Vidyamandir Classes. JEE Advanced Paper 2 Code 5 25 May, :00 PM 5:00 PM. Answer key and solutions by Vidyamandir Classes

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1 JEE Advanced 04 Paper Code 5 5 May, 04 :00 PM 5:00 PM key and solutions by Vidyamandir Classes VMC/JEE Advanced-04 Paper-

2 PART-A PHYSICS. A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross section of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its cross section is b. If the surface tension of water is S, its density is ρ, and its contact angle with glass is θ, the value of h will be : (g is the acceleration due to gravity) S cos bρ g S cos bρ g ( θ α) ( θ+ α) S α cos θ bρ g P s + ρ gh= P r 0 0 s h= rρ g From triangle AOP α b= r cos θ + b r = α cos θ + A θ + r θ α α b O P b S α cos θ+ bρ g r B s α h= cos θ + bρg. A planet of radius R R= (radius of Earth) has the same mass density as Earth. Scientists dig a 0 R well of depth on it and lower a wire of the same length and of linear mass density 0 kgm 5 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is : (Take the radius of Earth = m and the acceleration due to gravity of Earth is 0ms ) 96 N 08 N 0 N 50 N 4π Gρx E = = kx df = λ Edx R λk 6R F = λ.kxdx = R 5 4R 5 R 5 x O dx 4R 5 (centre) VMC/JEE Advanced-04 Paper-

3 9λ KR F = 5 9 4πGP R0 R 0 = radius of earch = M 0 = Mass of earth = 08 N. Charges Q, Q and 4Q are uniformly distributed in three dielectric solid spheres, and of radii R/, R and R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres, and are E, E and E respectively, then : E > E > E E > E > E E > E > E E > E > E Q E = 4π 0 R Q E = 4π 0 R E ρr 4Q R Q Q = = = = π( R) π 0 R 4π 0 R Hence E > E > E 4. If λcu is the wavelength of K α X-ray line of copper (atomic number 9) and λmo is the wavelength of the K α X-ray line of molybdenum (atomic number 4), then the ratio λcu / λmo is close to : λ ( z ) λ λ cu mo Zmo 4 4 = = = =. 4 zcu 9 8 VMC/JEE Advanced-04 Paper-

4 5. A wire, which passes through the hold in a small bead is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and its sides along the wire without friction. As the bead moves from A to B, the force it applies on the wire : always radially outwards. always radially inwards. radially outwards initially and radially inwards later. radially inwards initially and radially outwards later. After some time bead will tend to leave the wire. 6. A metal surface is illuminated by light of two different wavelength 48 mm and 0 mm. The maximum speeds of the photoelectrons corresponding to these wavelength are u and u, respectively. If the ratio u : u = : and hc= 40eV nm the work function of the metal is nearly.7 ev. ev.8 ev.5 ev hc w Kmax λ = + hc w K λ 4hc = = 4 K hc λ w λ hc 4hc w= 4w λ λ 4 w= hc λ λ 40 4 w= 0 48 w. 7 ev 7. A tennis ball is dropped on a horizontal smooth surface it bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which of the following sketches described the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale. VMC/JEE Advanced-04 4 Paper-

5 8. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90Ω, as shown in the figure. The least count of the scale used in the metre bridge is mm. The unknown resistance is : 60± 0. 5Ω 5± 0. 56Ω 60± 0. 5Ω 5± 0. Ω R 60= R= = 60Ω 60 Let at null point length of right part is x and left is y. Rx= 90y 90y R= x R y x = + R y x R = + R = R = 60± Parallel rays of light of intensity I = 9Wm are incident on a spherical black body kept in surroundings of temperature 00 K. Take Stefan-Boltzmann constant 8 4 σ = Wm K and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to : 00 K 660 K 990 K 550 Let steady state temp is T. Power gain = Power lost VMC/JEE Advanced-04 5 Paper-

6 I( π R ) = 6l ( 4π R )( T 4 T 4 ) 0 9= ( T ) T 00 = T = 0 T = K 0. A point source S is placed a the bottom of a transparent block of height 0 mm and refractive index.7. It is immersed in a lower refractive index liquid as shown in the figure. I is found that the light emerging from the block to the liquid forms a circular bright spot of diameter.54 mm on the top of the block. The refractive index of the liquids is : µ µ sinc = = µ. 7 r 0mm = h tanc = h. 54 / = = = sinc = (. ) µ 0. 5= µ = s C r C Paragraph for Questions - In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with moles of an ideal monatomic gas at 700 K and the upper compartment is filled with moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monatomic gas are CV = R, CP = R, and those for an ideal diatomic gas are CV = R, CP = R. VMC/JEE Advanced-04 6 Paper-

