1 of 55 www.aieeepage.com AIEEE -2004 Question Paper PHYSICS & CHEMISTRY PART -I 1. Which one of the following represents the correct dimensions of the coefficient of viscosity 1) 2 1T ML− 2) 1 MLT− 3) 1 1T ML − − 4) 2 2T ML − − Sol: 3) dx dv A η F = = 2 MLT F − → 2 L A − 1 LT dV − − dx -L 2. A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetics energy for any displacement x is proportional to 1) 2 x 2) x e 3) x 4) x loge Sol: 1) a x -α kx dr dv V − = = intergraty both sides m 21 ∫ = vx km 21 dr dv V [ ] ∫ ∫ = − 2 2 2 2 x0 ∆KEαx 2 x 2 km x V m 21 dx; -x 3. A ball is released from the top of height h meters. It takes T seconds to reach the ground. What is the position of the ball at 3T second ? 1) 9h meters from the groun 2) 9 7h meters from the ground 3) 9 8h meters from the ground 4) 18 17h meters from the ground Sol: 3) The highly from ground = h -h h 9 T 21 2 g 2 21 gt h T t = O h = 18 gT 2 gT 2 2 − = 18 ] 2 18 [ gT2 − = h 98 gT 98 2 = =2 of 55 www.aieeepage.com 4. If A B B A × = × , then the angle between A and B is : 1) π b) 3π c) 2π d) 4π Sol: 1) B A A B B A × − = × = × B A× AB B . A ∴= are antiparalled to opposite B , A ∴= are π angle apard 5. A projectivle can have the same range R' ' for two angles of projection. If ' T ' 1 and ' T ' 2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to : 1) 2 R1 b) R1 c) R d) 2 R Sol: 3) g u T u θ) -θ(90 sin 2 ; g θ sin 2 T 2 1 = = 2 2 2 1 g θ Cos Sinθ 2u 2 T T × = × g θ .ucos Sinθ 2u g2 × = But range = R = g θ ucos -Sinθ 2u = αR T T g 2R T T 2 1 2 1 ⇒ = ∴ 6. Which of the following statements is FALSE for a particle moving in a cirlce wiht a constant angular speed ? 1) The velocity vector is tangent to the circle 2) The acceleration vector is tangent to the circle 3) The acceleration vector points to the centre of the cirlce 4) The velocity and acceleration vectors are perpendicular to each other Sol: 2) When the particle moves with constant angular spped, it will have only centripetal acceleraties dinated two ands the centre. But the auleration vector can never be tangentical if the circular motin is uniform N a v 7. An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20 m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will be : 1) 20 m 2) 40 m 3) 60 m 4) 80 m Sol: 4) let αX v α 2 ∆ x x ⇒ = 2 α(2V) becomes 4 lines i.e. 80m 4 20 = ×3 of 55 www.aieeepage.com 8. A machine gun fires a bullet of mass 40 g with a velocity 1200 1 ms− . The man holding it can exert a maximum force of 144 N on the gun. Hown many bullets can be fire per second at the most 1) One 2) Four 3) Two 4) Three Sol: 4) (p) dt d n. F = 1200 10 40 144 3 × × × = − n 3 48 144 = = n 9. Two masses kg 5 1 m = and kg 4.8 2 m = tied to a string are hanging over a light frictionless pulley. What is the acceleration of the masses when left free to move ? ) 2 9.8m/s (g = 1 m 2 m 1) 2 m/s 0.2 2) 2 m/s 9.8 3) 2 m/s 5 4) 2 m/s 4.8 Sol: 1) for a m T g m m 1 1 1 = − for a m m -T m 2 2 2 = g T 5T 4.8 g 1 m g 2 m ) m a(m ) m g(m 2 1 2 1 + = − 2 1 2 1 m m m (m g a +− = 2 Sm 0.2 4.8 5 4.8 9.8(5 = + − =4 of 55 www.aieeepage.com 10. A uniform chain of length 2m is kept on a table such that a length of 60 cm hangs freels from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table ? 1) 7.2 J 2) 3.6 J 3) 120 J 4) 1200 J Sol: 2) lot g having part = linear dairty × 0.6 1.4m 0.6m 10 6 . 0 2m 4kg × × change in P.E = work done 3 . 0 10 6 . 0 24 × × × = = 3.6 J 11. A block rests on a rough inclined plane makign an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block ( in kg) is (Take g = 10 m/s²) 1) 2.0 2) 4.0 3) 1.6 4) 2.5 Sol: 1) F = fictional force F = mg sin 30° ° 30 ° 30 mg F ° = 30 cos mg N ° 30 sin mg ° 30 cos mg f 21 10 10 × × = m m = 2 12. A force 2k) j 3 i (5 F + + N is applied over a particle which displaces it from its origin to the point ) j i (2 r − = m. The work done on the particle in joules is 1) -7 2) + 7 3) + 10 4) + 13 Sol: 2) ) j -i ).(2 k 2 j 3 i (5 π , F w + + = = ) 1 ( 3 2 5 − + × = = 10 -3 = 7 joule5 of 55 www.aieeepage.com 13. A body of mass ‘m’, accelerates uniformly from rest to ' v ' 1 in time ' t ' 1 . The intantaneous power delivered to the body as a function of time t' ' is 1) 11 t t mv 2) 21 21 t t v m 3) 1 2 1 t t v m 4) 121 t t mv Sol: 2) 11 1 1 tV a ; at u = + = v Instantauouses power = dp = F.dv at dv .dv tv m 11 = × Insantauoues power as a function of f is .t tv t mv .at t mv 11 11 11 × = = t t mv21 21 × = 14. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that 1) its velocity is constant 2) its acceleration is constant 3) its kinetic energy is constant 4) it moves in a straight line Sol: 3) It takes a circular path with magnitude of velocity remaing constant ∴ let remain constant as it is a scalar quantity 15. A solid spher is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected ? 1) Moment of inertia 2) Angular momentum 3) Angular velocity 4) Rotational kinetic energy Sol: 3) Momnet of intertia angular and Rotationa KE, all depend on radius ecept Angular velocity 16. A ball is thrown from a point with a speed ' v ' 0 at an elevation angle of θ. From the same point and at the same instant a person starts running with a constant speed ' 2 /v ' θ to catch the ball, will the person be able to catch the ball ? If yes, what should be the angle of projection θ ? 1) Yes, 60° 2) Yes, 30° 3) No 4) Yes, 45° Sol: 1) The person has to catch the ball as both are moving as starting from same point & same instant cosθ V 2 V 0 0 = θ cosθ θ V 2 Vθ 21 cosθ =° = 60 θ6 of 55 www.aieeepage.com 17. One solid spher A and another hollow sphere B are of same mas and same outer radii. Their moment of inertial about their diameters are respectively A I and B I such that 1) B A I I = 2) B A I I > 3) B A I I < 4) BA BA dd II = Sol: 3) As A I (solid sphere) = 2 MR 52 B I ( hollow sphere) = 2 MR 32 B A I I < 18. A statellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is 1) gx 2) x -RgR 3) x RgR2 + 4) 2 1 2x R R g + Sol: 4) orbital velocity π GM ) ( = n V x x R π + = R but GM = 2 gR and = π height of satellite from coutre of earth x RgR V 2 0 + = ∴ 19. The time period of an earth satellite in circular orbit is independent of 1) the mass of the satellite 2) radius of its orbit 3) both the mass and radius of the orbit 4) neither the mass of the satellite nor the radius of its orbit Sol: 1) GM π π 2 3 = T M is mass of earth not that of satellite ∴ T does not depend on the mass of the satellite7 of 55 www.aieeepage.com 20. If ‘g’ is the acceleration due to gravity on the earth’s surface, the gain in the potential energy of an object of mass ' 'm raised from the surface of the earth to a height equal to the radius ' 'R of the earth is 1) 2 mgR 2) mgR 21 3) mgR 41 4) mgR Sol: 2) 2R GMm R RGMm PE ; RGm PE 2 1 − = + − = − = 1 2 PE PE ∆PE − = R GMm 2R GMm + − = 2R GMm = but GM = 2 gR mgR 21 2R m gR ∆PE 2 = × = ∴ 21. Suppose the gravitational force varies inversely as the th n power of distance. Then the time period of a planet in circular orbit of radius ' 'R around the sun will be proportional to 1) +21 n R 2) −21 n R 3) n R 4) −22 n R Sol: 1) Gratstatrind force = F = 2 n mRw R GMm = 22 1 n T 4π R GM + = ∴ GM 4π R T 2 1 n 2 + = GM 4π2 is constant 1 n 2 R α T + ∴ + ∴ 2 1 n R Tα 22. A wire fixed at the upper end stretches by length l by applying a force F. The work done in stretching is 1) l F2 2) Fl 3) 2Fl 4) 2 Fl Sol: 4) .d.dx L YA F.dx dh = = x = eytausian 2L . L YA .x.dx L YA W = ∫ = 2 FL 2L .L. L YA =8 of 55 www.aieeepage.com 23. Spherical balls of radius ' 'R are falling in viscous fluid of viscosity ' η ' with a velocity v' ' . The retarding viscous force acting on the spherical ball is 1 ) directly proportional to R' ' but inversely proportional to v' ' 2) directly proportional to both radius R' ' and velocity v' ' 3) inversely proportional to both radius R' ' and velocity v' ' 4) inversely proportional to R' ' but directly propotional to velocity v' ' Sol: 2) Retandiry its cous force accerdiry to Stodes law is RV 6π F 2 = Fα RV ∵ 24. If two soap bubbles of different radii are connected by a tube, 1) air flows from the bigger bubble to the smaller bubble till the sizes become equal 2) air flows from bigger bubble to the smaller bubble tillo the sizes are interchanged 3) air flows from the smaller bubble to the bigger 4) there is no flow of air Sol: 3) r 4T P =r1 Pα ∴ air blows from smaller bubble to the bigger ∴ p of smaller is more then bigger 25. The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob is 0 t in air. Neglecting frictional force of water and given that the density of the bob is 3 kg/m 1000 34 × . What relationship between t and 0 t is true ? 1) 0 t t = 2) 2 /t t 0 = 3) 0 t 2 t = 4) 0 t 4 t = Sol: 3) to gL π 2 ) ρσ -g(1L π 2 gL π 2 1 = = t In water ρ -denisty of = 3 1000kg/m 34 × -σ decisty of water = 3 kg/m 1000 − = × = ∴ 43 1 g L 2π 1000 341000 -1 g L 2π t g 4L π 2 41 g L π 2 = × = gL π 2 2× = 0 2 t t × =9 of 55 www.aieeepage.com 26. A particle at the end of a spring executes S.H.M. with a period 1 t while the corresponding period for another spring is 2 t . If the period of oscillation with the two springs in series is t, then 1) 2 1 t t T + = 2) 22 21 2 t t T + = 3) -1 2 -1 1 -1 t t T + = 4) -2 1 -2 1 -2 t t T + = Sol: 4) 2 1 k1 k1 k1 + = 1 2 21 km . 