SOLUTIONSA long horizontal rod has a bead which can slide along its length, and initially placed at...

SOL UTI ON S

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ANSWER KEY

AITS-FT # 03 [SET-A]

DROPPER ENGINEERING

(PHYSCIS, CHEMISTRY & MATHS)

TARGET : JEE (MAIN + ADVANCED) 2018-19

Exam. Date : 27-01-2019

AITS-FT # 03 (Engineering Dropper) (Solutions SET-A) - 2018-19


Potential & Concept Educations Page. 2

ANSWER KEYS FOR AITS-FT # 03 DROPPER ENGINEERING [SET # A]

DATE : 27-01-2019

ANSWERS [CHEMISTRY]

31. D 32. A 33. D 34. B 35. D 36. A 37. A 38. B 39. A 40. D41. C 42. A 43. A 44. B 45. D 46. A 47. A 48. A 49. C 50. C51. A 52. A 53. B 54. A 55. D 56. D 57. D 58. B 59. A 60. D

ANSWERS [PHYSICS]

1. C 2. A 3. A 4. B 5. D 6. A 7. B 8. B 9. B 10. C11. B 12. D 13. A 14. D 15. A 16. A 17. A 18. C 19. C 20. B21. B 22. A 23. D 24. B 25. D 26. B 27. B 28. C 29. A 30. B

ANSWERS [MATHS]

61. C 62. D 63. D 64. D 65. D 66. A 67. C 68. D 69. A 70. C71. C 72. A 73. A 74. B 75. B 76. A 77. B 78. B 79. A 80. D81. C 82. A 83. B 84. D 85. A 86. C 87. D 88. B 89. A 90. C



PHYSICS AITS-FT-03Engineering

Dropper Batch

1. Two masses are connected by a spring as shown in the figure. One of the masses was given velocity = 2k, as shown in figure where ‘k’ is the spring constant. Then maximum extension in the spring will be

(A) 2 m (B) m (C) 2mk (D) 3mkSolution :The answer is (C).

2. The centre of mass of a non uniform rod of length L whose mass per unit length varies as 2k.x

L

where k is a constant & x is the distance of any point from one end, is (from the same end)

(A) 3 L4

(B) 1 L4

(C) kL

(D) 3kL

Solution :

Hence the answer is (A).3. A person P of mass 50 kg stands at the middle of a boat of mass 100 kg moving at a constant velocity 10

m/s with no friction between water and boat and also the engine of the boat is shut off. With what velocity(relative to the boat surface) should the person move so that the boat comes to rest.

(A) 30 m/s towards right (B) 20 m/s towards right(C) 30 m/s towards left (D) 20 m/s towards leftSolution :Momentum of the system remains conserved as no external force is acting on the system in horizontaldirection. (50 + 100) 10 = 50 × V + 100 × 0 V = 30 m/s towards right, as boat is at rest. VPboat = 30m/sHence the answer is (A).

4. A ball of mass m approaches a moving wall of infinite mass with a speed ‘v’ along the normal to the wall.The speed of the wall is ‘u’ toward the ball. The speed of the ball after ‘elastic’ collision with wall is :(A) u + v away from the wall (B) 2u + v away from the wall(C) |u – v| away from the wall (D) |v – 2u| away from the wallSolution :The answer is (B).



5. A force of constant magnitude F acts on a particle moving in a plane such that it is perpendicular to thevelocity v(| v | v)

of the body, and the force is always directed towards a fixed point. Then the angleturned by the velocity vector of the particle as it covers a distance S is :(take mass of the particle as m)

(A) 2

FS2mv (B) 2

2FSmv

(C) 2FS

mv(D) 2

FSmv

Solution :

Hence the answer is (D).6. A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from

one end of A of the rod. The rod is set in angular motion about A with constant angular acceleration . Ifthe coefficient of friction between the rod and the bead is µ and gravity is neglected, then the time afterwhich the bead starts slipping is :

(A)

(B)

(C) 1 (D) infinitesimal

Solution :The answer is (A).