7 . Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be : 550 K 55 K 5 K 490 K Let final temp be T. For lower compartment process is isochoric and for upper process is isobaric Q = Q lower upper ( ) ( ) C 700 T = C T 400 V R = 400 T = 490K. P ( T) R ( T ). Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. The total work done by the gases till the time they achieve equilibrium will be : 50 R 00 R 00 R 00R Let T be final temperature Qlower = Qupper C ( 700 T) = C (T 400) (Bothprocessareisobaric) p 5 7 R( 700 T) = R(T 400) T = 55K w = R( ) + R( ) = 50R+ 50R = 00R p Paragraph for Questions - 4 A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 0 mm and mm respectively. The upper end of the container is open to the atmosphere.. If the piston is pushed at a speed of 0. ms 5 mms, the air comes out of the nozzle with a speed of : ms ms 8 ms VMC/JEE Advanced-04 7 Paper-

8 AV = AV π( 0) 5 0 =π( ) v = m/s v 4. If the density of air is ρ a and that of the liquid ρ l, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to : ρ a ρaρ l ρ l For air P + ρ ava = P0 + 0 For liquid 4 P0 + 0= P + ρ lvl ρa Vl = V a. ρ l ρl ρl ρ a 4 Paragraph for Questions 5-6 The figure shows a circular loop of radius a with two long parallel wires (numbered and ) all in the plane of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above. 5. When d a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case : current in wire and wire is the direction PQ and RS, respectively and h a current in wire and wire is the direction PQ and SR, respectively and h a current in wire and wire is the direction PQ and SR, respectively and h. a current in wire and wire is the direction PQ and RS, respectively and h =. a 0 0 µ Ia = µ π Ia π ( a + h ) 4π ( a + h ) h. a / VMC/JEE Advanced-04 8 Paper-

9 6. Consider d >> a, and the loop is rotated its diameter parallel to the wires by 0 from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop) µ 0 I a d µ 0I B= d π ( 80 0) τ = MB sin ( π a I) µ 0I = π d µ = 0a I d µ 0I a d µ 0I a d µ 0I a d 7. Four changes Q, Q, Q and Q 4 of same magnitude are fixed along the x axis at x= a, a, + a and + a, respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists. List I (P) Q, Q, Q, Q 4 all positive. +x List II (Q) Q, Q positive; Q, Q 4 negative. x (R) Q, Q 4 positive; Q, Q negative. + y (S) Q, Q positive; Q, Q 4 negative 4. y P Q R S P Q R S Option P 8. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists. VMC/JEE Advanced-04 9 Paper-

10 List I List II (P). r (Q). r/ (R). r (S) 4. r P Q R S P Q R S = f R 0 5 =. f R 0 5 =. f R = + = f f f R f = R = + = f f f R f = R = + = f f f R f = R 0 5 = + =. f f f R f = R VMC/JEE Advanced-04 0 Paper-

11 9. A block of mass m = kg another mass m = kg, are placed together (see figure) on an inclined plane with angle of inclinationθ. Various value of θ are given in List I. The coefficient or friction between the block m and the plane is always zero. The coefficient of static dynamic friction between the block m and the plane are equal to µ= 0.. In List II expressions for the friction on block m are given. Match the correct expression of the friction in List II with the angles in list I, and choose the correct option. The acceleration due to gravity is denoted by g. List I List II P. θ = 5. mg sinθ θ =. ( + ) Q. 0 m m g sinθ R. θ = 5. µ mg cosθ θ = 4. ( + ) S. 0 µ m m g cosθ [Useful information : ( ) ( ) ( ) tan ; tan ; tan ] Code : P, Q, R, S P, Q, R, S P, Q, R, S 4 P, Q, R, S For No slipping µ m g cosθ > m g sinθ + m g sinθ µ tanθ < tan ( θ ) < 0. θ <. 5 For P θ = 5 No slipping f = m+ m g sinθ Hence Option f = µ m g cosθ slipping. f = µ m g cosθ Hence option For Q ( ) For R For S θ 0 slipping f µ m g cos( θ) = Hence Option VMC/JEE Advanced-04 Paper-

