FIITJEE CM Monthly Test-1

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PHYSICS, CHEMISTRY & MATHEMATICS

Time Allotted: 3 Hours

Maximum Marks: 183

Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose. You are not allowed to leave the Examination Hall befo re 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 Sections.

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

4. All the section can be filled in PART-A of OMR.

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 Blue/Black Ball Point Pen 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 Only One Part.

(i) Part-A (01-07) – Contains seven (07) multiple choice questions which have One or More correct answer. For each question in the group Q. 01 – 06 of PART – A you will be awarded Full Marks: +4 If only the bubble(s) corresponding to all the correct options(s) is (are) darkened. Partial Marks: +1 For darkening a bubble corresponding to each correct option, provided NO incorrect option is darkened.

Zero Marks: 0 If none of the bubbles is darkened. Negative Marks: 1 In all other cases.

(i) Part-A (08-14) – Contains seven (07) multiple choice questions which have ONLY ONE CORRECT answer Each question carries +3 marks for correct answer and -1 marks for wrong answer.

(i) Part-A (15-18) - This section contains Two paragraphs. Based on each paragraph, there are Two multiple

choice questions. Each question has only one correct answer and carries +3 marks for the correct answer and -1 marks for wrong answer.

Name of the Candidate :____________________________________________

Batch :____________________ Date of Examination :___________________

Enrolment Number :_______________________________________________

BA

TC

HE

S –

21

23

FIITJEE – CM Monthly Test-1

Pattern -1

QP Code:

PAPER - 1

CM Test-PT-I-(Paper-1)-PCM-IITJEE_XI-2123

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

2

PPPAAARRRTTT ––– III ::: PPPHHHYYYSSSCCCIIISSS

SECTION – A

(Multiple Choice Type) This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct.

1. The velocity versus time of two particles moving along the x-axis varies as shown in the following

two plots. Both started from the same point.

(A) Maximum separation between the two particles is 2 m.

(B) Maximum separation between the two particles is 2.5 m.

(C) Maximum separation between the two particles occurs after time t = 2 s

(D) Maximum separation between the two particles occurs after time t = 3s

2. In the given figure, points A and C are on the horizontal ground. Points A, B and C are in the same

vertical plane. Bullets are fired simultaneously from A, B and C. The bullet at B is fired horizontally

with a speed of 172kmh

5

and the bullet at C is projected vertically upward at a velocity of

154kmh

5

. Choose correct option(s)

(A)Collision of all the three bullets at the same time is not possible

(B) The three bullets can collide at D if velocity of the bullet projected from A is 5 ms-1

.

(C) If velocity of the bullet projected from A is 5 ms-1

then all three will collide irrespective of the

height from which B is projected.

(D) If bullets are fired only from A and C, then there is a unique value of projection speed at A for

which it collides with the bullet fired from C.



3

3. Four vectors A,B,C,D

all have the same magnitude and lie in a plane. The angle between

adjacent vectors is 45º as shown. Which of the following equation(s) is/are correct?

(A) A C 2D

(B) B D 2 C 0

(C) A B B D

(D) A C / 2 B

4. Two balls A and B are thrown with the same velocity u from the top of a tower. Ball A is thrown

vertically upwards and ball B is thrown vertically downwards. Choose the correct statement(s).

(A) Both the balls reach the ground with same velocity.

(B) If tA and tB are the respective times taken by the balls A and B to reach the ground, then

A Bt t

(C) If tA = 6s and tB = 2s, then the height of the tower is 60 m

(D) If tA = 6s and tB = 2s, then velocity of each ball is 20 ms-1

.

5. Ball I is thrown towards a tower at an angle of 60º with the horizontal with unknown speed (u). At

the same moment ball II is released from the top of the tower as shown. Balls collide after 2 s, and

at the moment of collision, velocity of ball I is horizontal.

(A) Speed u is 140 / 3 ms

(B) Distance of point of projection of ball I from the base of the tower (x) is 40/ 3m

(C) Height of the tower (h) is 40 m

(D) The two balls cannot collide while the first one is moving horizontally.