7. A non-ideal voltmeter and a non-ideal ammeter are connected as shown in the figure. The reading of thevoltmeter is 20 V and that of the ammeter is 4 A. The value of R is :

(A) 5 (B) less than 5 (C) greater than 5 (D) may be less than 5 Solution :

Hence the answer is (B).8. Two circular rings of identical radii and resistance of 36 each are placed in such a way that they cross

each others centre C1 and C2 as shown in figure. Conducting joints are made at intersection points A andB of the rings. An ideal cell of emf 20 volts is connected across AB. The power delivered by cell is

(A) 80 watt (B) 100 watt (C) 120 watt (D) 200 watt



Solution :

Hence the answer is (B).9. A battery of internal resistor ‘r’ and e.m.f. is connected to a variable external resistance AB. If the

sliding contact is moved from A to B, then terminal potential difference of battery will :

(A) remain constant & is independent of value of external resistance(B) increase continuously(C)decrease continuously(D)first increase and then will decreaseSolution :The answer is (B).

10. A uniform wire of resistance R is stretched uniformly n times & then cut to form five identical wires.These wires are arranged as shown in the figure. The effective resistance between A & B will be :

(A) nR5

(B) 2R

5n(C)

2n R5

(D) 2n R2

Solution :

Hence the answer is (C).11. In a practical wheat stone bridge circuit as shown, when one more resistance of 100 is connected in

parallel with unknown resistance ‘x’, then ratio 1/2 become ‘2’. 1 is balance length. AB is a uniformwire. Then value of ‘x’ must be :



(A) 50 (B) 100 (C) 200 (D) 400 Solution :

Hence the answer is (B).12. Figure above shows a closed Gaussian surface in the shape of a cube of edge length 3.0 m. There exists

an electric field given by E [(2.0x 4.0)i 8.0 j 3.0k]N/C

, where x is in metres, in the region in whichit lies. The net charge in coulombs enclosed by the cube is equal to

(A) –54 0 (B) 6 0 (C) –6 0 (D) 54 0

Solution :

Hence the answer is (D).



13. Circuit for the measurement of resistance by potentiometer is shown. The galvanometer is first connectedat point A and zero deflection is observed at length PJ = 10 cm. In second case it is connect at point C andzero deflection is observed at a length 30 cm from P. Then the unknown resistance X is

(A) 2R (B) R2

(C) R3

(D) 3R

Solution :

Hence the answer is (A).

14. A ring of radius R having a linear charge density moves towards a solid imaginary sphere of radius R2

,

so that the centre of ring passes through the centre of sphere. The axis of the ring is perpendicular to theline joining the centres of the ring and the sphere. The maximum flux through the sphere in this process is:

(A) 0

R (B)

0

R2 (C)

0

R4 (D)

0

R3

Solution :

Hence the answer is (D).15. The given figure gives electric lines of force due to two charges q1 and q2. What are the signs of the two

charges?



(A) Both are negative (B) Both are positive(C) q1 is positive but q2 is negative (D) q1 is negative but q2 is positiveSolution :Since the lines of forces are terminating on the charges both have to be negative.Hence the answer is (A).

16. Figure shows three circular arcs, each of radius R and total charge as indicated. The net elecric potentialat the centre of curvature is :

(A) 0

Q2 R (B)

0

Q4 R (C)

0

2QR (D)

0

QR

Solution :

Hence the answer is (A).17. The figure shows a charge q placed inside a cavity in an uncharged conductor. Now if an external electric

field is switched on :

(A) only induced charge on outer surface will redistribute.(B) only induced charge on inner surface will redistribute.(C) both induced charge on outer and inner surface will redistribute.(D) force on charge q placed inside the cavity will change.Solution :The distribution of charge on the outer surface, depends only on the charges outside, and it distributesitself such that the net, electric field inside the outer surface due to the charge on outer surface and all theouter charges is zero. Similarly the distribution of charge on the inner surface, depends only on thecharges inside the inner surface, and it distributes itself such that the net, electric field outside the innersurface due to the charge on inner surface and all the inner charges is zero.Also the force on charge inside the cavity is due to the charge on the inner surface.Hence the answer is (A).