4π t = 22 2 2 21 2 1 t1 m. 4π k ; t1 m. 4π k = = 1 k m m 2 k m 2 k 1 k 1 1 km π 2 t = 2 2 um π 2 t = 2 1 2 1k k ) k m(k 2π T + = 2 1 k1 k1 k1 + = 22 2 21 2 2 2 t1 m. 4π t1 m. 4π T1 m. 4π + = 2 2 2 1 2 t t T − − − + = ⇒ 27. The total energy of a particle executing simple harmonic motion is 1) x ∝ 2) 2 x ∝ 3) independent of x 4) 1/2 x ∝ Sol: 3) Total energy during SHM is constant and is grieus by, 2 2A mCo 21 E = Where → A Amptitide ∴ E does not depend on x 28. The displacement y of a particle in a medium can be expressed as where t is in second and x in meter. The speed of the wave is 1) 2000 m/s 2) 5 m/s 3) 20 m/s 4) m/s π 5 Sol: 2) Comparing is the standred of eq for displacement φ) + + = Kx Asin(cot y w = 100 = π 2 100 η : η π 2 = λ 2π 20 k = = ; 20π 2 λ = wave speed = V = 5m/s 20 Lπ 2π 100 nλ = × =10 of 55 www.aieeepage.com 29. A particle of mass m is attached to a spring (of spring constant) and has a natural angular frequency . ω0 An external force F(t) proportional to cos ) ω t(ω ω 0 ≠ is applied to the oscillator. The time displacement of the oscillator will be proportional to 1) ) ω (ω m 2 20 − 2) ) ω m(ω 1 2 20 − 3) ) ω m(ω 1 2 20 + 4) ) ω (ω m 2 20 + Sol: 3) mk ωo = 2o mω k = F = cos t ω 30. In forced oscillation of a particle the amplitude is maximum for a frequency 1 ω of the force, while the energy is maximum for a frequency 2 ω of the force; then 1) 2 1 ω ω = 2) 2 1 ω ω > 3) 2 1 ω ω < when damping is small and 2 1 ω ω > when damping is large 4) 2 1 ω ω < Sol: 1) Both amplitude & energy get maximised when the frequency is equal to natural frequency i.e. the condition of resonance. 31. One mole of ideal monoatomic gas 5/3) γ ( = is mixed with one mole of diatomic gas 7/5) γ ( = . What is γ for the mixture ? γ denotes the ratio of specific heat at constant pressure, to that at constant volume. 1) 3/2 2) 23/15 3) 35/23 4) 4/3 Sol: 1) 1 -γ n n 1 γ n 1 γ n 2 1 2 2 1 1 + = − + − 1 -γ2 1 57 1 1 351 = − + − ∴ 23 γ = ∴ 32. If the temperature of the sum were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be 1) 4 2) 16 3) 35 4) 64 Sol: 4) ; σ.e.A.T E 4 = .T σ.e.4A.2 E' 4 = 2 R 4π A'= 2 2R) ( 4π = 2T T' A 4 = = × = = 64 E. 33. Which of the following statements is correct for any thermodynamic system ? 1) The internal energy changes in all processes 2) Internal energy and entropy are state functions 3) The change in entropy can never be zero 4) The work done in an adiabatic process is always zero Sol: 2) Internal energy and entropy are state functions.11 of 55 www.aieeepage.com 34. Two thermally insulated vessels 1 and 2 are filled with air at temperatures ), T , (T 2 1 volume ) V , (V 2 1 and pressure ) P , (P 2 1 respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be 1) 2 1 T T + 2) 2 /) T (T 2 1 + 3) ) T V P T V )/(P V P V (P T T 1 2 2 2 1 1 2 2 1 1 2 1 + + 4) ) T V P T V )/(P V P V (P T T 2 2 2 1 1 1 2 2 1 1 2 1 + + Sol: 3) 1 1 1 1 RT n V P = 1 1 1 1 RT n V P = After solving them n = 2 1 n n + 2 2 2 11 1 TV P TV P T PV + = 1 2 2 2 1 1 2 1 2 1 1 2 2 2 1 1 T V P T V P ) T PV(T T T T T V P T V P T PV + = = + = 1 2 2 2 1 1 2 1 2 2 1 1 T V P T V P T )T V P V (P + + = 35. A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is 1) E/c 2) 2E/c 3) Ec 4) 2 E/c Sol: 2) CE P = i CE Pf − = (for perfectly reflecting surface) CE P P P i f 2 − = − = ∆ CE P Psurface 2 = ∆ − = ∆ 36. The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and Ax, respectively, are 2 T and ) T (T T 1 2 1 > . The rate of heat transfer through the slab, in a steady state is x )K T A(T 1 2 − f, with f equal to K 2K 1 T x 4x 2 T 1) 1 2) 21 3) 32 4) 3112 of 55 www.aieeepage.com Sol: 4) Equivalent coefficient of thermal conductivity 2K 4x Kx 4x x KL KL L L K 22 11 2 1 ++ = ++ = = 3 5K 6x2K 5x 4K4x 2x5x = × = + = 3 5K KQ = 5x )A T (T K t Q H 1 2 Q − = = 31 . x )A T K(T 5x )A T (T 3 5K 1 2 1 2 − = − 31 f = ∴ 37. A light ray is incident perpendicular to one face of a 90° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45°, we conclude that the refractive index ? ° 45 ° 45 1) 2 1 n < 2) 2 n > 3) 2 1 n > 4) 2 n < Sol: 2) C Sin sinθ > sinC 1 sinθ 1 < = sinC 1 µ µ sinθ 1 < 2 µ sin45 1 µ sinθ 1 µ > > ⇒ > 38. A plano convex lens of refractive index 1.5 and radius of curvature 30 cm, is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object 1) 20 cm 2) 30 cm 3) 60 cm 4) 80 cm13 of 55 www.aieeepage.com Sol: 1) cm 10 1.5 2 30 2µ R F − = × − = = x Q C l It should be placed at 2F i.e., 20 cm 39. The angle of incidence at which reflected light in totally polarized for reflection from air to glass (refracting index n), is 1) (n) sin 1 − 2) (1/n) sin 1 − 3) (1/n) tan 1 − 4) (n) tan 1 − Sol: 4) n θ tan p = n -1 p tan θ = = 40. The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in young’s double-slit experiement is 1) infinite 2) five 3) three 4) zero Sol: 1) sinθ d nλ = 2λ 2d ; sinθ d. 2 2nλ = = given sinθ . 2λ 2nλ = n θ sin = (n = 0, 1, 2, ....∝ ) 41. An electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into a dielectric medium with permitivity . 0 . 4 ∈= Then 1) wave length is double and the frequency remains unchaged 2) wave length is double and frequency becomes half 3) wave length is halved and frequency remains unchaged 4) wave length and frequency both remain unchaged Sol: 3) 21 ε 1 C → ∝ C as 4 ε → nλ C = ∴λ becomes 2 1 n frequency remains unchanged 42. Two spherical conductors B and C having equal radil and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is 1) F/4 2) 3F/4 3) F/8 4) 3 F/814 of 55 www.aieeepage.com Sol: 4) When third spherical conductor in contact wih B change on both will be 2Q Now third conductor of charge 2Q is kept in contact with C. Therefore charge C will 4 3Q Now force between B and C 22 1 r Q K f F = = = Before contact 22 2 r Q K 83 F = after contact F 83 F2 = 43. A charged particle q' ' is shot towards another charged particle , Q' ' which is fixed, with a speed v'. ' It approaches Q' ' upto a closest distance r and them returns. If q were given a speed of , '2v' the closest distances of approach would be q Q v r 1) r 2) 2 r 3) r/2 4) r/4 Sol: 4) .....(1) EQ K. q 4π1 r mv 21 r qQ 4π1 0 2 0 ε = ⇒ = ε If v = 2v Hence, 41 EQ Kq 4π1 r 0 1 × ε = .........(2) From (1) and (2) 4r r1 = 44. Four charges equal to -Q are placed at the four corners of a square and a charge q is at its centre. If the system is in equilibrium the value of q is 1) ) 2 2 (1 4Q + − 2) ) 2 2 (1 4Q + 3) ) 2 2 (1 2Q + − 4) ) 2 2 (1 2Q + Sol: 2) OB DB 2BC 2AB F F F F = + + -Q q D B A AB F 2BC 2 AB F F + -Q-Q -Q O DB F qQ K ) 2 (KQ KQ 2 2 2 2 2 = + or 3qQ 21 2 2Q = + q 1 2 2 41 = + ∴15 of 55 www.aieeepage.com 45. Alternating currect can not be measured by D.C. ammeter because 1) 1 A.C. can not pass thorugh D.C. Ammeter 2) A.C. changes direction 3) Averaga value of current for complete cycle is zero 4) D.C Ammeter will get damaged Sol: 2) 46. The total current supplied to the circuit by the battery is Ω 2 Ω 3 Ω 5 . 1 Ω 6 6V 1) 1 A 2) 2 A 3) 4 A 4) 6 A Sol: 3) Equivalent Resistance of the circuit 3 6 2 6 2 23 3 6 2 6 2 23 Req + +× + + +× + = 1.5 3 6 2 23 69 3 23 23 3 23 23 Req = = + ++ = 4Amp 3/2 6 RV I = = = ∴ 47. The resistance of the series combination of two resistances is S. When they are joined in parallel the total resistance is P. If S = n P then the Minimum possible value of n is 1) 4 2) 3 3) 2 4) 1 Sol: 1) 48. An electric current is passed through a circuit containint two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of 34 and , 32 then the ratio of the currents passing thorugh the wires will be 1) 3 2) 1/3 3) 8/9 4) 2 Sol: 2) 2 1 2 2 r r Ir I + = and 2 1 1 2 r r 1r I + = 13 49 34 AA 11 rr 12 21 21 = × = × = 3 /1 rr 11 12 21 = = ∴16 of 55 www.aieeepage.com 49. In a meter bridge experiment null points is obtained at 20 cm. from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4 X against Y ? 1) 50 cm 2) 80 cm 3) 40 cm 4) 70 cm Sol: 1) 50. The thermistors are usually made of 1) metals iwth low temperature coefficient of resistivity 2) metals with high temperature coefficient of resistivity 3) metal oxides with high temperature coefficient of resistivity 4) semiconducting material having low temperature coefficient of resistivity Sol: 4) Thermistors are usually made of semiconducting material having low temperature coefficient of resistiveity 51. Time taken by a 836 W heater to heat one litre of water from 10°C to 40°C is 1) 50 s 2) 100 s 3) 150 s 4) 200 s Sol: 3) 1 litre of water = 1000 gm T m.s. q ∆ = ∴= (1000) (30) C) (cal/gm0 × q = 30, 000 calorie ⇒ 1 calorie = 4.2 joule ⇒ 4.2 30,000 q × = = 126000 joule Power of the heater = 8.36 W i.e. taken = sec 150 836 126000 = 52. The thermo emf of a thermocouple varieis with the temperature θ of the hot junction as 2 bθ aθ E + = involts where the ratio b a is 700°C. If the cold junction is kept at 0°C, then the neutral temperature is 1) 700°C 2) 350°C 3) 1400°C 4) NO neutral temperature is possible for this thermocouple Sol: 2) 2 bθ aθ E + = For neutral temperature 0 dθ dE = 2bθ a + ⇒ C 350 700 a1 ba 21 2b -a θ ° = = − = = ⇒ 53. The electrochemical equivalent of a metal is -7 10 3.3× kg per Coulomb. The mass of the metal liberated at the cathode when a 3A current is passed for 2 seconds will be 1) kg 10 19.8 7 − × 2) kg 10 9.9 7 − × 3) kg 10 6.6 7 − × 4) kg 10 1.1 7 − × Sol: 1) m = z.q 2 3 10 3.3 z.L.t 7 × × × = = − kg 10 19.8 7 − × =17 of 55 www.aieeepage.com 54. A current i ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is 1) infinite 2) zero 3) ri 2 . 