physics, chemistry & mathematics jee - mains 2015 paper class 11.pdf · 2017. 9. 19. · jee -...

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PHYSICS, CHEMISTRY & MATHEMATICS

JEE - MAINS 2015 PHASE – I

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 specific ally for this purpose.

Important Instructions:

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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 parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30

questions in each part 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.

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FIITJEE - JEE (Mains)


Useful Data Chemistry:

Gas Constant R = 8.314 J K1

mol1

= 0.0821 Lit atm K1

mol1

= 1.987 2 Cal K1

mol1

Avogadro's Number Na = 6.023 1023

Planck‘s Constant h = 6.626 10–34

Js

= 6.25 x 10-27

erg.s

1 Faraday = 96500 Coulomb

1 calorie = 4.2 Joule

1 amu = 1.66 x 10-27

kg

1 eV = 1.6 x 10-19

J

Atomic No : H=1, D=1, Li=3, Na=11, K=19, Rb=37, Cs=55, F=9, Ca=20, He=2, O=8,

Au=79.

Atomic Masses: He=4, Mg=24, C=12, O=16, N=14, P=31, Br=80, Cu=63.5, Fe=56, Mn=55,

Pb=207,

Au=197, Ag=108, F=19, H=2, Cl=35.5, Sn=118.6

Useful Data Physics:

Acceleration due to gravity g = 10 2m/ s


Section – I (Physics)

1. Three vectors ,P Q

and R

are such that Q

= 2A and the angles between P

and Q

, Q

and R

, R

and

P

are 900, 150

0 and 120

0 respectively. Find the value of P

.

(A) 2

A (B)

2

3

A (C)

2

3

A (D)

2

A

2. The velocity and acceleration of a particle at time t = 0 are ˆ ˆ2 2 /u a i a j m s

and 0

ˆ ˆa ai aj

respectively. Find the angle made by the velocity of the particle at t = 2sec with initial velocity.

(A) 1tan 2 (B) 1tan 2 (C) 1tan 1 (D) 1 1tan2

3. When a man walks at the rate of 3 km/hr, rain appears to fall vertically. The speed of rain is 3 2 km/hr. At what speed man should walk so that the rain appears to fall at an angle of 45

0 with vertical.

(A) 3 km/hr (B) 4 km/hr (C) 3 2 km/hr (D) 6 km/hr

4. The trajectory of a projectile in a vertical plane is 2 ,y ax bx where a and b are positive constants and x and

y are respectively the horizontal and vertical distances of the projectile from the point of projection. The

maximum height attained by the projectile is

(A)

22a

b (B)

2a

b (C)

2

2

a

b (D)

2

4

a

b

5. A car accelerates from rest at a constant rate for some time after which it decelerates at a constant rate to come to rest. If the total time elapsed is t, the distance travelled by the car is given by

(A) 21

2t

(B)

21

2t

(C) 2 2

21

2t

(D)

2 221

2t

6. The upper half of an inclined plane of inclination is perfectly smooth while the lower half is rough. A block starting from rest from the top of the plane will again come to rest at the bottom if the coefficient of friction

between the block and the lower half of the plane is given by

(A) 2tan (B) tan (C) 2

tan

(D)

1

tan

space for rough work


7. A ball of mass m is moving towards a batsman at a speed v. The batsman strikes the ball and deflects it by an

angle without changing its speed. The impulse imparted to the ball is given by

(A) cosmv (B) sinmv (C) 2 cos2

mv

(D) 2 sin2

mv

8. An insect starts crawling up a hemispherical bowl of radius R from its lowest point. If the coefficient of friction is

1

3, the insect will be able to go up to height h equal to (take

30.95

10 )

(A) 5

R (B)

10

R (C)

20

R (D)

30

R

9. A simple pendulum of length 1 m and bob of mass 100 gm is swinging with an angular amplitude of 600. What is

the tension in the string when the bob passes through the equilibrium position? Take g = 10 ms-2

.

(A) 1 N (B) 2 N (C) 3 N (D) 4 N

10. Which one of the following statements is NOT true about the motion of a projectile?

(A) The time of flight of a projectile is proportional to the speed with which it is projected

(B) The horizontal range of a projectile is maximum when angle of projection is 450 for a given speed

(C) The average acceleration for any time interval is varying.

(D) At maximum height the acceleration due to gravity is perpendicular to the velocity of the projectile.

11. A point P moves in counter- clockwise direction in a circular path

as shown in the figure. The movement of P is such that it sweeps

out a length 3 5S t , where S is in metres and t is in sec. The

radius of the path is 20 m. The acceleration of ‗P‘ when t = 2s is

nearly.

(A) 14 m/s2 (B) 13 m/s

2

(C) 12 m/s2 (D) 7.2 m/s

2

A

B

O

20 m

P(x, y)

12. A person standing in a stationary lift drops a coin from a certain height h. It takes time ‗t‘ to reach the floor of the

lift. If the lift is rising up with a uniform acceleration a, the time taken by the coin, dropped from the same height

h, to reach the floor will be

(A) t (B) a

tg

(C)

1

t

a

g

(D)

1

t

a

g



13. A block is placed on the top of an inclined plane of inclination kept on the floor of a lift which is moving down

with an acceleration a a g . The coefficient of friction between the block and the incline so that the block just remains stationary with respect to incline plane.

(A) tan (B) tana

g (C) 1 tan

a

g

(D) 1 tana

g

14. Force F acting on a body moving in a straight line varies with the velocity v of the body as k

Fv

where k is a

constant. The work done by the force in time t is proportional to

(A) t (B) 3/2t (C) 1/2t (D) 3/2t

15. A bob of mass m is suspended with a string from a fixed point, when it is projected with a velocity which is just

required to loop the circle completely. At what angle with the horizontal, tension in the string will be equal to 2

mg.

(A) 1 1sin

3

(B) 1 2sin

3

(C) 1 1cos

3

(D) 1 1tan

3

16. With what force must a man pull on the rope to hold the plank in the

position as shown in the figure. If the man weighs 60 kg and plank

weighs 40 kg. The rope and pulley are massless

(A) 100 N (B) 150 N

(C) 125 N (D) 250 N

17. The work done by the force

2 2ˆ ˆF x i y j

around the path

shown in the figure is

(A) 32

3a (B) zero

(C) 3a (D) 3

4

3a

y

xO A

BC(0, a)

(0, 0) (a, 0)

(a, a)



18. A particle is moving with velocity ˆ ˆv k yi xj

where k is a constant. The trajectory equation of the particle

is

(A) 2y x constant (B) 2y x constant

(C) xy constant (D) 2 2y x constant

19. Two fixed frictionless inclined planes making an angle 300

and 600 with the horizontal are shown in the figure. Two

blocks A and B are placed on the two planes. What is the

relative horizontal acceleration of A with respect to B?

(A) 4.9 ms-2

in horizontal direction

(B) 9.8 ms-2

in horizontal direction

(C) zero

(D) 4.9 ms-2

in vertical direction

A

B

060 030

20. A system of wedge and block is as shown in figure. If wedge is fixed

and block is released from rest with spring in its natural length, find

maximum elongation in the spring. All surfaces are frictionless

(A) 2 sinmg

k

(B)

sinmg

k

(C) 4 sinmg

k

(D)

sin

2

mg

k

k

m

21. A particle of mass m is located in a one dimensional potential field where potential energy of the particle has the

form 2a b

U xx x

where a and b are positive constants. The position of equilibrium is

(A) 2

b

a (B)

2b

a (C)

a

b (D)

2a

b

22. In the shown figure mass of A is m and that of B is 2m. All the

surface are smooth. System is released from rest with spring

unstretched. Then, the maximum extension (xm) in the spring will

be

(A) mg

k (B)

2mg

k

(C) 3mg

k (D)

4mg

k

k

A

B



23. In the figure shown all the surfaces are frictionless, and mass of the

block, m = 1kg. The block and wedge are held initially at rest. Now

wedge is given a horizontal acceleration of 10 m/s2 by applying a

force on the wedge, so that the block does not slip on the wedge.

Then work done by the normal force in ground frame on the block in

3s is (A) 30 J (B) 60 J

(C) 150 J (D) 100 3 J

m10 m/s2

M

24. Three stones A, B and C are simultaneously projected from same point with same speed. A is thrown upwards, B

is thrown horizontally and C is thrown downwards from a building. When the distance between stone A and C

becomes 10 m, then distance between A and B will be

(A) 10 m (B) 5 m (C) 5 2 (D) 10 2m

25. The coefficient of friction between 4 kg and 5 kg blocks is 0.2 and

between 5 kg block and ground is 0.1 respectively. Choose the correct

statements.

(A) minimum force needed to cause system to move on ground is 17 N

(B) When force F = 4N, static friction at all surfaces I 4 N to keep

system at rest.

(C) Maximum acceleration of 4 kg block is 2 m/s2

(D) Slipping between 4 kg and 5 kg block starts when F is 17 N.