18. A diatomic ideal gas undergoes a thermodynamic change according to the P–V diagram shown in thefigure. The total heat given of the gas is nearly :



(A) 2.5 P0V0 (B) 1.4 P0V0 (C) 3.9 P0V0 (D) 1.1 P0V0

Solution :

Hence the answer is (C).19. Logarithms of readings of pressure and volume for an ideal gas were plotted on a graph as shown in

Figure. By measurring the gradient, it can be shown that the gas may be

(A) monoatomic and undergoing an adiabatic change(B) monoatomic and undergoing an isothermal change(C) diatomic and undergoing an adiabatic change(D) triatomic and undergoing an isothermal change.Solution :

Hence the answer is (C).20. A particle moves along the parabolic path y = ax2 in such a way that the x component of the velocity remains

constant, say c. The acceleration of the particle is

(A) âck (B) 2ˆ2ac j (C) 2 âc k (D) 2 â cjSolution :

Hence the answer is (B).21. The rate of disintegration of a radioactive substance falls from 800 decay/min to 100 decay/min in 6

hours. The half-life of the radioactive substance is :(A) 6/7 hour (B) 2 hrs. (C) 3 hrs (D) 1 hr.Solution :

Hence the answer is (B).



22. When a metallic surface is illuminated with monochromatic light of wavelength , the stopping potential is5 V0. When the same surface is illuminated with light of wavelength 3, the stopping potential is V0. Thenthe work function of the metallic surface is :

(A) hc6

(B) hc5

(C) hc4

(D) 2hc4

Solution :


23. A bead of mass m is attached to one end of a spring of natural length R and spring constant ( 3 1)mgK

R

.

The other end of the spring is fixed at point A on a smooth vertical ring of radius R as shown in figure. Thenormal reaction at B just after it is released is :

(A) mg2

(B) 3mg (C) 3 3mg (D) 3 3mg2

Solution :

Hence the answer is (D).24. In which of the following process the number of protons in the nucleus increases

(A) – decay (B) – decay (C) + decay (D) k – captureSolution :




25. In the figure the variation of potential energy of a particle of mass m = 2kg is represented w.r.t. itsxcoordinate. The particle moves under the effect of this conservative force along the x-axis. Which of thefollowing statements is incorrect about the particle :

(A) If it is released at the origin it will move in negative x-axis.(B) If it is released at x = 2 + where 0 then its maximum speed will be 5 m/s and it will performoscillatory motion

(C) If initially x = – 10 and i. ˆu 6i then it will cross x = 10

(D) x = –5 and x = + 5 are unstable equilibrium positions of the particleSolution :

Hence the answer is (D).26. A block of mass 10 kg is released on a fixed wedge inside a cart which is moved with constant velocity 10

m/s towards right. Take initial velocity of block with respect to cart zero. Then work done by normalreaction (with respect to ground )on block in two second will be : (g = 10 m/s2).

(A) zero (B) 960 J (C) 1200 J (D) None of theseSolution :




27. The diagram shows a solenoid and a loop such that the solenoid’s axis lies in the plane of the loop. Boththe solenoid and the loop carry constant currents in the directions as shown in the diagram. If the systemis released from rest, the loop

(A) move towards the solenoid, rotates clockwise(B) move towards the solenoid, rotates anticlockwise(C) move away the solenoid, rotates clockwise(D) move away the solenoid, rotates anticlockwiseSolution :The loop rotates anticlockwise due to magnetic torque and as a result north and south poles will attracteach other.Hence the answer is (B).

28. The magnetic field at the origin due to the current flowing in the wire is

(A) 0I ˆ ˆ– (i k)8 a

(B) 0I ˆ ˆ(i k)2 a

(C) 0I ˆ ˆ(–i k)8 a

(D) 0I ˆ ˆ(i – k)4 a 2

Solution :

Hence the answer is (C).