4π µ0 Tesla 4) ri 2 Tesla Sol: 2) Magnetic field inside the tube = zero 55. A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is B. It is then bent into a circular loop of n turns. The magnetic field at the centre of the coil will be : 1) n B 2) B n2 3) 2n B 4) B n2 2 Sol: 2) 2r ron.i B' , 2R roi B = = nr R n.2π. 2Ar = ⇒ = ∴ 2 0 n 2R µoi n 2Rni µ nR 2 µoni B' = = = B n B' 2 = 56. The magnetic field due to a current carrying circular loop of radius 3 cm at an point on the axis at a distance of 4 cm from the centre is µT. 54 What will be its value at the centre of the loop ? 1) µT 250 2) µT 150 3) µT 125 4) µT 75 Sol: 1) 57. Two long conductors, separated by a distance d carry current 1 I and 2 I in the same direction. The exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3d. The new value of the force between them is 1) -2 F 2) 3F 3) 3 2F − 4) 3F − Sol: 3) 1 2 1 o d 2π .d1 1 1 µ F = 1 2 1 o d 2π .d1 1 1 µ F = given 2 1q 2l l = 2 2l d'= 3d d'= 3d 2π )dl (2l l µ F' 2 1 0 = ; 23 F' F = 3 2F F'= Since the current is reversed 32F F' − =18 of 55 www.aieeepage.com 58. The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is 2 s. The magnet is cut along its length into three equal parts and these parts are then placed on each other with their like poles together. The time period of this combination will be 1) 2 s 2) s 32 3) s 3 2 4) s 3 2 Sol: 1) As magnetic moment remains some 59. The materials suitable for making electromagnets should have 1) high retentivity and high coercivity 2) low retentivity and low coercivity 3) high retentiviy and low coercivity 4) low retenivity and high coercivity Sol: 3) Material should have low reactivity and low coercivity 60. In a LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50 V. The voltage across the LC combination will be 1) 50 V 2) V 2 50 3) 100 V 4) 0 V (zero) Sol: 4) There will be phase angle of 180° between L and C circuit ∴and C circuit ∴ voltage across to LC combination will be zero 61. A coil having n turns and resistance Ω R is connected with a galvanometer of resistance Ω 4R . This combination is moving in time t seconds from a magnetic field 1 W weber to 2 W weber. The induced current in the circuit is 1) Rnt 5 ) W (W 1 2 − − 2) Rt 5 ) W n(W 1 2 − − 3) Rnt ) W (W 1 2 − − 4) Rt ) W n(W 1 2 − − Sol: 2) dt d , R1 I φ − = −− = 1 2 1 2 t t W W n R1 I t ) W n(W 4R) (R 1 I 1 2 − +− = 5Rt ) W n(W I 1 2 − = 62. In a uniform magnetic field of induction B a wire in the form of a semicircle of radius r rotates about the diameter of the cirlce with an angular frequency ω. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is R the mean power generated per pediod of rotation is 1) R 2 ω r Bπ 2 2) R 8 ω) r (Bπ 2 2 3) R 2 ) rω (Bπ 2 4) R 8 ) rω (Bπ 2 Sol: 2) 8Rω) r (Bπ 2 2 63. In a LCR circuit capacitance is changed from C to 2 C. For the resonant frequency to remain unchanged, the inductance should be changed from L to 1) 4 L 2) 2 L 3) L/2 4) L/4 Sol: 3) LC π 2 1 W0 =19 of 55 www.aieeepage.com 64. A metal conductor of length 1 m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth’s magnetic field is , 10 2 . 0 4T − × then the e.m.f developed between the two ends of the conductor is 1) V µ 5 2) V µ 50 3) mV 5 4) 50 mV Sol: 2) Induced e.m.f E = 2l Bω 2V 50µ 2 1 5 10 0.2 2 4 = × × × − 65. According to Einstein’s photoelectric equation the plot of the kinetic energy of the emitted photo electrons from a metal vs the frequency, of the incident ratiation gives a straght line whose slope 1) depends on the naturue of the metal used 2) depends on the intensity oif the radiation 3) depends both on the intensity of the radiation and the metal used 4) is the same for all metals and independent of the intensity of the radiation Sol: 4) Einstein’s Photoelectric equation is φ − = hv E max K The slope of the line is plank’s constant 66. The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approiximately. 1) 540 nm 2) 400 nm 3) 310 nm 4) 220 nm Sol: 3) 4eV 0 = φ Threshould wavelength 4eV 1242evnm hc λ 0 φ = 310 nm 67. A charged oil drop is suspended in a uniform field of 4 10 3× v/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 15 10 9 . 9 − × kg and ) m/s 10 2 = g 1) C 18 10 3 . 3 − × 2) C 18 10 2 . 3 − × 3) C 18 10 6 . 1 − × 4) C 18 10 8 . 4 − × Sol: 1) Sincne ball is hanging force by gravity is balanced by electric force E g m q mg qE × = ⇒ = 4 -15 10 3 10 10 9.9 q × × × = ⇒ C 10 3.3 q 18 − × =20 of 55 www.aieeepage.com 68. A nucleus disintegrates into two nuclear parts which have their velocities in the ratio of 2 : 1. The ratio of their nuclear sizes will be 1) 1 : 2 3 1 2) 3 1 2 : 1 3) 1 : 3 2 1 4) 2 1 3 : 1 Sol: 2) 2 2 1 1 v m V m = 21 vv mm 12 21 = = 21 πr 34 ρ πr 34 ρ 3 23 1 = ×× 21 rr3 23 1 = 3 /1 21 21 rr = 69. The binding energy per nucleon of deuteron ( ) H 21 and helium nucleus ( ) He 42 is 1.1 MeV and 7 MeV respectively. If two deuteron nuclei react to form a single hellium nucleus, then the energy released is 1) 13.9 MeV 2) 26.9 MeV 3) 23.6 MeV 4) 19.2 MeV Sol: 3) BEN ∴ for H 21 is 1.1 ∴ B. E = 2.2 MeV and BEN for He 42 is 7 BE = 28 MeV Energy released = Total B.E of product -Total B.E of reactants 2.2 2 -28 × = 4.4 -28 = 6 . 23 = MeV 70. An -α particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of 1) 1 Å 2) cm 10-10 3) cm 10-12 4) cm 10-15 Sol: 3) Energy = r e K.ze× r e K.ze eV 10 5 6 × = × ⇒ m 10 VK 10 5 V 10 1.6 2 K 10 5 V Kze r 14 6 19 6 − − = × × × × × = × × = ⇒ cm 12 10 VK − = 71. When npn transistor is used as an amplifier 1) electrons move from base to collector 2) holes move from emitter to base 3) electrons move from collector to base 4) holes move from base to emitter Sol: 3) When npn transistor is used, majority charge carrier electron of n type emitter move from emitter to base and than base to collector.21 of 55 www.aieeepage.com 72. For a transistor amplifier in common emitter configuration for loas impedence of Ω k 1 ) 25 h and 50 ( oe = = fe h the current gain is 1) -5.2 2) -15.7 3) 24.8 4) -48.78 Sol: 4) 73. A piece of copper and another of germanium are cooled from room temperature ot 77 K, the resistance of 1) each of tehm increases 2) each of them decreases 3) copper increases and germanium decreases 4) copper decrases and germanium increases Sol: 4) Resistance of the metal decreased with decrease in temperature and resistance of semiconductor incrased with decrease in temperature. 74. The manifestation of band structure in solids is due to 1) Heisenberg’s uncertainty principle 2) Pauli’s exclusion principle 3) Bohr’s correcpondence principle 4) Boltzmann’s law Sol: 2) Pauli’s exclusion principle 75. When p-n junction diode is forward biased, then : 1) the depletion region is reduced and barrier height is increased 2) the depletion region is widened and barrier height is reduced 3) both the depletion region and barrier height are reduced 4) both the depletion region and barrier height are increased Sol: 3) P-n junction diode is forward biased then both deplation region and barrier height are reduced 76. Which of the following sets of quantum numbers is correct for an electron in 4 t orbital ? 1) n = 4, I = 3, m = + 4, s = + 21 2) n = 4, l = 4, m = -4, s = 21 − 3) n = 4, I = 3, m = + 1, s = + 21 4) n = 4, l = 2, m = -2, s = 21 + Sol: 3) Fro l = 3, value of m is -l to +l including zero So, m = +1 possible but m = + 4 (option 1) not possible 77. Consider the ground states of Cr atom (Z = 24). The numbers of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively : 1) 12 and 4 2) 12 and 5 3) 16 and 4 4) 16 and 5 Sol: 2) Cr = 1 5,4s 3d Full configuration = 1 5 2 6 2 9 2 2 4 , 3 , 4 , 3 , 2 , 2 , 2 , 1s s d s p s p s 12 6 6 1 = + = = = p l 5 2 = = = d l 78. Which one of the following lons has the highest value of ionic radius ? 1) + Li 2) + 3 B 3) − 2 O 4) − F Sol: 3)22 of 55 www.aieeepage.com 79. The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = ) 10 097 . 1 1 7 − × m 1) 91 nm 2) 192 nm 3) 406 nm 4) nm 10 9.1 8 − × Sol: 1) 22 11 n1 n1 R λ1 − = 7 10 097 . 1 λ1 × = nm 91 λ = 80. The correct order of bond angles (smallest first) in 3 3 2 , , BF NH S H and 4 SiH is 1) 3 3 4 2 BF NH SiH S H < < < 2) 3 3 2 3 BF SiH S H NH < < < 3) 3 4 3 2 BF SiH NH S H < < < 4) 4 3 3 2 SiH BF NH S H < < < Sol: 3) 3 4 3 2 BF SiH NH S H < < < 92.5 < 107 < 109° < 120 Bond angles in the aboves case depend on lp lp − repoulsion in first two compounds and bp-bp repulsion and their no. of later two compounds. 81. Which one of the following sets of ions represents the collection of isoelectronic species ? 1) − + + + Cl Sc Ca K , , , 3 2 2) − + + + F Sc Ca Na , , , 3 2 3) + + − + 3 2 , , , Sc Mg Cl K 4) 4 3 3 2 SiH BF NH S H < < < Sol: 1) − + + + Cl , Sc , Ca , K 3 2 represent collection of isoelectronic species. All have − e 18 and 19, 20, 21 and 17 no. of protons respectively. 