4

6. A body moves in a circular path of radius R with deceleration so that at any moment of time its

tangential and normal accelerations are equal in magnitude. At time t = 0, the velocity of the body is

v0. The velocity of the body at some later instant can be expressed as

(A) 0

0

vv

v t1

R

at time t

(B)

s

R0

v v e

after it has moved a distance s.

(C) sR0

v v e after it has moved a distance s.

(D) None of these

7. Two swimmers A and B start swimming from different positions on the same bank as shown in

figure. The swimmer A swims at angle 90º with respect to the river to reach point P. He takes 120

seconds to cross the river of width 10 m. The swimmer B also takes the same time to reach the

point P. Now choose from the following the correct alternative(s).

(A)velocity of A with respect to river is 1

m/ s6

(B) river flow velocity is 1

m/ s4

(C) velocity of B along y-axis with respect to earth is 1

m/ s3

(D) velocity of B along x-axis with respect to earth is 5

m/ s24

SECTION – A : (Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

8. The velocity of a particle is given as 1ˆv i 6t j ms

. The particle is moving in the xy plane. At t =

0, the particle is at the origin. Find the radius of curvature of its path at 2 2

m, m3 3

.

(A) 9 m (B) 4.5 m (C) 13.5 m (D) 2.25 m



5

9. If 1e

and 2e

are two unit vectors and is the angle between them, then sin2

is

(A) 1 2

1e e

2

(B)

1 2

1e e

2

(C) 1 2

e e

2

(D) 1 2

1 2

e e

2 e e

10. A small marble is projected with a velocity of 10 ms-1

in a direction 45º from the y-direction

(horizontal) on a smooth inclined plane. The magnitude of its velocity after 2s will be

(take g= 10 ms-2

)

(A) 5 1ms (B) 6.5 1ms (C) 10 1ms (D) 12.5 1ms

11. A stone projected at an angle of 60º from the ground level strikes the roof of a building h meter

high, making an angle of 30º with the roof. Then the speed of projection of the stone is

(A) 2gh (B) 6gh (C) 3gh (D) gh

12. A particle starts moving with velocity 10 ms-1

in a straight line with an acceleration varying linearly

with time. Its velocity-time graph is shown in the figure. Its velocity is maximum at t = 3 seconds.

The time (in seconds) when the particle stops, is (tan 37º = 3/4)

(A) 6.41 sec (B) 8.18 sec (C) 7s (D) 12.43 sec



6

13. A particle having a velocity v = v0 at t = 0 is decelerated at a rate of |a| = v , where is a positive

constant. After what time will the particle come to rest?

(A) 0v

(B)

02 v

(C) 02 v (D)

04 v

14. A car is at rest at the foot of a = 37º incline. The driver finds that it is raining and the raindrops

which are falling make an angle of 37º to the vertical. The car begins to climb while moving at a

uniform acceleration of a = 2 ms-2

.

After 4 s, he finds that the raindrops are falling vertically. Find the actual speed of the rainfall.

(A) 18.67ms (B) 19.5 ms (C) 110.67ms (D) 112.35 ms

Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Questions 15 and 16

Projectile inside a box sliding on an incline.

A large heavy box is sliding without friction down an inclined plane of inclination . From a point P,

at the bottom of the box, a particle is projected inside the box. The initial speed of projection of the

particle relative to the box is u and direction of projection makes an angle with the bottom surface

as shown.

15. The distance along the bottom of the box between the point of projection P and the point Q where

the particle lands is

(A) 2u sin 2

gsin

(B)

22u sin 2

gcos

(C)

2u sin 2

gcos

(D)

2 2u sin

gcos

16. If the horizontal displacement of the particle as seen by an observer on the ground is zero, then the

speed of the box with respect to the ground at the instant when the particle was projected is

(A) usin( )

sin

(B)

ucos( )

sin

(C)

usin( )

cos

(D)

ucos( )

cos



7


Have you ever seen the trajectory of a projectile. If you watch the trajectories of projectiles thrown

at different angles, you would find, there are some projectiles, whose distance from point of projection first

increases and then decreases (figure 1), while there may be certain projectiles whose distance from point

of projection always increases (figure 2)

A projectile is thrown from O on an inclined plane as shown in (figure 3). It hits the plane

perpendicularly.

