axis, then the angle is:
(A) 1 1tan2
(B) 1 1tan4
(C) 1tan 1 (D) 1tan 0
A 1,1P
B
q2
q1
y
x O *
* *
22. A Sol.
y
x O
BE
AE
P
1 / 2q
1
0 0
/ 2,4 4A B
q qE E
1tan2
B
A
EE
1 1tan2
23. Three point charges 4 ,q Q and q are placed in straight line of length l at points distant 0, ,2l l
respectively. If the net force on charge q is zero, the magnitude of the force on charge 4q is
24. The capacitance of a sphere of radius R1 is increased 3 times when it enclosed by an earthed sphere of radius R2. The ratio R2/R1 is
A 2 B 32
C 43
D 3
24. B
Sol. 1 0 1
4C R ,
0 1 2
22 1
4 R RC
R R
Given 2 1
3C C Hence
0 1 2
0 12 1
43 4
R RR
R R
Which given 2
1
3 ,2
RR
Which is choice B
25. The network shown in the following figure is part of a circuit. What is the potential difference B AV V . When current I is 5A and is decreasing
at a rate of 310 / ?A s (A) 5 V (B) 10 V (C) 15 V (D) zero
1 5H 15V A
I
25. B Sol. Kirchhoff’s 2nd law
A BV V = diRi E Ldt
3 3 31 5 15 5 10 10 10 /di A sdt
10A BV V V 10B AV V V 26. A particle of mass m and charge q moves with a constant velocity v along the positive x direction.
It enters a region containing a uniform magnetic field B directed along the negative z direction, extending from x a to x b . The minimum value of v required so that the particle can just enter the region x > b is
26. B Sol. The radius r of the circular path is given by (see Fig.)
2mv qBqvB or v r
r m
min min,qB qBv r b a
m m which is choice B.
27. A square loop is placed in a uniform magnetic field B
as shown in figure. The power needed to pull it out of the field with a constant velocity v is proportional to (A) 1/2v (B) v (C) 2v (D) 3/2v
x x x x
x x x x
x x x x
x x x xB
v
Loop
27. C Sol. Induced emf Blv
BlvIR
2 2
magB L vF BIL
R
maxextF F
Power = 2 2 2B L vFvR
28. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. when the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is:
A 52
B 52
C 32
D 32
h
3h
2h
2 h
28. B
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x
Sol. When beaker is filled upto a height 2h, the bottom Q of the rod will be visible if the ray QD travelling in the liquid refracts along DB in air. It follows from Fig. that D is the mid-point of diagonal PB of square ABPR. Hence DE = PQ = h. Also
045BDF since 045 .DPE From
Snell’s law, we have (since the object is in a denser medium)
0
1 sin / / 5 2sin 51/ 2sin 45
i QG QD h hr
or 5 ,2
which
is choice B.
29. A U shaped wire is placed in front of a concave mirror of radius of
curvature 20 cm as shown in the figure. The total length of the image of the wire ABCD is nearly: (A) 2.5 cm (B) 6 cm (C) 12.5 cm (D) 15 cm
Section – II (Chemistry) 1. For a hypothetical reaction
2 2 31 ;2g g gAB B AB H x kJ
More 3AB could be produced by
(A) using a catalyst (B) removing some of 2B (C) increasing the temperature (D) increasing the pressure 1. D Sol. On increasing the pressure, equilibrium shifts towards lesser number of gaseous moles 2. The bpK of CN is 4.7. The pH of solution prepared by mixing 2.5 mole of KCN and 2.5 moles
of HCN in water and making the total volume upto 500 ml, is (A) 10.3 (B) 9.3 (C) 4.7 (D)8.3 2. B Sol. If forms basic buffer
2.5 / 500log2.5 / 500bpOH pK
= 4.7 pH = 14 – 4.7 = 9.3 3. The pH of a solution of weak base at half neutralization with strong base is 8, bK for base is
(A) 41 10 (B) 61 10 (C) 81 10 (D) 71 10 3. B Sol. At half equi. point, ;B BOH Salt Base
logb
SaltpOH pK
Base
6bpOH pK
66, 1 10b bpK K 4. HBr and HI reduce sulphuric acid, HCl can reduce KMnO4 and HF can reduce (A) H2SO4 (B) KMnO4 (C) K2Cr2O7 (D) None of these 4. D Sol. HF does not behave as a reducing agent. Therefore, it cannot be oxidized by any of these 5. The correct IUPAC name of the compound given below is:
CH3CH3
CH3
CH3
(A) 4-Ethyl-3-methyloctane (B) 3-Methyl-4-ethyloctane (C) 2, 3-Dimethylheptane (D) 5-Ethyl-6-methyloctance. 5. A Sol. Substituents must be written in alphabetical order.