82. Among 3 2 2 3 2 , , O P SiO O Al and 2 SO the correct order of acid strength is : 1) 3 2 2 3 2 2 O Al SiO O P SO < < < 2) 3 2 3 2 2 2 O P O Al SO SiO < < < 3) 3 2 2 2 3 2 O P SO SiO O Al < < < 4) 2 3 2 2 3 2 SO O P SiO O Al < < < Sol: 4) 2 3 2 2 3 2 SO O P SiO O Al < < < As we move to right in a period, acidic character of oxides increases, 3 2O Al is amphoteric oxide while others are acidic with increasing nature towards group 14 and 16. 83. The bond order in NO is 2.5 while that in + NO is 3. Which of the following statements is true for these two species ? 1) Bond length in + NO is greater than in NO 2) Bond length in NO is greater than in + NO 3) Bond length in + NO is equal to that in NO 4) Bond length is unpredictable Sol: 2) Bond length in No(2.5) is greater than in ) 0 . 3 ( NO+ because bond length is inversely related to bond order.23 of 55 www.aieeepage.com 84. The formation of the oxide ion − 2 ) (g O requires first an exothermic and then an endotherms step as shown below : 1 (g) (g) 142kJmol ∆H O e O − − − − = ° = + ; 1 2(g) (g) -kJmol 844 ∆H O e O − − − = ° = + This is because 1) Oxygen is more electronegative 2) Oxygen has high electron affinity 3) − O ion will tend to resist the addition of another electron 4) − O ion has comparatively larger size than oxygen atom Sol: 3)− O will tend to resist the addition of another electron because of electrostatic repulsion. 85. The states of hybridization of boron and oxygen atoms in boric acid ) ( 3 3BO H are respectively : 1) 2 2 sp and sp 2) 3 2 sp and sp 3) 2 3 sp and sp 4) 3 3 sp and sp Sol: 2) Boron of group 13 with at no. 5 has electronic configuration 1 1 1 2 2py , 2px , 2s , 1s in excited state and shows 2 sp hybridization while oxygen in covalent stage 4 2 2 2p , 2s , 1s shows 3 sp hybridization. 86. Which one of the following has the regular tetrahedral structure ? 1) − 4 XeF 2) 4 SF 3) − 4 BF 4) − 2 4 ] ) ( [ CN Ni Sol: 3) -4 BF has regular tetrahedral structure as it doesn’t contain any lone pair of electron 3 2 4 3 sp sp BF F BF − − → + * In 2 XeF4 − lone pairs are present hence distored geometry. * In ) 4s , (3d Ni ] [Ni(CN) 8 2 2 4 ° = + − After pairing of electrons shows sq. dsp2 planer geometry (Atomic nos. : B = 5, S = 16, NI = 28, Xe = 54) 87. Of the following outer electronic configurations of atoms, the highehst oxidtion state is achieved by which one of them ? 1) 2 8 ) 1 ( ns d n − 2) 1 5 ) 1 ( ns d n − 3) 2 3 ) 1 ( ns d n − 4) 2 5 ) 1 ( ns d n − Sol: 4) Highest oxidation state is shown by involvement of maximum no. of electrons of 2 5 ns & d 1) -(n state. The electrons which are unpaired take part in bond formation. In , ns 1)d -(n 2 8 only 2e of (n-1) d participate along with 2e of ns. 88. As the temperature is raised from 20° C to 40° C, the average kinetic energy of neon atoms changes by a factor of which of the following ? 1) 2 1 2) ) 293 /313 ( 3) 293 313 4) 2 Sol: 3) 1 2 21 KE 293 313 KE 313 R 23 293 R 23 KE KE = ⇒ ×× = 89. The maximum number of 90° angles between bond pair-bond pair of electrons is observed in : 1) 3 dsp hybridization 2) d sp3 hybridization 3) 2 dsp hybridization 4) 2 3d sp hybridization Sol: 4) , d sp 2 3 six 90° angles in octahedral structure observed.24 of 55 www.aieeepage.com 90. Which of the following aqueous solutions will exhibit highest boiling point ? 1) 0.01 M 4 2SO Na 2) 0.01 M 3 KNO 3) 0.015 M urea 4) 0.015 M glucose Sol: 1) 0.01 M . SO Na 4 2 It furnishes 3 ions ( ) − + + 24 1SO Na 2 in aqueous solution and shown maximum elevation in boiling point. 91. Which among the following factors is the most important in making fluorine the strongest oxidizing halogen ? 1) Electron affinity 2) ionization enthalpy 3) Hydration enthalpy 4) Bond dissociation energy Sol: 4) Bond dissociation energy of 2 F is least ( 151 kJ/mole) & therefore most important factor in making fluorine the strongest oxidizing halogen. 92. In van der waals equation of state of the gas law, the constant ‘b’ is a measure of : 1) intermolecular repulsions 2) intermolecular attraction 3) volume occupied by the molecules 4) intermolecular collisions per unit volume Sol: 3) ‘b’ the vander walls constnat is a measure of volume occupied by the molecules also called co-volume or excluded volume 4N πr 34 b 3 × = It is four times the actual volume of gas molecules its unit . mol m 1 3 − 93. The conjugate base of −4 2 H PO is 1) − 34 PO 2) 5 2O P 3) 4 3PO H 4) − 24 HPO Sol: 4) The conjugate base of −42PO H is : HPO24 − base conj. Acid HPO H PO H 24 -4 2 − + + → 94. 20 10 02 . 6 × molecules of urea are present in 100 ml of its solution. The concentration of urea solution is 1) 0.001 M 2) 0.01 M 3) 0.02 M 4) 0.1 M (Avogadro constant, ) mol 10 6.02 N 1 23 A − × = Sol: 2) 0.01M -6.02 21 10 × molecules/litre urea sol. 23 10 02 . 6 × ∵ molecules ≡ 1 M soluiton 21 10 02 . 6 × ∴ molecules ≡ 0.1 M solution 95. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid ), ( 3 3PO H the volume of 0.1 M aqueous KOH solution required is : 1) 10 mL 2) 20 mL 3) 40 mL 4) 60 mL Sol: 3) 40 ml ( 3 3PO H is dibasic acid) Base Acid V N V N 2 2 1 1 = 2 V 0.1 20 0.1 2 × = × × 40 ml = 2 V25 of 55 www.aieeepage.com 96. For which of the following parameters the structural isomers OH H C 5 2 and 3 3OCH CH would be expected to have the same values ? (Assume ideal behavious ) 1) Heat of vaporization 2) Vapour pressure at the same temperature 3) Boiling points 4) Gaseous densities at the same temperature and pressure Sol: 4) , Hvap ∆ B. point & V.P. of OH H C 5 2 and 3 3OCH CH would be different. 97. Which of the following liquid pairs shows a positive deviation from Raoult’s law ? 1) Water -hydrochloric acid 2) Benzene -methanol 3) Water -nitric acid 4) Acetone -chloroform Sol: 2) -OH CH H C 3 6 6 − shows +ve deviation all others are example of negative deviation show strong solvent -solvent interaction compared to solute -solute interaction. 98. Which of the following statements is FALSE ? 1) Raoult’s law states that the vapour pressure of a component over a solution 2) The osmotic pressure ) (π of a solution 3) The correct order of osmotic pressure for 0.01 M aqueous solution of each compound is > > > COOH CH KCl BaCl 3 2 sucrose 4) Two sucrose solutions of same molality prepared in different solvents will have the same freezing point depressiion Sol: 4) 99. What type of crystal defect is indicated in the diagram below ? − + − + − + Cl Na Cl Na Cl Na − Cl − Cl + Na + Na + Na − Cl − + − Cl Na Cl + − + − Na Cl Na Cl + Na 1) Frenkel defect 2) Schottky defect 3) Interstitial defect 4) Frenkel and Schottky defects Sol: 2) Schottky defect -equal no of cations and anion are missing from lattice points. 100. An ideal gas expands in volume from 3 3 10 1 m − × to 3 m 2 10 1 − × at 300 K against a constant pressure of . 10 1 2 5 − × Nm The work done is : 1) -900 J 2) -900 kJ 3) 270 kJ 4) 900 kJ Sol: 1) By formula w = V P∆ − ) 10 1 10 1 ( 105 1 3 2 − − × − × × × − = = -900 J 101. In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to 1) generate heat 2) create potentiaol difference between the two electrodes 3) produce high purity water 4) remove adsorbed oxygen from electrode surfaces Sol: 2) 102. In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15n minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is 1) 30 minutes 2) 15 minutes 3) 7.5 minutes 4) 60 minutes Sol: 1) Conc. reduced to 1/2 in 15 ml so in 30 minutes it reduces to 1/426 of 55 www.aieeepage.com 103. What is the equilibrium expression for the reaction ) ( 2 ) ( 4 5 g s O P + ) ( 10 4 s O P ? 1) 5 2 4 10 4 ] [ ] /[ ] [ O P O P Kc = 2) ] [ ] [ 5 /] [ 2 4 10 4 O P O P Kc = 3) 5 2 ] [O Kc = 4) 5 2 ] /[ 1 O Kc = Sol: 4) (s) P4 and -(s) O P 10 4 concentrations are taken as unity. Therefore eq.constant expression [ ]5 2 c O1 K = ∴ 104. For the reaction, ) ( 2 ) ( g g Cl CO + ) ( 2 g COCl the c p K K is equal to 1) RT 1 2) RT 3) RT 4) 1.0 Sol: 1) ∆n(g) c p RT K K = r p n n ∆ng − = = -1 (g) Cl (g) [Co 2 + (g)] CoCl2 1 cp RT KK − = = 1/RT 105. The equilibrium constant for the reaction ) ( 2 2(g) N g O + ) ( 2 g NO at temperature T is 4 10 4 − × . The value of c K for the reaction ) (g NO ) ( 2 ) ( 2 21 21 g g O N + at the same temperature is 1) 2 10 5 . 2 × 2) 50 3) 4 10 4 − × 4) 0.02 Sol: 2) The equation reversed & reduced to 1/2, so c 'c K1 K = 106. The rate equation for the reaction 2A + B → C is found to be rate k [A] [B]. The correct statement in relation to this reaction is that the 1) unit of k must be 1 − s . 2) 2 1 t is a constant 3) rate of formation of C is twice the rate of disappearance of A 4) value of k is independen of the initial concentrations of A and B Sol: 4) This is characteristic feature of eq.const K. I is independent of binitial conc. of reactant27 of 55 www.aieeepage.com 107. Consider the following E° values V 77 . 0 2 3 + = + + FE Fe o E ; V 14 . 