17. Range of the projectile will be

(A) 20 m (B) 40 m (C) 70 m (D) 80 m

18. If projectile is thrown at time t = 0 then time when projectile will the inclined plane.

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



8

PPPAAARRRTTT ––– IIIIII ::: CCCHHHEEEMMMIIISSSTTTRRRYYY SECTION – A

(Multiple Choice Type) This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONE OR MORE THAN ONE is/are correct.

1. The van der Waa‟s equation for a real gas is

(A) 2

aP V b RT

V

(B)

2

aP V nb nRT

V

(C) 2n a

P V nb nRTV

(D) 2

2

n aP V nb nRT

V

2. In the following reaction (unbalanced), equivalent wt. of As2S3 is related to molecular wt. M By:

3 22 3 3 2 4 4As S HNO NO H O AsO SO

(A) M/2 (B) M/4 (C) M/28 (D) M/24 3. The kinetic energy of one mole of a gas is given by the expression :

3PV

K.E.2

and 3

K.E. RT2

(A) K.E. P at constant temperature

(B) K.E. T at constant pressure (C) K.E. is not directly proportional to volume at constant temperature

(D) K.E. V at constant temperature

4. A photon having = 854 Å causes the ionization of a nitrogen atom. Given the I.E. per mole of nitrogen in kJ.

(A) 2.21 × 103 kJ/mole (B) 9.67 × 10

8 kJ/mole

(C) 1.39 × 103 kJ/mole (D) 5.32 × 10

4 kJ/mole

5. The longest wavelength of He

+ in Paschen series is “m”, then stortest wavelength of Be

3+ in

Paschen series is (in terms of m)

(A) 5

m36

(B) 64

m7

(C) 53

m8

(D) 7

m64

6. 22 7Cr O is reduced to Cr

3+ by 2Fe . Identify the incorrect statement from the following :

(A) 6 moles of 2Fe are oxidised to 3Fe ions

(B) The solution becomes yellow (C) The solution becomes green

(D) 3 moles of 2Fe get oxidised to 3Fe

7. The graph of P vs V is given at different temperatures and number of rule curves, n1, n2, n3 are

number of moles the correct relationship are

(A)

1 2 3T T T (B)

1 2 3T T T (C)

3 2 1n n n (D)

1 2 3n n n



9

SECTION – A : (Single Correct Answer Type) This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

8. The third line in Balmer series corresponds to an electronic transition between which Bohr‟s orbits in hydrogen :

(A) 5 3 (B) 5 2 (C) 4 3 (D) 4 2 9. The ratio of difference in wavelengths of 1

st and 2

nd lines of Lyman series in H-like atom to

difference in wavelength for 2nd

and 3rd

lines of same series is : (A) 2.5 : 1 (B) 3.5 : 1 (C) 4.5 : 1 (D) 5.5 : 1 10. One mole of N2H4 loses ten moles of electrons to from a new Compound y. Assuming that all the

nitrogen appears in the new compound. What is the oxidation state of nitrogen in y. There is no change in the oxidation state of hydrogen :

(A) - 1 (B) – 3 (C) + 5 (D) + 3 11. An equimolar mixture of NaHC2O4 and H2C2O4 consumes 20 ml 0.3 M NaOH solution for complete

neutralization. The same mixture requires V ml. 0.05 M KMnO4 solution in acidic medium for oxidation.

