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

Transcription

1 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. () of the bigger disc. The value of α is () () 3 (3) () 6 Let the mass of the larger disc be M. Then mass of the smaller disc M = = M 3M Mass of remaining part = If the position of the centre of mass of the remaining disc be (x, ) with respect to the centre of the larger disc, then 3M M ( x) + ( ) = M( ) x = 3 If the centre of mass of the new disc is α from the centre of the bigger disc, then α= 3. A round uniform body of radius, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is gsinθ () M /I gsinθ (3) + M /I gsinθ () + I/M gsinθ () I/ M Career Launcher 8 / Solutions AIEEE Test Paper 7 Page

2 Sol. () N f θ Mg Mgsinθ f = Ma Ia and f =α= I Ia f = Ia Mgsinθ = Ma + gsinθ a = I + M 3. Angular momentum of the particle rotating with a central force is constant due to () Constant Torque () Constant Force (3) Constant linear momentum () Zero Torque Sol. () If the force is central, then r r r τ= r F = Angular momentum is constant.. A kg block slides on a horizontal floor with a speed of m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 5 N and spring constant, N/m. The spring compresses by () 8.5 cm () 5.5 cm (3).5 cm (). cm Sol. () mv = fx + kx 6 = 5x + 5x 5x + 5x 6 = 5 ± ± x = = = 55 m = 5.5cm Career Launcher 8 / Solutions AIEEE Test Paper 7 Page

3 5. A particle is projected at 6 to the horizontal with a kinetic energy K. The kinetic energy at highest point is () K / () K (3) Zero () K / Sol. () If v is the velocity of projection, then the velocity at the highest point is k = mv At the highest point v K K.E. = m = v v cos6 = 6. In the 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 I. If I denotes the maximum intensity, I I is equal to () 3 () (3) 3 () Sol. () x π Phase difference, φ= π= λ 3 π I = I + I + I.I cos = 3I 3 ut I = I ( φ = ) o I 3 = I o 7. Two springs, of force constant k and k, are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k and k are made four times their original values, the frequency of oscillation becomes k m k () f () f / (3) f / () f Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 3

4 Sol. () k k + k f = = m m eff k+ k k+ k f' = = = f m m 8. 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 a f Sol. () i () cal. () 6 cal. (3) 6 cal. () 66 cal. U Uiaf for both the paths will be the same. = Uibf Qiaf Wiaf = Qibf Wibf or 5 = 36 W ibf w = 6 cal ibf 9. 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 () π ma ν (3) ma ν () π ma ν Sol. () Let y = asinωt dy v = = aωcosωt dt k = mv = ma ω cos ωt Average KE: Career Launcher 8 / Solutions AIEEE Test Paper 7 Page b

5 K T/ avg T/ kdt = = dt T/ ma ω cos tdt ω T/ T/ T/ ma ω ma ω sinωt = ( cos t) dt t T + ω = + T ω T ma ω T sin ω = + = ma ω ω T = π T ω ma πν =π ma ν = ( ) ( Q ) 5. The displacement of an object attached to a spring and executing simple harmonic motion is given by x = cos π t metres. The time at which the maximum speed first occurs is ().5 s ().5 s (3).75 s ().5 s Sol. () ( ) x cos t = π ; SHM with particle at extreme position at t =. Velocity is maximum for the st time when it crosses the origin i.e. or ω=π π =π T T t = =.5s T t = 5. In an a.c. circuit the voltage applied is E = E sin ωt. The resulting current in the circuit is π I= I sin( ω ). The power consumption in the circuit is given by E I E () P = E I () P = I (3) P = zero () P = EI P = cosϕ ϕ=π/ EI p = cos π / = Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 5

6 5. An electric charge 3 µ C is placed at the origin (, ) of X Y co-ordinate system. Two points A and are situated at (, ) and (, ) respectively. The potential difference between the points A and will be ().5 volt () 9 volt (3) zero () volt A (, ) q (, ) r r (, ) V V A k q = r k q = r kq kq 9 9 VA V = = 9 = r r 53. 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) () Sol. () Energy shared in the battery = QV Work done by the battery = QV QV ratio = = QV Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 6

