part i : physics - eenadupratibha.netconsider a disc rotating in the horizontal plane with a...

SRIGAYATRI EUDCATIONAL INSTITUTIONS

Hyderabad Nalgonda Karimnagar Vijayawada Guntur Vizag Kadapa Tirupati


PART I : PHYSICS

SECTION I : Single Correct Answer TypeThis section contains 8 multiple choice questions. Each question has fourchoices (A), (B), (C) and (D) out of which ONLY ONE is correct.

1. A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector k̂ is coming out of theplane of the paper. The magnetic momentof the current loop is

(A) 2 ˆa Ik (B) 2 ˆ1

2a Ik

(C)

2 ˆ12

a Ik

(D) 2 ˆ2 1 a Ik

Sol. Key - B

M NiA

2 ˆ1. . 12

I a k

2. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating verticallyinwater in half-submerged state. If c is the relative density of the material of the shell with respect towater, then the coprrect statement is that the shell is(A) More than half-filled is c is less than 0.5.

(B) More than half-filled is c is more than 1.0.

(C) Half-filled if c is more than 0.5

(D) Less than half-filled if c is less than 0.5.Sol. Key - (A)

1in L B B WV d g V d g V d g

1

2 2W W WV Vd d V d

12 1V V

1 12VV

If 10.52VV



SRIGAYATRI EUDCATIONAL INSTITUTIONS3. An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform

current density along its length. The magnitude of the magnetic field, B

as a function of the radial distance

r from the axis is best represented by

(A) (B)

(C) (D)

Sol. Key - D

/ 2| 0in r RB

2 20/ 2| . / 4

2R r RB j r Rr

20

2 4Rj r

r

0|2r R

iBr

4. Consider a disc rotating in the horizontal plane with a constant angular speed about its centre O. Thedisc has a shaded region on one side of the diameter and an unshaded region on the other side as shownin the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneouslyprojected at an angle towards R. The velocity of projection is in the y-z plane and is same for bothpebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed18 rotation, (ii) their range is less than half the disc radius, and (iii) remains consant throughout. Then

(A) P lands in the shaded region and Q in the unshaded region(B) P lands in the unshaded region and Q in the shaded region(C) Both P and Q land in the unshaded region(D) Both P and Q land in the shaded region



SRIGAYATRI EUDCATIONAL INSTITUTIONSSol. Key - A

In 18 rotation the disc rotates by 4

and occupies the position as shown.

Since, the velocity of pebble is constant w.r.t. disc. Both the pebbles will move in y direction.Hence, P lands in the shaded region and Q in the unshaded region.

5. A student is performing the experiment of Resonance Column. The diameter of the column tube is 4 cm.The frequency of the tuning fork is 512 Hz. The air temperature is 038 C in which the speedof sound is336 m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. Whenthe first resonance occurs, the reading of the water level in the column is(A) 14.0 cm (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm

Sol. Key - B

0 1.24

l

336 84 21512 4 512 128

021 100 1.2

128l

15.2

6. In the given circuit, a charged of 80 C is given to the upper plate of the 4 F capacitor. Then in thesteady state, the charge on the upper plate of the 3 F capacitor is

(A) 32 C (B) 40 C (C) 48 C (D) 80 CSol. Key - C

80 9 36 volt20

V

1 2 180 4 5 20V V V volt

2 16V V

3| 16 3 48Fq C



SRIGAYATRI EUDCATIONAL INSTITUTIONS7. Two identical discs of same radius R are rotating about their axes in opposite directions with the same

constant angular speed . The discs are in the same horizontal plane. At time 0t , the points P and Qare facing each opther as shown in the figure. The relative speed between the two points P and Q is rv . In

one time period (T) of rotation of the discs, rv as a function of time is best represented by

(A) (B)

(C) (D)

Sol. Key -A

2 sinr PQV V R t i

2 sinrV R t

8. Two moles of ideal helium gas are in a rubber balloon at 030 C . The balloon is fully expandable and canbe assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowlychanged to 035 C . The amount of heat required in raising the temperature is nearly (take

