CM21-MT-2-1921-1 FIITJEE MONTHLY ASSESSMENT TEST Monthly test - II CLASS - 11.pdf · CM21-MT-2-1921-1 FIITJEE KOLKATA NORTH CENTRE: VIP TOWER, Gr. Floor, 80, VIP Road, Golaghata,

CM21-MT-2-1921-1

FIITJEE KOLKATA NORTH CENTRE: VIP TOWER, Gr. Floor, 80, VIP Road, Golaghata, Ultadanga, Kolkata–7000 48,Tel.40221300 (16 lines) E-Mail: [email protected] URL: www.fiitjeekolkatanorth.com

PHYSICS, CHEMISTRY & MATHEMATICS

Time Allotted: 3 Hours

Maximum Marks: 240

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 Section.

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

4. Each section is further divided into two parts: Part-A & Part-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) Part-A (01 – 08) contains 8 multiple choice questions which have only one correct answer. Each question carries +3 marks for correct answer and – 2 mark for wrong answer.

(09 – 12) contains 4 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) Part -B (01 – 02) contains 2 Matrix Match Type questions containing statements given in 2 columns.

Statements in the first column have to be matched with statements in the second column. Each question carries +8 marks for all correct answer. For each correct row +2 marks will be awarded and –1 mark for each row matched incorrectly.

(iii) Part-C (01 – 06) contains 6 Numerical based questions with single digit integer as answer, ranging from 0

to 9 and each question carries +4 marks for correct answer and -1 mark for wrong answer.

Name of the Candidate :__________________________________________

Batch :___________________ Date of Examination :___________________

Enrolment Number :______________________________________________

BA

TC

HE

S –

19

21

FIITJEEFIITJEEFIITJEEFIITJEE MONTHLY ASSESSMENT TEST

CM – TEST – 2

CODE:

CM21-MT-2-1921-2


PHYSICS SECTION – A

(Single Correct Choice Type) 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. In the setup shown, blocks of masses m, 3m and 4m are placed on a frictionless horizontal surface and the free end ‘P’ of the thread is being pulled by a constant force 2F. Find the acceleration of the free end P (a) 8 F/m (b) 4 F/m (c) 16 F/m (d) 2 F/m

2. The system of springs and masses are in equilibrium. If the spring 2 is cut then the initial acceleration of block B will be

(a) 7

g2

downward (b) 2g downward

(c) 2g upward (d) 72

g upward

3. A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1kg/s and at a speed of 5 m/s. The initial acceleration of the block is

(a) 25m / s

3 (b) 225

m / s4

(c) 225m / s

8 (d) 25

m / s2

Space for rough work

CM21-MT-2-1921-3


4. Two smooth cylindrical bars weighing W N each lie next to each other in contact. A similar third bar is placed over the two bars as shown in figure. Neglecting friction, the minimum horizontal force on each lower bar necessary to keep them together is (a) W/2 (b) W

(c) W / 3 (d) ( )W / 2 3

5. The kinetic energy K of a particle of mass 1kg moving along a circle of radius R depends upon the distance travelled s, as K = as2. The force acting on the particle is

(a) 2s

2aR

(b)

1/22

2s

2as 1R

+

(c) 2as (d)

1/22s2as 1

R

+

6. You wish to lift a heavy block through a height h be attaching a string of

negligible mass to it and pulling so that it moves at a constant velocity. You have the choice of lifting it either by pulling the string (i) vertically upward or (ii) along a frictionless inclined plane. Which one of the following statements is true? (a) The magnitude of the tension force in the string is smaller in case (i) than in case (ii) (b) The work done on the block by the tension force is the same in both cases. (c) The work done on the block by the tension force is smaller in case (ii) than is case (i) (d) The work done on the block by the gravitational force is smaller in case (ii) than in case (i)

7. A particle moving along the x-axis is acted upon by a single force F = F0e-kx, where F0 and k are constants. The particle is released from rest at x = 0. It will attain a maximum kinetic energy of: (a) F0/k (b) F0/e2 (c) kF0 (d) 1/2 (kF0)2

8. The work done by the force 2 2ˆ ˆF x i y j= +�

around the path shown in the figure is

(a) 32a

3 (b) zero

(c) a3 (d) 34a

3


CM21-MT-2-1921-4


(Multi Correct Choice Type) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE OR MORE may be correct.