29. A vertical wall is in North South direction. A current carrying wire is kept in the wall such that to the westof the wall magnetic field is towards South. Then the wire should be(A) vertical and current is upwards (B) vertical and current is downwards(C) Horizontal and current is towards west (D) Horizontal and current is towards east.Solution :The answer is (A).

30. Magnetic field strength at the centre of regular pentagon made of a conducting wire of uniform crosssection area as shown in figure is :

(I) o

05 i 722sin4 a 2

(II) 0

(III) Not zero if current 'i' leaves D point instead of E(IV) Zero even if the current ‘i’ leaves point D instead of point E(A) I & III (B) II & IV (C) I, II & III (D) None of theseSolution :




CHEMISTRY AITS-FT-03Engineering

Dropper Batch

31. For the two moving particles A and B, following values are givenParticle Velocity MassA 0.1 ms–1 m kgB 0.05 ms–1 5m kgRatio of de-Broglie wavelength associated with the particles A and B is(A) 2 : 5 (B) 3 : 4 (C) 6 : 4 (D) 5 : 2

Solution :


32. The wave number of the spectral line in the emission spectrum of H-atom will be equal to 89

times theRydberg's constant if the electron jumps from(A) n = 3 to n = l (B) n = 10 to n = 1 (C) n = 9 to n = 1 (D) n = 2 to n = 1

Solution :

Hence the answer is (A).33. In the redox reaction,

xKMnO4 + NH3 yKNO3 + MnO2 + KOH + H2O(A) x = 4, y = 6 (B) x = 3, y = 8 (C) x = 8, y = 6 (D) x = 8, y = 3

Solution :

Hence the answer is (D).34. The correct decreasing order of oxidation number of oxygen in compounds BaO2, O3, KO2 and OF2 is :

(A) BaO2 > KO2 > O3 > OF2 (B) OF2 > O3 > KO2 > BaO2

(C) KO2 > OF2 > O3 > BaO2 (D) BaO2 > O3 > OF2 > KO2

Solution :Oxidation no.of O and +2, 0, –1/2 and –1 respectivelyHence the answer is (B).

35. The pair of amphoteric hydroxides is(A) LiOHOHAl ,)( 3 (B) 22 )(,)( OHMgOHBe (C) 23 )(,)( OHBeOHB (D) 22 )(,)( OHZnOHBe



Solution :Both Be(OH)2 and Zn(OH)2 are amphoteric in nature.Hence the answer is (D).

36. If 500 mL of gas A at 400 torr and 666.6 mL of B at 600 torr are placed in a 3 litre flask, the pressure ofthe system will be(A) 200 torr (B) 100 torr (C) 550 torr (D) 366 torr

Solution :

Hence the answer is (A).37. Volume of an ideal gas is to be decreased by 10% by increase of pressure by x% under isothermal

condition. Thus, x is

(A) 1009

(B) 9100

(C) 10 (D) 1

10Solution :

Hence the answer is (A).38. An ideal gas is at 300 K and 1 atm. Then volume remains constant if

(A) temperature is doubled and pressure is halved(B) temperature is halved and pressure is halved(C) only temperature is doubled(D) None of these

Solution :




39. Which of the following is least ionic ?(A) AgCl (B) KCl (C) BaCl2 (D) CaCl2

Solution :

Hence the answer is (A).40. The number and type of bonds between carbon atoms in calcium carbide are

(A) Two sigma, one pi (B) Two sigma, two pi(C) One sigma, one pi (D) One sigma, two pi

Solution :

Hence the answer is (D).41. Which of the following is non-existent according to molecular orbital theory ?

(A) 2H (B) 2O (C) He2 (D) 2O

Solution :


42. The molecular formula of diphenyl methane,

How many structural isomers are possible when one of the hydrogen is replaced by chlorine atom(A) 4 (B) 8 (C) 7 (D) 6

Solution :




43. The structures given below are :

(A) Enantiomers (B) Diastereomers (C) Geometrical isomers (D) Homomers

Solution :

Hence the answer is (A).44. Maximum enol content is in :

(A) (B) (C) (D)

Solution :Hence the answer is (B).