0 2 − = + Sn Sn o E Under standard conditions the potential for the reaction ) ( ) ( 2 ) ( 2 2 2 3 ) ( aq Sn aq Fe aq Fe Sn s + + + + → + is 1) 1.68 V 2) 1.40 V 3) 0.91 V 4) 0.63 V Sol: 3) Anode Cathode ocel E E E = /Sn Sn /Fe Fe 2 2 3 E E + + + − = = 0.77 -( 0.14) = 0.91 V 108. The molar solubility (in mol ) L 1 − of a sparingly soluble salt 4 MX is ‘s’. The corresponding solubility product is sp K . ‘s’ is given in terms of sp K by the relation : 1) 4 1 ) 128 /( sp K s = 2) 4 1 ) 128 ( sp K s = 3) 5 1 ) 256 ( sp K s = 4) 5 1 ) 256 /( sp K s = Sol: 4) 4 Mx − + + 4X M4 (s) 4 (4s) 4 (4s) (s) Ksp = 5 S 256 Ksp = S KsP/256 5 = 109. The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is ( F = 500 , 96 C ; mol 1 − R = 8.314 1 JK− 1 mol− ) 1) 1 10 0 . 1 × 2) 5 10 0 . 1 × 3) 10 10 0 . 1 × 4) 30 10 0 . 1 × Sol: 3) c ocel logK n 0.059 E = c logK 1 0.0591 0.591 = c logK 0591 . 00.591 = Antilog 10 = c K 110. The enthalples of combustion of carbon and carbon monoxide are -393.5 and -283 kJ 1 mol− respectively. The enthalpy of formation of carbon monoxide per mole is 1) 110.5 kJ 2) 676.5 kJ 3) -676.5 kJ 4) -110.5 kJ Sol: 4) (8); CO (s) O C(s) 2 2 → + 1 c 393.5KJmol ∆H − − = 1 c 2 2 mol 283KJ ∆H (8); CO (s) O 21 O(s) − − = → + ; ) ( 21 ) ( 2 ) ( g s CO g O C → + 1 c mol 110.5KJ ∆H − − =28 of 55 www.aieeepage.com 111. The limiting molar conductivities A° for NaCl, KBr and KCl are 126, 152 and 150 S 1 2 mol cm − respectively. The A° for NaBr is 1) 128 S 1 2 mol cm − 2) 176 S 1 2 mol cm − 3) 278 S 1 2 mol cm − 4) 302 S 1 2 mol cm − Sol: 1) KBr KCl NaCl NaBr Λ Λ Λ Λ + − = 15 150 126 ΛNaBr + − = = 128 S 1 2 mole cm − 112. In a cell that utilises the reaction ) ( 2 2 ) ( ) ( ) ( 2 g s H aq Zn aq H Zn + → + + + addition of 4 2SO H to cathode compartment, will 1) lower the E and shift equilibrium to he left 2) lower the E and shift equilibrium to the right 3) increase the E and shift equilibrium to the right 4) increase the E and shift equilibrium to the left Sol: 3) Addition of 4 2SO H increase + H ion concentration and increase E. The equilibrium shift to right. 113. Which one of the following statements regarding helium is incorrect ? 1) It is used to fill gas balloons instead of inflammable 2) it is used as a cryogenic agent for carrying out experiments at low temperature 3) It is used to produce and sustain powerful superconducting magnets 4) It is used in gas-cooled nuclear reactions Sol: 3) 114. Identify the correct statement regarding enzymes : 1) Enzymes are specific biological catalysts that can normally function at very high temperature ( T -1000 K) 2) Enzymers are normally heterogeneous catalysts that are very specific in their action 3) Enzymes are specific biological catalysts that cannot be poisoned 4) Enzymes are specific biological catalysts that possess well-defined active sites Sol: 4) Enzymes are biocatalyst which are temperature & pH sensitive. They have active site for binding to sunstrate molecules. 115. One mole of magnesium nitride on the reaction with an excess of water gives : 1) one mole of ammonia 2) one mole of nitric acid 3) two moles of ammonia 4) two moles of nitric acid Sol: 3) 3 2 2 3 2NH 3MgO O 3H N Mg + → + 116. Which one of the following ores is best concentrated by froth-flotation method ? 1) Magnetite 2) Cassiterite 3) Galena 4) Malachite Sol: 3) Galena is the only sulphite ore which is concentration by froath floatation 117. Beryllium and aluminium exhibit many properties which are similar. But the two elements differ in : 1) exhibiting maximum covelency in compounds 2) forming polymeric hybrides 3) forming covalent halides 4) exhibiting amphoteric nature in their oxides Sol: 1) Maximum valency of Be is 14 & that of Al is 629 of 55 www.aieeepage.com 118. Aluminium chloride exists as dimer, 6 2Cl Al in solid state as well as in solution of non-polalr solvents such as benzene. When dissolved in water, it gives 1) − + + 3Cl Al3 2) − + + 3Cl ] O) [Al(H 3 6 2 3) 3HCl ] [Al(OH) 3 6 + − 4) HCl O Al 6 3 2 + Sol: 2) 6 2Cl Al is ionic in aqueous solution as exist as a coordination compound [ ] − + + 3Cl O) Al(H 3 6 2 119. The soldiers of Napolean army whiel at Alps during freezing winder suffered a serious problem as regard to the tin buttons of thier uniforms. While metallic tin buttons got converted to grey powder. This transformation is related to 1) an interaction with nitrogen of the air at very low temperature 2) a change in the crystalline structure of tin 3) a change in the partial pressure of oxygen in the air 4) an interaction with water vapour contained in the humid air Sol: 2) At low temperature, crystaline tin changes to amorphous form 120. The + + 2 3 M M o E values for Cr, Mn, Fe and Co are -0.41, + 1.57, + 0.77 and +1.9 7V respectively for which one of these metals the change in oxidation state from + 2 to + 3 is easiest ? 1) Cr 2) Mn 3) Fe 4) Co Sol: 1) More negative the vlaue of reduction potential more in the tendency to get oxidize & act as reducing agent 121. Excess of KI reacts with 4 CuSO solution and then 3 2 2 O S Na solution is added to it. Which of the statements is incorrect for this reaction ? 1) 2 2I Cu is formed 2) 2 CuI is formed 3) 3 2 2 O S Na is oxidised 4) Evolved 2 I is reduced Sol: 2) 4 2 2 2 2 4 SO 2K I I Cu 4KI 2CuSO + + → + 6 4 2 3 2 2 2 O SO Na 2Nal O S 2Na I + → + 122. Among the properties (a) reducing (b) oxidising (c) complexing, the set of properties shown by − CN ion towards metal species is 1) a, b 2) b, c 3) c, a 4) a, b, c Sol: 2) − CN acts as oxidizing & complexation con towards metals. It form NaCN as well as [ ]2 Ag(CN) Na type of compounds. 123. The coordination number of a central metal atom in a complex is determined by 1) the number of ligands around a metal ion bonded by sigma bonds 2) the number of ligands around a metal ion bonded by pi-bonds 3) the number of ligands around a metal ion bonded by sigma and pi-bonds both 4) the number of only anionic ligands bonded to the metal ion Sol: 1) Coordination no.of metal = No of s bonds formed by metal with ligands. 124. Which one of the following complexes is an outer orbital complex ? 1) [ ]− 4 6 ) (CN Fe 2) [ ]− 4 6 ) (CN Mn 3) [ ]+ 3 6 3) (NH Co 4) [ ]+ 2 6 3) (NH Ni Sol: 4)[ ]+ 2 6 3) Ni(NH is 2 3d sp hybridized compled and outer d orbital type.30 of 55 www.aieeepage.com 125. Coordination compounds have great importance in biological systems. In this context which of the following statements is incorrect ? 1) Chlorophylls are green pigments in plants and contain calcium 2) Haemoglobin is the red pigment of blood and contains iron 3) Cyanocobalamin is 12 B and contains cobalt 4) Carboxypeptidase -A is an enzyme and contains zinc Sol: 1) Chlorophylls are magnesium complexes. 126. Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statements about cerium is incorrect ? 1) The common oxidation states of cerium are +3 and + 4 2) The +3 oxidation state of cerium is more stable than the +4 oxidation state 3) The +4 oxidation state of cerium is not known in solutions 4) Cerium (IV) acts as an oxidizing agent Sol: 3) The +4 oxidation state of cerium is now know in solutions. 127. Which one of the following has largest number of isomers ? 1) [ ]2 4 3) ( Cl NH Ru 2) [ ]+ 2 5 3) ( Cl NH Co 3) [ ]+ 2 2 3 ) ( ) ( CO H PR Ir 4) [ ]+ 2 2 ) ( Cl en CO Sol: 4)[ ]+ 2 2Cl Co(en) has 2 geometrical isomerms cis & trans. The its cis-geometrical isomer has 2 optical isomers d & l. The same comp.also exists in a meso form . 128. The correct order of magneitc moments (spin only value in B.M.) among is 1) − − − > > 4 6 2 4 2 4 ] ) ( [ ] [ ] [ CN Fe CoCl MnCl 2) − − − > > 2 4 4 6 2 4 ] [ ] ) ( [ ] [ CoCl CN Fe MnCl 3) − − > > 2 4 2 4 4 6 ] [ ] [ ] ) ( [ CoCl MnCl CN Fe 4) − − − > > 2 4 2 4 4 6 ] [ ] [ ] ) ( [ MnCl CoCl CN Fe Sol: 1) The no.of unpaired electrons are respectively in order − − − > > 4 6 2 4 2 4 ] [Fe(CN) ] [CoCl ] [MnCl 129. Consider the following nuclear reactions : He N M 42 238 92 2 + → ; + + → β 2 L N AB xy The number of neutrons in the element L is : 1) 142 2) 144 3) 140 4) 146 Sol: 2) He 2 N M 3042 82 88 22 92 + → + + → β 2 L N 3 02 86 23 88 No. of neutrons in L = 230 -86 = 144 130. The half -life of a radioisotope is four hours. If the initial mass of the isotope was 200 g, the mass remaining after 24 hours undecayed is : 1) 1.042 g 2) 2.084 g 3) 3.125 g 4) 4.167 g Sol: 3 n t 21 No N = g 125 . 3 21 200 6 = =31 of 55 www.aieeepage.com 131. The compound formed in the positive test for nitrogen with the Lassalgne solution of an organic compound is 1) [ ]3 6 4 ) (CN Fe Fe 2) [ ]6 3 ) (CN Fe Na 3) ( )3 CN Fe 4) [ ] NOS CN Fe Na 5 4 ) ( Sol: 1) 132. The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hrdroxide solution for complete neutralization. The organic compound is 1) acetamide 2) benzamide 3) urea 4) thiourea Sol: 3) 133. Which one of the following has the minimum boiling point ? 1) n-Butane 2) 1-Butane 3) 1-Butene 4) isobutane Sol: 4) In hydrocarbon molecular force exist is weak vander waals force of attraction which depend upon the contact surface area, it decreases with branching. Therefore the minimum b.p. is of Isobutane. 134. The IUPAC name of the compound HO is 1) 3, 3-dimethyl-1-hydroxy cyclohexane 2) 1, 1-dimethyl-3-hydroxy cyclohexane 3) 3, 3-dimethyl-1-cyclohexanol 4) 1, 1-dimethyl-3-cyclohexanol Sol: 3. 135. Which one of the following does not have 2 sp hybridized carbon ? 1) Acetone 2) Acetic acid 3) Acetonitrile 4) Acetamide Sol: 3) No 2 sp hybridized carbon in Acetonitrile 136. Which of the following will have a meso-isomer also ? 