9. On the figure shown, a stone tied to a light string is oscillating between extreme points A and C in a

vertical plane. Acceleration of stone has magnitude aA, aB and aC at the respective points then: (Given: sin 4 / 5)θ=

θ

CB

A

(a) aA = aB (b) aB = 2aA (c) 2aB = aA + aC (d) aA = aC

10. A small sphere of mass m is connected by a string to a nail at O and moves in a circle of radius r on the

smooth plane inclined at an angle θ with the horizontal. If the sphere has a velocity u at the top position A. Mark the correct option(s)

o90B

C

Au

θ = 3

7o

o

(a) Minimum velocity at A so that string does not get slack instantaneously is 3

gr5

(b) Tension at B if sphere has required velocity to just complete circle A is 11

mg5

(c) Tension at C if sphere has required velocity to just complete circle is 21

mg5

(d) Centripetal force at point A is 3

mg5

if it just get slack instantaneously

11 In figure, a block of mass m is released from rest when spring was in its natural length. The pulley also has mass m but it is frictionless. Suppose the value of m is such that finally it is just able to left the block M up after releasing it

(a) The weight of m required to just lift M is M

g2

(b) The tension in the rod, when m is in has zero acceleration M

g2

(c) The normal force acting on M when m has zero acceleration M

g2

(d) The tension in the string when displacement of m is maximum possible is Mg

M

Rod

mString


CM21-MT-2-1921-5


12. As shown in figure BEF is a fixed vertical circular tube. A block of mass m starts moving in the tube at point B with velocity V towards E. It is just able to complete the vertical circle, then

(a) velocity at B must be 3Rg

(b) velocity at F must be 2Rg

(c) Normal reaction at point F is 2 mg (d) The normal reaction at point E is 6mg

60ο R

C F

E

B

PART– B (Matrix-Match Type)

This Section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.

1. In column-I, a situation is depicted each of which is in vertical plane. The surfaces are frictionless. Match with appropriate entries in column-II

Column – I Column – II (a) Bead is threaded on a

circular fixed wire and is projected from the lowest point

R 1m=

5m / s

(P) Normal force is zero at top most point of its trajectory

(b) Block loosely fits inside the fixed small tube and is projected from lowest point.

R 1m=

20m / s

(Q) Velocity of the body is zero at top most point of its trajectory.

(c) Block is projected horizontally from lowest point of a smooth fixed cylinder.

R 1m=

6m / s

(R) Acceleration of the body is zero at the top most point of its trajectory.

(d) Block is projected on a fixed hemisphere from angular position .θ

θ

cos 2 / 3θ=

R 1m=

20m / s

3

(S) Normal force is radially outward at top most point of trajectory.


CM21-MT-2-1921-6


2. Find extension of spring in equilibrium in Column-I and match them with Column-II. [Assume ideal spring, light string and frictionless, pulley

Column – I Column – II

(a)

m

k

(P) mg

k

(b)

k

m

(Q) 4mg

3k

(c) k

m

(R) 2mg

k

(d) k

m

(S) mg

4k


CM21-MT-2-1921-7


SECTION – C (Integer Type)

This section contains 6 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 small block of mass m is placed in a groove carved inside a disc. The disc is placed on smooth horizontal surface and pulled with an acceleration of magnitude 25 m/s2 as shown. Find half of the acceleration of block with respect to the disc in m/s2?

θ2A 25m / s=

(Given 3

sin ,cos5

θ= 24,g 10m / s

5θ= = and co-efficient of friction between groove and the block is

2

5µ = )

2. In the diagram shown, no relative motion takes place between the wedge and the block placed on it.

The rod slides downwards over the wedge and pushes the wedge to move in horizontal direction. The

mass of wedge is same as that of the block and is equal to M = 1 kg. If 1

tan3

θ= , find the mass of

rod.

θ θ

Rod Smoo

th

Smooth

3. A system consists of two identical cubes, each of mass 3kg, linked together by a compressed weightless

spring of force constant 1000 N/m. The cubes are also connected by a thread which is burnt at a certain moment. At what minimum value of initial compression, x0 (in cm) of the spring will the lower cube bounce up after thread is burnt through?

3kg

3kg

k 1000 N / m=


CM21-MT-2-1921-8


4. Two identical beads of m = 100 g are connected by an inextensible massless string that can slide along the two arms AC and BC of a rigid smooth wireframe in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of 0.1 m is 16 x × 10-3 J. Find the value of x. (g = 10 m/s2)

C0.3m

2

B

1A 0.4 m

5. For the given figure. Find the tension in the string

5kg 10kg 18 N0.02µ =

6. Two blocks are connected by a string passing over a pulley as shown. Wedge is fixed. M is sliding

down with speed 11 m/s at an instant and suddenly the string breaks. The co-efficient of friction

between the blocks and wedge is 1

2 for both the blocks. Find the velocity of block M with respect to

m after t = 15 seconds from the moment the string breaks

mM

o30o30

[Taking g = 10 ms-1 and 3

1.22

= and assuming that the length of inclined is sufficiently long]


CM21-MT-2-1921-9


CHEMISTRY SECTION – A


1. Shape of NH3 is not similar to

(a) 3SeO− (b) 3CH− (c) 3CF•

(d) 3Ph C•

2. In which case H-bonding will not be observed

(a) 3 2H O− (b) H2O (c) 5 2H O+ (d) 3H O+

3. Which of the following combination result in the formation of π ( π anti bonding)?

(a)

