FIITJEE Solutions to AIEEE PHYSICS

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1 FTJEE Solutions to AEEE - 7 -PHYSCS Physics Code-O 4. The displacement of an object attached to a spring and executing simple harmonic motion is given by x = cos πt metres. The time at which the maximum speed first occurs is ().5 s ().75 s (3).5 s (4).5 s x = cos πt v =.π sin πt v is maximum at t = =.5sec 4. n an a.c. circuit the voltage applied is E = E sin ωt. The resulting current in the circuit is π = sin ωt. The power consumption in the circuit is given by E () P = () P = zero E (3) P = (4) P E Sol. () cos φ = So power = 43. An electric charge 3 C is placed at the origin (, ) of X Y co-ordinate system. Two points A and B are situated at (, ) and (, ) respectively. The potential difference between the points A and B will be () 9 volt () zero (3) volt (4) 4.5 volt Sol. () Both points are at same distance from the charge 44. A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be () () (3) 4 (4) qv = qv 45. An ideal coil of H is connected in series with a resistance of 5 Ω and a battery of 5V. second after the connection is made the current flowing in amperes in the circuit is () ( e) () e (3) e (4) ( e ) FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

2 FTJEE Solutions to AEEE - 7 -PHYSCS i = i e = ( e ) Rt L 46. A long straight wire of radius a caries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a and a is () 4 () 4 (3) (4) a i πa Bπ = a π 4 i B = (i) 4 π a B(a) π = i i B = (ii) 4 π a B B = 47. A current flows along the length of an infinitely long, straight, thin walled pipe. Then () the magnetic field is zero only on the axis of the pipe () the magnetic field is different at different points inside the pipe (3) the magnetic field at any point inside the pipe is zero (4) the magnetic field at all points inside the pipe is the same, but not zero Use Ampere s law f M O is the mass of an oxygen isotope O, M 8 p and M N are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is () (MO 8M P)C () (MO 8MP 9M N)C (3) MC (4) (M 7M )C O Sol. () Binding energy = (M O 8M P 9 M N )C 49. n gamma ray emission from a nucleus () both the neutron number and the proton number change () there is no change in the proton number and the neutron number. (3) only the neutron number changes (4) only the proton number changes Sol. () O N FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

3 FTJEE Solutions to AEEE - 7 -PHYSCS 3 5. f in a p-n junction diode, a square input signal of V is applied as shown Then the output signal across R L will be 5V 5V R L () V () (3) 5V V (4) +5V 5. Photon of frequency ν has a momentum associated with it. f c is the velocity of light, the momentum is () ν/c () hνc (3) hν/c (4) hν/c h hν P = = λ c 5. The velocity of a particle is v = v + gt + ft. f its position is x = at t =, then its displacement after unit time (t = ) is () v + g + 3f () v + g/ + f/3 (3) v + g + f (4) v + g/ + f Sol. () x dx = (V + gt + ft )dt x = v + g + f For the given uniform square lamina ABCD, whose centre is O, = () AC EF () AD = 3 EF = (3) AC EF A E B (4) AC = EF D F C = (from rd axis theorem) AC EF 54. A point mass oscillates along the x-axis according to the law x = x cos (ωt π/4). f the acceleration of the particle is written as a = A cos(ωt + δ) () A = x, δ = π/4 () A = x ω, δ = π/4 (3) A = x ω, δ = π/4 (4) A = x ω, δ = 3π/4 FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

4 FTJEE Solutions to AEEE - 7 -PHYSCS 4 v = x ω sin (ωt π/4) a = x ω π cos ω+π t 4 a = A cos(ωt + δ) A = x ω 3π ; δ= Charges are placed on the vertices of a square as shown. Let E be the electric field and V the potential at the centre. f the charges on A and B are interchanged with those on D and C respectively, then () E remains unchanged, V changes () Both E and V change (3) E and V remains unchanged (4) E changes, V remains unchanged q q A D q B C q As E is a vector quantity 56. The half-life period of a radio-active element X is same as the mean life time of another radioactive element Y. nitially they have the same number of atoms. Then () X will decay faster than Y () Y will decay faster than X (3) X and Y have same decay rate initially (4) X and Y decay at same rate always. Sol. () ln t / = λ x τ mean = ; dn = λ N λ dt x y y ln = λ λ λ x = λ y(.693) λ y >λ x 57. A Carnot engine, having an efficiency of η = / as heat engine, is used as a refrigerator. f the work done on the system is J, the amount of energy absorbed from the reservoir at lower temperature is () 99 J () 9 J (3) J (4) J Sol. () T T W = Q η= T T T T 9 = Q 9 = = = T T T = Q 9 = T 9 Q = 9 J 58. Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate? () The number of free conduction electrons is significant in C but small in Si and Ge. () The number of free conduction electrons is negligible small in all the three. (3) The number of free electrons for conduction is significant in all the three. (4) The number of free electrons for conduction is significant only in Si and Ge but small in C. FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