The value of V is : (A) 160 ml (B) 32 ml (C) 24 ml (D) None of these 12. n moles of Hellum gas are placed in a vessel of volume V liter. At TK, If V1 is free volume of Hellum

then diameter of He atom is

(A) n

1

31

A

V3

2 N

(B)

n

1

31

A

3 V V

2 N

(C)

n

1

31

A

6 V V

N

(D) n

1

31

A

6V

N

13. Temperature at which most probable speed of O2 becomes equal to root mean square speed of N2

is [Given : N2 at 427ºC] (A) 732 K (B) 1200 K (C) 927 K (D) 800 K 14. Correct option regarding a container 1 mol of a gas in 22.4 litre container at 273 K is (A) If compressibility factor (z) > 1 then „P‟ will be less than 1 atm. (B) If compressibility factor (z) > then „p‟ will be greater than 1 atm (C) If „b‟ dominates, pressure will be less than 1 atm (D) If „b‟ dominates, pressure will be greater than 1 atm.



10

Comprehension Type

This section contains 2 paragraphs. Based upon each paragraph, 2 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.


25 ml from a stock solution containingNaHCO3 and Na2CO3 was diluted to 250 ml with CO2 free distilled water.25 ml of the diluted solution when titrated with 0.12 M HCl required 8 ml., when phenolphthalein was used as an indicator.

HPh

2 3 3Na CO HCl NaHCO

When 20 ml of diluted solution was titrated with same acid it required 18 ml when methlyorange was used as an indicator.

MeOH2 3 2 2Na CO 2HCl 2NaCl H O CO

MeOH3 2 2NaHCO HCl NaCl H O CO

15. Concentration of NaHCO3 in gm/lit. (A) 0.312 (B) 2.62 (C) 3.12 (D) 26.208 16. Amount of NaOH that should be added to convert all bicarbonate into carbonate in 100 ml stock

solution (A) 1.248 gm (B) 0.312 gm (C) 3.12 × 10

–2 gm (D) 7.8 × 10

–2 gm


On the recently discovered 10th planet it has been found that the gaseous follow the relationship

PeV/2

= nCT where C is constant other notation are as usual (V is atm and T iin Kelvin). A curve is plotted between P & V at 500 K & 2 moles of gas as shown in figure

17. The value of constant C is (A) 0.01 (B) 0.001 (C) 0.005 (D) 0.002

18. Find the slope of the curve plotted between P Vs T for closed container of volume 2 lit, having

same moles of gas

(A) e

2000 (B) 2000 e (C) 500 e (D)

2

1000e



11

PPPAAARRRTTT ––– IIIIIIIII ::: MMMAAATTTHHHEEEMMMAAATTTIIICCCSSS

(Multi Correct Choice Type) This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.

1. Which of the following is/are correct?

(A)

ln sinx ln sinx

tanx cot x , x 0, / 4 (B) lncosec x lncosec x4 5 , x 0, / 2

(C)

ln cosx ln cosx

1/2 1/3 , x 0, /2 (D) ln tanx ln sinx2 2 , x 0, / 2

2. Let f : (-1, 1) R be such that 2

2f cos4

2 sec

for 0, ,

4 4 2

. Then the value(s) of

1f

3

is(are)

(A) 3

12

(B) 3

12

(C) 2

13

(D) 2

13

3. If 3

x = 4

x-1, then x =

(A) 3

3

2log 2

2log 2 1 (B)

2

2

2 log 3 (C)

4

1

1 log 3 (D) 2

2

2log 3

2log 3 1

4. Which of the following is/are correct?

(A) n 1

n xx dx c n R

n 1

(B) sec xdx ln sec x tanx c

(C) x

x

dxln c e 1

1 e

(D) xd

x ,x 0dx x

5. Which of the following is INCORRECT ?

(A) If 2 4

2

d 1 x xax b

dx 1 x x

, then a + b = 3

(B) tanxy e , then x 0

dy0

dx

(C) 2x 1

x 1lim 0

2x 7x 5

(D) x 0

x h x 1lim

h x

6. Let fn() = n

4 r

rr 0

1sin 2

4

. Then which of the following alternative(s) is/are correct?