(b)

(c)

(d)

4. The normal C – O bond length in CO is 1.128 Å, so the C – O bond length the CO+ will be (a) 1.128 Å (b) 1.130 Å (c) 1.115 Å (d) 1.132 Å

5. Considering the modern periodic table where A,B,C …………… are the element in concern period and respective groups. Select the incorrect statement

Period Group

1 2 14 15 17 2 A B C D E 3 F G H I J 4 K L M N O

(a) Covalent character of compounds in increasing order is AE > FE > KE (b) Ionization potential in increasing order A > F > K (c) M+4 compounds act as good oxidizing agent whereas C+2 compounds act as good reducing agent (d) Element ‘H’ is used to prepare interstitial hydrides

6. Electron affinity of X would be equal to (a) electron affinity of X– (b) ionization energy of X (c) ionization energy of X– with sign reversed (d) none of these


CM21-MT-2-1921-10


7. Which of the following can not be oxidized by H2O2? (a) KI + HCl (b) O3 (c) PbS (d) Na2SO3.

8. Acidified solution of chromic acid on treatment with H2O2 yield (a) CrO3 + H2O2 + O2 (b) Cr2O3 + H2O + O2 (c) CrO5 + H2O (d) H2Cr2O7 + H2O + O2.


9. Report the incorrect statement(s)?

(a) 3 identical S – O linkages are found in 23SO − ion

(b) NF3 has same no. of lone pairs as in 3I−

(c) Structure of PBr5 can be drawn without dative bond and without expanding octet of ‘P’ (d) PCl5, PBr5 (separately) both exist as ionic compound in solid form and their anionic part is

6PX− type

10. Choose the incorrect statement(s) (a) All S-F bond length are identical in SF4 (b) All Cl-F bond length are identical in ClF3

(c) All �F Cl F− − angles are identical in ClF3 (d) In PCl5 molecule equatorial P-Cl bond lengths are different from the axial P – Cl bond lengths.

11. For the group − I alkali metal, select the correct statement(s) (a) Lithium is the only element in group−1 which form nitride Li3N (b) Li acts as strongest reducing agent in aqueous medium (c) Li reacts with water slowly as compare to Na & K because of its comparatively high melting point (d) Li can form monoxides (major amount) as well as peroxides (minor amount) when react with O2

12. Which is/are interstitial hydride (a) CH4 (b) ScH2 (c) NaH (d) ZrH1.4

PART– B (Matrix-Match Type)

This Section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.

1. Match the following


(a) CsI3 (p) Upon dissociation it gives two products (b) PCl5(s) (q) Number of p-orbitals involved in hybridization of central

covalently bonded atom of the gives compound 2≥ (c) N2O5(s) (r) Hydrolysis produces an acidic solution (d) Li2CO3 (s) Oxidation number of central atom 1≥

(t) The dissociation product can acquire a linear shape


CM21-MT-2-1921-11


2.

Column – I (Alkali metal) Column – II (Oxide)

(a) Na (p) Generally forms super oxide (b) Cs (q) Shows golden yellow flame in Bunsen burner (c) K (r) Reacts with water to form H2 (d) Li (s) Shows diagonal relationship

(t) Forms basic oxide



1. How many of the following compounds do not contain unpaired electrons KO2, NO2, SO2, BaO2, ClO2,

O2, NO, N2O, O3, CO

2. The number of species out of the following which undergoes hydrolysis in water is XeF6, SF6, ScF6, TeF6, XeF4, SF4, CCl4, SiCl4, P4O8, SOCl2, NCl3, PH3

3. How many of the following has square planar geometry according to the VSEPR theory XeO2F2, SF4,

4BF− , XeF4, [ ] [ ] ( ) [ ]2 2 24 4 44PtCl , NiCl , Ni CO , PdCl ,− − −

( )2

4 4BrF , Co Cl ,−−

( )3 2Rh pph Cl

4. How many hydrides can be classified as ionic hydrides among following example?

KH, LiH, NaH, CaH2, BH3, NH3, CH4, H2O, TiH

5. The electronegativity value of Cl according to Pauling’s scale __

6. How many type of gaseous products are formed when LiNO3 is strongly heated?


CM21-MT-2-1921-12


MATHEMATICS SECTION – A


1. If the circumference of the circle x2 + y2 + 8x + 8y – b = 0 is bisected by x2 + y2 – 2x + 4y + a = 0, then a + b is equal to (a) – 56 (b) 2 (c) – 64 (d) – 14