5 FTJEE Solutions to AEEE - 7 -PHYSCS A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields E and B with a velocity v perpendicular to both E and B, and comes out without any change in magnitude or direction of v. Then () v = E B/B () v = B E/B (3) v = E B/E (4) v = B E/E v B = E 6. The potential at a point x (measured in m) due to some charges situated on the x-axis is given by V(x) = /(x 4) Volts. The electric field E at x = 4 m is given by () 5/3 Volt/m and in the ve x direction () 5/3 Volt/m and in the +ve x direction. (3) /9 Volt /m and in the ve x direction (4) /9 Volt/m and in the +ve x direction. Vx = x 4 dv E = = (x ) dx (x 4) 6 = = Which of the following transitions in hydrogen atoms emit photons of highest frequency? () n = to n = 6 () n = 6 to n = (3) n = to n = (4) n = to n = h ν= Rhcz n n 6. A block of mass m is connected to another block of mass M by a spring (massless) of spring constant k. The blocks are kept on a smooth horizontal plane. nitially the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Find the force on the block of mass m () mf (M + m)f () M m (3) mf (m + M) (4) MF (m + M) Kx = ma = mf m+ M 63. Two lenses of power -5 D and + 5D are in contact with each other. The focal length of the combination is () cm () cm (3) + cm (4) + cm Sol. () P = P + P = f =. m cm P FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

6 FTJEE Solutions to AEEE - 7 -PHYSCS One end of a thermally insulated rod is kept at a T temperature T and the other at T. The rod is composed of two sections of lengths and and k thermal conductivities k and k respectively. The temperature at the interface of the two sections is () (k T + k T ) / (k + k ) () (k T + k T ) / (k + k ) (3) (k T + k T ) / (k + k ) (4) (k T + k T ) / (k + k ) k T (T T)k (T T )k = Tk l + Tk T = k + k 65. A sound absorber attenuates the sound level by db. The intensity decreases by a factor of () () (3) (4) B = log ' B = log given B B = ' = log ' = 66. f C p and C v denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then () C p C v = R/8 () C p C v = R/4 (3) C p C v = R (4) C p C v = 8R Mayer Formula 67. A charged particle moves through a magnetic field perpendicular to its direction. Then () the momentum changes but the kinetic energy is constant () both momentum and kinetic energy of the particle are not constant (3) both, momentum and kinetic energy of the particle are constant (4) kinetic energy changes but the momentum is constant 68. Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current and COD carries a current. The magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by () + π d πd () ( + ) πd (3) ( + ) πd (4) ( + ) FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

7 FTJEE Solutions to AEEE - 7 -PHYSCS 7 Sol. () πd ( + ) 69. The resistance of a wire is 5 ohm at 5 C and 6 ohm at C. The resistance of the wire at C will be () ohm () ohm (3) 4 ohm (4) 3 ohm 5 + 5α = 6 + α 5 = R (+ α 5) R O = 4 7. A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volts. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is () ½ (K )CV () CV (K )/K (3) (K )CV (4) zero 7. f g E and g m are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan s oil drop experiment could be performed on the two surfaces, one will electronic charge on the moon find the ratio to be electronic charge on the earth () () (3) g E /g m (4) g m /g E 7. A circular disc of radius R is removed from a bigger circular disc of radius R such that the circumferences of the discs coincide. The centre of mass of the new disc is α/r from the centre of the bigger disc. The value of α is () /3 () / (3) /6 (4) /4 n this question distance of centre of mass of new disc is αr not R α. 3M M α R+ R = 4 4 α= 3 FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

8 FTJEE Solutions to AEEE - 7 -PHYSCS A round uniform body of radius R, mass M and moment of inertia, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is () (3) + () + (4) Mg sin θ f = Ma N fr = a R a = + θ f Mg 74. Angular momentum of the particle rotating with a central force is constant due to () Constant Force () Constant linear momentum. (3) Zero Torque (4) Constant Torque 75. A kg block slides on a horizontal floor with a speed of 4 m/s. t strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 5 N and spring constant is,. N/m. The spring compresses by () 5.5 cm ().5 cm (3). cm (4) 8.5 cm 76. A particle is projected at 6 to the horizontal with a kinetic energy K. The kinetic energy at the highest point is () K () Zero (3) K/ (4) K/4 77. n a Young s double slit experiment the intensity at a point where the path difference is 6 λ (λ being the wavelength of the light used) is. f denotes the maximum intensity, 3 () () (3) / (4) 3/4 is equal to FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

9 FTJEE Solutions to AEEE - 7 -PHYSCS 9 φ = cos max 78. Two springs, of force constants k and k, are connected to a mass m as shown. The frequency of oscillation of the mass is f. f both k and k are made four times their original values, the frequency of oscillation becomes k m k () f/ () f/4 (3) 4f (4) f k + k f = π m k + k = = π m f' f 79. When a system is taken from state i to state f along the path iaf, it is found that Q = 5 cal and W = cal. Along the path ibf Q = 36 cal. W along the path ibf is () 6 cal () 6 cal. (3) 66 cal. (4) 4 cal. a i f b 8. A particle of mass m executes simple harmonic motion with amplitude a and frequency ν. The average kinetic energy during its motion from the position of equilibrium to the end is () π ma ν () 4 π ma ν (3) 4π ma ν (4) π ma ν ma ω = π f ma 4 ************* FTJEE Ltd., CES House, 9 A, Kalu Sarai, Sarvapriya Vihar, New Delhi -6, Ph , , Fax:

Physics. 41. A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the

Physics. 41. A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the Physics. A circular disc of radius is removed from a bigger circular disc of radius such that the circumferences of the discs coincide. The centre of mass of the the new disc is α from the center Sol.