1) 2-chlorobutane 2) 2, 3-Dichlorobutane 3) 2, 3-Dichloropentane 4) 2-Hydroxypropanoic acid Sol: 2) 137. Rate of the reaction C -R Nu + O Z C -R ONu Z + is fastest when Z is 1) Cl 2) 2 NH 3) 5 2H OC 4) 3 OCOCH Sol: 1) C -R Nu + O Z C -R ONu Z + The rate is fastest when Z is Cl because of its weak basicity Cl is a better leaving gr. Secondly least stabilization by Resonance due to ineffective overlapping between the 3p orbital of Cl and 2p orbital of carbon Z is Cl. 138. Amongst the following compounds, the optically active alkane having lowest molecular mass is 1) 3 2 2 3 CH CH CH CH − − − 2) 3 2 3 CH CH CH CH − − − 3 CH 3) − − C CH3 H 5 2H C 4) CH C CH CH ≡ − − 2 3 Sol: 3)32 of 55 www.aieeepage.com 139. Consider the acidity of the carhoxylic acids : 1) PhCOOH 2) COOH H C NO o 4 6 2 − 3) COOH H C NO p 4 6 2 − 4) COOH H C NO m 4 6 2 − Sol: 4) The Acidity of carboxylic acid COO COO 2 NO COO 2 NO COO 2 NO (1) (2) (3) (4) E.W.G increase the acidity of benzoic acid, o-isomer will have higher acidity then corresponding and p isomer due to ortho-effect. As -M gr 9 (ie 2 NO ) at p-position have more pronounced electron with drawing effect than As -2 NO gr at m -position (I-effect) \ Correct order of acidity is 2 > 3 > 4 > 1 140. Which of the following is the strongest base ? 1) 2 NH 2) 3 NHCH 3) 2 NH 3 CH 4) 2 2NH CH Sol: 2) 141. Which base is present in RNA but not in DNA ? 1) Uracil 2) Cytosine 3) Guanine 4) Thymine Sol: 1) Uracil base is present in RNA but not in DNA 142. The compound formed on heating chlorobenzene with chloral in presence of concentrated sulphuric acid, is 1) gammexene 2) DDT 3) freon 4) hexachloroethane Sol: 2) 143. On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is 1) NaCl H COOC + 5 2 3 CH 2) OH H C COONa 5 2 3 CH + 3) COONa H C COCl 5 2 3 CH + 4) COONa H C Cl 5 2 3 CH + Sol: 2) No reaction 144. Acetyl bromide reacts with excess of MgI CH3 followed by treatment with a saturated solution of Cl NH4 gives : 1) acetone 2) acetamide 3) 2-methyl-2-propanol 4) acetyl iodide Sol: 3) 145. Which one of the following is reduced with zinc and hydrochloric acid to given the corresponding hydrocarbon ? 1) Ethyl acetate 2) Acetic acid 3) Acetamide 4) Butan-2-one Sol: 4) Butane -2 -one Zn + HCl reduces carbonyl compounds. Whereas Zn | HCl do not reduces esters, acids and amides. 146. Which one of the following undergoes reaction with 50% sodium hydroxide solution to given the corresposnding alcohol and acid ? 1) Phenol 2) Benzaldehyde 3) Butanal 4) Benzoic acid Sol: 2.33 of 55 www.aieeepage.com 147. Among the following compounds which can be dehydrated very easily is 1) OH CH CH CH CH CH 2 2 2 2 3 2) 3 2 2 3 CH CH CH CH CH OH 3) 3 2 2 3 CH CH C CH CH 3 CH OH 4) OH CH CH CH CH CH 2 2 2 3 3 CH Sol: 3) The more stable carbocation is generated thus more easily it will be the dehydrated. CH CH C CH CH CH CH C CH CH 2 2 3 H 3 2 2 3 − − − − → − − − − + 3 CH OH 3 CH The more stable carbocation is generaged thus more easily it will be dehydrated. 148. Which of the following compounds is not chiral ? 1) 1-chloropentane 2) 2-chloropentane 3) 1-chloro-2-methyl pentane 4) 3-chloro-2-methyl pentane Sol: 1) 149. Insulin production and its action in human body are responsible for the level of diabetes. This compound belongs to which of the following categories ? 1) A co-enzyme 2) A hormone 3) An enzyme 4) An antibiotic Sol: 2) 150. The smog is essentially caused by the presence of 1) 2 O and 3 O 2) 2 O and 2 N 3) Oxides of sulphur and nitrogen 4) 3 O and 2 N Sol: 3)34 of 55 www.aieeepage.com 1. Let R = {(1, 3), (4, 2), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is : 1) reflexive 2) transitive 3) transitive 4) a function Sol: 3) ⇒ ≠ −1 R R not symmetric 2. The range of the function f(x) = 3 -x x -7 P is : 1) {1, 2, 3, 4, 5} 2) {1, 2, 3, 4, 5, 6} 3) {1, 2, 3, 4} 4) {1, 2, 3} Sol: 1) Domain = {1, 2, 3, 4, 5} ) 3 7 ( − ≥ − x x ∵ Range = } 3 , 2 , 1 { } 2 , 3 , 4 { 2 1 0 = P P P 3. Let z, w be complex numbers such that 0 w i z = + and are zw = π . The arg z equals : 1) 4π 5 2) 2π 3) 4 3π 4) 4π Sol: 3) w i z = iw z = ⇒arg ⇒ + 2π arg w z arg -π 2π arg + = ⇒ z 4 3π arg = ⇒ z 4. If z = x -iy and , 31 iq p z + = then ) ( 2 2 q p qy px + + is equal to 1) 1 2) -1 3) 2 4) -2 Sol: 4) 3 ) ( iq p z + = ) 3 ( ) 3 ( 2 3 2 3 q p q i pq p − − − but z = x -iy q p q y pq p x 2 3 2 3 3 & 3 − = − = ∴ 2 2 2 2 3 3 p q qy q p px − = − = ⇒ 2 2 2 2 2 q p qy px − − = = or 2 1 2 2 − = + + qy px q p MATHEMATICS PART -II35 of 55 www.aieeepage.com 5. If , 1 1 2 2 + = − z z then z lies on 1) the real axis 2) the imaginary axis 3) a circle 4) an ellipse Sol: 2) Let z = x + iy 1 | | | 1 | 2 2 + = − z z 1 4 ) 1 ( 2 2 2 2 2 2 2 + + = + − − ⇒ y x y x y x 0 0 2 = ⇒ = ⇒ x x 6. Let . 0 0 1 0 1 0 1 0 0 − − = A The only correct statement about the matrix A is 1) A is a zero matrix 2) A = (-1)I, where I is a unit matrix 3) 1 − A does not exist 4) I A = 2 Sol: 4) I A2 = 7. Let 1 1 1 3 -1 2 1 1 -1 A = and 3 2 -1 0 5 2 2 4 10B α − = . If B is the inverse of matrix A, then α is : 1) 5 2) -1 c) 2 4) -2 Sol: 4) 3 2 -1 α 0 5 2 2 4 1 1 1 3 1 2 1 1 1 10 1 − − − = = − A A On comparing, 5 α = 8. If ..... ......, , , a 3 2 1 n a a a are in G.P., then the value of the determinant 8 7 6 5 4 3 2 1 log log log log log log log log log + − + + + + + − n n n n n n n n n a a a a a a a a a is : 1) -2 2) 1 3) 2 4) 0 Sol: 1) since n a a a a ...... 3 2 , 1 are in G.P log 1 a n a a a log .... log , log 3 2 2 are in A.P. ∴ 2 log 2 1 log log + + + = x x x a a a 2 log 5 3 4 log log + + + + = x x x a a a 2 log 8 6 7 log log + + + + = x x x a a a Apply 2 3 1 2 2c c c C − + → then the value of the determinat = 0 9. Let two numbers have arithmetic mean 9 and geometric mean 4. Then these number are the roots of the quadratic equation : 1) 0 16 18 2 = − − x x 2) 0 16 18 2 = + − x x 3) 0 16 18 2 = − + x x 4) 0 16 18 2 = + + x x36 of 55 www.aieeepage.com Sol: 2) 18 = β + α ⇒ = 2 β + α 9 16 = αβ ⇒ = αβ 4 0 αβ β)x α ( 2 = + + − ∴x 0 16 18 2 = + = ⇒ x x 10. If (1 -p) is a root of quadratic equation 0 ) 1 ( 2 = − + + p px x , then its roots are : 1) -1, 2 2) -1, 1 3) 0, -1 4) 0, 1 Sol: 3) Sum of the roots = p o β = + α if p -1 α = 1 -p + -p β = -1 β = ⇒∴ other root 0 α = 11. Let S(K) = 1 + 3 + 5 + .....+(2K -1) = 3 + . K2 Then which of the following is true ? 1) Principle of mathematical induction can be used to prove the formula 2) S(K) ⇒ S(K+1) 3) S(K) S(K+1) 4) S(1) is correct Sol: 2) S(K) = 1 + 3 + 5 + ..... + (2k + 1) = 3 = 2 k S(K + 1) = 1 + 3 = 5 + ......+(2k + 1) -1 = 1 + 3 + 5 + ......+ 2k + 1 = 1 + 3 + 5 + .....+ (2x -1) + (2k + 1) 1 2 k 3 2 + + + = k 1 2 3 2 + + + = k k = 3 + 2 ) 1 ( + k ) 1 ( ) ( + ⇒ ∴ k S k S 12. How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order ? 1) 480 2) 240 3) 360 4) 120 Sol: 3) AE always together ways of arrangemtn = 12 5 = ∠ *G * R * D * N* Two places select can be 5 places way of arrangemnt 240 C 4 2 5 = ∠ Total way of arrangement = 120 + 240 = 360 13. The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes empty is : 1) 3 8C 2) 21 3) 8 3 4) 5 Sol: 2) Required No. of ways 1 3 1 8 C − − = = 21 14. If one root of the equation 0 12 2 = + + px x is 4, while the equation 0 2 = + + q px x has equal roots, then the the value of a' ' is 1) 4 2) 12 3) 3 4) 49/437 of 55 www.aieeepage.com Sol: 1) 4 in a root of 0 12 2 = + + px x Let the other root be β 12 β 4 = × 3 β = ⇒ P 3 4 − = + ∴ 7 P − = ⇒ 0 9 2 = + + q x x then equal roots 0 4 2 = − ⇒ q p 0 4 4 = − ⇒ q p4 49 = p 15. The coefficient of the middle terms in the binomial expansion in powers of x of 4 x) (1 α + and of 6 x) (1 α − is the same if α equals : 1) 53 2) 3 10 3) 103 − 4) 35 − Sol: 3) 3 3 2 2 α) ( 6 α 4 − = c c 3 2 α 20 α 6 − = ⇒ 10 3 α − = ⇒ 16. The coefficient of n x in expansion of (1 + x) n x x ) ( − is : 1) n n 1 ) 1 ( − − 2) ) 1 ( ) 1 ( n n − − 3) 2 1 ) 1 ( ) 1 ( − − − n n 4) (n -1) Sol: 2) When x is odd (1 + n) ) .... 1 )( 1 ( ) 1 ( 2 2 n x x x nc nx x n − + + − + = − when x is given ) .... 1 )( 1 ( ) 1 )( 1 ( 2 2 n x x x nc nx x n n − + + − + = + − coeffecient of ) 1 ( ) 1 ( n x n n − − = 17. If r n n0 r n C1 Σ s = = and , Cr Σ t r n n0 r n = = then nn st is equal to : 1) 2 1 -2n 2) 1 21 − n 3) n -1 4) n 21 Sol: 1) r n n r C Sn 1 0 = Σ= 1 2 1 0 1 ..... 1 1 1 C C C C Sn n n n n + + = r n n r n Cr t 0 = Σ= n n n n n Cn C C t + + + + = ..... 2 1 0 2 1 Then 2n Sn Tn =38 of 55 www.aieeepage.com 18. Let r T be the rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, n, m ≠ n Tm 1 = and , 1m Tn = then a-d equals : 1) n m 1 1 + 2) 1 3) mn 1 4) 0 Sol: 1) n d x a Tm 1 ) 1 ( = − + = m d x a Tn 1 ) 1 ( = − + = mn d mn a 1 1 = = ⇒ 0 = − ∴ d a 19. The sum of the first n terms of the sum 2 ) 1 ( .... 6 . 2 5 4 . 2 3 2 . 2 1 2 2 2 2 2 2 2 + + + + + + + n n when n is even. When n is odd sum is : 1) 2 2 ) 1 ( + n n 2) 2 ) 1 ( 2 + n n 3) 4 12 + n n 4) 2 1 3 + n n Sol: 2) If can be easily verified from options Put n = 3, 18 3 2 . 2 1 2 2 2 = + + 18 2 4 9 2 ) 1 ( 2 = × = + n n 2 ) 1 ( 2 + ∴ n n 20. The sum of series ..... ! 61 ! 41 ! 21 + + + is : 1) e e 2 2 − 2) e e 2 1 2 − 3) e e 2 1 2 − 4) 2 1 2 − e Sol: 2) ∞ + + ∠ + ∠ + = ..... 6 2 1 1 3 2 x x x ex we know that2 2 1 1 0 − ∞= + = ∠ Σ e e n n 1 2 .... 