(A) 2

1f

4 2

(B) 3

2 2f

8 4

(C) 4

3f 1

2

(D) 5f 0



12

7. If sin sin a and cos cos b, a b,a 0,b 0 , then

(A) 2 2 2

8abtan tan

a b 4b

(B)

22 2 2

2 2

a b 4acos .cos

4 a b

(C) 2 2

2 2

b acos

b a

(D)

2 2 2

4absin

a b 2b

SECTION – A : (Single Correct Answer Type) This section contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

8. A circle is drawn in a sector of a larger circle of radius r, as shown in figure. The smaller circle is

tangent to the two bounding radii and the arc of the sector. The radius of the smaller circle is

60º

r

(A) r

2 (B)

r

3 (C)

2 3r

5 (D)

r

2

9. The maximum value of the expression 2 2 2 2sin x 2a 2a 1 cos x , where a and x are real

numbers, is

(A) 3 (B) 2 (C) 1 (D) 5

10. If , , , are the smallest positive angles in ascending order of magnitude which have their sines

equal to the positive quantity k, then the value of 4sin 3sin 2sin sin2 2 2 2

is equal to

(A) 2 1 k (B) 2 1 k (C) 1 k

2

(D) None of these

11. The solution set of the equality

x 2e 1 2x 3 x x 20

sinx 2 x x 1

(A) 3

,2

(B)

3, 1 ,

2

(C) 3

1,0 ,2

(D) R 0, 1

12. The value of the integral 2

0

cosxdx

cosx sinx

is

(A) 1 (B) 2

(C) 0 (D)

4



13

13. The largest number among the following numbers is

(A) tan 47º + cos 47º (B) cos 47º + 2 sin 47º

(C) 2 cos 47º + sin 47º (D) tan 47º + cot 47º

14. The equation 2x x

| x |x 1 | x 1|

will be always true for x belonging to

(A) (1, ) (B) (1, ) {0}

(C) (1, 1) (D) (, )

(Paragraph Type) This section contains 2 paragraphs. Based upon the paragraphs 2 multiple choice questions have to be answered. Each of these questions has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is

correct.

Paragraph for Question no. 15 to 16

In ABC, BC = 1, sinA

2= x1, sin

B

2 = x2, cos

A

2= x3 and cos

B

2= x4 with

2007 2006

31

2 4

xx0

x x

.

15. Length of side AC is equal to (A) 1/2 (B) 1 (C) 2 (D) Can‟t be determined

16. If C = 90º, then area of ABC is (A) 1/2 sq. units (B) 1/3 sq. units (C) 1 sq. units (D) 2 sq. units

Paragraph for Question no. 17 to 18

Consider a function 2

3x af(x)

x 3

which has greatest value equal to 3/2.

17. The value of the constant number a is equal to (A) 1 (B) 2 (C) 3 (D) 4 18. The minimum value of f(x) is equal to

(A) tan3

(B) sin6

(C) cos3

(D) cot2



14

FIITJEE MONTHLY ASSESSMENT TEST-1 PHYSICS, CHEMISTRY & MATHEMATICS

IIT-JEE, 2022-23 QP CODE:

ANSWERS PHYSICS C.CODE CHEMISTRY C.CODE MATHEMATICS C.CODE

1 BD P110308 1 AD C111204 1 ABCD M111401 2 B P110304 2 C C111101 2 AB M111403 3 ABD P110202 3 BC C111206 3 ABC M111401 4 ACD P110301 4 C C110105 4 BCD M111701 5 ABC P110304 5 D C110103 5 ABCD M111803 6 AB P110307 6 BD C110101 6 CD M111402 7 BD P110310 7 AD C111201 7 BC M120307 8 B P110304 8 B C110102 8 B M111402 9 B P110307 9 B C110103 9 B M111401 10 C P110305 10 C C111101 10 B M111601 11 C P110304 11 B C111104 11 B M120701 12 D P110302 12 B C111204 12 D M110604 13 B P110302 13 B C111206 13 D M111404 14 C P110309 14 B C111203 14 B M111401 15 C P110308,

P110305 15 D

C111103 15 B

M111402

16 D P110308, P110305

16 A C111103

16 A M111402

17 B P110305 17 B C111103 17 C M111304 18 D P110305 18 D C111103 18 B M111304



15

HINTS OR SOLUTION

Physics 1. BD

1. At the instant of maximum separation both have same velocity

Maximum separation

2 2 1 1

1 1

2 2 1 12 2

= 5 – 2.5 = 2.5 m

2 B

2. 54 5

3m/ s5 18

72 5

4m/ s5 18

For A and C colide at D

u sin = 3 ….… (1)