8.31 / .R J mol K )(A) 62 J (B) 104 J (C) 124 J (D) 208 J

Sol. Key - DAs balloon is expanding under isobaric conditions

52 5 25 8.312PQ nC T R

207.85 208J




SECTION II : Paragraph TypeThis section contains 6 multiple choice questions relating to 3 paragraphs withtwo questions on each paragraph. Each question has four choices (A), (B), (C)and (D) out of which ONLY ONE is correct

Paragraph for Questions 9 and 10The decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory,,

a proton (p) and an electron e are observed as the decay products of the neutron. Therefore, consideringthe decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energyof the electron should be a constant. But experimentally, it was observed that the electron kinetic energyhas a continuous spectrum. Considering a three-body decay process, i.e. en p e v , around 1930,

Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ev to be masslessand possessing negligible energy, and the neutron to be at rest, momentum and energy conservationprinciples are applied. From the calculation, the maximum kinetic energy of the electron is 60.8 10 eV .The kinetic energy carried by the proton is only the recoil energy.

9. What is the maximum energy of the anti-neutrino?(A) Zero (B) Much less than 60.8 10 eV(C) Nearly 60.8 10 eV (D) Much larger than 60.8 10 eV

Sol. Key - C

pe

p e

mKEKE m

ep e

p

mKE KEm

31

627

9 100.8 101.6 10

450eV

PKE of proton is negligible and constant Energy is distributed between anti-neutrino and electron KE of anti-neutrino 60.8 10 eV

10. If the anti-neutrino had a mass of 23 /eV c (where c is the speed of light) instead of zero mass, whatshould be the range of the kinetic energy, K, of the electron?(A) 60 0.8 10K eV (B) 63.0 0.8 10eV K eV (C) 63.0 0.8 10eV K eV (D) 60 0.8 10K eV

Sol. Key - DRest mass energy of anti neutrino = 3eV KE of electron is

60 0.8 10K eV



SRIGAYATRI EUDCATIONAL INSTITUTIONSParagraph for Questions 11 and 12

Most materials have the refractive index, 1n . So, when a light ray from air enters a naturally occuring

material, then by Snell’s law, 1 2

2 1

sinsin

nn

, it is understood that the refracted ray bends towards the

normal. Bjut it never emerges on the same side of the normal as the incident ray. According to

electromagnetism, the refractive index of the medium is given by the relation. r rcnv

, where

c is the speed of electromagnetic waves in vacuum, v its speed in the medium, r and r are the relative

permittivity and permeability of the medium respectively. In normal materials, both r and r are positive,

implying positive n for the medium. When both r and r are negative, one must choose the negativeroot of n . Such negative refractive index materials can now be artificially prepared and are called meta-materials. they exhibit significantly different optical behavior, without violating any physical laws. Since nis negative, it results in a change in the direction of propagation of the refracted light. However, similar tonormal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

11. Choose the correct statement

(A) The speed of light in the meta-material is v c n

(B) The speed of light in the meta-mateiral is | |cvn

(C) The speed of light in the meta-material is v c

(D) The wavelength of the light in the meta-material m is given by m air n , where air is thewavelength of the light in air.

Sol. Key - B

constantf and cnV

.air mm air airmm

mm air mm mm

n V n V cVn V n n

12. For light incident from air on a meta-material, the appropriate ray diagram is

(A) (B)

(C) (D)



SRIGAYATRI EUDCATIONAL INSTITUTIONSSol. Key - C

metamaterialn ve and 1mm airn n

1 2

2 1

sinsin

nn

sinsin

a mm

mm air

n ven

and Greater than one

mm a if a ve the mm ve Hence ‘C’ is correct answer

Paragraph for Questions 13 and 14The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre ofmass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.These axes need not bestationary. Consider, for example, a thin uniform disc welded (rigidly fixed)horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotatedabout the origin on a horizontal frictionless plane with angular speed , the motion at any instant can betaken as a combination of (i) a rotation of the centre of massof the disc about the z-axis, and (ii) a rotationof the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from thechanged orientation of points P and Q). Both these motions have the same angular speed in this case.

Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical andparallel to x-z plane; Case (b) the disc with its face making an angle of 045 with x-y plane and itshorizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems arerotated with constant angular speed about the z-axis.

13. Which of the following statements regarding the angular speed about the instantaneous axis (passingthrough the centre of mass) is correct?(A) It is 2 for both the cases

(B) It is for case (a); and 2

for caseb (b).

(C) It is for case (a); and 2 for case (b).(D) It is for both the cases



SRIGAYATRI EUDCATIONAL INSTITUTIONSSol. Key - D

As point P is welded It has no relative motion with respective rod. The angular velocity will remain same in both the cases

14. Which of the following statements about the instantaneous axis (passing through the centre of mass) iscorrect?(A) It is vertically for both the cases (a) and (b).(B) It is vertical for case (a); and is at 045 to the x-z plane and lies in the plane of the disc for case (b)(C) It is horizontal for case (a); and is at 045 to the x-z plane and is normal to the plane of the disc for case (b).(D) It is vertical for case (a); and is at 045 to the x-z plane and is normal tothe plane of the disc for case (b).

Sol. Key - CThe instantaneous axis of rotation is the axis of rotation of disc about is centre of mass.Hence (C) is correct answer

SECTION III : Multiple Correct Answer(s) TypeThis section contains 6 multiple choice questions. Each question has fourchoices (A), (B), (C) and (D) out of which ONE or MORE are correct

15. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane fromthe same height at the same time. Cylilnder P has most of its mass concentrated near its surface, while Qhas most of its mass concentrated near the axis. Which statement(s) is(are) correct?(A) Both cylinders P and Q reach the ground at the same time(B) Cylinder P has larger linear acceleration then cylilnder Q(C) Both cylinders reach the ground with same translational kinetic energy(D) Cylinder Q reaches the ground with larger angular speed

Sol. Key - DMoment of inertia of cyllinder P is greater than the moment of inertia of cylinder Qi.e. P QI I

Lilnear acceleration P QI I

Now 2 2P Q P QI I MK MK P QK K (K = radius of gyration)

P Qa a

Cylinder P has lesser linear acceleration then cylinder Q. Therefore option (B) is wrong.Option (A) is also wrong as P Qa a .Total mechanical energy is conserved.

2 21 12 2cmMgh MV I

putting 2I MK2

22

1 12 cm

Kmgh MVR

P QK K Translational KE of cylinder P is less than that of .Hence option (C) is wrong.

cmV for cylinder P is less than for cylinder Q .

cmV R . But both the cylinders have the same radius.

P Q

Only option (D) is correct




16. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it.The correct statement(s) is (are)(A) The emf induced in the loop is zero if the current is constant(B) The emf induced in the loop is finite if the current is constant(C) The emf induced in the loop is zero if the current decreases at a steady rate(D) The emf induced in the loop is finite if the current decreases at a steady rate

Sol. Key - A,C

If the circuit is constant in the wire, then zerodBdt

through the circular wire loop.

Therefore, emf induced is zero.Hence option (A) is correct.

If the current in the wire decreases at a steady rate, dBdt would decrease, but the flux through the upper

half of the loop is opposite to the flux through the lower half of the loop.Therefore, the induced emf is zero.Hence option (C) is correct.

17. In the given circuit, the AC source has 100 /rad s . Considering the inductor and capacitor to beideal, the correct choice(s) is(are)

(A) The current through the circuit, I is 0.3 A.(B) The current through the circuit, I is 0.3 2 A.

(C) The voltage across 100 resistor 10 2V(D) The voltage across 50 resistor = 10V

Sol. Key - ACThe impedance of R - C circuit equals to 1z

1 100 2 ohmz

Impedance of L - r circuit = 2 50 2z ohm

As current in C - R circuit leads the emf by phase 1 such that 1100tan

100 2

01 45

Similarly in L - R circuit,Current lags the emf by phase 2 such that

02 2

50tan 4550 2




15 2ii A

22

5i A

2 2 01 2 22 cos90ii i i i i

1 0.310

i A A

(A) is correct

Voltage across 100 resistor 1 100

5 2

10 2V (C) is correct

18. Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the

figure. Given that 20

14

qKL

, which of the following statement(s) is(are) correct?