F

4 kg

5 Kg

26. A spring of natural length l is compressed vertically downward against the floor so that its compressed length

becomes 2

l. On releasing, the spring attains its natural length. If k is the stiffness constant of spring then work

done by the spring on the floor is

(A) zero (B) 2

2kl

(C)

21

8kl (D) 2

1

4kl

27. A particle is given an initial speed u inside a smooth spherical

shell of radius R so that it is just able to complete the circle.

Acceleration of the particle, when its velocity is vertical, is

(A) 3g (B) 2g

(C) g (D) 10g u



28. A block is sliding along a smooth incline as shown in figure. If the acceleration

of chamber is ‗a‘ as shown. The time required to cover a distance L along

incline is

(A) 2

sin cos

L

g a (B)

2

sin sin

L

g a

(C) 2

sin cos

L

g a (D)

2

sin

L

g

am

29. Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory

is (u = projection speed, = angle of projection from horizontal).

(A) cosu (B) 21 3cos2

u (C) 22 cos

2

u (D) 21 cos

2

u

30. A particle moves along the curve

2

2

xy . Here x varies with time as

2

2

tx . Where x and y are measured in

metre and t in second. At 2sec.t the velocity of the particle (in ms-1) is

(A) ˆ ˆ2 4i j (B) ˆ ˆ2 4i j (C) ˆ ˆ4 2i j (D) ˆ ˆ4 2i j



Section – II (Chemistry)

1. The dehydration yield of cyclohexanol 6 11C H OH (m.wt = 100) to cyclohexene 6 10C H (m.wt = 82) is 75%.

What would be the yield if 100 gm of cyclohexanol is dehydrated?

(A) 61.5 gm (B) 16.5 gm (C) 6.15 gm (D) 615 gm

2. Four particles have speed 2, 3, 4 and 5 cm/sec respectively. Their RMS speed is

(A) 3.5 cm/sec (B) (27/2) cm/sec (C) ( 45 / 2 ) cm/sec (D) ( 54 / 2 ) cm/sec

3. Identify the paramagnetic species

(A) K2O (B) KO2 (C) CaO (D) BaO

4. The orbital angular momentum for 11th

electron of Na atom is

(A) 2

h

(B) 2

2

h

(C)

4

h

(D) zero

5. Under identical conditions, helium will diffuse through a pin hole(At. wt. of argon = 40 and helium = 4)

(A) 3.16 times as fast as argon (B) 7.32 times as fast as argon

(C) 1.58 times as fast as argon (D) 10 times as fast as argon

6. Velocity of photoelectron is

(A)

1

20

0

2hc

m

(B)

2

0

0

2hc

m

(C) 2hc

m (D)

2

2

m

hc

Where, m = mass of photoelectron, h = planck‘s constant, 0 Threshold wavelength,

wavelength incident, c velocity of light

7. The normality of 0.3 M boric acid 3 3H BO solution is (A) 0.3 (B) 0.15 (C) 0.6 (D) 0.9

8. Degree of hardness of a sample of water containing 6 mg of 4MgSO per kg of water is

(m. wt of MgSO4 = 120)

(A) 5 ppm (B) 7 ppm (C) 3 ppm (D) 7.8 ppm

9. Consider a titration of potassium dichromate solution with acidified Mohr‘s salt

4 4 4 22. .6FeSO NH SO H O solution using diphenylamine as indicator. The number of moles of Mohr‘s salt required per mole of dichromate is

(A) 3 (B) 4 (C) 5 (D) 6



10. Choose the correct formula

(A) Critical pressure = 8

27

a

b (B) Inversion temperature =

2a

Rb

(C) Critical temperature = 227

a

b (D) All the above

11. The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na will be (A) - 10.2 eV (B) - 2. 55 eV (C) - 5. 1 eV (D) + 2.55 eV

12. Choose the correct statement

(A) In 7IF molecules, total number of bonds with angle 720 is 5

(B) Number of F – Se – F arrangement with 1800 in

6SeF is 6

(C) F should be placed at equatorial position in 3 2PCl F

(D) Chlorine is 3sp d hybridized in 4HClO molecule

13. Correct order of paramagnetic property is

(A) 2 2 2 2O N N N

(B) 2 2 2 2N N N O

(C) 2 2 2N N O N (D) 2 2 2 2O N N N

14. Choose the correct order of periodic property

(A) P > N (electron affinity) (B) S > O (electronegativity)

(C) F < Cl (Ionisation energy) (D) N > C (electron affinity)


(A) Both Li and Mg are solid (B) Both Li and Mg forms nitride

(C) Both Li2O and MgO are basic (D) All the above

16. Bond length of x – y bond is 100 pm and its observed dipole moment is 2 D. It is % covalent character is

approximately

(A) 41.67 % (B) 58.33 % (C) 47.1 % (D) 61.3 %

17. Boiling point of glycol 2 2HO CH CH OH is much higher than that of propyl alcohol

3 2 2CH CH CH OH due to (A) Presence of dipole moment (B) Presence of two bonds (C) Presence of back bonding (D) Presence of hydrogen bonding

18. Momentum of particle ‗A‘ is thrice to the momentum of particle ‗B‘. The ratio of wavelength associated with B to

A is

(A) 3 : 1 (B) 1 : 3 (C) 1 : 2 (D) 3 : 2



19. Distance of electron of hydrogen atom from nucleus is .o

Ax The distance of shell in which 3rd electron of Li atom is present is

(A) 3

4

o

Ax

(B) 9

2

o

Ax

(C) 4

3

o

Ax

(D) 2

9

o

Ax

20. Oleum sample is identified as 102.25 %. The % free SO3 present in this sample is

(A) 20 % (B) 10% (C) 40 % (D) 80 %

21. Number of peroxylinkage (O – O) present in 2 8K CrO is(Given: Oxidation state of Cr = 6)

(A) 2 (B) 8 (C) 4 (D) 2

22. During preparation of H2, by electrolysis of acidic water, correct statement is

(A) O2 is liberated at anode (B) O2 is liberated at cathode

(C) H2 is liberated at anode (D) Both H2 and O2 are liberated at anode

23. The correct order of basic nature is

(A) LiOH NaOH KOH RbOH CsOH

(B) NaOH KOH RbOH CsOH LiOH

(C) LiOH KOH NaOH RbOH CsOH

(D) CsOH RbOH LiOH KOH NaOH


(A) BeO is amphoteric (B) Be forms polymeric halide

(C) Be forms polymeric hydride (D) All the above

25. Which of the following element shows hydride gap?

(A) Sc (B) Mg (C) Fe (D) Zn

26. Inert pair effect is shown by

(A) Pb (B) Mg (C) Si (D) I

27. Element which gives negative flame test is

(A) Na (B) Rb (C) Ca (D) Mg

28. Two vessels having equal volumes contains H2 and He at 1 and 2 atm respectively at the same temperature. Select

the correct statement.

(A) 2rms H rms HeU U (B) 2 2rms H rms HeU U

(C) 23

rms H rms HeU U (D)

22

rms H rms HeU U

29. The maximum number of electrons that can have principal quantum number, n = 3 and spin quantum number,

1

2s is

(A) 3 (B) 6 (C) 9 (D) 18

30. The equivalent weight of an element is 4. It‘s chloride has a vapour density 59.25. The valency of element is

(A) 2 (B) 3 (C) 4 (D) 5

pace for rough work


Section – III (Mathematics)

1.

2

0

cos 2

cos sin

xI dx

x x

is equals to

(A) 2 (B) 1 (C) 0 (D) None of these

2. 2lim 2x

x x x

is equals to

(A) 1 (B) -1 (C) -2 (D) 1

2

3. The number of solution of equation 16 4log log 2x x is equals to (A) 0 (B) 1 (C) 2 (D) 3

4. The value of 3 5 9 11 13

sin sin sin sin sin sin14 14 14 14 14 14

is equal to

(A) 1

32 (B)

1

64 (C)

1

16 (D)

1

128

5. Total number of lines of the form 1ax by , 0a b which intersect the circle 2 2 50x y at two integral points is/are equals to

(A) 66 (B) 60 (C) 56 (D) 78

6. If , lie on the circle 2 2 4x y then maximum value of 3 4 is equal to (A) 4 (B) 6 (C) 8 (D) 10

7. Let points A, B, C form a acute angled triangle and P is a point inside the ABC such that

2 2 2

PA PB PC is maximum then for ,ABC P is …………

(A) ortho centre (B) circum centre (C) centroid (D) Incentre

8. Let p and q be two statements, then ~p q p is (A) tautology (B) contradiction

(C) Both tautology & contradiction (D) neither tautology nor contradiction



9. The solution set of

3

42

1 20

5

x x

x x

is

(A) 1,2 0x (B) , 5 5, 1 2,x

(C) 1,2x (D) , 1 2,x

10. The complete set of values of x satisfying the inequality

10

3

x

x

is

(A) 1,3x (B) 3, 1 1,3x

(C) 1,3x (D) 3, 1 1,3x

11. Tangents 1PT and 2PT are drawn from a point P lying on the line 0ax by c to the circle 2 2 2 ,x y r then the locus of circumcentre of 1 2PTT is