For A and B collide at D

u cos = 4 ……. (2)

2 2 2 2 2(1) (2) u 3 4

u = 5 m/s

3 ABD

3. Let A = B = C = D = 2

Â 2 j

ˆC 2 i

ˆ ˆB i j

ˆ ˆD i j

ˆ Â C 2 j 2 i

2D

b. B D 2C

ˆ ˆ ˆ ˆ î j i j 2 2 i



16

0

ˆ Â C 2 j 2 i

2 2

ˆ î j B

4. ACD

4.

21

h u(6) g 62

21

u(2) g 22

u 20m/s

h 60m

5. ABC

5. (a) usin60º

2g

u 3 40

2 u m/ s20 3

(b)

2

2

40sin120º

u sin2 3

2g 20

40 40 3 40

m20 3 2 3

(c) 2

21 usinh gt

2 2g

2 2

21 40 sin 60º10 2

2 2g3

240 3 1

202 203

= 20 + 20 = 40 m



17

6. AB

2dv v

dt R

0

v t2

v 0

dtv dv

R

0

v2 1

v

v t 0

2 1 R

0

1 1 t

v v R

0

0 0

R v t1 1 t

v v R Rv

0

0

Rvv

R v t

2dv ds dv v

dt dt ds R

2vdv v

ds R

v s

0 0

dv ds

v R

0

v Sln

v R

S/R0

V V e

7. BD

8. B

8. ˆ ˆv i 6t j

x

2x v t t sec

3

dv â 0 6jdt



18

22v 1 6t

2v 1 36t

t

dva

dt

1

12 2

11 36t (72t)

2

2

36t

1 36t

at 2

236

3

21 36

3

at 12 2

1 8

at = 4 2 m/s2

2 2 2c t

a a a

ca 36 32 2m/ s

2

c

va

R

22

1 363

R2

1 8

R 4.5m2

9. B

9. 2 21 2e e 1 1 2 1 1 cos

1 1 2cos

2 1 cos

22 2sin / 2

1 2

1sin e e

2 2

10. C

10. xv 10sin45º 10sin45º 2



19

= 10 sin 45º

yv 10cos45º

2 2x yv v v 10m/ s

11 C

11. u cos60º = v cos30º

V = u

3

2 2

vsin30º usin60º 2gh

223u u

2gh4 2 3

2 29u u

2gh12

u 3gh

12 D

12. 1 1

a k k t

at t = 0, a = 1

3k

4

2

3a k t

4

0 2

3k 3

4

2

1k

4

dv 3 t

adt 4 4

t

2

100

3 tv t

4 8

23 t

10 t4 8

2t 6t 80 0

6 36 4 80

t2

= 12.43 sec



20

13 B

13. 1/ 2a v

0

0 t1/ 2

v 0v dv dt

0

01

12

0

v

vt

11

2

0

0

1/ 2

v

vt

11

2

t = 02 v

14. C

After t = 4s, velocity of the car up the incline is

VC = u + at = 0 + 2 × 4 = 8 ms-1

RC R Cv v v

RCv

cannot be vertical if R

v

is in the direction shown in (a) above. In this case, we cannot have

RCv

in the vertical direction.

Hence, the correct direction of rainfall is as shown in (b) above.

In the diagram shown, RCv

is vertical and R

v

makes = 37º with it. Hence, R

v

is normal to the

incline.