45. Which is the correct matched forthe following reactions

(A) Free Radical reaction

(B) Electrophilic add ition

(C) Electrophilic substitution

(D) Nucreophitic substitution

Solution :Hence the answer is (D).

46. Which one of the following anions is NOT stabilised by delocalisation

(A) (B) (C) (D)

Solution :

Hence the answer is (A).47. An ideal gas expands in volume from 1 × 10–3 m3 to 1 × 10–2 m3 at 300 K against a constant pressure of

1 × 105 Nm–2. The work done is(A) – 900 J (B) – 900 kJ (C) 270 kJ (D) + 900 kJ



Solution :

Hence the answer is (A).48. Given that, Ka(HCN) = 4.9 × 10–10, Ka(HNO2) = 4.5 × 10–4

Thus, which of the following reaction is favoured in forward direction ?(A) HNO2(aq ) + CN– (aq ) 2NO (aq) + HCN(aq )

(B) 2NO (aq )+ HCN(aq) HNO2(aq ) + CN–(aq )(C) Both are favoured in forward direction(D) None is favoured in for-ward direction

Solution :

Hence the answer is (A).49. Given,

H3PO4 + H2O 2 4H PO + H3O+, 1apK = x 2 4H PO + H2O 2

4HPO + H3O+,2apK = y

24HPO + H2O 3

4PO + H3O+, 3apK = z

Thus, pH of 0.1 M K2HPO4 solution is

(A) 2.0 (B) x y

2

(C) y z2

(D) x y z2

Solution :

Hence the answer is (C).50. The pKa of a weak acid HA is 4.80 The pKb of a weak base BOH is 4.78. The pH of an aqueous solution

of the corresponding salt BA will be(A) 9.58 (B) 4.79 (C) 7.01 (D) 9.22



Solution :

Hence the answer is (C).51. Which one of the following is most reactive for hydrogenation reaction ?

(A) (B) CH3 – CH = CH2 (C) CH3 – CH = CH – CH3 (D)

Solution :

Hence the answer is (A).52. What is the Product (A) of following reaction

(A) (B) (C) (D)

Solution :

Hence the answer is (A).53. In the give reaction (A) is :

(A) (B)

(C) Both (a) and (b) (D) None of these




54. In the structure of NaCl given below, ratio Na Clr / r is

(A) 0.414 (B) 0.681 (C) 0.745 (D) 0.520

Solution :

Hence the answer is (A).55. There is formation of Prussian blue when Fe3+ reacts with [Fe(CN)6]4–. Two solutions as shown are

separated by a semipermeable membrane AB. Due to osmosis, there is

(A) blue colour formation in side X(B) blue colour formation in side Y(C) blue colour formation in both the sides X and Y(D) no blue colour forrnation

Solution :

Hence the answer is (D).56.

The reagent [X] may be(A) aq.NaOH (B) alc. KOH (C) (C2H5)3N (D) Both (B) and (C)

Solution :




57. What is the major product of the following reaction ?

(A) (B) (C) (D)

Solution :Hence the answer is (D).

58. Product (A) is

(A) (B) (C) (D)


59. Compare rate of reaction with OH– :

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

Solution :Hence the answer is (A).

60. Given, (Scm2 mol–1) for different electrolytes

Thus, of CH3COOH is(A) 516.5 (B) 420.3 (C) 172.9 (D) 390.7

Solution :




61. A survey shows that 63% of the Americans like cheese whereas 76% like apples. If x% of the Americanslike both cheese and apples, then we have(A) x 39 (B) x 63 (C) 39 x 63 (D) None of these

Solution :

Hence the answer is (C).62. Four numbers are in arithmetic progression. The sum of first and last term is 8 and the product of both

middle terms is 15. The least number of the sequenc is(A) 4 (B) 3 (C) 2 (D) 1

Solution :