(A) 02

cbx ay (B) 0

2

cbx ay

(C) 2 2 0ax by c (D) 2 2 0ax by c

12. Let 0 0 0tan18 cot18 tan18

0 0 0

1 2 3tan18 , tan18 , cot18t t t and 0cot18

0

4 cot18 ,t then

(A) 1 2 3 4t t t t (B) 4 3 1 2t t t t (C) 3 1 2 4t t t t (D) 4 3 2 1t t t t

13. Let the coordinate axes is shifted as well as rotated in such a way that new x-axes is along the line

3 4 10 0x y and new y –axes is along the line 4 3 20 0x y . If the old coordinates of point P is (5,

5), then new coordinate of P can be

(A) (11, 1) (B) (1, 11) (C) (11, -1) (-1, 11)

14. The number of points equidistant from the lines 0, 0x y x y and 2 0y is/are

(A) 1 (B) 2 (C) 3 (D) 4

15. Total number of lines touching two circles of the family of circles defined as 2 2 4 4 0x y x y is

(A) 8 (B) 10 (C) 12 (D) 14

16. The acute angle between the medians drawn from the acute angle of a right angles isosceles triangle is

(A) 1 2cos

3

(B)

1 3cos4

(C)

1 4cos5

(D)

1 5cos6



17. If A = 2 4sin cos , then

(A) 1 2A (B) 3

14

A (C) 13

116

A (D) 3 13

4 16A

18. Given the family of lines 3 4 6 2 0a x y b x y . The line of the family, situated at the greatest

distance from the point 2,3P has equation (A) 4 3 8 0x y (B) 5 3 10 0x y (C) 15 8 30 0x y (D) None

19. The contrapositive of ―if two triangles are identical, then these are similar‖ is

(A) if two triangle are not similar, then these are not identical

(B) if two triangles are not identical then these are not similar

(C) if two triangles are not identical then these are similar

(D) none of these

20. Let .....y x x x then dy

dx at x = 2 is equal to

(A) 1 (B) 3 (C) 1

2 (D)

1

3

21. Let in ABC coordinates of A is (0, 0). Internal angle bisector of ABC is 1 0x y and mid point of

BC is (1, 3). Then ordinate of ‗C‘ is

(A) 2 (B) 4 (C) 5 (D) 6

22. If sin 2 0.44x then, sin cosx x is equal to

(A) 4

3 (B)

5

4 (C)

6

5 (D)

7

6

23. If 0 0 02 2 2log 1 tan1 log 1 tan 2 ......... log 1 tan 45 is a two digit number, whose sum of digits equals to

(A) 3 (B) 5 (C) 7 (D) 9

24. If 20, 15n A n B , max 5n A B , 30n C , then min n A B C equals to (A) 40 (B) 35 (C) 45 (D) 30



25. If circle 2 2 0x y ax c is completely inside the circle 2 2 0x y bx c , then

(A) 0c (B) 0c (C) 0ab (D) None of these

26. The value of ‗k‘ for which circles 2 2 81 0x y and 2 2 4 6 0x y x y k are orthogonal is

(A) 81 (B) -81 (C) 68 (D) None of these

27. The portion of the line 1ax by intercepted between the lines 1 0ax y and 0x by subtends a

right angle at origin then

(A) 22 2 0a b b (B) 22 0a b b (C) 2 2 0a a b (D) 2 0a b ab

28. Let line 2 7 0x y cuts the circle 2 2 16 0x y at points A and B. If P is (6, 5) then PA PB is

equals to


29. The number of values of ‗a‘ for which lines 1 0, 2 1 0x y ax y and 4 2 7 0x ay are

concurrent is equal to


30. Lines L1 and L2 are rotating in anticlockwise direction about points (-2, 0) and (2, 0) respectively in such a way

that angle of rotation of line L2 is double that of L1. If initially equation of lines are y = 0 and angle of rotation of

line L2 varies between 0 to 2

, then locus of point of intersection of L1 and L2 is part of circle with radius equals

to

(A) 2 (B) 4 (C) 6 (D) 8




JEE - MAINS 2015

PHASE – I

SET - A ANSWERS

PHYSICS

1. B 2. B 3. D 4. D

5. A 6. A 7. C 8. C

9. B 10. C 11. A 12. C

13. A 14. A 15. A 16. D

17. B 18. D 19. C 20. A

21. D 22. D 23. C 24. C

25. C 26. A 27. D 28. C

29. B 30. B

CHEMISTRY

1. A 2. D 3. B 4. D

5. A 6. A 7. A 8. A

9. D 10. B 11. C 12. A

13. D 14. A 15. D 16. B

17. D 18. A 19. C 20. B

21. C 22. A 23. A 24. D

25. C 26. A 27. D 28. D

29. C 30. B

MATHEMATICS

1. C 2. D 3. B 4. B

5. B 6. D 7. C 8. A

9. A 10. D 11. C 12. B

13. C 14. D 15. D 16. C

17. B 18. A 19. A 20. D

21. D 22. C 23. B 24. D

25. B 26. D 27. B 28. A

29. B 30. B

FIITJEE - JEE (Mains)


HINTS & SOLUTIONS

PHYSICS 1. B

Sol. 0tan 30

P

Q

2

3

AP

P

Q

R

090

0150

0120

030

2. B

Sol. u

and at

are perpendicular vectors

Also v u at

At 2sect , 2 , 2 2 2 2u a at a a

1/22 2

12v u at a

tanat

u

1tan 2

at

u

v

3. D

Sol. Let ˆ ˆRV ai bj

Case I ˆ3MV i

ˆ ˆ3RM R MV V V a i bj

Now 3 0a as RMV

is vertical

Also 2

2 2

RV a b

2

2 23 2 3 b

b = 3

Case II ˆMV ki

ˆ ˆ ˆ ˆ3 3 3RMV a k i j k i j

For angle to be 450, ˆ ˆ3 3RMV i j

6k 4. D

Sol. When x = R, y = 0

0 = aR – bR2

a

Rb

When max,

2

Rx y H

2y ax bx

2 2

max2 2 4

a a aH a b

b b b

5. A


Sol. max 1 2V t t

2 1t t

1 2t t t

1t t

Distance = Area under Vt graph

= 1

1

2t t

21

2D t

V

Vmax

t1 t1 + t2

0

6. A

Sol. For the block to come to rest at the bottom of the inclined plane, the acceleration in the first half must be equal to

the retardation in the second half

sin cos sing g 2tan 7. C

Sol. 1 2P P mv

2 1P P P

1/2

2 2

2 1 2 12 cos 180P P P P P

1/22 2 22 cosP P P P

2 cos2

P

2 cos2

mv

P

2P

1P

0180

8. C

Sol. sinf mg

cos sinmg mg

1tan

3

3

cos 1 1 0.9510

h R R R R

0.05R

20

R

cosmg mgsinmg

N

hf

9. B

Sol. Speed at equilibrium position

2 2 1 cosv gR gR 2

2 2mv

T mg mg NR

m

v

060

10. C

Sol. avga g

always

11. A


Sol. 3 5S t

23

dsv t

dt

6tdv

a tdt

, At t = 2sec, v = 12 ms-1

212ta ms

2 2212 144 7.2

20 20n

va ms

R

2 2

t na a a

214a ms

12. C

Sol. 21

2h gt when lift is stationary

When lift is accelerating upwards.

21

'2

h g a t

22 'gt g a t

1

1

g tt t

g a a

g

13. A

Sol. FBD of block wrt to inclined plane

/maxsin s sm g a f N

cosN m g a

sin cossm g a m g a

tans

N

ma

mg a

14. A

Sol. k

P Fv v kv

0

t

w Pdt Kt

15. A

Sol. 2

sin Pmv

T mgr

2

sin 2Pmv

T mg mgr

2 2 sinPv gr gr …..(1)

From w – E theorem

cosmg

sinmg

mg

O

P

B

2 21 1

1 sin2 2

P Bmv mv mgr

Also 5Bv gr to loop circle.