7 5. An ideal coil of H is connected in series with a resistance of 5 Ω and a battery of 5 V. seconds after the connection is made, the current flowing in amperes in the circuit is () ( e ) () ( e) (3) e () e Sol. () t x5 E L 5 i= e = e = e 5 ( ) 55. A long straight wire of radius a carries a steady current i. The current is uniformly distributed across its cross section. The ratio of the magnetic field at a and a is () () (3) () Sol. () Let field at a/ be and at a be. Then µ ia µ i = = πa π a µ i = = π a 56. A current I flows along the length of an infinitely long, thin walled pipe. Then () the magnetic field at all points inside the pipe is the same, but not zero () the magnetic field is zero only on the axis of the pipe (3) the magnetic field is different at different points inside the pipe () the magnetic field at any point inside the pipe is zero Sol. () r P ur ur dl = µ o i enclosed For any loop inside the hollow tube i enclosed is zero hence = Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 7

8 57. If M is the mass of an oxygen isotope 7 8O,M P and M N are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is () (M 7 M )C () (M 8M )C (3) (M 8M 9M )C () M C O y definition N O P O P N 58. In gamma ray emission from a nucleus () only the proton number changes () both the neutron number and the proton number change (3) there is no change in the proton number and the neutron number () only the neutron number changes Gamma ray are EM waves produced, when daughter nuclei make a transition from higher energy state to lower energy state. There is a no change in mass number or atomic number. 59. If in a p-n junction diode, a square input signal of V is applied as shown 5V O L 5V Then the output signal across L will be +5V V () () (3) V () 5V Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 8

9 Sol. () Half wave rectification 6. Photon of frequency ν has a momentum associated with it. If c is the velocity of light, the momentum is () h ν /c () ν /c (3) hν c () h ν /c Sol. () h h ν P = = λ c 6. The velocity of a particle is v = v + gt + ft. If its position is x = at t =, then its displacement after unit time (t = ) is () v + g/ + f () v + g + 3f (3) v + g/ + f/3 () v + g + f v = v + gt + ft 3 ft x = x + vt+ gt + 3 x t= = x = g f x v t= = For the given uniform square lamina ACD, whose centre is O, D F C O A E () I AC = I EF () IAC = IEF I (3) AD EF = 3I () I AC = I EF Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 9

10 Sol. () D F C O A E I AC = ( axis theorem) I I IEF = IAC = IEF 63. A point mass oscillates along the x-axis according to the law x = x cos ( ωt π /). If the acceleration of the particle is written as a = Acos( ω t + δ ), then Sol. () 3 π π () A = x ω, δ = () A = x, δ = π π (3) A = x ω, δ = () A = x ω, δ = x π x cos cot = dx π = xω sin t dt ω dx = ω ω dt π x cos t π A x ; = ω δ = Career Launcher 8 / Solutions AIEEE Test Paper 7 Page

11 r 6. Charges are placed on the vertices of a square as shown. Let E be the electric field and V the potential at the centre. If the charges on A and are interchanged with those on D and C respectively, then q q A Sol. () D C q q r r () E changes, V remains unchanged () E remains unchanged, V changes r r (3) oth E and V change () E and V remain unchanged q A q q q O O D C q q q q (I) (II) V = V = ur ur Direction of E is as indicated Direction of E is as indicated 65. The half-life period of a radio-active element X is same as the mean life time of another radio-active element Y. Initially they have the same number of atoms. Then () X and Y decay at same rate always () X will decay faster than Y (3) Y will decay faster than X () X and Y have same decay rate initially ().693 t / = λ τ () = λ () () t / =τ λ =.693 λ or λ < λ (same initial amount). Y decays faster. 66. A Carnot engine, having an efficiency of η= as heat engine, is used as a refrigerator. If the work done on the system is J, the amount of energy absorbed from the reservoir at lower temperature is () J () 99 J (3) 9 J () J Career Launcher 8 / Solutions AIEEE Test Paper 7 Page

12 For a heat engine W η= Q H W QH = = J η QH = QL + W QL = 9 J When the same engine is used as a refrigerator, heat will be absorbed from the system at lower temperature. Hence heat absorbed at lower temperature = Q L = 9 J 67. 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 electrons for conduction is significant only in Si and Ge but small in C. () The number of free conduction electrons is significant in C but small in Si and Ge. (3) The number of free conduction electrons is negligibly small in all the three. () The number of free electrons for conduction is significant in all the three. Sol. () The bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out an electron from these atoms (i.e. ionisation energy E g ) will be least for Ge, followed by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C. (NCET ook XII, Page 5) 68. A charge particle with charge q enters a region of constant, uniform and mutually orthogonal fields E r r and with a velocity v r r r perpendicular to both E and, and comes out without any change in magnitude or direction of v r. Then () r v = r E/E ur r ur ur () v = E / r ur ur (3) r ur ur v = E/ () v = E /E Career Launcher 8 / Solutions AIEEE Test Paper 7 Page