(A) The electric field at O is 6K along OD(B) The potential at O is zero(C) The potential at all points on the line PR is same(D) The potential at all points on the line ST is same

Sol. Key - A, B, CThe electric fields of positive charges are in the same direction as electric fields of negative charges at 0.

Electric field of (at B) 204

qq KL

along OE

Electric field of q at E K along OEElectric field of +q (at E) = K along OCELectric field of -q (at C) = K along OCNet electric field along OE = 2KNet electric field along OC = 2KSimilarly, neet electric field along OD = 4K.Resultant of electric field along OE and OC = 2K along ODHence not electric field of system of charges = 6K along ODTherefore option (B) is correct



SRIGAYATRI EUDCATIONAL INSTITUTIONSThe system can be through of as three electric dipoles.Potential on the perpendicular bisector of a dipole is zero.Therefore option (C) is correctOption (D) is wrong as potential at O is zero.But as point S A , potential tends to infinity

19. Two spherical planets P and Q have the same uniform density , masses pM and QM , and surface

areas A and 4A, respectively. A spherical planet R also has uniform density and its mass is P QM M .

The escape velocities from the planets P, Q and R, are ,P QV V and RV respecgtively. Then

(A) Q R PV V V (B) R Q PV V V (C) / 3R PV V (D) 1/2P QV V

Sol. Key - B, DPlanet P Planet QDensity Density

Mass PM Mass QM

Surface area 24 PR A Surface area = 24 4QR A

2Q PR R (radius of planets)

Given R P QM M M

or 3 34 4

3 3R P QM R R

or 3 34 49 . 93 3R P PM R R

3 1/ 34 93 PR R RR R

Escape velocities

3422 3 P

PP

P P

G RGMVR R

or

83P PV G R

342 . .2 . 3 QQ

QQ Q

G RG MV

R R

8 23Q Q Q PV G R V V

3422 3 RR

RR R

G RGMVR R

or 1/38 93R R R PV G R V V




R QV V

R Q PV V V

and 12

P

Q

VV

Option (B) and (D) are correct

20. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping ona horizontal surface with angular speed and (ii) an inner disc of radius 2R rotating anti-clockwise withangular speed / 2 . The ring and disc are separated by frictionless ball bearings. The system is in the x-z plane. The point P on the inner disc is at a distance r from the origin, where OP makes an angle of 030with the horizontal. Then with respect to the horizontal surface.

(A) The point O has a linear velocity ˆ3R i

(B) The point P has a linear velocity 11 3 ˆˆ4 4

R i R k

(C) The point P has a linear velocity 13 3 ˆˆ4 4

R i R k

(D) The point P has a linear velocity 3 1 ˆˆ3

4 4R i R k

Sol. Key - A,B

Since the outer ring rolls without slipping on the horizontal surface,Linear velocity of point O

3V R

Velocity of point ˆ3O R i Option (A) is correctThe point (P) has two velocitiesTranslational velocity ˆ3V R i Hence option (A) is correctThe point (P) has two velocities

Translational velocity ˆ3V R i

Tangential velocity = 2R

as shown.

Resolving the tangential velocity of P, linear velocity of point P.

0 0ˆ ˆ3 cos 60 sin 602 2R RR i j



SRIGAYATRI EUDCATIONAL INSTITUTIONSPART - II : CHEMISTRY

SECTION - I : Single Correct Answer TypeThis section contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and(D) out of which ONLY ONE is correct.