Hence the answer is (D).63. The sum of 10 terms of the sequence 1, 6, 27, 108, ... is

(A) 280481 (B) 280482 (C) 280484 (D) 280483

Solution :

MATHS AITS-FT-03Engineering

Dropper Batch



Hence the answer is (D).64. Let a, b and c be real numbers, a 0. If is a root of a2x2 + bx + c = 0, is the root of

a2x2 – bx – c = 0 and 0 < < , then the equation a2x2 + 2bx + 2c = 0 has a root that alwayssatisfies

(A) 2

(B)

2

(C) (D)

Solution :

Hence the answer is (D).65. If b > a, then the equation (x–a)(x–b)–1=0 has

(A) both roots in (a, b) (B) both roots in ( , a)

(C) both roots in (b ) (D) one root in ( , a) and the other in (b, )

Solution :




66. If the angles of elevation of the top of a tower from three collinear points A, B and C on a line leading tothe foot of the tower are 30o, 45o and 60o respectively, then the ration AB : BC is

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

Solution :

Hence the answer is (A).67. A line L intersects the three sides BC, CA and AB of a ABC at P, Q and R respectively..

Then, BP CQ AR. .PC QA RB

is equal to

(A) 1 (B) 0 (C) –1 (D) None of the above



Solution :


68. A variable straight line drawn through the point of intersection of lines x y 1a b and

x y 1,b a meets

the coordinate axes in A and B. Then, the locus of the mid-point of AB is(A) xy (a + b) = ab (x + y) (B) 2xy (a – b) = ab (x + y)(C) 2xy (a – b) = ab (x – y) (D) 2xy (a + b) = ab (x + y)

Solution :



Hence the answer is (D).69. Tangents PA and PB drawn to x2 + y2 = 9 from any arbitrary point P on the line x + y = 25. Locus of mid-

point of chord AB is(A) 25 (x2 + y2) = 9 (x + y) (B) 25 (x2 + y2) = 3 (x + y)(C) 5 (x2 + y2) = 3 (x + y) (D) None of these

Solution :

Hence the answer is (A).70. Radius of the largest circle that can be drawn to pass through the point (0, 1), (0, 6) and touching X-axis,

is

(A) 52

(B) 32

(C) 72

(D) 92

Solution :

Hence the answer is (C).71. If RR is given by f(x) = x3 + 3, then f –1(x) is equal to :

(A) 13x 3 (B) 1/3x 3 (C) 1/3(x 3) (D) 1/3x 3

Solution :




72. Find the interval of 'x' satisfying the equation 1 [sin x [sin x [sin x ....n

upto n terms] = [cos x] where

x (0, 2 ). (where [ ] denotes G.I.F.)

(A) 0,2

(B) ,2

(C)

3 , 22

(D) None of these

Solution :

Hence the answer is (A).73. Let tan(x + y) = a and y = tan–1 b, then find the value of x.

(A) 1 a btan

1 ab

(B) 1 a btan

1 ab

(C) 1 a btan

ab

(D) None of these

Solution :




74.

nn n

n

p qlim ,

2

p, q > 0 equals :

(A) 1 (B) pq (C) pq (D) pq2

Solution :

Hence the answer is (B).75. The following functions are continuous on (0, ) :

(A) tan x (B) x

0

1t sin dtt

(C)

xsin x,0 x2

sin ( x), x2 2

(D) None of these

Solution :

f(x) = x

0

1t sin dtt

1f (x) x sinx

which is well defined on (0, ) f(x) being differentiable and continuous on (0, )Hence the answer is (B).