21

C

R

vtan37º

v

1R

R

8 3 32v 10.67ms

v 4 3

15. C

16. D

Sol. 15-16

(i) Consider the x and y-directions as shown in the figure below. Acceleration of the particle is g()

and can be resolved as :

x ya gsin ,a gcos

Relative to the box, the componenets of acceleration are

xba gsin gsin 0

yba gcos 0 cos

Relative to the box, the componenets of initial velocity of the particle are

xb

u ucos

Yb

u usin

In RF of the box, the time of flight can be calculated using



22

2yb yby u t a t

210 usin T gcos T

2

2usin

Tgcos

PQ = Range of the projectile in RF of the box

2xb xb

1u .T a T

2

ucos T sba 0

2usin

ucosgcos

(ii) Taking x and y as horizontal and vertical,

Velocity of the box can be written as

bˆ ˆv vcos i v sin j

Initial velocity of the particle relative to the box is

Pbˆ û ucos i usin j

P bˆ û v ucos i usin j

Pˆ û ucos i usin j

ˆ ˆvcos i vsin j

ˆ ûcos vcos i usin v sin i

If the particle does not move horizontally, it means that the x componenet of the

above velocity is zero.

ucos vcos

ucos

vcos



23

17. B

Vx = ux + axt

80 – 40 = 40 m

Ans. D

CHEMISTRY 1. AD

2

2

anP V nb nRT

V

2

aP V b RT

V

2. C

22 3 4As S SO

3 5 2 6As As S S;

4 24

n-factor = 28 for 2 3As S

3. BC

3PV 3

K.E. RT2 2

K.E. P at constant volume

K.E. T at constant – pressure

Ans. C

34 8

10

hc 6.6 10 3 10E

854 10

For 1 mole



24

34 18 23

mole

6.6 10 10 3 6.022 10E

854

76.6 10 3 6.022

854

= 0.139 0 × 107 J/mole

= 1.39 × 103 KJ/mole

5. D

6. BD

7. AD

8. B

Balmer means transition

to n = 2

1. line 3 to 2

2. line 4 to 2

3. line 5 to 2

9. B

H1

1 1R 1

4



25

H1

1 3R

4

1H

4

3R

2H

9

8R

1 2H

1 9 16

R 8 15

H

1 32 27

R 24

H

1 5

R 24

2H

9 / 8

R

3H

16

15R

2 3H

1 4 9

R 3 8

H

1 135 128

R 120

H

1 7

R 120

Ratio = 5 / 24

3.57 /120

10. C

2

10e2 4

N H Y : N

should be in +3 or - 7

2 3N N 5e

(N can‟t be in – 7, because it should be oxidation)

11. B

m. mole of NaHC2O4 = 2; m.mole of H2C2O4 = 2

m.eq. of KMnO4 = 0.05 × V × 5 V

0.05 × V × 5 = (2 × 2) + (2 × 2) = 8

8

V 32ml0.05 5

12. B

b = 4 Vm

b = 4 × A 3

4N r

3



26

1

V V nb

1

nb V V

31A

V V 4b 4 N r

h 3

1

A

v V3r

16 nN

diamater =

1/ 3

1

A

8 3 V V2r

16 nN

13. B

M.P.S. =

2 2O N

2 R T 3R 700R.M.S

M M

On solving T = 1200 K

= 927ºC

14. B

If z> 1 then PV

1nRT

P 22.4

11 273 0.0821

P > 1.

15. D

16. A

Sol. 15-16

When HPh is used : 2 3 3Na CO NaHCO

Eq. of Na2CO3 = Eq. of HCl

mole of Na2CO3 × 1 = mole of HCl × 1

mole of Na2CO3 = 0.12 × 8 × 10–3

= 9.6 × 10–4

mole

9.6 × 10–4

mole of NaCO3 present in 25 ml diluted solution

Orignal mole of Na2CO3 present = 9.6 × 10-3

in 250 ml of diluted solution.

When MeOH is used (20 ml soln is used)

Eq. of Na2CO3 + Eq. of NaHCO3 = Eq. of HCl

9.6 × 10-3

20

250× 2 + mole of NaHCO3 = 0.12 × 18 × 10

-3

mole of NaHCO3 6.24× 10-4

6.24 × 10-4

mole NaHCO3 present in 20 ml of solution

In 250 ml NaHCO3 present = 6.24 × 10-4

× 250

20



27

7.8 × 10-3

mole

In original 25 ml stock the mole of compounds are

NaHCO3 = 7.8× 10-3

mole = 0.6552 g.