6. The transition metal ion with least magnetic moment has the electronic configuration (A) 43d (B) 93d (C) 23d (D) 83d 6. B Sol. 93d configuration has only 1 unpaired electron. So magnetic moment is least. 7. Potassium ferricyanide on ionization produces (A) 2 ions (B) 1 ions (C) 5 ions (D) 4 ions 7. D Sol.
3
3 6 63 .K Fe CN K Fe CN
Total 4 ions 8. Arrange the following free radicals in order of stability: Benzyl (I), Allyl (II), Methyl (III), Vinyl (IV) (A) IV > III > II > I (B) I > II > III > IV (C) I > III > IV > II (D) IV > III > I > II 8. B
Sol. Benzyl radical
CH2
is most stable due to greater number of resonating structures.
Allyl radical 2 2C H CH CH
is less stable than benzyl due to lesser no. of resonating
structures. Vinyl radical is least stable of these because of sp hybridization . 9. A hydrocarbon ‘A’ with molecular mass 84 gives a single monochloride but four dichlorides on
photochemical chlorination. The hydrocarbon ‘A’ is (A) Cyclopentane (B) Cyclohexane (C) 2, 3-Dimethylbutane (D) Methyl cyclopentane 9. B Sol.
Cl
2 ,Cl hv 2 ,Cl hv
ClCl Cl
Cl
Cl
Cl Cl
Cl
+ + +
10. What would be the expected formula of a compound which is formed of element X containing 2
electrons in the valence shell and element Y contains 7 electrons in the valence shell? (A) X2Y (B) XY2 (C) XY (D) X2Y2 10. B Sol. X belongs to group 2 and Y belongs to Group 17 11. If one million atoms of silver weight 161.79 10 ,g the gram atomic mass of silver is (A) 107 g (B) 107.2 g (C) 107.8 g (D) 108.2 g 11. C Sol. 1 million i.e. 106 atoms of Ag have mass = 1.79 1610 g
13. The blue colour developed when Lassaigne’s extract is heated with fresh 4FeSO in presence of
alkali, cooled and acidified with dil. H2SO4 indicates…. and is due to the formation of… (A) 4 6 3
,N Fe Fe CN (B) 4 6,N Na Fe CN
(C) 4 5,S Na Fe CN NOS (D) 3
,N S Fe CNS 13 A Sol. Blue colour in Lassaigne’s test indicates nitrogen and is due to the formation of Prussian blue i.e.,
4 6 3Fe Fe CN .
14. If the critical temperature of the gas be 827C
aTRb
and BT is the Boyle’s temperature, then which
of the following, is the correct relation between TC and TB?
(A) 427C BT T (B) 27
4C BT T (C) 827C BT T (D) 27
8C BT T
14. C
Sol. 8 and 27B C
a aT TRb Rb
827
C
B
TT
15. The composition of a sample of wustite is 0.93 .Fe O What is the percentage of iron present as 3 ?Fe (A) 15% (B) 25% (C) 35% (D) 45% 15. A Sol. Let, no. of O-2 ions in 0.93 100Fe O Total no. of 2 3Fe and Fe ions = 93 Let no. of 2Fe ions in the sample = x and No. of 3Fe ions in the sample = y 93x y …(i) Total –ve charge carried by 100 O-2 ions = 200 Total +ve charge carried by 2xFe ions = 2x Total charge carried by 3yFe ions = 3y Thus 2 3 200x y …(ii) Solving (i) and (ii), we get x = 79 and y = 14 No. of Fe2+ ions, x = 79, and No. of Fe3+ ions, y = 14 % of iron present as Fe(III)
14 100 15.05%79 14
16. The uncertainty in the position of a dust particle with mass equal to 1 mg (if uncertainty in its
20 1 65.5 10 , 1 10v ms m mg kg 34 2 13.143; 6.626 10 .h kgm s
34 2 1
6 20 1
6.626 10
4 3.143 10 5.5 10
kgm sx
kg ms
o
109.58 10 9.58A.m 17. Which of the following has highest bond dissociation energy? (A) O2 (B) 2O (C) 2O (D) 2
2O 17. B Sol. 2O has higher bond order and highest bond dissociation energy.