2. Given a circle (x + 4)2 + (y – 2)2 = 25. Another circle is drawn passing through (- 4, 2) and touching the given circle internally at the point A (-4, 7). AB is the chord of length 8 units of the larger circle intersecting the other circle at the point C. The AC will be

(a) 4 units (b) 17 units (c) 5 units (d) 3 units

3. A rectangle PQRS joins the points P, Q, R, S. Co-ordinates of P and R be (2, 3) and (8, 11) respectively. The line QS is known to be parallel to the y – axis. Then coordinates of Q and S are (a) (0, 7) and (10, 7) (b) (5, 2) and (5, 12) (c) (7, 6) and (7, 10) (d) (7, 2) and (7, 12)

4. The sum of the squares of the lengths of the chords intercepted on the circle, x2 + y2 = 16, by the lines, x + y = n, n N∈ , where N is the set of all natural numbers, is (a) 320 (b) 105 (c) 160 (d) 210

5. An incident ray (L1) is reflected by the mirror (L2) 3x + 4y = 5 and equation of reflected ray (L3) is x + y = 1, the equation of L1 is (a) x – y = - 3 (b) 17x + 31y = 45 (c) 17x – 31y = - 77 (d) 31x + 17y = - 3

6. Let A (5, 12), B (-13 cos θ , 13 sin θ ) and ( )C 13sin , 13cosθ − θ , where θ is real, be the vertices of a

ABC∆ . Then, the orthocenter of ABC∆ lies on the line (a) x – y – 7 = 0 (b) x + y – 7 = 0 (c) x – y + 7 = 0 (d) x + y + 7 = 0

7. Which one of the following statements is a tautology?

(a) ( ) ( )( )p q ~ p q∨ ∧ ∧ (b) ( ) ( )( )p q ~ p q∧ ∨ ∨

(c) ( )( )q ~ p q∨ ∧ (d) ( )( )p ~ p q∧ ∧

8. A point on the straight line, 3x + 5y = 15 which is equidistant from the coordinate axes will lie only in

(a) 4th quadrant (b) 1st quadrant (c) 1st and 2nd quadrants (d) 1st, 2nd and 4th quadrants


CM21-MT-2-1921-13



9. Which of the following statements is/are correct

(a) The truth value of p ~ q ~ r∧ ∧ is T then the truth values of p, q, r are respectively T, F, F (b) If the truth value of the statement p ~ q r∨ ∨ is F, then the truth values of p q r∧ → and p r q∧ → are T, T respectively (c) The negation of the statement p q↔ is logically equivalent to ~ p q↔

(d) The converse of the statement ( ) ( )p ~ q ~ p q∧ → ∧ is logically equivalent to p ~ q∨

10. Equations of bisectors of angle between intersecting lines x 3 y 5 x 3 y 5

,cos sin cos sin

− + − += =

θ θ φ φ are

x 3 y 5 x 3 y 5and , then

cos sin− + − +

= =α α β γ

(a) 2

α + φα = (b) 2 r2β + can be 1 (c) tan /α = −β γ (d) tan /α =β γ

11. Consider the circles

S1: x2 + y2 + 3x + 2y + 1 = 0 S2: x2 + y2 – x + 6y + 5 = 0 S3: x2 + y2 + 5x – 8y + 15 = 0 Also let S4 be a circle which cuts S1, S2, S3 orthogonally. Then identify the correct statements from the following (a) Point from which length of tangents to S1, S2, S3 are equal is (3, 2)

(b) Radius of S4 is 3 3

(c) Radical centre of S1, S2, S4 is 3 8

,5 5

− −

(d) Centre of S4 is (3, 2)

12. Suppose line x – 2y + 1 = 0, mx + y + 3 = 0 meet co-ordinate axes in concyclic point, then (a) m = - 2

(b) the circle cuts a chord of length 52

on x – axis

(c) radius of circle is 2 (d) a parabola can not pass through above four points


CM21-MT-2-1921-14


PART– B (Matrix-Match Type) This Section contains 2 questions. Each question has four statements (A, B, C and D) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the bubbles corresponding to q and r in the ORS.

1. Match the following


(a) The number of lines passing through (2, 3) each at distance of 5 units from the point (7, 8) is

(p) 0

(b) The orthocenter of the triangle formed by x – y = 0 and 2x2 + 3xy – 2y2 + 7x – y + 3 = 0 lies on x + y + k = 0 for k =

(q) 1

(c) If a, b, c are 3 consecutive positive multiples of 5, then the line ax – by + c = 0 passes through a fixed point whose ordinate is

(r) 2

(d) Mirror image of (1, 4) in the tangent at (0, 0) on the circle x2 + y2 – x + y = 0 has ordinate

(s) 3

(t) 5

2. Let Lp be pair of lines 7x2 – 18xy + 7y2 = 0 and C1 be the circle x2 + y2 – 8x – 8y = 0. The circle C2

touches both line of Lp and touches C1 internally. Let R be the region between the line represented by Lp in which the circle lies