Na2CO3 = 9.6 × 10-3

mole = 1.0176 g.

NaHCO3 (g/lt) = 1000

0.6552 26.20825

* In 100ml stok

mole of NaHCO3 = 3 31007.8 10 31.2 10

25

reaction would be

NaOH required = 31.2 × 10-3

mole

= 31.2 × 10-3

× 40 g

= 1.248 g

17. B

v /2Pe nCT

T = 500 K

N = 2 moles

P = 1 atm

On solving

C = 01 e

0.001500 2

18. D

v /2P.e nCT

v / 2

nCP T

e

Slope v / 2

nC

e

C = 0.001 & V = 2L

N = 2

12P e

1000



28

MATHEMATICS

1. ABCD

A. For x 0,4

, tanx < cotx

Also ln(sinx) < 0

(tanx)ln(sinx)

> (cotx)ln(sinx)

B. For x 0,2

, cosecx 1

ln(cosecx) 0

4ln(cosecx)

< 5ln(cosecx)

C. x 0,2

cos x (0, 1)

ln(cosx) < 0

D. For x 0,2

Since sinx < tanx, we get ln(sinx) < ln(tanx)

2ln(sinx)

< 2ln(tanx)

2. AB

For 0, ,4 4 2

Let 1

cos43

1 cos4 2

cos22 3

2

2 2

1 2 2cos 1f 1

3 2 sec 2cos 1 cos2

1 3 3

f 1 or 13 2 2

3. ABC

3x = 4

x-1 log2 3

x = (x – 1)log2 4 = 2(x – 1)

or xlog2 3 = 2x – 2

or x = 2

2

2 log 3

Rearranging, we get

3

3

3

2log 22x

1 2log 2 12

log 2

4. BCD 5. ABCD 6. CD 7. BC Square and add



29

8. B

Let the radius of the smaller circle be x. Here OP = x cosec 30º And OQ = r = x + x cosec 30º

x = r

3

9. B For maximum value,

cos2x = 2a

2 – 1

sin2x = 1 – cos

2 x = (2 – 2a

2)

2a2 + sin

2x = 2

Maximum value of 2 2 2 2sin x 2a 2a 1 cos x

= 2 0 2

10. B

Since < < < and sin = sin = sin = sin = k, we have

= - , = 2 + , = 3 -

4sin 3sin 2sin sin2 2 2 2

= 4sin 3cos 2sin cos2 2 2 2

= 2sin 2cos 2 1 sin 2 1 k2 2

11. B 12. D

13. D

14. B 15. B 16. A



30

15-16

Sol. In given ABC both A

2and

B

2lie strictly in 0,

2

and sinx is always increasing in 0,2

whereas

cosx is always decreasing in 0,2

.

So, if A B

2 2

A B

sin sin2 2

or x1 > x2 and x3 < x4 or 3 4

1 1

x x

So, 2007 2006 2007 2006

1 4 2 3x x x x is not valid

Similarly for A B

2 2

A B

sin sin2 2

x1 < x2 and 3 4

1 1

x x

For this also 2007 2006 2007 2006

1 4 2 3x x x x is not valid.

So

2007 2006

31

2 4

xx0

x x

is possible only when

A B

2 2 .

x1 = x2 and 3 4

1 1

x x

Hence, ABC is isosceles with ABC = CAB

BC = AC = 1 unit

If C = 90º

Area, A = 1 1

BC AC2 2

sq. units

17. C 18. B

(i) We have 2

3x af(x)

x 3

Now, 2

2

3 3x a3x 9 6x 2a

2 x 3

23x 6x 9 2a 0

This must be a perfect square, so

D 0 36 12(9 2a) a 3

(ii) 2

2

3x 3y f(x) x y 3y 3x 3

x 3

2x y 3x 3 3 0

As x R, so D 0

9 4y(3y 3) 0

(2y 3)(2y 1) 0

Hence 1 3

y ,2 2

Hence minimum value of 1

f(x) is sin2 6

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