6 1 4 1 2 1 1 − + = + ∠ + ∠ + ∠ ∴ − e e e e 2 ) 1 ( 2 − =39 of 55 www.aieeepage.com 21. Let β α, be such that π. 3 < β − α < π If 65 21 sin sin − = β + α and , cos 65 27 = β + α then the value of 2 cos β + α is : 1) 656 − 2) 130 3 3) 65 6 4) 130 3 − Sol: 1) 2 2 ) cosβ α (cos β) sin (sin + + + α 2 2 265 27 21 + = 65 65 130 9 β) -α cos( 1 × × = + ⇒ 130 3 2β -α cos = ⇒ 2π 3 2β -α 2π < < ∵ 22. If θ + θ + θ + θ = 2 2 2 2 2 2 2 2 cos b sin sin b cos a a u , then the difference between the maximum and minimum values of 2 u is given by 1) 2 b) -(a 2) 2 2 b -a 2 3) 2 b) (a + 4) ) b 2(a 2 2 + Sol: 4) 2 2 2 2 4 4 2 2 θsin ) ( 2 a µ x a v ax b a b + + + + = θ 2 sin 2 2 a 2 2 2 2 2 2 2 − + + + = b a b a b when 0, θ = 2 2 ) ( µ b a + = when ) ( 2 µ 4π θ 2 2 2 b a + = = ab b a b a b a d 2 ) ( ) ( 2 2 2 2 2 2 − + = + − + = 2 ) ( b a − = 23. The sides of a triangle are sin α , cos α and α α + 1 cos sin for some . 0 2π < α < Then the greatest angle of the triangle is : 1) 150° 2) 90° 3) 120° 4) 60° Sol: 3) cosα α. sin 2 cos) sinα (1 -α cos α sin θ cos 2 2 + + = 24. A person standing on the bank of a rive observes that the angle of elevation of the vop of a tree on the opposite bank of the rive is 60° and when he retires 40 meters away from the tree the angle of elevation becomes 30°. The breadth of the river is : 1) 60 m 2) 30 m 3) 40 m 4) 20 m40 of 55 www.aieeepage.com Sol: 1) x h 3 = 3 40 x h + = h x ° 60 ° 3040 3 40 3 x x + = x x + = ⇒ 40 3 20 = ⇒ x 25. If R : S, R → defined by f(x) = sin x -3 cos x + 1, is onto, then the interval of S is : 1) [-1, 3] 2) [-1, 1] 3) [0, 1] 4) [0, 3] Sol: 4) [ ] 3 1 1 , 3 1 1 + + + − = S[ ]2 1 , 2 1 + − =[ ]3 , 1 − = 26. The graph of the function y = f(x) is symmetrical about the line x = 2, then : 1) f(x) = -f(-x) 2) f(2+x) = f(2-x) 3) f(x) = f(-x) 4) f(x+2) = f(x-2) Sol: 2) Graph is symmetrical about line x = 2 i.e. the value of the function will be same at the points equidistant from x = 2 on eithe side or it ) 2 ( ) 2 ( x f x f − = + ⇒ 27. The domain of the function 2 1 9 ) 3 ( sin f(x) x x − − = − is : 1) [1, 2) 2) [2, 3) 3) [1, 2] 4) [2, 3] Sol: 2) solving the inequalities 1 3 1 ≤ − ≤ − x and 0 9 2 > − x we get ) 3 , 2 [ ∈ x 28. If , 1 lim 2 2 e xb xa x = + + ∞ → then the values of a and be are : 1) a = 1 and b = 2 2) a = 1, R b∈ 3) 2 b R, a = ∈ 4) R b R, a ∈ ∈41 of 55 www.aieeepage.com Sol: 2) 2 2 2 1 lim e xb xa x x = + + ∞ → 2 2 lim 2 = + ⇒ ∞ → x xb xa x 2 2 lim = + ⇒ ∞ → xb a x ⇒ a = 1 and R b∈ 29. Let f(x) = . 2π , 0 , 4π , π 4 tan 1 ∈ ≠ − − x x x x If f(x) is continuous in , 2π , 0 then 4π is : 1) -1 2) 21 3) 21 − 4) 1 Sol: 3) π/4) ( π 4 tan 1 limπ/4 f x x x = − − → Let 4x -π/4 4Y x y π + = ⇒ = π/4) ( π/4) 4 /tan( 1 lim0 f y y y = + − ⇒ → π/4) ( 21 f = ⇒ 30. If , .... ∞ + + = to y e y e x x > 0, then dx dy is : 1) x x 1+ 2) x1 3) xx -1 4) x 1 x + Sol: 3) x y e x + = log x = ( 3 + x) 1 1 + = ⇒ dx dy x x x x dx dy − = − = ⇒ 1 1 1 31. A point on the parabola x y 18 2 = at which the ordinate increases at twice the rate of the abscissa is : 1) 29 , 89 2) (2, -4) 3) 29 , 89 − 4) (2, 4) Sol: 4) From options dt dx dt dy y 18 2 = 29 = ⇒ y 89 = ⇒ x42 of 55 www.aieeepage.com 32. A function y = f(x) has a second order derivative ) 1 ( 6 ) ( " − = x x f . If its graph passes through the point (2, 1) and at that point that tangent to the graph is y = 3x -5, then the function is : 1) 2 ) 1 ( + x 2) 3 ) 1 ( − x 3) 3 ) 1 ( + x 4) 2 ) 1 ( − x Sol: 2) From options ; 3 ) 1 ( ) ( − = = x y x f 33. The normal to the curve x = a(1 = cos θ ), y = a sin θ at θ' ' always passes through the fixed point : 1) (a, a) 2) (0, a) 3) (0, 0) 4) (a, 0) Sol: 1) normal passes through the centre of the circle = (a, 0) 34. If 2a + 3b + 6c = 0, then at least one root of the equation 0 c bx ax2 = + + lies in the interval 1) (1, 3) 2) (1, 2) 3) (2, 3) 4) (0, 1) Sol: 1) consider f(x) = cx x b x a + + 2 3 2 3 0 ) 0 ( = f c b a + + = 2 3 ) 1 ( f = 2c + 3b + bc = 0 0 f(1) f(0) = = 0 ) ( 2 1 = + + = − c bx ax x f ∴ By Radius is the atleast one value of ) 1 , 0 ( ∈ x 35. ∑= ∞ → n r n r e n 1 /n 1 Lim is : 1) e + 1 2) c -1 3) 1 -e 4) e Sol: 2) ∫ − = = Σ − − ∞ → 10 1 ) 1 ( 1 lim e dx e e n x n r n r n 36. If B Ax dx x x + = α − ∫sinsin log sin C, α) ( + − x than value of (A, B) is : 1) α) sin α, cos (− 2) α) sin α, (cos 3) α) cos α, sin (− 4) α) cos α, (sin Sol: 2) c α) sin( log α) sin( sin + − + = − ∫ x B Ax dx x x Put t α = − x dx = dt c α) -(x sin log sinα x α cos sin ) sin( + + = + ∫ dt t t d sinαi α, (cos ∴43 of 55 www.aieeepage.com 37. ∫ − x xdxsin cos is equal to : 1) C x + + 8π 3 2 tan log 2 1 2) C x + 2 cot log 2 1 3) C x + 8 3π − 2 tan log 2 1 4) C x + 8π − 2 tan log 2 1 Sol: 1) ∫ − x xdxsin cos ∫ − = x xdx sin 2 1 cos 2 1 2 1 ∫ − = sinx π/4 cos cos π/4 sin 2 1 xdx ∫ − = dx x ec 4π cos 2 1 C x + − = 8π 2 tan log 2 1 38. The value of ∫− − 32 2 1 dx x is : 1) 31 2) 3 14 3) 37 4) 3 28 Sol: 1)∫ ∫ ∫ −− − − + − + − 12 11 31 2 2 2 ) 1 ( ) 1 ( ) 1 ( dx x dx x dx x 31 3 11 3 12 3 3 3 3 − + − + − = − −− x x x x x x 3 28 = 39. The value of ∫ π/2 ++ ( = 0 2 2 sin 1 x) cos sin I dx x is : 1) 3 2) 1 3) 2 4) 0 Sol: 3) ∫ + = π/2 0 cosx sinx I [ ] cos sin 2 sin 1 x x x + = + ∵ π/2 0 ] sin cos [ x x + − = = 244 of 55 www.aieeepage.com 40. If ∫ ∫ π π/2 = 0 0 x)dx, f(sin ) (sin A dx x xf then A is : 1) π 2 2) π 3) 4π 4) 0 Sol: 2) 1 ) 1 . 1 ( 21 ) 1 )( 1 ( 21 = + 0 1 1 2 3 ALITER Area = ∫ ∫ = − + − 21 32 1 2 2 x x 41. If ∫− − = + = ) ( ) ( 1 x x )} 1 {( I , e 1 e f(x) a f a f dx x xg and ∫− − = ) ( ) ( 2 , )} 1 ( { a f a f dx x x g I then the value of 12 II is : 1) 1 2) -3 3) -1 4) 2 Sol: 3) 42. The area of the region bounded by the curves y = | x -2|, x = 1, x = 3 and the x-axis is : 1) 4 2) 2 3) 3 4) 1 Sol: 4) 43. The differential equation for the family of curves , 0 2 2 2 = − + ay y x where a is an arbitrary constant is : 1) xy y y x 2 ' ) ( 2 2 = + 2) xy y y x = + ' ) ( 2 2 2 3) xy y y x 2 ' ) ( 2 2 = − 4) xy y y x = − ' ) ( 2 2 2 Sol: 3) yx x a 2 2 2 + = 0 2 2 2 1 1 = − − + y yy x 1 1 yyy x a + = ⇒ 0 2 1 1 2 2 = + − + yyy x xy y x xy y y x 2 ' ) ( 2 2 = − 44. The solution of the differential equation ydx + dy y x x ) ( 2 + = 0 is : 1) log y = Cx 2) C y log 1 = + − xy 3) C y log xy 1 = + 4) C xy 1 = −45 of 55 www.aieeepage.com Sol: 2) y.dx + 0 ) ( 2 = + dy y x x 0 . 2 = + + ⇒ y x x dy dx y 2 x yx dy dx − = + ⇒ 1 1 12 = − − ⇒ xy dy dx x dy dt dy dx x t x = − ⇒ = ⇒ 2 1 1 Put 1 1 = − ⇒ t y dy dt y F I 1 . = ∴ The solution is t.I.F = dy F I. . 1 ∫ dy y y t 1 . 1 1 . ∫ = ⇒ c y xy + = log 1 45. Let A(2, -3) and B(-2, 1) be vertices of a triangle ABC. If the centroid of this triangle moves on the line 2x + 3y = 1, then the locus of the vertex C is the line : 1) 3x -2y = 3 2) 2x -3y = 7 3) 3x + 2y = 5 4) 2x + 3y = 9 Sol: 1) Let (h, k) he C centrod G = −3 2 , 3 k h −3 2 , 3 k h lies in 2x + 3x = 1 9 3 2 = + ⇒ k h 9 3 2 = + ⇒ y x 46. The equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinate axes whose sum is -1 is : 1) 1 1 2 -x and 1 3 2 = + = − y y x 2) 1 1 2 -x and 1 3 2 − = + − = − y y x 3) 1 1 2 -x and 1 3 2 = + = + y y x 4) 1 1 2 -x and 1 3 2 − = + − = + y y x Sol: 4) solve by options 1 3 3 2 = − + x sum of intercepts = 2 -3 = -1 1 13 2 = + −x sum of intercept = -2 + 1 = -1 Both are satisfied by (4, 3)46 of 55 www.aieeepage.com 47. If the sum of the slopes of the lines gives by 0 7 2 2 2 = − − y xy x is four times their product, then C has the value : 1) -2 2) -1 3) 2 4) 1 Sol: 3) 2 1 2 1 4 m m m m = + 7 4 7 2 − = − ⇒ c 2 = ⇒c 48. If one of the lines given 0 4 6 2 2 = + − cy xy x is , 0 4 3 = + y x then c equal to : 1) -3 2) -1 3) 3 4) 1 Sol: 4) Put 43x y − = in 0 4 6 2 2 = + − cy xy x we get 0 4 9 4 27 2 − + c x 9c = -27 c = -3 49. If a circle passes through the point (a, b) and cuts the circle 4 2 2 = + y x orthogonally, then the locus of its centre is :1) 2ax -2by -0 ) 4 ( 2 2 = + + b a 2) 2ax + 2by -0 ) 4 ( 2 2 = + + b a 3) 2ax -2by + 0 ) 4 ( 2 2 = + + b a 4) 2ax + 2by + 0 ) 4 ( 2 2 = + + b a Sol: 2) 2 2 2 2 ) ( ) ( ) 3 ( ) ( b k a h k h x − + − = − + − centre (h, k) 2 2 2 ) ( ) ( ( b k a h r − + − = ∵ ky k y hx h x 2 2 2 2 2 2 − + + − + ⇒ bk b k ah a h 2 2 2 2 2 2 − + + − + = 0 ) 2 2 ( 2 2 2 2 2 2 = − − + + − − + ⇒ b a bk ah ky hx y x 0 4 2 2 = − + y x 1 1 1 c c 2ff 2gg + = + ∵ ( condition for orthogonality) 0 1 = + ⇒ c c 0 4 2 2 2 2 = − − − + ⇒ b a bk ah 0 ) 4 ( 2 2 2 2 = + + − + ⇒ b a by ax 50. A variable circles passes through the fixed point A (p, q) and touches x-axis. The locus of the other and of the diameter through A is : 1) px q y 4 ) ( 2 = − 2) py q x 4 ) ( 2 = − 3) px p y 4 ) ( 2 = − 4) qy p x 4 ) ( 2 = − Sol: 1) Equation of the circle with end-point of diameters as (p, q) & (h, k) is 0 ) ( ) ( ) ( 2 2 = + + + − + − + kq hp q k y p h x y x since it touche x -axis kq hp p h + = + ∴ 2 2 or qy p x 4 ) ( 2 = −47 of 55 www.aieeepage.com 51. If the lines 2x + 3y = 1 = 0 and 3x -y -4 = 0 lie along diameter of a circle of circumference π, 10 then the equation of the circle is : 1) 0 23 2 2 2 2 = − − + + y x y x 2) 0 23 2 2 2 2 = − − − + y x y x 3) 0 23 2 2 2 2 = − + + + y x y x 4) 0 23 2 2 2 2 = − + − + y x y x Sol: 1) 2x + 3y + 1 = 0 3x -y -4 = 0 we get x = 1, y = -1 centre = (1, -1) 10π πr 2 = 5 = ⇒ r 25 ) 1 ( ) 1 ( 2 