13 Sol. () V E r ur Fe = qe r r ur F = qv ur r ur ur ur r ur ur qe + qv = E + v = ur ur ur r ur ur ur r E + ( v) ( ) v = ur ur r E or v = 69. The potential at a point x (measure in µ m ) due to some charges situated on the x-axis is given by V(x) = / (x ) Volts The electric field E at x = µ m is given by () /9 Volt/ µ m in the +ve x direction () 5/3 Volt/ µ m and in the ve x direction (3) 5/3 Volt/ µ m and in the +ve x direction () /9 Volt/ µ m and in the ve x direction Sol. () V( x) = = x x x+ = 5 x x + dv 5 5 E( x ) = = dx x x+ E (x = ) = V = 9 µ m ( ) ( ) 5 5 V 8 5 V 36 = µ m 36 µ m Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 3

14 7. Which of the following transitions in hydrogen atoms emit photons of highest frequency? () n = to n = () n = to n = 6 (3) n = 6 to n = () n = to n = Sol. () < 6 7. 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. Initially 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) () mf M F Sol. () Common acceleration of two blocks a = m+ M mf Force on mass m, F' = ma = M+ m ( ) (3) M + m F m mf () ( m+ M) 7. Two lenses of power 5D and +5D are in contact with each other. The focal length of the combination is () + cm () cm ( 3) cm () + cm p = p + p = = D P = = D F F = m = cm 73. One end of a thermally insulated rod is kept at a temperature T and the other at T. The rod is composed of two sections of lengths l and l and thermal conductivities k and k respectively. The temperature at the interface of the two sections is T l l T k k () ( k l T + k l T ) / (k l + k l ) () (k l T + K l T ) / ( k l + k l ) (3) (k l T + k l T ) / (k l + k l ) () (k l T + k l T ) / ( k l + k l ) Career Launcher 8 / Solutions AIEEE Test Paper 7 Page

15 Sol. () Let the temperature at the interface is T T T T T = l l ka ka k k Tk T k T + = + l l l l kl T T = + k lt kl + k l 7. A sound absorber attenuates the sound level by d. The intensity decreases by a factor of () () (3) () Sol. () I β= ( d) log I I β β = log I I = log I I log = I I = = I 75. If 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 = 8 () C p C v = / 8 (3) C p C v = / () C p C v = Sol. () C P and C v are specific heats per unit mole, C P C V = ( ) MC C = P V CP CV = = M 8 Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 5

16 76. A charge particle move through a magnetic field perpendicular to its direction. Then () kinetic energy changes but the momentum is constant () the momentum changes but he kinetic energy is constant (3) both momentum and kinetic energy of the particle are not constant () both, momentum and kinetic energy of the particle are constant Sol. () Magnitude of velocity remains same where as direction changes. 77. Two identical conducting wires AO and COD are placed at right angles to each other. The wire AO carries an electric current I and COD carries a current I. The magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wires AO and COD, will be given by µ () ( I + I ) πd µ (3) ( I + I ) / πd µ I µ I =, = πd πd Angle between and is 9 µ / = (I + I ) πd / µ I + I () π d () ( I + I ) 78. The resistance of a wire is 5 ohm at 5 C and 6 ohm at C. The resistance of the wire at C. The resistance of the wire at C will be () 3 ohm () ohm (3) ohm () ohm Sol. () From, = ( +α T) t 5 α =...() ( 5 ) 6 α =...() ( ) 6 = 5 = ohm Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 6 µ πd

17 79. 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 () zero () / (K ) CV (3) CV (K ) / K () (K ) CV Sol. () work done (insert) = work done (removed) 8. If 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 find the ratio electronic charge on the moon electronic charge on the earth to be () g M / g E () (3) () g E / g M Sol. () Charge is independent of gravity. Career Launcher 8 / Solutions AIEEE Test Paper 7 Page 7