2 5 2 1 sinPv gr gr

2 3 2 sinPv gr gr …….(2)

From (1) and (2)

2 sin 3 2 singr gr gr gr

1

sin3

1 1sin

3

16. D

Sol. For the (Man + Plank) system to be at rest

4 100T g

250T N

17. B

Sol. .W F dr

x yW F dx F dy

0

0

A B C

x y x y y x y x y

A B C

W F dx F dy F dx F d F dx F dy F dx F dy

0W 18. D

Sol. ˆ ˆv k yi xj Hence ,x yv ky v kx

dx dy

ky kxdt dt

/

/

dy dy dt kx x

dx dx dt ky y

ydy xdx

Integrating

2 2

2 2

y xc

2 2y x constant

19. C

Sol. Acceleration along the plane is sina g

Horizontal component of a is cos sin cos sin 22

ga g

For Block A, horizontal acceleration is 03 3

sin 2 602 2 2 4

g g g

cosa

sina g

For Block B, horizontal acceleration is 03 3

sin 2 302 2 2 4

g g g

Relation horizontal acceleration of A wrt B = 0

20. A

Sol. W – E theorem

2T T

T

(60 g + 40 g)


g sk w w

210 sin

2mgx kx

2 sinmg

xk

21. D

Sol. dU

Fdx

3 2

2a bF

x x

For equilibrium F= 0

3 2

2a b

x x

2a

xb

22. D

Sol. From conservation of energy

212 0

2mgx kx

4mg

xk

23. C

Sol. Work done by the normal force on the block relative to ground

frame = K

21

2mv

21 300

1 10 3 1502 2

J

m

M

90

N

a

mg

24. C

Sol. Initial velocity of C wrt A

CA C Au u u

ˆ ˆ ˆ2uj uj u j 0CA C Aa a a g g

CA CAS u t

CA CAS u t

10 2ut

5t

u

Initial velocity of B wrt A

ˆ ˆBA B Au u u ui uj

0BA B Aa a a g g

ˆ ˆ5 5BA BAs u t i j

5 2BAs m

25. C


Sol. /maxsf between 4 kg and 5 kg = 1 8f N

/maxsf between 5 kg and ground =

2 9f N

Minimum force needed to move the system

is 9 N

When F = 4N friction between the blocks is

zero

When /maxsf acts on 4 kg it will have

maximum acceleration

21max

82 /

4

fa m s

m

When slipping between 4 kg and 5 kg starts

max. friction acts between them and their

accelerations are just same.

8 9 5 2F

27F N

4 N

4 kg

5 Kg

0f

4f N

F

4 kg

5 Kg

8f N

2 9f N

1 8f N

22 /a m s

22 /a m s

26. A

Sol. Point of application of force is at rest.

27. D

Sol. At lowest point A, 5u gR

When the velocity is vertical, at point B

3v gR

2

3 ,n tv

a g a gR

2 2 10n ta a a g

u

naO

v

ta

B

A 28. C

Sol. acceleration of block wrt chamber

2

( cos sin )

1

2

2

sin cos

BC

BC BC

a a g

s a t

Lt

g a

from the chamber frame

N

ma

cos sinma mg cosmg

sinma BCa

29. B

Sol. 22

2

Rs H

2 2 2

2

sin sin cos sin, ,

2 2

1 3cos2

Time of ascent,T

av

u R u uH

g g g

s uv

T

2

R

s

A

H

O

30. B

Sol.

2 2 4

2 2 8

t x tx y

3

2x y

dx tv t v

dt

3

ˆ ˆ2

tv ti j

at t = 2 sec


ˆ ˆ2 4v i j

(m/s)

Chemistry

1. A

Sol. 26 11 6 10

H OC H OH C H

m.wt. = 100 m.wt. = 82

100 gm cyclohexanol = 82 gm C6H10

100gm cyclohexanol = 6 1082 100

100gm C H

Also % yield is 75% wt. of 6 1082 75

61.5100

C H

2. D

Sol.

2 2 2 22 3 4 5 54 54

4 4 2RMS

cm/sec

3. B

Sol. Presence of 1

2O

ion which has one unpaired electron.

4. D

Sol. orbital angular momentum = 12

hl l

for 11th

electron, 0l 5. A

Sol. 40

10 3.164

He

Ar

r

r

6. A

Sol.

0

1 1KE hc

2

0

1 1 1

2mv hc

2

0

2 1 1hcv

m

or

1

2

0

2 1 1hcv

m

7. A

Sol. n.f. of boric acid = 1

8. A

Sol. 120 gm 4 3100MgSO gm CaCO

3 3

4 36 10 5 10gm MgSO gm CaCO

36

3

5 1010

1000ppm of CaCO

5 ppm

9. D

Sol. 3 3

2 7 26 14 6 2 7Fe Cr O H Fe Cr H O

21 6 Fen

10. B

Sol. 8

27C

aT

Rb ,

227C

aP

b ,

2aTi

Rb


11. C

Sol. Electron gain enthalpy of ion = - I.E. of atom = - 5.1 eV

12. A

Sol. In 0

7 , 72 5IF ; 090 10 ; 0180 1

13. D

Sol. 2 22 2

8, 3 , 0O NN N

14. A

15. D

Sol. Diagonal relationship

16. B

Sol. 10 10

100% 4.8 10 100 10

20480 10

184.8 10 esu cm

18

100% 18

4.8 104.8

10D

% ionic = 2

100 41.674.8

% Covalent = 100 41.67 58.33% 17. D

Sol. Glycol has more number of hydrogen bonding as compare to propyl alcohol hence Boiling point of glycol is much

higher than that of propyl alcohol.

18. A

Sol. h

P

1

3 3

A B

B A

P x

P x

3:1B

A

19. C

Sol.

2

0.529o

An

rZ

distance of 1st electron =

20.529 1 x

distance of 3rd

electron = 0.529 4 4

3 3

o

Ax

20. B

Sol. 18 gm H2O reacts with 80 gm SO3

2.25 gm water reacts with = 380

2.2518

gm SO

% 3

80 2.25100 10%

18 100SO

21. C

Sol.

O

O

Cr

O O

O

O

O

O

22. A


Sol. H is reduced at cathode and OH is oxidized at anode. 23. A

Sol. Basic nature increases down the group

24. D

Sol. Abnormal behavior of Be

25. C

Sol. Hydride gap = No hydride formation

26. A

Sol. 4, 2Pb (common)

27. D

Sol. Due to very high I.E.

28. D

Sol. 3RT

UM

29. C

Sol. Number of orbitals = n2 = 9

Number of electrons with 1

92

s

30. B

Sol. Mole. wt. of 2 . . 2 59.25 118.5nMCl V D

Now 35.5 118.5a n 35.5 118.5 4 35.5 118.5E n n n n

3n

MATHEMATICS 1. C

Sol. /2

/2 /2

0 0

0

cos sin sin cos 0I x x dx x x

2. D

Sol. 22

2

21

2 1lim 2 lim lim

21 221 1

x x x

x xx x xx x x

x x

3. B

Sol. 2

16 4log log 2 2 1,4x x x x x

But 1 is rejected

4. B

Sol. 3 5 9 11 13

sin sin sin sin sin sin14 14 14 14 14 14

=

2

2 2 23 5 2 4sin sin sin cos cos cos14 14 14 7 7 7

2 23

3

8sin 2 sin17 7

82 sin sin

7 7

1

64

5. B

Sol. Total integral points on 2 2 50x y equals to 12

Total lines passing through these points are 11 10 .... 2 1 66 in which 6 lines are diameter Answer is 60

6. D

Sol. Let 3 4x y k is tangent to 2 2 4x y then k equals to 10 Maximum value of 3 4 is 10


7. C

Sol. Let A be 1 1, ,x y B be 2 2,x y C be 3 3,x y and P be ,x y then 2 2 2

PA PB PC is

2 2 2 2 2 2

1 1 2 2 3 3x x y y x x y y x x y y

2 21 2 32 2 1 2 3 1 123 2

3 3 3 3

x x x y y y x yx x y y

2 2

2 2 2 21 133 3

x yx x y y x y

The value is maximum when &x x y y

8. A

Sol.

p q p q ~ p ~p q p T T T F T

T F T F T

F T T T T

F F F T T

9. A

Sol.

1,2 0x 10. D

Sol. 1,3 3, 1 1,3x x 11. C

Sol. Because 1 2, , , 0,0 &P T O T are

concyclic points, so circumcentre of 1 2PTT is

mid – point of OP which is Q(h, k)

2 , 2h k

Because , lies on line 0ax by c so

2 2 0a h b k c locus of Q is 2 2 0ax by c

O (0, 0)

,P

2T

1T

,Q h k

12. B

Sol. 4 3t t because 0 0 0cot18 1 & cot18 tan18

3 1t t because 3 11 & 1t t

1 2t t because 0 0 0tan18 0,1 & cot18 tan18

13. C

Sol. Distance of (5, 5) from 3 4 10 0x y is 1 which must be magnitude of new y coordinate similarly distance

of (5, 5) from 4 3 20 0x y is 11 which will be magnitude of new x coordinate.

Now point P be either in 2nd

or IVth quadrant according to new coordinate system.

14. D

Sol. One incentre and 3 excentres

15. D

Sol. Let 1c be circle where equation is 2 2 4 4 0x y x y

2c be circle where equation is 2 2 4 4 0x y x y

Not defined 0 Not defined

3

42

1 2

5

x x

x x

x -5 - 1 0 2

0 + + +


3c be circle where equation is

2 2 4 4 0x y x y

4c be circle where equation is 2 2 4 4 0x y x y

Now new common tangent between 1c & 2c is 2, 1 3&c c is 2, 1 4&c c is 3, 2 3&c c is 3, 2 4&c c is 2 ,

3 4&c c is 2.