12 0. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of. m from the person. In the following, state of the lift s motion is given in List-I and the distance where the water jet hits the floor of the lift is given in List-II. Match the statements from List-I with those in List-II and select the correct answer using the code given below the lists. List-I List-II (P) Lift is accelerating vertically up. d =. m (Q) Lift is acceleration vertically down with an acceleration less than the gravitational acceleration.. d >. m (R) Lift is moving vertically up with constant speed.. d <. m (S) Lift is falling freely. 4. No water leaks out of the jar Code: P -, Q -, R, S - 4 P, Q, R, S - 4 P, Q, R, S 4 P, Q, R, S t= h g+ a d = vt = 4h h Which is independent of a Hence P I, Q, R For S, V = 0 (as ( a= g) VMC/JEE Advanced-04 Paper-

13 PART-B CHEMISTRY. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure. The correct order of their boiling point is : I > II > III III > II > I II > III > I III > I > II III > II > I, Branching leads to decrease in B.P.. The major product in the following reaction is :. The acidic hydrolysis of ether (X) shown below is fastest when : One phenyl group is replaced by a methyl group One phenyl group is replaced by a para-methoxyphenyl group Two phenyl groups are replaced by two para-methoxyphenyl group No structural change is made to X Methoxy group stabilizes the carbocation. 4. Hydrogen peroxide in its reaction with KIO 4 and NH OH respectively, is acting as a : reducing agent, oxidising agent reducing agent, reducing agent VMC/JEE Advanced-04 Paper-

14 oxidising agent, oxidising agent oxidising agent, reducing agent Θ HO + IO4 O + I (Re ducing Agent) HO + N HOH HO + N (Oxidisin g Agent) 5. The product formed in the reaction of SOCl with white phosphorus is : PCl SO Cl SCl POCl P4 ++ 8SOCl 4PCl + 4SO + 4SCl 6. For the identification of β -naphthol using dye test, it is necessary to use : Dichloromethane solution of β -naphthol Acidic solution of β -naphthol Neutral solution of β -naphthol Alkaline solution of β -naphthol 7. For process HO( l ) HO(g) at T= 00 C and atmosphere pressure, the correct choice is : Ssystem > 0 and Ssurroundings > 0 Ssystem > 0 and Ssurroundings < 0 Ssystem < 0 and Ssurroundings > 0 Ssystem < 0 and Ssurroundings < 0 Ssystem because HO change its state from liquid to gas and Ssurrounding because heat is supplied to the system from the surroundings. Heat is lost by the surrounding. 8. For the elementary reaction M N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is : 4 d[m] d[n] r = dt = + dt, as [M] [M] r [M] them r 8r r = k[m] Order of reaction w.r.t M = VMC/JEE Advanced-04 4 Paper-

15 9. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is : 0 H _ O OH / HO disproportionation gas gas XeF 6HF XeO HXeO XeO Xe O 0. Assuming s-p mixing is NOT operative, the paramagnetic species among the following is : Be B C N If s-p mixing is NOT operative them even the specises with 4 or less 4 electrons will have the normal molecular orbital energy levels as shown below. Clearly, Only C among the given options will be paramagnetic if the following diagram is followed for fill electrons. Paragraph for Questions - Schemes and describe sequential transformation of alkynes M and N. Consider only the major products formed in each step for both the schemes.. The product X is : VMC/JEE Advanced-04 5 Paper-

16 . The correct statement with respect to product Y is : It gives a positive Tollens test and is a functional isomer of X It gives a positive Tollens test and is a geometrical isomer of X It gives a positive iodoform test and is a functional isomer of X It gives a positive iodoform test and is a geometrical isomer of X VMC/JEE Advanced-04 6 Paper-

17 Paragraph for Question - 4 An aqueous solution of metal ion M reacts separately with regents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M always forms tetrahedral complexes with these reagents. Aqueous solution of M on reaction with reagent S gives white precipitate which dissolves in express of S. The reactions are summarized in the scheme given below : SCHEME :. M, Q and R respectively are : Zn Cd + +, KCN and HCl, KCN and HCl Ni Co + +, HCl and KCN, HCl and KCN 4. Reagent S is : K Fe( CN) NaHPO 4 KCrO 4 KOH 4 6 Paragraph for Question 5-6 X and Y are two volatile liquids with molar weights of 0 g mol and 40 g mol respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 4 cm, as shown in the figure. The tube is filled with an inert gas at atmosphere pressure and a temperature of 00 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameter and assume ideal behaviour for the inert gas and the two vapours. VMC/JEE Advanced-04 7 Paper-

18 5. The value of d in cm (shown in the figure), as estimated from Graham s law is : The experimental value of d is found to be smaller than the estimate obtained using Graham s law. This is due to : Larger mean free path for X as compared to that of Y Larger mean free path for Y as compared to that of X Increased collision frequency of Y with the inert gas as compared to that of X with the inert gas Increased collision frequency of X with the inert gas as compared to that of Y with the inert gas 7. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway from List with an appropriate structure from List and select the correct answer using the code given below the lists. VMC/JEE Advanced-04 8 Paper-