2 = + + − ∴ y x 0 23 2 2 2 2 = − + − + ⇒ y x y x 52. The intercept on the lines y = x by the circle 0 2 2 2 = − + x y x is AB. Equation of the circle on AB as a diameter is 1) 0 2 2 = − + + y x y x 2) 0 2 2 = + − + y x y x 3) 0 2 2 = + + + y x y x 4) 0 2 2 = − − + y x y x Sol: 1) Hence equation of the circle is y x' ) 0 , 0 ( ' y ) 1 , 1 ( 1/2) , centre(1/2 X x ( x -1) + y ( y -1) = 0 0 2 2 = − − + ⇒ y x y x 53. If 0 ≠ a and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas ax y 4 2 = and , 4 2 ay x = then : 1) 0 ) 2 3 ( 2 2 = − + c b d 2) 0 ) 2 3 ( 2 2 = + + c b d 3) 0 ) 3 2 ( 2 2 = − + c b d 4) 0 ) 3 2 ( 2 2 = + + c b d Sol: 1) Put a = θ then equation of the plarabolar are x 4 y2 = and y x 4 2 = Points as of intersection of the parabolar are (0, 0) and (4, 4) put b = 1, d = 0 ⇒ The line is 2x + 3cy = 0 ⇒ c = -2/3 54. The eccentricity of an ellipse, with its center at the origin, is 21 . If one of the directrices is x = 4, then the equation of the ellipse is : 1) 1 3 4 2 2 = + y x 2) 12 4 3 2 2 = + y x 3) 12 3 4 2 2 = + y x 4) 1 4 3 2 2 = + y x48 of 55 www.aieeepage.com Sol: 2) Given 21 = e 4 = ea ∵2a = 4 ∵ a= 2 ) 1 ( 4 2 2 e b − = ) 41 1 ( 4 2 − = b 3 2 = b equation of ellipse 1 3 4 2 2 = + y x 12 4 3 2 2 = + y x 55. A line makes the smae angle , θ with each of the x and z axis. If the angle β, which it makes with y-axis, is such that θ, sin 3 sin 2 2 = β then θ cos2 equals : 1) 52 2) 51 3) 53 4) 32 Sol: 3) 1 γ cos β cos α cos 2 2 2 = + + 1 β cos θ cos 2 2 2 = + β sin θ cos 2 2 2 = θ sin 3 θ cos 2 2 2 = 32 θ tan2 = ∴or 53 θ cos2 = 56. Distance between two parallel planes 2x + y + 2x = 8 and 4x + 2y + 4z + 5 = 0 is : 1) 29 2) 25 3) 27 4) 23 Sol: 3) 2x + y + 2z = 8 2x + y + 2z = -5/2 distance = 2 2 2 2 1 c b a d d + +− 27 3 25 8 = + = 57. A line with direction cosines proportinal to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The coordinates of each of the points of intersection are given by : 1) (2a, 3a, 3a), (2a, a, a) 2) (3a, 2a, 3a), (a, a, a) 3) (3a, 2a, 3a), (a, a, 2a) 4) (3a, 3a, 3a), (a, a, a)49 of 55 www.aieeepage.com Sol: 2) x = y + a = Z passing through (0, -a, 0) or’s 1, 1, 1 x + a = 2y = 2z passing (-a, 0, 0) 0.r’s 21 , 21 , 1 points 3a) 2a, (3a, and (a, a, a) satisfy condition 58. If the straight lines x = 1 + s, y = -3 -5, λ z = 1 λs and x = , 2t y = 1 + t, z = 2 -t, with parameters s and t respectively, are co-planar, then λ equals : 1) 0 2) -1 3) 21 − 4) -2 Sol: 1) lines are λ 1 λ3 1 1 -x − = −+ = z y 1 -2 1 1 21 x− = − = z y 0 1 4 1 1 1 21 λ λ 1 = − − − ∴ -2 λ = ∴ 59. The intersection of the spheres 13 2 7 2 2 2 = − − + + + z y x z y x and 8 4 3 3 2 2 2 = + + − + + x y x z y x is the same as the intersection of one of the sphere and the plane 1) 2x -y -z = 1 2) x -2y -z = 1 3) x -y -2z = 1 4) x -y -z = 1 Sol: 4) The equation of surface of intersection is 13 2 7 2 2 2 = − − + + + z y x z y x ------(1) ) 8 4 3 3 ( 2 2 2 = + + − + + − z y x z y x -----(2) i.e. 2x -y -z = 1 60. Let b a , and c be three non-zero vectors such that no two of these are collinean. If the vector b a 2 + is collinear with c and c b 3 + is collinear with a λ ( being some non-zero scalar) then c b a 6 2 + + equals : 1) 0 2) b λ 3) c λ 4) a λ Sol: 4) Let c x b a = + 2 and a y c b = + 3 Then c x c b a ) 6 ( 6 2 + = + + and also a y c b a ) 2 1 ( 6 2 + = + + so c x a y ) 6 ( ) 2 1 ( + = + ⇒ Since a and c are non zero and no collinear we have x = 6 = 0 ⇒ x = /06 and 1 + 2y = 0 i.e., x = -6 y = -1/2 0 6 2 = + + c b a 50 of 55 www.aieeepage.com 61. A paracticle is acted upon by constant forces k ˆ3 -jˆ i ˆ4 + and k ˆ3 -jˆ i ˆ3 + which displace it from a point k ˆ3 -jˆ 2 i ˆ+ to the point k ˆ4 jˆ 4 i ˆ5 + + . The work done in standard units by the forces is given by : 1) 15 2) 30 3) 25 4) 40 Sol: 1) 2 1 net F F F + = k j i ˆ 4 ˆ 2 ˆ 7 − + = k j i s ˆ 4 ˆ 2 ˆ 4 − + = Wd = S . F = 28 + 4 + 8 = 40 62. If c b a , , are non-coplanar vectors and λ is a real number, then the vectors , c 3 b 2 a + + c 4 b λ + and c 1) -(2λ are non-coplanar for : 1) no value of λ 2) all except one value of λ 3) all except two values of λ 4) all values of λ Sol: 3) They are coplaner is 0 1 -2λ 0 0 4 λ 0 3 2 1 = 0 3(0) 2(0) -1) -λ(2λ = + = if 0 λ = and 21 λ = 63. If w , v , u be such that . 3 w 2, v 1, u = = = if the projection v and u is equal to that of w along u and v , w are perpendicular to each other then | w v -u | + equals : 1) 14 2) 7 3) 14 4) 2 Sol: 3) Given uu w uu v . . = and 0 . = w v Now let w v u y + − = w u w v v u w v u y . 2 . 2 . 2 2 2 2 2 + − − + + = = 1 + 4 + 9 -2u.v -0 + 2u.v = 14 14 = y 64. Let b a, and c be non-zero vectors such that . | | | | 31 ) ( a c b c b a = × × If θ is the acute angle between the vectors b and c , then sinθ euqals : 1) 32 2 2) 32 3) 32 4) 3151 of 55 www.aieeepage.com Sol: 4) a c b c b a | || | 31 ) ( = × × a c b a c b b c a | || | 31 ) . ( ) . ( = − ⇒ a c b b c a 31 θ cos | | | | ) . ( + − ⇒ 32 2 θ sin 31 -θ cos = ⇒ = ⇒ 65. Consider the following statements : a) Mode can be computed from histrogram b) Median is not independent of change of scale c) Variance is independent of change of origin and scale Which of these is/are correct ? 1) (a), (b) and (c) 2) only (b) 3) Only (a) and (b) 4) Only (a) Sol: 3) 66. In a series of 2 n observations, half of them equal and remaining half equal -a. If the standard deviation of the observations is 2 then | q | equals : 1) n2 2) 2 3) 2 4) n1 Sol: 2) S.D = xx x 2 ) ( − 0 2 = − = xna na x x times x a a a x D S 2 2 ..... . 2 2 2 + + + = xx a2 2 2 2 = ⇒ 2 2 = ⇒ a 2 | | = ⇒ a 67. The probability that A speaks truth is , 54 while this probability for B is . 43 The probability that they contradict each other when asked to speak on a fact is 1) 54 2) 51 3) 20 7 4) 20 3 Sol: 3) 51 , 54 ) ( = = A P A P 41 , 43 ) ( = = B P B P then ) ( ) ( ) ( ) ( B P A P B P A P +20 7 43 51 41 54 = × + ×52 of 55 www.aieeepage.com 68. A random variable X has the probability distribution : : X P(X) 1 2 3 4 5 6 7 8 0.15 0.23 0.12 0.10 0.20 0.08 0.07 0.05 For the events E = {X is a prime number} and F= X < 4, the probability P = F E∪ is : 1) 0.50 2) 0.77 3) 0.35 4) 0.87 Sol: 2) ) ( ) ( ) ( ) ( F E P F P E P F E P ∩ − + = ∪ P(E) = 0.12 + 0.23 + 0.20 + 0.20 + 0.07 = 0.62 P(F) = 0.15 + 0.23 + 0.12 = 0.35 77 . 0 35 . 0 50 . 0 62 . 0 ) ( = − + = ∪ F E P 69. The mean and the variance of a binomial distribution are 4 and 2 respepctively. Then the probability of 2 successes is : 1) 256 28 2) 256 219 3) 256 128 4) 256 37 Sol: 4) mean 4 = ⇒ np variance 2 = ⇒ npq 4q = 2 q = 21 21 = p then n = 8 Then the probability of 2 successes 256 28 21 , 21 8 2 6 2 = c 70. With two forces acting at a point, the maximum effect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the force are : 1) 3 21 2 + N and 3 21 2 − N 2) 3 2 + N and 3 2 − N 3) 2 21 2 + N and 2 21 2 − N 4) 2 2 + N and 2 2 − N Sol: 3) θ cos 2 2 1 2 2 2 1 F F F F R + + = given 4 2 1 = + F F ....(1) 9 & 2 2 2 1 = + F F ...(2) solving (1) & (2) we get 22 2 1 + = F 22 2 2 − = F 71. In a right angles ∆ ABC, ° = ∠ 90 A and sides a, b, c are respectively, 5 cm, 4cm and 3 cm. If a force F has moments 0, 9 and 16 in N cm. units respectively about verticles A, B and C, then magnitude of F is : 1) 9 2) 4 3) 5 4) 353 of 55 www.aieeepage.com Sol: 3) Clearly F acts at pt A and its components perpendicular to 4 A C5 5 3 B 4 AC & AB are 4 16 ( = 4) & ) 3 ( 39 = 5 4 3 | | 2 2 = + = ∴ F } & { r are AB Ac ⊥ ∵ 72. Three forces Q , P and R acting along IA, IB and IC, where I is the incentre of a ∆ ABC, are in equilibrium. Then R Q P : : is : 1) cosec 2C cosec : 2B cosec 2A 2) 2C sin : 2B sin : 2A sin 3) 2C sec : 2B sec : 2A sec 4) 2C cos : 2B cos : 2A cos Sol: 1) A A/2 A/2 B/2 B/2Q B R C/2 C/2 C α β P γ γ sin β sin sinα P R Q = = 2 B A 180 sin R 2 C A 180 sin Q 2 C B 180 sin R + − ° = + − ° = + − ° ⇒ 2 B A sin R 2 C A sin Q 2 C B sin P + = + = + ⇒ 2C cosR 2B cosQ 2A cosP = = ⇒ 73. A particle moves towards east from a point A to a point B at the rate of 4 km/h and then towards north from B to C at the rate of 5 km/h. If AB = 12 km and BC = 5 km, then its average speed fro its journery from A to C and resultant average velocity direct from A to C are respectively : 1) km/h 9 17 and km/h 9 13 2) km/h 4 17 and km/h 4 13 3) km/h 9 13 and km/h 9 17 4) km/h 4 13 and km/h 4 1754 of 55 www.aieeepage.com Sol: 1) Average speed = time Total distance Total 55 4 12 5 12 + = + N C A km 12km/h 4 km 5km/h 5B 4 17 = km/h Average velocity = time Total nt Displaceme Total 55 4 12 5 12 2 2++ = 4 13 = km/h.from A to C 74. A velocity m/s 41 is resolved into two components along OA and OB making angles 30° and 45° respectively with the given velocity. Then the component along OB is : 1) m/s 2 6 81 − 2) m/s 1 3 41 − 3) m/s 41 4) m/s 81 Sol: 4) s m/2 6 81 21 2 1 23 2 1 21 41 75 sin 1/4sin30 u − = + × × = ° ° =O B° 45 ° 30 v wu 75. If 1 t and 2 t are the times of flight of two particles having the same initial velocity u and range R on the horizontta then 22 21 t t + is equal to : 1) 1 2) 2 2/g 4u 3) /2g u2 4) /g u255 of 55 www.aieeepage.com Sol: 2) The only distinct case will be when angles of projection are θ -90 & θ Now g 2usinθ Nt = 2 2 22 2 1 g θ cos 2 g θ sin 2 + = + ∴ u u t t 22 4gu = .