16. C

Sol. Let vertices of triangle are (0, 0), (a, 0) and (0, a), then slope of medians through (a, 0) and (0, a) are 1

2 and -2

respectively

12

3 42tan cos1 1 4 5

17. B

Sol. 2

22 2 2 1 3sin 1 sin sin

2 4A

3

14

A

18. A

Sol. Family of lines concurrent at (-2, 0)

Now the required line is the line passing through (-2, 0) and perpendicular to line segment joining (2, 3) and (-2,

0)

Line is 0 2 2

4 3 8 02 3 0

yx y

x

19. A

Sol. Contrapositive of p q is ~ ~q p

20. D

Sol. 2 2 1

dxy x y x y y y

dy

Now at x = 2, y is also equal to 2

Hence dy

dx at (2, 2) is

1 1

2 2 1 3

21. D

Sol. Reflection of A about 1 0x y is (1,1) which lies on side BC equation of BC is 1x

B is (1, 0) and C is (1, 6)

22. C

Sol. sin cos 1 sin 2x x x

If sin 2 0.44x then sin cos 1.2x x

23. B

Sol. Because 1 tan 1 tan 45 2 , so the two digit number is 23. 24. D

Sol. min 20 15 5 30n A B

If min n A B n C then minimum n A B C equals to 30 25. B

Sol. radical axis of circles is x = 0

y axis goes outside both the circle 0ab

Now on solving x = 0 and 2 2 0x y ax c simultaneously we does not get any point c is positive

26. D

Sol. If we apply the condition of orthogonally 1 2 1 2 1 22 2g g f f c c then we get k = 81


but at k = 81 circle 2 2 4 6 81 0x y x y becomes imaginary

27. B

Sol. Combined equation of 1 0ax y and 0x by is 2 2 1 0ax by ab xy x by After making it homogeneous equation of degree two with the help of 1ax by we get

2 2 1ax by ab xy x ax by by ax by = 0

Now coefficient of 2x coefficient of

2y must be equal to zero 22 0a b b

28. A

Sol. Let PT is tangent on 2 2 16 0x y from point (6, 5)

Now PA PB = PT2 2 21 6 5 16 45for 6,5PA PB S 29. B

Sol.

1 1 1

2 1 0

4 2 7

a

a

13

,22

rejecteda

a = 2 is rejected because at a = 2 lines are parallel

30. B

Sol.

2 2,0A

2,0C

B

6,0D

1L 2L

2BCD BAD Locus of B is circle with centre (2, 0) and radius 4 units.



Time Allotted: 3 Hours

Maximum Marks: 210

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

You are not allowed to leave the Examination Hall before the end of the test.

INSTRUCTIONS

Caution: Question Paper CODE as given above MUST be correctly marked in the answer OMR sheet before attempting the paper. Wrong CODE or no CODE will give wrong results.

A. General Instructions 1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.

2. This question paper contains Three Parts.

3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.

4. Each part is further divided into two sections: Section-A & Section-C

5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be provided for rough work.

6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices, in any form, are not allowed.

B. Filling of OMR Sheet 1. Ensure matching of OMR sheet with the Question paper before you start marking your answers on OMR

sheet. 2. On the OMR sheet, darken the appropriate bubble with HB pencil for each character of your Enrolment No.

and write in ink your Name, Test Centre and other details at the designated places. 3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Three Parts. (i) Section-A (01 – 10) contains 10 multiple choice questions which have only one correct answer. Each

question carries +3 marks for correct answer and – 1 mark for wrong answer. Section-A (11 – 15) contains 5 multiple choice questions which have one or more than one correct

answer. Each question carries +4 marks for correct answer. There is no negative marking. (ii) Section-C (01 – 05) contains 5 Numerical based questions with single digit integer as answer, ranging from

0 to 9 and each question carries +4 marks for correct answer. There is no negative marking.

Name of the Candidate :__________________________________________

Batch :___________________ Date of Examination :___________________

Enrolment Number :______________________________________________

BA

TC

HE

S –

Tw

o Y

ea

r C

RP

(1

31

5)-

Ad

va

nc

e (

B L

ot)

FIITJEE

CPT1 - 1

CODE: SET-A

PAPER - 1



Gas Constant R = 8.314 J K1 mol1

= 0.0821 Lit atm K1 mol1

= 1.987 2 Cal K1 mol1


Planck‘s Constant h = 6.626 10–34 Js

= 6.25 x 10-27 erg.s



1 amu = 1.66 x 10-27 kg

1 eV = 1.6 x 10-19 J


Au=79.

Atomic Masses: He=4, Mg=24, C=12, O=16, N=14, P=31, Br=80, Cu=63.5, Fe=56,

Mn=55, Pb=207, Au=197, Ag=108, F=19, H=2, Cl=35.5, Sn=118.6




PPPAAARRRTTT ––– III ::: PPPHHHYYYSSSIIICCCSSS SECTION – A

(Single Correct Choice Type)

This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. Which of the following is a unit vector

(A) ĵî (B) cos î - sin ĵ (C) sin ĵcos2î (D) ĵî3

1

2. A force ˆ ˆ ˆF 5i 3j 2k N

is applied over a particle which displaces it from its origin to the

point ˆ ˆr 2i j m.

The work done (in J) on the particle is :

(A) + 13 (B) + 10 (C) + 7 (D) – 7

3. A uniform chain of length 2m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table ?

(A) 3.6 J (B) 7.2 J (C) 1200 J (D) 120 J

4. A projectile is thrown with velocity u at an angle above the horizontal. Find the average velocity during the time of ascent

(A) u cos (B) usin

2 (C) 2

u1 3cos

2 (D) None of these

5. A block of mass m is attached with a spring in its natural length, of spring constant k. The other end A of spring is moved with a constant acceleration ‗a‘ away from the block as

mAa

shown in the figure. Find the maximum extension in the spring. Assume that initially block and spring is at rest w.r.t ground frame

(A) ma

k (B)

1 ma

2 k (C)

2ma

k (D)

4ma

k.

6. A balloon B is moving vertically upward and viewed by a

telescope T. At a particular angular position = 53° measured

parameters are r = 1 km, dr

3m / sdt

and d

0.02 rad / s.dt

The

magnitude of the linear velocity of the balloon at this instant is

(A) 1.2 m/s (B) 2.4 m/s

(C) 3.6 m/s (D) 4.8 m/s

= 53°

B

r

T

Space For Rough Work


7. Width of a river is 60 m. A swimmer wants to

cross the river such that he reaches from A to B directly. Point B is 45 m ahead of line AC (perpendicular to river) Assume speed of river and speed of swimmer as equal. Swimmer must

try to swim at angle with line AC. Value of is A

BC

River Flow

(A) 37º (B) 53º (C) 30º (D) 16º

8. Find minimum value of the angle so that block of mass m does not move on rough surface, whatever may be the value of applied force F.

The coefficient of state friction between the block and surface is .

F m

() Rough Surface

(A) tan1() (B) 11

tan ( )2

(C) cot1() (D) 11

cot ( )2

9. At time t = 0, a bullet is fired vertically upwards with a speed of 98 ms1. At time t = 5 s (i.e., 5 seconds later) a second bullet is fired vertically upwards with the same speed. If the air resistance is neglected, which of the following statements will be true ?

(A) The two bullets will be at the same height above the ground at t = 12.5 s (B) The two bullets will reach back their starting points at the same time (C) The two bullets will have the same speed at t = 20 s (D) The two bullets will attain the different maximum height 10. Figure shows the changes in speed of a marble as it rolls down

an inclined plane P1, travels on a flat horizontal surface and then up another inclined plane P2. What can you say about the steepness of P1 and P2 from the information given in the figure ?

(A) P1 is steeper than P2 (B) P2 is steeper than P1 (C) P1 and P2 are equally steep (D) Nothing can be said about the relative steepness of P1 and P2

as the information given is insufficient

C

P2

BA20

0

10

ED

P1

20 50 100

Time (s)

Sp

eed

(m

s)

-1

(Multi Correct Choice Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.

11. A spring and block is placed on a fixed smooth wedge as shown. Following conclusion can be drawn about block.

(i) magnitude of its momentum will be max when Fnet on block is zero

(ii) its kinetic energy will be max when Fnet on block is zero (iii) KE of block is max when block just touches the spring. (iv) net force on block is maximum when KE = 0

m

Block

Spring

Fix Wedge

(A) (i) (B) (ii) (C) (iii) (D) (iv)



12. In the figure, if F = 4 N, m = 2kg, M = 4 kg then

(A) The acceleration of m w.r.t. ground is 22

m / s3

(B) The acceleration of m w.r.t. ground is 1.2 m/s2 (C) Acceleration of M is 0.4 m/s2

F

s=0.1=0m

M

k = 0.08

Ground

z

(D) Acceleration of m w.r.t. ground is 22

m / s3

13. A particle moves along positive branch of the curve x

y2

where 3t

x ,3

x and y are

measured in metres and t in seconds, then :

(A) The velocity of particle at t = 1 s is 1ˆ î j2

(B) The velocity of particle at t = 1 s is 1 ˆ î j2

(C) The acceleration of particle at t = 1 s is ˆ ˆ2i j

(D) The acceleration of particle at t = 2 s is ˆ î 2 j

14. Two blocks of masses m1 and m2 are connected through a massless inextensible string. Block of mass m1 is placed at the fixed rigid inclined surface while the block of mass m2 hanging at the other end of the string, which is passing through a fixed massless frictionless pulley shown in figure. The coefficient of static friction between the block and the inclined plane is 0.8. The system of masses m1 and m2 is released from rest.

m=4kg1 m=2kg2

30º Fixed

g=10m/s2

=0.8

(A) The tension in the string is 20 N after releasing the system (B) The contact force by the inclined surface on the block is along normal to the inclined

surface

(C) The magnitude of contact force by the inclined surface on the block m1 is 20 3N

(D) None of these

15. A particle ‗P‘ of mass ‗m‘ is rotating in horizontal circle about vertical axis AB with the help of two strings each of length ‗L‘ as shown in

figure. The separation AB = L, and ‗P‘ rotates with angular velocity ‗‘ about axis AB. Tension in the upper and lower strings are T1 and T2

respectively, then :

(A) T2 will be zero for 2g

L

(B) T1 will always be greater than T2 for any ‗‘

(C) T1 = 3T2, for 4g

L

(D) 21

T mL for 2g

L

L

L

P

L

T1

A

B

T2



SECTION–C (Integer Type)

This section contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The correct digit below the question number in the ORS is to be bubbled.