(a) A point with least abscissa among the given options within region R is

(p) (1, 5)

(b) A point within C1 but not within R is (q) (2, 3) (c) A point within C1 but outside C2 having

greatest abscissa is (r) (6, 8)

(d) A point on C2 is (s) (8, 4) (t) (5, 6)


CM21-MT-2-1921-15




1. Consider the family of lines (x – y – 6) + λ (2x + y + 3) = 0 and (x + 2y – 4) + µ (3x – 2y – 4) = 0. If

the line of these 2 families are at right angle to each other the locus of their point of intersection is ax2 + by2 + cx + dy + e = 0 then a + b + c + d + e is

2. The centre of variable circle x2 + y2 + 2gx + 2fy + c = 0 lies on the line 2x – 2y + 9 = 0 and the variable circle cuts the circle x2 + y2 = 4 orthogonally if the variable circle passes through two fixed points (a, b) and (c, d) where (b < d) then find the value of 2b + d

3. In triangle ABC, if y = x + 1 and y = 2x + 3 are altitude through A and angle bisector of B respectively. If vertex ( )C 3,4≡ equation of median through C is x y 5 0λ +µ + δ = then find λ +µ + δ

4. Two circles of unequal radii have four common tangents. A transverse common tangent meets the

direct common tangents at the points P & Q. If length of direct tangent (between the point of contacts) is 8 then length of PQ is.

5. If A B C+ + = π and sin 2A sin 2B sin 2C A B C

cos .cos .coscos A cos B cosC 1 2 2 2

+ +=λ

+ + −then λ equals

6. The value of tan 1o tan 2o + tan 2o tan 3o +…..+ tan 88o tan 89o is equal to cot2 1o – n. where n is a two

digit number ab. Then the value of b – a is equal to.


CM21-MT-2-1921-16


FFFFIIIIIIIITTTTJEEJEEJEEJEE MONTHLY ASSESSMENT TEST – 2

ANSWER KEY

Q.N Physics CC Chemistry CC Mathematics CC

1. B P110409 D C110305 A M110820

2. A P110420 D C110309 A M110813

3. D P110413 D C110302 B M110715

4. D P110420 C C110302 D M110827

5. B P110413 D C110702 B M110731

6. B P110507 C C110704 C M110734

7. A P110502 B C111006 C M112603

8. B P110501 C C111006 C M110718

9. ACD P110413 BCD C110302 ABCD M112602

10. AD P110319 ABC C110302 ABC M110726

M110727

11. ABCD P110420 ABCD C110808 ABCD M110821

M110819

12. ABC P110316 BD C111003 AB M110802

1.

A – QS

B – PQ

C – P

D – QRS

P110316

A – PQT

B – PQRS

C – PQST

D – PQST

C110305

A – R

B – R

C – R

D – Q

M110723

M110706

M110723

M110722

M110806

2.

A – R

B – S

C – Q

D – P

P110420

A – QRT

B – PRT

C – PRT

D – RST

C110802

C110803

A – QT

B – P

C – R

D – S

M110730

M110731

M110802

1. 5 P110408 5 C110302 0 M110817

2. 3 P110409 9 C110305 5 M110827

3. 1 P110503 4 C110305 2 M110707

4. 4 P110503 5 C111003 8 M110820

5. 6 P110412 3 C110706 8 M111412

6. 0 P110411 2 C110808 2 M111406

CM21-MT-2-1921-17


Solution PHYSICS

1. Ans. B

Sol. Magnitude of acceleration of each block is F/m. By using constraint ap = 4 F/m

2. Ans. A

3. Ans. D

Sol.

dmForce V

dt=

4. Ans. D

Sol.

For upper cylinder 2N cos 30o = W …..(1)

For lower cylinder N cos 60o = F …..(2)

From (1) and (2) o

o2cos30 W

Fcos 60=

WF

2 3 =

5. Ans. B

Sol.

k = as2 2

21 asmv

2 m=

2av s

m

=

Centripetal acceleration = 2 2v 2as

R mR=

Tangential acceleration = dv

v 2asds

=

Net acceleration = 2

22as s

1m R

+

Net force 2

2s

2as 1R

= +

CM21-MT-2-1921-18


6. Ans. B

7. Ans. A

Sol.

kxe

0

KE work done = F e dx∞

−∆ =

kx0 0

0

F Fe

k k

∞−−

= =

8. Ans. B

Sol.

Force is conservative and displacement zero so work done is zero.

9. Ans. ACD

Sol.

A C 0υ = υ =

So, A Ca a gsin 4g / 5= = θ =

According to Conservation of Energy

( )2B

1mv mgl 1 cos

2= − θ

( )2Bmv 2mgl 1 3 / 5 = −

2Bv 4gl / 5=

2B

Bv 4g

al 5

= =

A B Ca a a∴ = =

And B A C2a a a= +

θ

CAg sin θ

gg sin θ

l

10. Ans. AD

Sol.