19 List I List II (P) Pathway P. (Q) Pathway Q. (R) Pathway R. (S) Pathway S 4. P Q R S P Q R S VMC/JEE Advanced-04 9 Paper-

20 8. Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I, II, III, IV) provided in List II and select the correct answer using the code given below the lists : List I List II (P) Scheme I (Q) Scheme II (i) KMnO 4, HO, heat (ii) H +, HO (iii) SOCl (iv) NH? C H N O (i) Sn / HCl (ii) CHCOCl (iii) conc. HSO4 (iv) HNO (v) dil. HSO 4, heat (vi) HO 7 6? C H N O 6 6 (R) Scheme III (i) red hot iron, 87 K (ii) fu min g HNO, HSO 4,heat (iii) HS.NH (iv) NaNO, HSO 4 (v) hydrolysis? C H NO 6 5 (S) 4 Scheme IV (i) conc. HSO 4, 60 C (ii) conc. HNO, conc.hso 4 (iii) dil. HSO 4, heat? C H NO P Q R S P Q R S VMC/JEE Advanced-04 0 Paper-

21 9. Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists. { en= HNCHCHNH ; atomic numbers : Ti= ;Cr= 4;Co= 7;Pt= 78} List I List II (P) [Cr(NH ) 4Cl ]Cl Paramagnetic and exhibits ionization isomerism (Q) [Ti(HO) 5Cl](NO ) Diamagnetic and exhibits cis-trans isomerism (R) [Pt(en)(NH )Cl]NO Paramagnetic and exhibits cis-trans isomerism (S) [Co(NH ) 4(NO ) ]NO 4 Diamagnetic and exhibits ionization isomerism P Q R S P Q R S VMC/JEE Advanced-04 Paper-

22 40. Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct answer using the code given below the lists. List I List II (P). p d π antibonding (Q). d d σ bonding (R). p d π bonding (S) 4. d d σ antibonding P Q R S P Q R S VMC/JEE Advanced-04 Paper-

23 PART-C MATHEMATICS 4. Coefficient of x in the expansion of ( + ) ( + ) ( + ) x x x is : Let f : [ 0, ] R be a function which is continuous on [0, ] and is differentiable on (0, ) with f ( 0) =. Let ( ) e x = for x [ 0, ]. If F ( x) = f ( x) for all x ( 0, ) F x f t dt 0 4 e e, then F() equals : 4 e 4. The function y= f ( x) is the solution of the differential equation dy xy x + x + = dx x x 4 in (, ) satisfying f ( 0) = 0. Then : f ( ) x dx is : π π π π VMC/JEE Advanced-04 Paper-

24 44. The following integral ( cosec x) ( ) 6 π 7 dx is equal to : π 4 log + u u log + u u ( e + e ) du ( + ) 0 ( ) 7 0 ( ) 7 e e du log + u u log + u u ( e e ) du ( ) 0 0 ( ) 6 e e du VMC/JEE Advanced-04 4 Paper-

25 45. The quadratic equation p( x ) = 0 with real coefficient has purely imaginary roots. Then the equation ( ( )) = 0 p p x has : only purely imaginary roots all real roots two real and two purely imaginary roots neither real nor purely imaginary roots 46. For x ( 0,π), the equation sin x+ sinx sinx= has : infinity many solutions three solutions one solution no solution 47. In a triangle the sum of two sides is x and the product of the same two sides is y. If x c = y, where c is the third side of the triangle, then the ratio of the in-radius to the circum-radius of the triangle is : y x x ( + c) y c x ( + c) y 4x x ( + c) y 4c x ( + c) VMC/JEE Advanced-04 5 Paper-

26 48. Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is : The common tangents to the circle x + y = and the parabola y P,Q and the parabola at the points R, S. Then the area of the quadrilateral PQRS is : = 8x touch the circle at the points VMC/JEE Advanced-04 6 Paper-

27 50. Six cards and six envelopes are numbered,,, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered is always placed in envelope numbered. Then the number of ways it can be done is : Paragraph for Questions 5-5 Let a, r, s, t be nonzero real numbers. Let P (at, at), Q, R (ar, ar) and S (as, as) be distinct points on the parabola y = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (a, 0). 5. The value of r is : t t + t t t t VMC/JEE Advanced-04 7 Paper-