21. 2 2 5 6 52 2NiCl P C H C H exhibits temperature dependent magnetic behaviour (paramagnetic/

diamagnetic). The coordination geometries of 2Ni in the paramagnetic and diamagnetic states arerespectively.A) tetrahedral and tetrahedral B) square planar and square planarC) tetrahedral and square planar D) square planar and tetrahedral

Key: CSol. tetrahedral (paramagnetic) & square planar (Diamagnetic)22. The reaction of white phosphorus with acqueous NaOH gives phosphine along with another phosphorus

containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the otherproduct are respectively.A) redox reaction; –3 and –5 B) redox reaction; +3 and +5C) disproportionation reaction; –3 and +5 D) disproportionation reaction; –3 and +3

Key: DSol. 4 2 3 2 2PH 3NaOH 3H O PH 3NaH PO

2 2NaH PO may undergo further disproportionation23. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are

A) O2 and CO respectively B) O2 and Zn dust respectivelyC) HNO3 and Zn dust respectively D) HNO3 and CO respectively

Key: B

Sol: 2 22Ag S 4NaCl 2Na Ag CN Na S

2 2 2 4Na S 2O Na SO

22 42Na Ag CN Zn Na Zn CN 2Ag

24. The compound that undergoes decarboxylation most readily under mild condition is

A)

COOH CH COOH2

B)

COOH

OC)

COOH

COOH D)

CH COOH2

O

Key: BSol. Keto carboxylic acid undergoes decarboxylation most readily..25. Using the data provided, calculate the multiple bond energy (kJ mol–1) of a C C bond in C2H2. That

energy is (take the bond energy of a C – H bond as 350 kJ mol–1)

2 2 22C s H g C H g 1225H kJ mol

2 2C s C g 11410H kJmol

2 2H g H g 1330H kJ mol A) 1165 B) 837 C) 865 D) 815

Key: D




Sol. 2 Heat of sub lim ation of C H H Bond dissociation energy 2 C H bond energy C C bond energy = 225

1410 330 2 350 x 225 1410 300 700 x 225

x 81526. The shape of XeO2F2 molecule is

A) trigonal bipyramidal B) square palnarC) tetrahedral D) see - saw

Key: DSol: XeO2F2 has 4 bond pair & 1 lone pair hence destorted trigonal bipyramidal (or) sea saw

Xe OF

FO

27. The major product H of the given reaction sequence is2 495%

3 2 3H SOCNCH CH CO CH G H

heat

A) 3CH CH C COOH

3CH B) 3CH CH C CN

3CH

C) 3 2CH CH C COOH

3CH

OH

D) 3 2CH CH C CO NH

3CH

Key: A

Sol.3 2 3CH CH C CH

CN2 495% H SO

H Heat3 2CH CH C C N 3 2CH CH C COOH

3CH CH C COOH

OH OH

3CH 3CH

3CH

O

28. For a dilute solution containing 2.5g of a non - volatile non - electrolyte solute in 100g of water, theelevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower thanthe concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb= 0.76 K kg mol–1)A) 724 B) 740 C) 736 D) 718

Key: A

Sol.1

b b1 2

w 1000T i kM w




1

2.5 1000 0.76 252 1 0.76 9.5M 100 2

0 s 1 2 1

s 2 1 0

p p w M 760 p 2.5 18p w M p 100 9.5

ss

760 p 0.0474 p 760 760 0.0474 724760

SECTION - II : Paragraph TypeThis section contains 6 Multiple choice questions relating to three paragraphs with two questions on eachparagraph. Each question has four choices (A), (B), (C) and (D) out of which only one is correct.

Paragraph for Questions 29 and 30The electrochemical cell shown below is a concentration cell.

2|M M (saturated solution of a sparingly soluble salt, 2MX ) 2 3|| 0.001 |M mol dm M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes.The emf of the cell at 298K is 0.059V

29. The solubility product 3 9spK ;mol dm of MX2 at 298K based on the information available for the given

concentratioin cell is (take 2.303×R×298/F = 0.059V)A) 1×10–15 B) 4×10–15 C) 1×10–12 D) 4×10–12

Key: B

Sol:

2

a2

c

M0.059E logn M

2

3

M0.0590.059 log2 10

3 32 2 5

2 2

10 10log 2 10 M 10M M

33 5 15pKS 4s 4 10 4 10

30. The value of G (kJ mol–1) for the given cell is (take 1F = 96500 C mol–1)A) – 5.7 B) 5.7 C) 11.4 D) – 11.4

Key: DSol. G nFE 2 96500 0.059 11387J 11.387kJ

Paragraph for Questions 31 and 32Bleaching powder and bleach solution are produced on a large scale and used in several house holdproducts. The effectiveness of bleach solution is often measured by iodometry.