1. A particle of mass 10 kg is in equilibrium with the help of two ideal

and identical strings. Now one string is cut then, find the ratio of tension in the other string just before cutting and just after cutting.

30°30°

10 kg 2. In a car race, car A takes 4 seconds less than car B to reach the finish line and passes the

finishing line with velocity v more than car B. Assume cars start from rest and travel with constant acceleration aA = 4 m/s

2 and aB = 1 m/s2. Find the value of v in m/s.

3. In the figure, find the velocity of m1 in ms

–1 when m2 falls by 9m. Given m1 = m m2 = 2m (take g = 10 ms–2)

m1=0.1

m2 4. A ball is projected from some height with initial horizontal speed

20 m/s. There is a wall at a horizontal separation of 100 m from

the building. If collision is perfectly elastic find the time in sec

after which it will hit the wall. (t = 0 is taken when ball is thrown).

All surfaces one smooth.

100 m

20 m/s

5. Figure shows a smooth cylindrical pulley of radius R with centre at origin

of co-ordinates. An ideal thread is thrown over it on the two parts of ideal

thread two identical masses are tied initially at rest with co-ordinates (R, 0)

and (-R, -R) respectively. If mass at x-axis is given a slight upward jerk, it

leaves contact with pulley at (R cos, Rsin). Then find /sin.

x

y

m

m



PPAARRTTIIII :: CCHHEEMMIISSTTRRYY SSEECCTTIIOONNAA

Single Correct Choice Type This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is

1. The distance between 3rd and 2nd orbit of hydrogen atom is

(A) 2.646108 cm (B) 2.116108 cm (C) 2.646 cm (D) 0.529 cm

2. H–B–H bond angle in 4BH is:

(A) 180° (B) 120° (C) 109° (D) 90° 3. Which of the following has maximum lattice energy? (A) CaO (B) Na2O (C) MgO (D) BaO 4. The atomic radii of F and Ne in angstrom unit are respectively given by (A) 0.72, 1.60 (B) 1.60, 1.60 (C) 0.72, 0.72 (D) 1.60, 0.72 5. The K.E. of N molecule of O2 is x Joules at –123°C. Another sample of O2 at 27°C has a KE of

2x Joules. The latter sample contains. (A) N molecules of O2 (B) 2N molecules of O2 (C) N/2 molecules of O2 (D) N/4 molecule of O2 6. Out of the following, which does not have zero dipole moment is (A) CO2 (B) CCl4 (C) BCl3 (D) NH3 7. The wave function for 1s orbital of hydrogen atom is given by

r /a01s e2

a0 = radius of Bohr orbit r = distance from nucleus What will be ratio of probability density of finding the electron at the nucleus to the first Bohr‘s

orbit (a0)? (A) e (B) e2 (C) 1/e (D) 0 8. The IP1, IP2, IP3, IP4 and IP5 of an element are 7.1, 14.3, 34.5, 46.8, 162.2 eV respectively.

The element is likely to be (IP ionization potential) (A) Na (B) Si (C) K (D) Ca

Space for rough work


9. 12.25 g KClO3 on heating gives enough O2 to react completely with H2 produced by the action of the Zn on dilute H2SO4.

3 2

2KClO 2KCl 3O , 2 4 4 2

H SO Zn ZnSO H , 2 2 2

2H O 2H O

The weight of Zn required for this is: [At.wt of Zn = 65.5] (A) 9.825 g (B) 19.65 g (C) 39.3 g (D) 8.5 g

10. 2 moles of 4

FeSO in acid medium are oxidised by x moles of 4

KMnO , whereas 2 moles of

2 4FeC O in acid medium are oxidised by y moles of

4KMnO . The ratio of x and y is:

(A) 1

3 (B)

1

2 (C)

1

4 (D)

1

5

Multiple Correct Answer(s) Type

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE or MORE are correct.

11. Which of the following statement is correct regarding H2O2?

(A) it has open booklike structure (B) it is both an oxidizing as well as reducing agent (C) it is a bleaching agent (D) it acts as only oxidizing agent 12. Which of the following will represent Boyle‘s law correctly?

(A) PV

P

(B) V

PV

(C) PV

1/P

(D) P

V 13. Which of the following pairs will not diffuse at the same rate through porous plug at same

conditions of temperature and pressure? (A) CO & NO2 (B) NO2 & CO2 (C) NH3 & PH3 (D) CO2 & N2O 14. A gas obeys the equation P(V-b) = RT. Which of the following is/are correct about the graphs

of gas?

(A) The isochoric curves have slope = R

V b

(B) The isobaric curves have slope = R

P and intercept b.

(C) For the gas compressibility factor = 1+Pb

RT

(D) For the gas compressibility factor = 1Pb

RT



15. Highly pure dilute solution of sodium in liquid ammonia: (A) Shows blue colour (B) Exhibits electrical conductivity (C) Shows reducing properties (D) Shows oxidizing properties

SECTIONC Integer Answer Type

This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive).

1. 1 g of an acid (Molar mass = 150 g/mol) is completely neutralized by 1.5 g KOH. Calculate the number of neutralizable protons in acid.

2. Find out the number of angular nodes in the orbital to which the last electron of Cr enter. 3. According to molecular orbital theory, the number of electrons present in the antibonding

molecular orbitals of N2 is (are)

4. A 17 gm sample of H2O2 contains a% H2O2 by weight and requires a mL of KMnO4 in acidic

medium for comlete oxidation. Thus what is the molarity of KMnO4? 5. The value of x+y+z in following redox reaction

xFeCl3 + yH2S zFeCl2 + S + HCl, is



PART – III : MATHEMATICS SECTION – A

(Single Correct Choice Type) This section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. In a triangle ABC, A(2, 4) and internal angular bisector of B & C are y = x & 2x + y = 3, then find the equation of BC

(A) x = 2 (B) y = 2 (C) x + y = 2 (D) none of these 2. Find the equation of minimum radius of that circle which contain all free circles S1, S2 & S3

where S1 x2 + y2 = 1, S2 (x – 2)

2 + y2 = 9, S3 (x – 2)2 + (y – 2)2 = 4

(A) x2 + y2 – 2x – 2y = 2 (B) x2 + y2 + 2x + 2y = 9

(C) x2 + y2 – 2x – 2y = 9 2 (D) none of these

3. The value of

20log 0.1 0.01 0.001 .............

0.05

is

(A) 81 (B) 1

81 (C) 20 (D)

1

20

4. The equation of the bisector of the acute angle between the lines 2x – y + 4 = 0 and x – 2y = 1

is : (A) x + y + 5 = 0 (B) x – y + 1 = 0 (C) x – y = 5 (D) x – y + 5 = 0

5. The maximum value of cos2x sin2x27 81 is

(A) 23 (B) 53 (C) 73 (D) 3



6. The pair of straight lines joining the origin to the common points of x2 + y2 = 4 and y = 3x + c

are perpendicular, if c2 = (A) – 1 (B) 6 (C) 13 (D) 20 7. The centre of circle inscribed in square formed by the lines x2 – 8x + 12 = 0 and

y2 – 14y + 45 = 0 is (A) (4, 7) (B) (7, 4) (C) (9, 4) (D) (4, 9) 8. |x + 1| + |x − 2| > 3

(A) (−, −1) (2, ) (B) (2, ) (C) (−1, 2) (D) none of these

9. In the equilateral ABC the side length is 8 unit, inscribe this another triangle is form through the

midpoints of vertices A,B and C is DEF. Inside

DEF another triangle is also form through the

midpoints of vertices D,E & F is PQR Find the

area of PQR.

(A) 2 3 (B) 3

(C) 3

2 (D) none of these

A B

C

D E

F

P

Q R

10. Consider the following statements P : Suman is brilliant Q : Suman is rich R : Suman is honest The negation of the statement ―Suman is brilliant and dishonest if and only if Suman is rich‖

can be expressed as

(A) ~ (Q (P ~ R) (B) ~ Q ~P R

(C) ~ (P ~ R) Q (D) ~ P (Q ~ R)



SECTION – A

(Multiple Correct Answers Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE or MORE may be correct.