28 5. If st =, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is. ( t + ) t ( + ) a t t ( + ) a t t ( + ) a t t Paragraph for Questions 5-54 h a Given that for each ( 0 ) ( ) + h 0 h a a,, lim t t dt exists. Let this limit be g (a). In addition, it is given that the function g (a) is differentiable on (0, ). 5. The value of g is : π π π π 4 VMC/JEE Advanced-04 8 Paper-

29 54. The value of π g' π π 0 Paragraph for Questions Box contains three cards bearing numbers,, ; box contains five cards bearing number,,, 4, 5 ; and box contains seven cards bearing numbers,,, 4, 5, 6, 7. A card is drawn from each of the boxes. Let x i be the number on the card drawn from the i th box, i =,,. 55. The probability that x + x+ x is odd, is : VMC/JEE Advanced-04 9 Paper-

30 56. The probability that x, x, x are in an arithmetic progression, is : Let kπ kπ Zk = cos + i sin ; k =,,..., List I List II (P) For each z k there exists a z j such that z z =. True k j (Q) There exists { 9} (R) a k,,..., such that z z = z k has no solution z in the set of complex numbers (S) 9 z z... z 9 k = 0 k cos equals 0 equals. False. 4. P Q R S P Q R S VMC/JEE Advanced-04 0 Paper-

31 58. MATCH THE FOLLOWING LISTS : (P) List I The number of polynomials f (x) with non-negative integer coefficients of degree, satisfying ( ) ( ) f 0 = 0 and f x dx =, is (Q) The number of points in the interval, (R) f (x) = sin (x ) + cos (x ) attains its maximum value, is x dx equals x ( + e ) (S) x + cos x log dx x equals + x cos x log dx 0 x 0 at which List II P Q R S P Q R S VMC/JEE Advanced-04 Paper-

32 59. MATCH THE FOLLOWING LISTS : (P) (Q) (R) List I Let ( ) ( ) [ ] ( x ) ( ) y x y x = cos cos x, x,, x ±. Then ( ) ( ) d y x dy x + x equals dx dx Let A, A,, A n (n > ) be the vertices of a angular polygon on n sides with its centre at the origin. Let a k be the position vector of the point A k, k =,,,n. If n n ( ak ak + ) = ( a k ak + ), then the minimum k= k = value of n is x y If the normal from the point P (h, ) on the ellipse + = 6 is perpendicular to the line x + y = 8, then the value of h is (S) Number of positive solutions satisfying the equation tan tan tan + = x 4x + + x List II VMC/JEE Advanced-04 Paper-

33 P Q R S P Q R S Let f : R R, f :[ 0, ) R, f : R R and f [, ) ( x) f ( x) f x if x< 0 = x e if x 0 sin x if x< 0 = x if x 0 ; f ( x) x and f ( x) 4 4 : R 0 be defined by : = ; ( ( )) ( ) f f x if x< 0 = f( f x ) if x 0 List I List II (P) f 4 is. Onto but not one-one (Q) f is. Neither continuous nor one-one (R) fof is. Differentiable but not one-one (S) f is 4. Continuous and one-one Codes : P Q R S P Q R S VMC/JEE Advanced-04 Paper-

34 VMC/JEE Advanced-04 4 Paper-

35 JEE Advanced 04 Paper 5 May 04 :00PM - 5:00 PM Code - 5 key PHYSICS CHEMISTRY MATHS Q.No. D Q.No. B Q.No.4 C Q.No. B Q.No. D Q.No.4 B Q.No. C Q.No. C Q.No.4 B Q.No. 4 B Q.No.4 A Q.No.44 A Q.No. 5 D Q.No.5 A Q.No.45 D Q.No. 6 A Q.No.6 D Q.No.46 D Q.No. 7 B Q.No.7 B Q.No.47 B Q.No. 8 C Q.No.8 B Q.No.48 A Q.No. 9 A Q.No.9 C Q.No.49 D Q.No. 0 C Q.No.0 C Q.No.50 C Q.No. D Q.No. A Q.No.5 D Q.No. D Q.No. C Q.No.5 B Q.No. C Q.No. B Q.No.5 A Q.No. 4 A Q.No.4 D Q.No.54 D Q.No. 5 C Q.No.5 C Q.No.55 B Q.No. 6 B Q.No.6 D Q.No.56 C Q.No. 7 A Q.No.7 A Q.No.57 C Q.No. 8 B Q.No.8 C Q.No.58 D Q.No. 9 D Q.No.9 B Q.No.59 A Q.No. 0 C Q.No.40 C Q.No.60 D