31. 25mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4N acetic acid. Inthe titration of the liberated iodine, 48mL of 0.25 N Na2S2O3 was used to reach the end point. Themolarity of the household bleach solution isA) 0.48M B) 0.96M C) 0.24M D) 0.024M

Key: C




Sol.

2 3 2 2 3

3 2 4 6 22

CaOCl 2CH COOH 2KI 2Na S O

CH COO Ca 2KCl 2NaI Na S O H O

Number of milli moles of CaOCl2 = 2 2 3numberof milli moles Na S O

2No.of milli equvalents Hypo = No.of milli moles of Hypo = 48×0.25=12 Number of milli moles of CaOCl2 =6

25 M 6

6M 0.2425

32. Bleaching powder contains a salt of an oxoacid as one of its component. The anhydride of that oxoacid isA) Cl2O B) Cl2O7 C) ClO2 D) Cl2O6

Key: ASol. 2 22HOCl Cl O H O

Paragraph for Questions 33 and 34In the following reaction sequence, the compound J is an intermediate

I J K (CH CO ) O3 2 2 (i)H ,Pd/C 2

CH COONa3 (ii) SOCl 2

(iii) anhyd. AlCl3

J (C9H8O2) gives effervescence on treatment with NaHCO3 and a positive Baeyer’s test33. The compound K is

A) B) C) D)

Key: C

Sol.

2 2CH CH

3AlClC 0

Cl O34. The compound I is

A) B) C) D)

Key: A




Sol.CHO

SECTION - III : Multiple Correct Answer(s) TypeThis section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which One or More are correct.

35. With respect of graphite and diamond, which of the statement(s) given below is (are) correct ?A) Graphite is harder than diamondB) Graphite has higher electrical conductivity than diamondC) Graphite has higher thermal conductivity than diamondD) Graphite has higher C - C bond order than diamond

Key: A & DSol: Conceptual36. The given graphs/data I, II, III and IV represent general trends observed for different physisorption and

chemisorption processes under mild conditions of temperature and pressure. Which of the followingchoice(s) about I, II, III and IV is (are) correct ?

A) I is physisorption and II is chemisorption B) I is physisorption and III is chemisorptionC) IV is chemisorption and II is chemisorption D) IV is chemisorption and III is chemisorption

Key: A & CSol: Conceptual37. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure.

Which of the following statement(s) is (are) correct ?

A) T1 = T2 B) T3 > T1

C) wisothermal > wadiabatic D) isothermal adiabaticU U Key : A,C,DSol: Conceptual



SRIGAYATRI EUDCATIONAL INSTITUTIONS38. For the given aqueous reactions, which of the statement(s) is (are) true ?

A) the first reaction is a redox reaction B) white precipitate is 3 6 2Zn Fe CN

C) addition of filtrate to starch solution gives blue colourD) white precipitate is soluble in NaOH solution

Key: A,C,DSol. Conceptual39. With reference to the scheme given, which of the given statement(s) about T,U,V and W is (are correct?

A) T is soluble in hot aqueous NaOH B) U is optically activeC) molecular formula of W is 10 18 4C H O

D) V gives effervescence on treatment with aqueous 3NaHCOKey: A,CSol. Conceptual40. Which of the given statement(s) about N,O,P and Q with respect to M is (are) correct ?

A) M and N are non - mirror image stereoisomersB) M and O are identical C) M and P are enantiomersD) M and Q are idential

Key: A,B,CSol. Conceptual



SRIGAYATRI EUDCATIONAL INSTITUTIONSMATHEMATICS

SECTION I : Single Correct Answer TypeThis section contains 8 multiple choice questions. Each question has fourchoices (A), (B), (C) and (D) out of which ONLY ONE is correct.