Species Bond order O2 2.0
2O 2.5
2O 1.5 22O 1.0
18. Given, that the abundance of isotopes 54 56,Fe Fe and 57 Fe are 5%, 90% and 5% respectively,
the atomic mass of Fe is (A) 55.85 (B) 55.95 (C) 55.75 (D) 55.05 18. B
Sol. Av. atomic mass = 54 5 56 90 5 57
100
270 5040 285 55.95
100
19. The volume strength of 1.5 N H2O2 solution is (A) 4.8 (B) 8.4 (C) 3 (D) 8 19. B Sol. volume strength = 5.6 N = 5.6 1.5 8.4 20. Which of the following has the maximum number of atoms? (A) 24 g of C(12) (B) 56 g of Fe(56) (C) 27 g of Al(27) (D) 108 g of Ag(108) 20. A Sol. No. of atoms of:
21. An aqueous solution of 6.3 g oxalic acid dihydrate (H2C2O4.2H2O) is made upto 250 ml. The
volume of 0.1 N NaOH required to completely neutralize 10 ml of this solution is (A) 40 ml (B) 10 ml (C) 20 ml (D) 4ml 21. A
Sol. 3
6.3 0.2126 250 10oxalicacid dihydrateM
neq NaOH neq oxalic acid dihydrate
30.1 0.2 2 10 10NaOHV
0.04 40NaOHV l ml 22. In the standardization of Na2S2O3 using 2 2 7K Cr O by iodometry, the equivalent weight of
2 2 7K Cr O is:
(A) 2
molecular weight (B)
6molecular weight
(C) 3
molecular weight (D)
1molecular weight
22. B Sol. 3
2 2 7 2I K Cr O I Cr
2 2 2 3 2 4 6I Na S O I Na S O
2 2 7.6
molecular weightE wt K Cr O
as, 12 326 2e Cr Cr
23. What volume of CO2 at STP will evolve when 1 gm of CaCO3 reacts with excess of dil HCl? (A) 224 ml (B) 112 ml (C) 58 ml (D) 448 ml 23. A Sol. 100 gm CaCO3 gives 22,400 ml CO2 1 gm CaCO3 will give 224 ml CO2 24. The vapour pressure of benzene at 280 K is 40 mbar. When urea is mixed at the same
temperature, the vapour pressure falls by 8 mbar. The mole fraction of benzene in the solution is (A) 0.20 (B) 0.25 (C) 0.70 (D) 0.80 24. D Sol. Vapour pressure of the solution = 40 mbar – 8 mbar = 32 mbar benzene benzene
oP P 32 mbar = 40 mbar × benzene benzene = 0.8 25. Ammonia gas is passed into water, yielding a solution of density 0.93 g/cm3 and containing
18.6% NH3 by weight. The mass of NH3 per cc of the solution is (A) 0.17 g/cm3 (B) 0.34 g/cm3 (C) 0.51 g/cm3 (D) 0.68 g/cm3
(II) In DMF (1) OH (2) 2H O (3) O O H (A) I, 3; II, 2 (B) I, 2; II, 1 (C) I, 1; II, 2 (D) I, 3; II, 3 29. D Sol. In MeOH solvation of anions occurs, hence MeS is least solvated and is the best nucleophile. Since in DMF, no solvation of anion occurs and greater are the LP of electrons near the
nucleophilic site, better is the nucleophilicity. 30. Which of the following statement is correct regarding the rate of hydrolysis of the compounds (x)
and (y) by 1NS reaction?
Br O Br
(x) (y)
(A) x reacts faster than y (B) y reacts faster than x (C) Both x and y reacts at the same rate (D) Neither x nor y reacts 30. B
7. The points (x1, y1), (x2, y2) , (x1, y2) and (x2 , y1) are always (A) Vertices of a rhombus (B) Vertices of a square (C) Con-cyclic (D) Collinear 7. C Sol.
ABCD Diagonal always intersect at 900 So ABCD is con-cyclic.