11. In a ABC

(A) sinA.sinB.sinC 3 3 / 8 (B) 2 2 2sin A sin B sin C 9/ 4

(C) sinAsinBsinC is always positive (D) 2 2sin A sin B 1 cosC

12.

n n

cos A cosB sinA sinB

sinA sinB cos A cosB

(n, even or odd) is equal to

(A) nA B

2tan2

(B) nA B

2cot2

(C) 0 (D) none of these

13. Common tangents to the circles x2 + y2 – 2x – 6y + 9 = 0 & x2 + y2 + 6x – 2y + 1 = 0 are. (A) 3x + 4y – 10 = 0 (B) 4x – 3y = 0 (C) y = 4 (D) x = 0

14. Two circles x2 + y2 + x = 0 & x2 + y2 = c2 touch each other if

(A) + c = 0 (B) c = 0 (C) 2 = c (D) none of these

15. If x satisfies x 1 x 12 2

log (9 7) 2 log (3 1) then

(A) x Q (B) x {x Q: x 0}

(C) x N (D) x Ne (set of even natural numbers)



SECTION – C

(Integer Answer Type)

This Section contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in the ORS.

1. A(0, 0), B(2, 1) and C(3, 0) are the vertices of a triangle ABC, and BD is its altitude. The line

through D parallel to the side AB intersects the side BC at a point K. If the product of the areas of the triangle ABC and BDK is k, then the value of 2k is

2. If 2sinx sin x 1 then the value of 2 4 4 2cos x cos x cot x cot x is equal to

3. If 1 2 3cos 2cos 3cos 6 then 1 2 3tan tan tan equals to

4. If 2 2 2log x log y log z

4 6 3k and x3y2z = 1

Then |k| is 5. Let the co-ordinates of the circumcentre of the triangle whose vertices are A(5, – 1), B(–1, 5)

and C(6, 6) is (a, b) then [a + b] is (where [.] denotes the greatest integer function)



FIITJEE COMMON TEST BATCHES: TWO YEAR CRP (1315)-Advance-(B LOT)

PHASE TEST-I (PAPER-1) ANSWER KEY

PART – I (PHYSICS) PART – II (CHEMISTRY) PART – III (MATHS)

SECTION-A

1. B

2. C 3. A 4. C 5. C 6. C 7. D 8. C 9. A 10. A 11. A, B, D 12. B, C 13. A, C 14. A, B, C 15. A, B, C

SECTION-C

1. 2 2. 8 3. 4 4. 5 5. 2

SECTION – A 1. A 2. C 3. C 4. A 5. A 6. D 7. B 8. B 9. B 10. A 11. A,B,C 12. A,B,D 13. A,B,C 14. ABC 15. A,B,C

SECTION–C 1. 4 2. 2 3. 5 4. 2 5. 5

SECTION-A

1. B 2. C 3. A 4. A 5. B 6. D 7. A 8. A 9. B 10. A 11. A,B,C,D 12. B,C 13. A,B,C,D 14. A,B,C 15. A,C

SECTION-C

1. 1 2. 2 3. 0 4. 8 5. 5

Paper Code

SET-A


HINT & SOLUTIONS

PPPAAARRRTTT ––– III ::: PPPHHHYYYSSSIIICCCSSS SECTION – A


1. B

2. C

F.r

= 10 – 3 = 7 3. A (1.2) (10) (0.3) 4. C

av

R ˆ ˆi Hj2

| v |T

2

5. C From work energy in the frame which is attached to point A

2 2 2xm m1 1 1

max k x m(0) m(0)2 2 2

max2ma

x .k

6. C

y = r sin

dy dr d

sin cos .r 3.6m / sdt dt dt

7. D

45

tan60

= 37°

since, VS = VR,

53° = + 37°

= 16°.

vR

vS

53º

37º

8. C Fcos N …..(A)

N Fsin mg …..(B)

By (A) and (B)

Fcos Fsin mg

Fcos Fsin mg


F(cos sin ) mg

if F is . Hence cos sin 0

cos sin 0 cot =

= cot1()


9. A

y1 = 9.8t 4.9t2

y2 = 9.8 (t5) 4.9 (t5)2

y1 = y2 gives t = 12.5 10. A

Acceleration = slope of v/t graph = g sin is greater for P1 where is angle of inclined plane.


11. A, B, D 12. B, C

Here sm

F mg 1M

For m

kF mg m.a

For M

kmg MA

2A 0.4m / s

13. A, C

2dx

tdt

…(i)

31 t

y2 3

2dy t

dt 2 …(ii)

t = 1, vx = 1, vy = 1

2

1ˆ ˆv i j2

2

2

d x2t

dt …(iii)

2

2

d yt

dt …(iv)

at t = 1 s ax = 2 and ay = 1

ˆ ˆa 2i j

.

14. A, B, C If the tendency of relative motion along the common tangent does not exist, then component

of contact force along common tangent will be zero. 15. A, B, C For particle ‗P‘

21 2(T T )cos 30 mL cos 30

1 2T sin 30 T sin 30 mg.

SECTION–C

(Integer Type) 1. 2


2. 8

B A

2S 2S4

a a

A B2a S 2a S v

v 8m / s.

3. 4 Applying work energy theorem Kf – Ki = w

2 22 11 1

m v m (4v)2 2

2 1m g y m g(4y)

v = 4 m/s.

m1

m2

mg

v

T

4. 5 Horizontal velocity of ball will not change 100 = 20 t t = 5 sec.

5. 2

mgR – mg R sin = ½ (2m)v2 …. (i)

mv2/R = mg sin … (ii)


PPPAAARRRTTT ––– III ::: CCCHHHEEEMMMIIISSSTTTRRRYYY

SECTION – A (Single Correct Choice Type)

1. A

r = r3r2 = r09 r04

r = 5r0 = 50.0529108 = 2.646108 cm

2. C

Hybridization in BH4 is sp3, hence bond angle in 109

3. C Latlic energy in inversely propotional to size 4. A Noble gases have more radii than halogen in respective periods 5. A

KE =3

RT2

; T = – 123 + 273 = + 150 K

3

R 1502

38.314 75

2 = xJ = 225 8.314 = xJ

At 27°C = 27+ 223 = 300K

KE for = 2x Joule = 3

8.314 3002

N molecules

x Joule = 3 8.314 75 In both the cases x Joules correspond to N molecules. 6. D NH3 does not have zero dipole moment 7. B

Probability density 2

02

(r 0) 2

22(r 1)

e2 e

e2

8. B There is large difference between 4th and 5th IP, hence element should contain four valence

electrons. 9. B

Moles of O2 produced = 3/2 moles of KClO3 = 3 12.25

0.152 122.5 mole

According to given equation One moles of O2 required 2 mole of H2 = 2 moles of Zn

Moles of Zn = 2moles of O2=0.152=0.3 mole

Mass of Zn require = 0.365.5 = 19.65 g 10. A

nfactor of 4FeSO 1

nfactor of 2 4FeC O 3

Hence geq. of 4 4FeSO g e.q of KMnO

12 = 5x (1)

geq. of FeC2O4 = geq of KMnO4

32 = 5y Hence x/y = 1/3



11. A,B,C H2O2 is both oxidising and reducing agent 12. A,B,D

According to Boyle‘s law 1

P or PV constantV

13. A,B,C

Rate of diffusion 1

M

CO2 and N2O have same molar mass = 44 g/mol 14. ABC

P(V b) RT

PV Pb RT

PV PbZ 1

RT RT

15. A,B,C Solution of alkali metal in liquid ammonia shows blue colour, exhibit electrical conductivity and

show reducing property SECTION–C

(Integer Type)

1. 4

geq. of acid = geq. of base

1 1.5

n150 56

n=4 2. 2

Last electron of Cr enters in dsubshell, hence l=2 3. 5

Electronic configuration of N2 is

1s2 1s2 2s2 2s22px22py

22pz

2 2px1

4. 2

geq. of H2O2 = geq. of KMnO4

17 2 a

a 5 M100 34 1000

M = 2 molar 5. 5 Balance equation is

2FeCl3 + H2S 2FeCl2 + S + 2HCl x + y + z = 5


SECTION – A


MATHEMATICS 1. B Take the reflection of A about internal angular bisector of B & C lie on the line BC. 2. C For centre - find the circumcentre of the centres S1, S2 & S3, For radius - find the circum radius from the centres S1, S2 & S3 and add the maximum radius

of the circles S1, S2 & S3 in the circum radius. 3. A Use the log properties 4. A

2x – y + 4 = ( x – 2y 1) For acute angle bisector use + sign 5. B 33cos2x+4sin2x Then maximum value is 35 6. D Use the homogenization 7. A Lines are x = 2 and x = 6, y = 5, and y = 9 Then centre is (4, 7) 8. A 9. B

Area = 23

24

10. A (Multi Correct Choice Type)

11. A,B,C,D

2 2 2 2 2 21 1

sin A sin B sin C (1 cos A) (1 cos B) sin C2 2

22 cos C cosC.cos(A B)

22 (cos C cosC)

2

9 1cosC

4 2

sin2A + sin2A+sin2C 9/4

Now 2 2 2

2 2 2 1/ 3sin A sin B sin C (sin A.sin B.sin C)3

2 / 39 / 4

(sinA.sinB.sinC)3

or 3 3

sinA.sinB.sinC8

also 2 2 2 2sin A sin B sin C 2 cos C cosC

sin2A + sin2B 1+ cos C. 12. B,C

n n

cos A cosB sinA sinB

sinA sinB cos A cosB

n n

A B B A2cot cot

2 2

If n even, nA B

2cot2

, if n odd, 0.