41. The equation of a plane passing through the line of intersection of the planes 2 3 2x y z and

3x y z and at a distance 23 from the point 3,1, 1 is

(A) 5 11 17x y z (B) 2 3 2 1x y (C) 3x y z (D)

2 1 2x y Key: A

Equation of required plane

1 2 0

The perpendicular distance from 3,1, 1 to this plane 23

7 2 equation of plane is 5 11 17.x y z

42. If a and b

are vectors such that 29a b

and ˆ ˆˆ ˆ ˆ ˆ2 3 4 2 3 4 ,a i j k i j k b

then a

possible value of ˆˆ ˆ. 7 2 3a b i j k

is(A) 0 (B) 3 (C) 4 (D) 8

Key: C

29a b . Let ˆˆ ˆ2 3 4c i j k

a c c b

0a c b c

0a b c

a b tc

29 29 1t t ˆˆ ˆ2 3 4a b c i j k

ˆˆ ˆ. 7 2 3a b i j k

14 6 12 4

43. Let PQR be a triangle of area with 72,2

a b and 5 ,2

c where ,a b and c are the lengths of

the sides of the triangle opposite to the angles at P, Q and R respectively. Then 2sin sin 22sin sin 2

P PP P

equals

(A) 3

4 (B) 454 (C)

234

(D) 245

4



SRIGAYATRI EUDCATIONAL INSTITUTIONSKey: C

7 52, , 4.2 2a b C S

2

22sin sin 2 tan 22sin sin 2S B S CP P P

P P

2

21 3

32 24

44. Four fair dice 1 2 3, ,D D D and 4 ,D each having six faces numbered 1,2,3,4,5 and 6, are rolled simulta-

neously. The probability that 4D shows a number appearing on one of 1 2,D D and 3D is

(A) 91

216 (B) 108216 (C)

125216 (D)

127216

Key: AOn the first 3 tosses 4th toss

Case (i) 3 same 6 1

216 6

(ii) 3 different 36 3216 6

P

(iii) Two alike 2 different 90 2216 6

Total probability 91

216

45. The value of the integral 2

2

2

ln cosxx x dxx

is

(A) 0 (B) 2

42

(C) 2

42

(D) 2

2

Key: B2

2

0

2 cos .x x dx

2/ 22

0sin 2 cos 2sin 4

2x x x x x

46. If P is a 3 3 matrix such that 2 ,TP P I where TP is the transpose of P and I is the 3 3 identity

matrix, then there exists a column matrix

000

xX y

z

such that

(A)

000

PX

(B) PX X (C) 2PX X (D) PX X

Key: D

2 2T TP P I P I 2 2P I I 4 3P I P I

.PX X




47. Let 1 2 3, , ,....a a a be in harmonic progression with 1 5a and 20 25.a The least positive integer nfor which 0na is(A) 22 (B) 23 (C) 24 (D) 25

Key: D1 1195 25

a a d

1 1 41925 5 25

d 4

19 25d

1 4 11 0.5 19 25n

na a n d

4 1

119 5

n

19 5 9914 4

n n 25n

48. The a and a be the roots of the equation 23 61 1 1 1 1 1 0a x a x a

where 1.a

Then 0lima a and 0lima a are

(A) 52

and 1 (B) 12

and 1 (C) 72

and 2 (D) 92

and 3

Key: B

Sum of the roots 3

111 1

aa aa

Product of roots 6

3

1 1.1 1

aa aa

0

3lim .2xa a

0

1lim . 2xa a

Only option (B) satisfies

SECTION II : Paragraph TypeThis section contains 6 multiple choice questions relating to 3 paragraphs withtwo questions on each paragraph. Each question has four choices (A), (B), (C)and (D) out of which ONLY ONE is correct

Let 2 2 21 sinf x x x x for all ,x and let 1

2 1ln

1

x tg x t f t dt

t

for all

1, .x 49. Which of the following is true ?