76. If x = t3 + t + 5 and y = sin t then 2

2d ydx

is equal to:

(A) 2

2 3

(3t 1)sin t 6t cos t(3t 1)

(B) 2

2 2(3t 1) sin t 6t cos t

(3t 1)

(C) 2

2 2(3t 1)sin t 6t sin t

(3t 1)

(D) 2

cos t(3t 1)



Solution :

Hence the answer is (A).77. The curve y = x3 + x2 – x has two horizontal tangents. The distance between these two horizontal lines,

is pq (where p,q I and H.C.F. of p and q is unity). Find (p + q) :

(A) 58 (B) 59 (C) 60 (D) 61

Solution :

Hence the answer is (B).78. The range of values of m for which the line

y = mx and the curve 2

xyx 1

encloses a region , is :

(A) (–1, 1) (B) (0, 1) (C) [0, 1] (D) 1,

Solution :



Hence the answer is (B).79. The bottom of the legs of a three legged table are the vertices of an isosceles triangle with sides 5, 5 and

6. The legs are to be traced at the bottom by three wires in the shape of a Y. The minimum length of thewire needed for this purpose, is :(A) 4 3 3 (B) 10 (C) 3 4 3 (D) 1 6 2

Solution :


80. Evaluate 2(2x 41x 91) dx.

(x 1)(x 3)(x 4)

(A) ln (x–1) + 3 ln (x + 3) + C (B) ln (x–1) + 2 ln (x + 3) + 3 ln (x – 4) + C(C) ln (x+1) + 4 ln (x – 4) + C (D) 4 ln (x–1) – 7 ln (x + 3) + 5 ln (x – 4) + C

Solution :



Hence the answer is (D).81. f is an odd function. It is also known that f(x) is continuous for all values of x and is periodic with period 2.

If x

0g(x) f (t) dt, then :

(A) g(x) is odd (B) g(n) = 0, n N (C) g(2n) = 0, n N (D) g(x) is non-periodic

Solution :


82. The absolute value of

/ 2 sin x

0/ 2 cos x

0

(x cos x 1)e dx

(xsin x 1)e dx

is equal to

(A) e (B) e (C) e2

(D) e

Solution :




83. Area bounded by the curves y = loge x and y = (loge x)2 is :(A) e – 2 sq. units (B) 3 – e sq. units (C) e sq. units (D) e – 1 sq. units

Solution :


84. The solution of the equation dy x(2 log x 1)dx (sin y ycos y

is :

(A) 2

2 xysin y x log x c2

(B) 2ycos y x (log x 1) c

(C) 2

2 xycos y x log x c2

(D) y sin y = x2 log x + c

Solution :

Hence the answer is (D).85. The differential equation of the family of curves y = ex (A cos x + B sin x), where A and B are arbitary

constants is :

(A) 2

2d y dy2 2y 0

dxdx (B)

2

2d y 2dy 2y 0

dxdx

(C) 22

2

d y dy y 0dxdx

(D) 2

2d y dy7 2y 0

dxdx



Solution :

Hence the answer is (A).86. If a, b

are the vectors forming consecutive sides of a regular hexagon ABCDEF, then the vectors

representing side CD is :

(A) a b

(B) a b

(C) b a

(D) a b

Solution :


87. If a

and b

are two unit vectors inclined at an angle such that a b 1,

then :

(A) 3

(B) 23

(C) 2

3 3 (D)

23

Solution :

Hence the answer is (D).88. If non-zero vectors a

and b

are equally inclined to vector c

, then c

is :

(A) a b

a ba 3 b a b

(B) b a

a ba b a b

(C) a b

a ba 3 b a 3 b

(D) None of these



Solution :

Hence the answer is (B).89. If from the point P(a, b, c) perpendicular PQ, PR be drawn to yZ and Zx-planes then, the equation to the

plane oQR is :

(A) x y z 0a b c (B)

x y z 0a b c (C)

x y z 0a b c (D) None of these

Solution :Hence the answer is (A).

90. Find the value of 't' for which the lines x 1 y 3 z 5

2 4 7

and x t y 2 z 1

1 3 1

intersect.

(A) 4

17(B)

1017

(C) 1417

(D) None of these

Solution :


SOLUTIONSA long horizontal rod has a bead which can slide along its length, and initially placed at...

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Transcript of SOLUTIONSA long horizontal rod has a bead which can slide along its length, and initially placed at...