(a)

T

u

mg cosθ 2

o mvT mg sin 37

r+ =

3gr

5∴ υ=

(c) mg2r sin 37o = 2 21 1mv mv

2 2−

2mv3mg

r∴ =

T – mg sin 37o = mar 18mg

T5

∴ =

CM21-MT-2-1921-19


ra

omg sin37

T

11. Ans. ABCD

12. Ans. ABC

Sol.

( )2 2 oEV V 2g R R cos60= + −

24gr V gR= +

or V 3gR=

2 2B EV V 2gR 2gR= − =

2PmV

N 2mgR

= =

2E

E EmV

N mg N 5mgR

= =

Matrix Match 1. Ans. A – QS, B – PQ, C – P, D – QRS

Sol.

(a) u = 5 m/s v2 = u2 – 2gR (1 – cos θ )

1min

1u u , v 0 at cos

4−

< = θ=

∵

θmg

N

v 0=

(b) minu 20 m / s u= <

mg

v 0=

θ = 9

0

R 1= m

(c) u = 6 m/s < umin

mg

6m / s

o37

After leaving the cylinder it will follow projectile path.

(d) ( )2 21 1mu mv mgR 1 cos

2 2= + − θ

v = 0

CM21-MT-2-1921-20


u

mg

θ

2. Ans. A – R, B – S, C – Q, D – P

Integer Type 1. Ans. 5

Sol.

mA cos mg mAsin ma

4 2 2 325 10 25 a

5 5 5 5

θ−µ −µ θ=

× − × − × × =

20 – 4 – 6 = a

20 – 10 = a 2a 10 m / s =

a5

2=

θ1N

2f1f

mA cosθ

mA

mAsin θ

FBD of block with respect to disc

2. Ans. 3

Sol.

Let M1 be the mass of the rod

θ

A

1N

θ1N

1M g 1A

N

M1g – N1 cos θ = M1A1 (1)

N1 ( )sin M M Aθ= + (2)

A = g tan θ (3) Relation between A1 and A

1A A tan= θ

Thus, by solving these equations, M1 = 3M = 3 kg

3. Ans. 1

4. Ans. 4

Sol.

Frictional force on 5 kg block = 1N 1Nµ =

Frictional force on 10 kg block = 2N 2Nµ =

Now, 18 – T – 2 = 10 a and T – 1 = 5a a = 1 m/s2 and T = 6 N

50

1

5 kg T

a1N

100

2

10 kg 18 N

a

2N

T

5. Ans. 6

Sol.

6. Ans. 0

Sol.

CM21-MT-2-1921-21


Retardation on m when it is moving up (after string breaks) o o

2mgsin 30 mg cos30 g 3a 1 11ms

m 2 2− + µ

= = + =

So it comes to rest at v 11

t 1sa 11

= = =

Now since mg cos mg sin 0µ θ− θ>

1 3 1m 10 m 10 0 m

2 2 2

× − × × >

will remain at rest afterwards.

[ ]o o

2Mg cos30 Mgsin 30 ga 1.2 1 1ms

M 2−µ −

= = − =

It comes to rest at v 11

t 11sa 1

= = =′

For M, also limiting friction > Mg sin 30o Hence it also remains at rest after t = 11 s ∴ Relative velocity at 1 2t 15s v v 0= = − =

� �

CHEMISTRY

1. Ans. D

Sol.

In 3Ph C•

, carbon is Sp2 – hybridized

2. Ans. D

3. Ans. D

Sol.

As symmetry not matches. 4. Ans. C

Sol.

B.O of CO+ is 3.5 5. Ans. D

Sol.

Element ‘H’ is formed covalent hydrides. 6. Ans. C

Sol.

Electron affinity of X is equal to ionization energy of X- with sign reversed. 7. Ans. B

Sol.

O3 can not be oxidized by H2O2 8. Ans. C

Sol.

H2CrO4 + 2H2O2 → CrO5 + 3H2O 9. Ans. BCD

Sol.

[ ]I I I−

− − R

NF

R

FR

CM21-MT-2-1921-22


10. Ans. ABC

Sol.

S

F

F

F

F

Cl

F

F

F Cl

Cl

Cl

P

Cl

Cl

11. Ans. ABCD

Sol.

All are correct. 12. Ans. BD

Sol.

Generally transition metals are formed interstitial hydrides. Matrix Match 1. Ans. A – PQT, B – PQRS, C – PQST, D – PQST

Sol.

2 3 2 2Li CO Li O CO

Linear

→ +

2 5 2 3N O H O 2HNO+ →

PCl5 (Sp3d hybridization) 2. Ans. A – QRT, B – PRT, C – PRT, D – RST


2. Ans. 9

3. Ans. 4

4. Ans. 5

Sol.