(A) g is increasing on 1, (B) g is decreasing on 1,

(C) g is increasing on x 1, 2 and decreasing on 2,

(D) g is decreasing on 1, 2 and increasing on 2,



SRIGAYATRI EUDCATIONAL INSTITUTIONSSol: B

1 2 1ln

1x

g x x f xx

2 2 21 sinf x x x x , 1 0g x for 1,x .

50. Consider the statements.

P : There exists some x such that 22 2 1f x x x

Q : There exists some x such that 2 1 2 1f x x x Then(A) both P and Q are true (B) P is true and Q is false(C) P is false and Q is true (D) both P and Q are false

Sol: C

2

2

1: sin 11

P xx

is solution false

2

2 2

1 2 1: sin 21 2 1

xQ x xx x

has a solution true

PASSAGE - 51 TO 52Let na denote the number of all n digit positive integers formed by the digits 0, 1 or both such that no

consecutive digits in them are 0. Let nb the number of such n digit integers ending with digit 1 and

nc the number of such n digit integers ending with digit 0.51. Which of the following is correct ?

(A) 17 16 15a a a (B) 17 16 15c c c (C) 17 16 16b b c (D) 17 17 16b c b Sol: (A)

Using Recurrence formula 1 2n n na a a

17 16 15a a a

52. The value of 6b is(A) 7 (B) 8 (C) 9 (D) 11

Sol. B

6b 4 places can be filled with 0,1 & within the given condition =g ways

PASSAGE - 53 TO 54

A tangent PT is drawn to the circle 2 2 4x y at the point 3,1 .P A straight line L, perpendicular to

PT is a tangent to the circle 2 23 1.x y 53. A common tangent of the two circles is

(A) 4x (B) 2y (C) 3 4x y (D) 2 2 6x y



SRIGAYATRI EUDCATIONAL INSTITUTIONSSol: D

1 20,0 3,0c c

1 22 1r r

1 2 1 23c c r r

2 2 6x y only satisfies is the direct common tangent

54. A possible equation of L is(A) 3 1x y (B) 3 1x y (C) 3 1x y (D) 3 5x y

Sol: AProve that equation is 3 4x y

3 0x y k

touches 2 23 1x y

1 5k or

3 1x y

SECTION III : Multiple Correct Answer(s) TypeThis section contains 6 multiple choice questions. Each question has fourchoices (A), (B), (C) and (D) out of which ONE or MORE are correct

55. Let X and Y be two events such that 1 1,2 3

P X Y P Y X and 1 .6

P X Y Which of the

following is (are) correct ?

(A) 23

P X Y (B) X and Y are independent

(C) X and Y are not independent (D) 13

CP X Y

Key: A,B

1P(x y) 1 1 16 P(y)

P(y) 2 p(y) 2 3

1P(y x) 1 1 16 P(x)

P(x) 3 p(x) 3 2

Option :A) P(x y) = P(x) + P(y) – P(x y)

1 1 13 2 6

5 1 4 26 6 6 3

....... (A)




B) P(x y) = 16 - P(x) . P(y)

x and y are independent ...... (B)C) WrongD) P(x y) p(y) P(x y)

1 1 2 1 13 6 6 6

Wrong(A) and (B) only.

56. If 2

02 3

x tf x e t t dt for all 0, ,x then

(A) f has a local maximum at 2x (B) f is decreasing on 2,3

(C) there exists some 0,c such that '' 0f c (D) f has a local minimum at 3x Key : A,B,D

x 2x

0

f (x) e (t 2) (t 3)dt

21 xf (x) e (x 2) (x 3) for Min or Max x = 2 or x == 3

57. For every integer ,n let na and nb be real numbers. Let function :f be given by

sin , 2 ,2 1,

cos , 2 1, 2n

n

a x for x n nf x

b x for x n n

for all integers .n

If f is continuous, then which of the following hold(s) for all ?n(A) 1 1 0n na b (B) 1n na b (C) 1 1n na b (D) 1 1n na b

Sol: B,D

0 [0,1)n 1, 0n na b

[ 1,0) 1n na b

1 [2,3)n 1 1n na b

[1,2) 10, 1n na b

part i : physics - eenadupratibha.netconsider a disc rotating in the horizontal plane with a...

Documents