8. A closet has 5 pair of shoes. The number of ways in which 4 shoes can be drawn such that there
will be no complete pair is (A) 80 (B) 160 (C) 200 (D) 240 8. A Sol. Total number of selection = 5 2 2 2 2
4 1 1 1 1 C C C C C = 80
9. Solution of the differential equation 2 2
2dy x ydx xy
is
(A) k (x2 + y2) = x (B) k (x2 – y2) + x = 0 (C) k (x2 – y2) = x (D) k (x2 + y2) + x = 0 9. C Sol. Put y = v x 10. The complex number z =1 + i is rotated through an angle 3 / 2 in anticlockwise direction about
the origin and stretched by additional 2 unit, then the new complex number is
(A) 2 2 i (B) 2 2 i (C) 2 2 i (D) None of these 10. D
(A) 4 (B) 5 (C) 6 (D) 9 11. D Sol. (3 4 ) 3 4 5 4 5 9.iz i iz i z 12. If 4a + 2b + c = 0, then the equation 3ax2 + 2bx + c = 0 has at least one real root lying between (A) 0 and 1 (B) 1 and 2 (C) 0 and 2 (D) none of these 12. C Sol. f(x) = ax3+ bx2+ cx f(0) = 0, f(2) = 8a + 4b + 2c = 2 (0) = 0 At least one c (0, 2) , such that f ‘ (c) = 0 13. 2 / 3 4 / 3sec x cos ec xdx is equal to
(A) 1 / 33 tanx (B) 1 / 33 cot x (C) 1 / 33 tan x (D) 1 / 33 cot x 13. C
Sol. 2
2 / 3 4 / 3 4 / 3dx sec x dx
cos x sin x tan x
Put tan x t
14. The value of 100
0
x dx (where {x} is the fractional part of x) is
(A) 50 (B) 1 (C) 100 (D) None of these 14. D Sol. { } [ ], [ .] . .x x x G I F put x = t2
100 1 2 10
00 0 1 9
2000 2 0. 1. ........ 9.3
x x dx dx dx dx
15. If f
2
2
cos( ) sin sec
tan 1 2
xx x ex x x x
x then the value of
/2
/2
( )f x dx is equal to
(A) 0 (B) 1 (C) 2 (D) None of these 15. A
Sol. f(x) is odd function so 2
2
( ) 0f x
16. The lines p(p2 +1)x – y + q = 0 and (p2 + 1)2 x + (p2 + 1) y + 2q = 0 are perpendicular to a common line for (A) No value of p (B) Exactly one value of p (C) Exactly two values of p (D) More than two values of p 16. B Sol. Lines must be parallel , m1 = m2 17. Tangents to the circle x2 + y2 = a2 cut the circle x2 + y2 = 2a2 at P and Q. The tangents at P and Q
to the circle x2 + y2 = 2a2 intersect at angle , then is equal to
(A) 4
(B) 6
(C) 3
(D) 2
17. D
Sol. PQ is chord of contact of circle of x2 + y2 = 2a2 w.r. to point A (n, k) , A is intersection point of circle which lies on director circle to x2 + y2 = 2a2. 18. The length of the latus rectum of parabola 4y2 + 2x – 20y + 17 = 0 is (A) 3 (B) 6 (C) ½ (D) 0
20. The tangent at any point P on the hyperbola 2 2
2 2 1x ya b
meets the straight lines bx – ay = 0 and
bx + ay = 0 in the points Q and R respectively. If C is the centre of the hyperbola, then CQ.CR is
(A) a2 – b2 (B) a2 + b2 (C) ab (D) ab
20. B
Sol. Point
,sec tan sec tan
,sec tan sec tan
a bQ
a bR
21. The number of points common to the line 3 4 5
2 3 2x y z
and the plane 4x – 2y – z =1 is
(A) 0 (B) 1 (C) infinite (D) none of these 21. A Sol. Given line is parallel to the plane 4x 2y z 1
22. 4 5 9 0, ( )If a b c then a b b c c a is equal to
(A) A vector perpendicular to plane of ,a b and c (B) A scalar quantity (C) 0
(D) None of these 22. C Sol. are collinearb c and c a
23. A unit vector in xy-plane which makes an angle of 450 with the vector i j
and an angle of 600
with the vector 3 4i j
is
(A) i (B)
2i j
(C) 2
i j
(D) None of these
23. D
Sol. Let ^ ^
2 2, 1r a i b j a b
24. If x + y = a + b, x2+ y2 = a2 + b2, then xn + yn = an + bn is true for (A) n N (B) 4n (C) 3n (D) None of these 24. A Sol. Check for any 3 Natural No. 5 25. If 1 + 2 3sin sin ......... 4 2 3,0 , / 2,x x sin x x x then
27. The centre and radius of the circle (1 ) (1 ) 7 0 zz i z i z is (A) - 1 – i , 3 (B) 1 + i , 3 (C) 2 – i, 4 (D) 2 + i , 4 27. A Sol. It represents a circle with centre - 1 – i. 28. 1 1,If x and y then the sum to infinity of the series (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + ………… to is