KH, LiH, NaH, CaH2 and TiH are ionic hydrides. 5. Ans. 3

Sol.

Electronegativity of Cl is 3.0. 6. Ans. 2

Sol.

( ) ( )3 2 2 24LiNO 2Li O 4NO g O g∆→ + +

MATHEMATICS 1. Ans. A

Sol. The common chord (radical axis) of the circles should pass through the centre of first circle be. (x2 + y2 + 8x + 8y – b = 0) Now, equation of common chord ≡ x2 + y2 + 8x + 8y – b – (x2 + y2 – 2x + 4y + a) = 0 10x + 4y – a – b = 0 a + b = 10 x + 4y ………(I) As common chord will pass through center of x2 + y2 + 8x + 8y – b = 0 i.e. from (- 4, - 4), a + b = 10 × - 4 + 4 × - 4 = - 56

CM21-MT-2-1921-23


2. Ans. A

Sol.

B

A

( )4, 2−

C

( )4, 7−

AC 4 units∴ =

3. Ans. B

Sol. As diagonals of a rectangle bisect each other and they are of same length, PD = DR = QD = DS

( ) ( )2 2PD 2 5 3 7 9 16 5= − + − = + =

( ) ( )Q 5, 2 S 5,12∴ ≡ ≡

R

P

D SQ

( )5,7

( )2,3

( )8,11

4. Ans. D

Sol. Chord will be intercepted by x + y = n on circle x2 + y2 = 16 if and only of length of perpendicular drawn from center of circle i.e. (0, 0) is less than radius.

So, 2 2

0 0 n4

1 1

+ −≤

+ n 4 2 ≤

So, possible values of n = 1, 2, 3, 4, 5

Now, length of chord 2 2

2 n n2 4 2 16

22

= − = −

Sum of squares of length of chords 5

2

n 1

64 2n=

−

= ( ) ( )5 5 1 2 5 1

320 26

× + × × +− × = 320 – 110 = 210

4

x y n+ =

5. Ans. B

Sol. Equation of incident ray ≡ 3x + 4y – 5 + λ (x + y – 1) = 0 ( ) ( ) ( )3 x 4 y 5 0 + λ + + λ − λ + =

As incident ray and reflected ray form same angle with mirror.

( )

( )( )

3 3 314 4 4

333 1144 4

− + λ+ − +

+ λ=

+ λ++

+ λ

12 4 12 3 1

16 4 9 3 7

− − λ + + λ − =

+ λ + + λ

1

7 25 7

−λ − =

λ +

2514

−λ =

∴Equation of incident ray ≡ 17x + 31y = 45 6. Ans. C

Sol. Circumcenter (c) ≡ (0, 0) (All the vertices are at equal distance from origin)

CM21-MT-2-1921-24


( )13sin 13cos 5 13sin 13cos 12

Centroid G ,3 3

θ− θ+ θ− θ+ ≡

2 1

GH C

Orthocenter ( )

13sin 13cos 5 13sin 13cos 122 0 3 2 0 3

3 3H ,2 3 2 3

θ− θ+ θ− θ+ × − × × − ×

≡ − −

( )13sin 13cos 5, 13sin 13cos 12≡ θ− θ+ θ− θ+

∴Equation of line on which orthocenter lies ≡ x – y + 7 = 0 7. Ans. C

Sol. ( )( )q ~ p q∨ ∧

= ( )q ~ p ~ q∨ ∨

= q ~ q ~ p∨ ∨

= Always true So, it is a tautology

8. Ans. C

Sol. As the line 3x + 5y = 14 passes through 1st, 2nd and 4th quadrants, point which is equidistant from both coordinate axes will lie in only these quadrants. Now, 1st quadrant Y X y x= =

So, 3x + 5x = 15 15

x8

=

15x y

8= = i.e. point

15 15,

8 8

≡

( )5,0

( )0,3 3x 5y 15+ =

2nd quadrant y x y x= = −

3x – 5x = 15 15

x2

− =

15y

2=

Point 15 15

,2 2

− ≡

4th quadrant y x y x y x= − = = −

3x – 5x = 15 15x 2 = −

But value of x in 4th quadrant should be positive. So, there will be no such point in 4 th quadrant. 9. Ans. ABCD

10. Ans. ABC Sol.

Inclinations of two lines are andθ φ . So inclination of angular bisector is .2 2