13. A,B,C,D


Points can be calculated by the internal and external section formula using centres of both the circles and slope can be calculated by using the condition of tangency.

14. A,B,C Use the internal and external touching condition of two circles. 15. A,C

x 1

2 x 1

9 7log 2

3 1

32x2 + 7 = 4(3x1 + 1)

x 1 x 1(3 1)(3 3) 0

x 1 = 0 or x 1 = 1 x = 1, 2.

SECTION–C (Integer Type)

1. 1 Calculate the area of ABC and BDK then multiply these two 2. 2 Given sin x + cos x + tan x + cot x + sec x + cosec x = 7

1 sinx cosx

sinx cosx 7sinx cosx sinx cosx

1 1

sinx cosx 1 7sinx cosx sinxcosx

2 2

2 21 sin2x 1 7

sin2x sinx

2 2

1 t t 2 7t 2 , where t = sin 2x

3 2t 44t 36t 0

2t 44t 36 0 [sin 2x 0]

244 44 4 36

t 22 8 72

sin2x 22 8 7

3. 0

Given 1 2 3cos 2cos 3cos 6

1 2 3cos cos cos 1

1 2 3 0

1 2 3tan tan tan 0

4. 8 Use the log property. 5. 5 Circum centre can be calculated by using the perpendicular bisectors of vertices of the

triangle.



Time Allotted: 3 Hours

Maximum Marks: 198

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.

You are not allowed to leave the Examination Hall before the end of the test.

INSTRUCTIONS

Caution: Question Paper CODE as given above MUST be correctly marked in the answer OMR sheet before

attempting the paper. Wrong CODE or no CODE will give wrong results.

C. General Instructions 1. Attempt ALL the questions. Answers have to be marked on the OMR sheets.

2. This question paper contains Three Parts.

3. Part-I is Physics, Part-II is Chemistry and Part-III is Mathematics.

4. Each part is further divided into one section: Section-A

5. Rough spaces are provided for rough work inside the question paper. No additional sheets will be provided for rough work.

6. Blank Papers, clip boards, log tables, slide rule, calculator, cellular phones, pagers and electronic devices, in any form, are not allowed.

D. Filling of OMR Sheet 1. Ensure matching of OMR sheet with the Question paper before you start marking your answers on OMR sheet.

2. On the OMR sheet, darken the appropriate bubble with HB pencil for each character of your Enrolment No. and

write in ink your Name, Test Centre and other details at the designated places.

3. OMR sheet contains alphabets, numerals & special characters for marking answers.

C. Marking Scheme For All Three Parts.

(i) Section-A (01 – 08) contains 8 multiple choice questions which have only one correct answer. Each question

carries +3 marks for correct answer and – 1 marks for wrong answer.

Section-A (09 – 14) contains 3 paragraphs. Based upon paragraph, 2 multiple choice questions have to be

answered. Each question has only one correct answer and carries +3 marks for correct answer and – 1 mark for

wrong answer.

Section-A (15 – 20) contains 6 multiple choice questions which have one or more than one correct

answer. Each question carries +4 marks for correct answer. There is no negative marking.

Name of the Candidate :____________________________________________

Batch :____________________ Date of Examination :___________________

Enrolment Number :_______________________________________________

BA

TC

HE

S –

Tw

o Y

ea

r C

RP

(1

31

5)-

Ad

van

ce

(B

Lo

t)

FIITJEE

CPT1 - 2

CODE:SET-A

PAPER - 2



Gas Constant R = 8.314 J K1

mol1

= 0.0821 Lit atm K1

mol1

= 1.987 2 Cal K1

mol1


Planck‟s Constant h = 6.626 10–34

Js

= 6.25 x 10-27

erg.s



1 amu = 1.66 x 10-27

kg

1 eV = 1.6 x 10-19

J


Au=79.

Atomic Masses: He=4, Mg=24, C=12, O=16, N=14, P=31, Br=80, Cu=63.5, Fe=56,

Mn=55, Pb=207, Au=197, Ag=108, F=19, H=2, Cl=35.5, Sn=118.6




PPPAAARRRTTT ––– III ::: PPPHHHYYYSSSIIICCCSSS

SECTION – A


This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONLY ONE is correct.

1. A body of mass „m‟ is placed on a plank which is tilted till the mass just slips. What is the

maximum horizontal force that can be applied on „m‟ at this position without causing

slipping? 1

.3

(A) mg (B) 2mg (C) 3 mg (D) 3

mg2

2. A stone is projected vertically upwards so as to reach a height h passes points P and Q

with velocities v

2 and

v,

3 where v is initial velocity with which the body is thrown. The

distance between P and Q in terms of h is :

(A) 7

h36

(B) 5

h36

(C) 9

h36

(D) 8

h36

3. A particle moves in a straight line with a velocity tv t 2 m / s, where t is time in

second. What is the distance covered by the particle in 4 s ?

(A) 2m (B) 4m (C) 1m (D) 8m

4. A body of mass 500 g is accelerated from a velocity 1ˆ ˆ3i 4j ms to 1ˆ ˆ6j 2k ms . Find the work done :

(A) 3.75 J (B) 0 (C) 4.75 J (D) 16 J

5. The minimum work done in moving a particle from a point (1, 1) to (2, 3) in a plane

having force field with potential U = (x + y) is :

(A) 0 (B) (C) 3 (D) – 3

6. A particle of mass m is fixed to one end of a light spring of force

constant K and unstretched length . The particle is rotated about

the other end of the spring with an angular velocity , in gravity free

space. The increase in length of the spring will be :

(A) 2m

K

(B)

2

2

m

K m

(C) 2

2

m

K m

(D) None of these

m K



7. The velocity of a particle moving in the positive direction of x-axis varies as v A x. The

graph of displacement versus time is :

(A)

x

t

(B)

x

t

(C)

x

t

(D)

x

t

8. The bob of a simple pendulum at rest, is given a sharp hit to impart a horizontal velocity

8g where is length of pendulum. The tension in the string :

(A) T = 6 mg when the string is horizontal

(B) T = 4 mg when the bob is at highest point

(C) T = 8 mg when the string is horizontal

(D) T = 6 mg when the bob is at highest point.

(Paragraph Type)

This section contains 3 paragraphs based upon paragraph 2 multiple choice questions have to be

answered. Each of these questions has four choices (A), (B), (C) and (D) out of WHICH ONLY

ONE is correct.

Paragraph for Question Nos. 9 to 10

From the top of a tower of height H, a stone of mass 2 kg is thrown vertically upwards

with a speed U and it hits the ground below in 28 sec. When same stone was thrown with

same speed vertically down from same position, the time taken to hit the ground is 7 sec.

P1 is the average power of gravity in first case and P2 in second case (g = 10 m/s2)

9. The kinetic energy with which the stone was thrown initially is :

(A) 9025 J (B) 11025 J (C) 13225 J (D) 11449 J

10. Taking potential energy at ground level as zero, the maximum potential energy attained

by the stone thrown vertically upwards is :

(A) 30625 J (B) 32825 J (C) 28625 J (D) 31049 J




In figure shown on the right, the spring constant is K. The mass of block is m.

The block is imparted a downward velocity = v0 at t = 0, at its equilibrium

position.

11. The value of v0 for which the block has zero velocity when spring is in its natural position

is

(A) m

gK

(B) m

2gK

(C) m

g2K

(D) m

4gK

12. The value of v0 for which the minimum pull force on ceiling is mg

,2

will be

(A) m

g2K

(B) m

gK

(C) m

2gK

(D) g m

2 K


In the system shown in the figure, the mass 30 kg is

pulled by a force of 210 N. Answer the following

questions at the instant when the 15 kg mass has

acceleration 6 m/s2. Assume the spring to be mass

less and spring constant is 100 N/m. The surface of

ground is smooth.

210 N

30 kg 15 kg

13. Find the acceleration of 30 kg mass

(A) 2 m/s2

(B) 3 m/s2 (C) 3.4 m/s

2 (D) 4 m/s

2

14. Find the elongation in the string at this instant

(A) 0.3 m (B) 0.6 m (C) 0.9 m (D) None of these

(Multiple Correct answers Type)

This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONE OR MORE may be correct.

15. Acceleration vs time graph is shown in the figure for a

particle moving along a straight line. The particle is

initially at rest. Find the time instant(s) when the

particle is at rest?

(A) t = 0

(B) t =

physics, chemistry & mathematics jee - mains 2015 paper class 11.pdf · 2017. 9. 19. · jee -...

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