θ+ φ θ+ φα =

( )tan 1 tan / .γ

α = − α = −β γβ

If 2 2sin , cos 1β = − α γ = αβ + γ =

CM21-MT-2-1921-25


11. Ans. ABCD

Sol. S1 – S2 = 0 2 3x y 1 0, S S 0 6x 14 y 10 0 − − = − = − + − =

x 3, y 2, = = radius of S4 = length of tangents from (3, 2) to 1S 27=

Equation of S4: (x – 3)2 + (y – 2)2 = 27 2 2x y 6x 4y 14 0 + − − − =

1 4 1 2S S 0 3x 2y 5 0,S S 0 x y 1 0− = + + = − = − − =

x 3 / 5, y 8 / 5 = − = −

12. Ans. AB Sol. Equation of circle must be (x – 2y + 1) (mx + y + 3) + xy 0λ = Co-eff of x2 = Co-eff of y2 m = - 2 Co-eff of xy must be zero 1 2m 0 5− + λ = λ = −

Circle : x2 + y2 x2

− + 3y - 32

= 0 since ( )2 52 g c B

2− = is correct

Radius ( )1 3 3

2 C16 4 2

+ + ≠ is not correct

(d) is not true since a parabola can always pass through four point if no three of them are collinear. Matrix Match

1. Ans. A – R, B – R, C – R, D – Q

Sol. (a) Distance between points is more than 5. So 2 lines, one on each side of perpendicular. These lines are x = 2, y = 3 (b) Lines 2x2 + 3xy – 2y2 + 7x – y + 3 = 0 are perpendicular to each other. So, orthocentre is their point of intersection i.e. (-1, -1), giving k = 2. (c) Line is (5n – 5)x – 5ny + (5 + 5n) = 0 n(x – y + 1) + (1 – x) = 0, which is a family of lines through (1, 2) (d) Tangent at (0, 0) is y = x. So mirror image is (4, 1)

2. Ans. A – QT, B – P, C – R, D – S

Sol. C1 has centre P1: (4, 4) and radius 1r 4 2= . Angular bisector of Lp are x2 – y2 = 0 i.e. x = y or x +

y = 0 So, C2 must lie in 1st quadrant with centre P2 : (h, h) and radius ( )2r 8 h 2= −

Also r2 = OP2 sin θ where 2 22 9 7 4 2

tan 214 7

−θ = =

Since cos 2 7 / 9, sin 1/ 3θ= θ= giving ( )8 h 2 h 2 / 3− =

h 6. = So P2: (6, 6), r2 = 2 2, C2 : x2 + y2 – 12x – 12y + 64 = 0 So, point within R if Lp < 0 and x, y > 0


Sol.

( ) ( )x y 6 2x y 3 0− − + λ + + = family of line passing through

(1, - 5) and

( ) ( )x 2y 4 3x 2y 4 0+ − +µ − − = passing through (2, 1)

Then y 5 y 1

1x 1 x 2

+ −= −

− −

( )x, y

( )2,1

( )1, 5−

2π

2. Ans. 5

Sol.

Let the circle x2 + y2 + 2gx + 2fy + c = 0 as centre lies on 2x – 2y + 9 = 0 ∴ - 2g + 2f + 9 = 0 cuts x2 + y2 = 4 orthogonally ∴ 2g × 0 + 2f × 0 = C – 4

CM21-MT-2-1921-26


Equation of circle x2 + y2 + (2f + 9) x + 2fy + 4 = 0 (x2 + y2 + 9x + 4) + 2f (x + y) = 0 passes through point of intersection of x2 + y2 + 9x + 4 = 0 & x + y = 0

∴ points ( )1 1

, and 4,42 2

− −

1 1a ,c 4,b ,d 4

2 2= − = − = =

12b d 2 4 5

2

+ = + =

3. Ans. 2

Sol.

Equation of AD y = x + 1 Equation of BE y = 2x + 3 Midian through C is (slope of BC) (slope of AD) = - 1 Equation of BC is y – 4 = - 1(x – 3)

17 4B ,

3 3

=

BE BCAB BE

AB BE BE BC

M MM M

1 M M 1 M M

−−=

+ +

We will get MAB

D

A

EF

B C( )h, 2h 3+

( )3,4

4. Ans. 8

Sol.

PA = PR = 8 – 9 So PQ = 8 – a + b Also QD = QR = 8 – b So QP = 8 – b + a 8 – a + b = 8 – b + a

a b = PQ = 8

C R

Qb 8 b− D

Sa

8 a−A

P B

5. Ans. 8

Sol.

A + B = Cπ−

A A B B C C2sin cos sin cos sin cossin Asin Bsin C 2 2 2 2 2 2

A B C A B Csin sin sin sin sin sin

2 2 2 2 2 2

=

= A B C

8cos cos cos2 2 2

6. Ans. 2

Sol.

o oo o

otan 2 tan1

1 tan1 tan 2tan1

−+ =

o oo o

otan 3 tan 2

1 tan 2 tan 3tan1

−+ =

o oo o

otan 89 tan1

1 tan88 tan 89tan1

−+ =

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