2. Alcance y límites de la teoría de la participación
2.6. Particularidades de la autoría y participación en delitos especiales
70. Farad = coulomb / volt and not joule/volt. Hence (c) is wrong. All others can be checked and found to be correct remembering L/R , CR and LC have dimensions of time. Ans.c
71. e = -dφ / dt = -(24t +4). At the time t = 1/4 s induced emf will have a magnitude (24 x 1/4) + 4 = 10 V. The induced current will be induced emf / resistance = 10/10 = 1 A. Ans.d
72. The average power in an A C circuit = Erms Irms cos φ, where φ is the phase difference between current and voltage. Here φ = 900. (Phase difference between cos ωt and sin ωt) therefore cos φ = 0. Ans.c
73. Voltmeters read rms value of A.C voltages. In AC circuit voltages add according to vector rules. There is no phase difference between current and voltage in a resistor. Current lags 900 in an inductor and leads 900 in a capacitor. The vector sum is of voltages will be
3 2 2 2
1 (V V )
V + − = V2. Ans.c
74. In an AC circuit instantaneous voltages follow Ohm’s law. So we can write just as for DC circuit, V1 + V2 + V3 = V. Ans.a
75. When the current is switched on the induced emf in the coil will oppose the growth of the current due to self induction. So the bulb will not glow immediately. When current becomes steady it glows dim because it gets only 6 V when it needs 10. Ans.b
76. During decay of current the induced emf in the inductor will oppose decay of current by Lenz’s law and momentarily shoots up current, and then goes off. Ans.d (Note:These two questions are drawn from experimental demonstration of of self induction).
77. Impedance Z = R2+(XL−Xc)2 where XL , Xc are inductive and capacitative reactance.
Here R = 60, XL Erms/Z = 200/100 = 2 A. Ans.a
78. Refer to theory note tabales.. Tan φ = XL – Xc/R = (180-100)/60 = 4/3. Ans.b 79. Power factor = cos φ = R/Z = 60/100 = 0.6. Ans.c
80. Average power Prms = Erms Irms cos φ = 200 x 2 x 0.6 = 240 W. Ans.d
81. An electro magnetic wave is produced by a tuning circuit which consists of an inductor and a cpacitor. Here is exchange of inductor’s magnetic energy 1/2 Li2 and condensor’s electrostatic energy 1/2 CV2 produces electro-magnetic waves. Ans.c
82. In an AC circuit, voltages add according to a vector rule. Voltages across inductance and capacitance oppose. Hence all that we can say is VR2 + (VL – Vc)2 = V2 where VR . VL, Vc
and V are the voltages across the resistor, inductor, capacitor and the supply voltage respectively. Hence it is possible for VL or Vc to exceed V. Ans.d
83. The energy of a photon is given by Planck’s law E = hv = hc/λ. The blue light has less wave length and hence gives photons of more energy. If P is the power of the source, n is no. of photons emitted per second, we have nhc/λ= P. Since nhc/λ is constant, the number of photons is more for light of larger wave length, i.e. red. Hence red light gives photons of less energy but more in number. Ans.b
84. To compare the phase difference between voltage and current, we should have both voltage and current either as sine functions or cosine function. So we rewrite E = 100 sin [ωt + π/2 + π/3] = 100 sin {ωt + 5π/6]. Comparing this with current I = 4 sin ωt, we find phase difference is 5π/6. Ans.b
85. The peak values of AC, E0 = 282 V. Rms value of AC Rrms = Eo/√2 = 282/1.41 = 200 V. Irms
= Erms / resistance = 200/20 = 10 A. AC instruments show rms value. Ans.c
86. The effective inductance between them is given by L = L1 + L2 ± 2M, where M is the mutual inductance between them. When the coils are far apart, mutual inductance between them becomes zero. Then only the self inductance becomes L1 + L2 . Ans.a
87. Scattered intensity is proportional to 1/λ4. Of the given electro-magnetic radiation radio waves have longest wavelength. Hence they are scattered least. Ans.d
88. Using the Rayleigh’s scattering law Iα 1/λ4 , λ (blue)/ λ (red) = 440/660 = 2/3. I(red)/I(blue)
= [λ(blue)/ λ(red)]4 = (2/3)4 = 16/81 = 1/5 . Ans.d
89. The maximum distance ground waves can reach by an antenna of height h = 2Rh , where R is radius of earth. Here h = 0.5 km. This gives the maximum distance as 80 km. Ans.a 90. The capacitative reactance Xc = 1/ωC = 1/2πf C. Thus Xc is inversely proportional to f. The
graph between them will be rectangular hyperbola. Ans.d
91. Any point on the Y-axis is on the equitorial line of the dipole. Hence field will be parallel to the axis. (a is correct). Electric field at a point between the charges will be along positive X-axis, but outside along negative X-axis (b is wrong). Dipole moment = -2pd i, as it is negative to positive charge. (c is wrong) work = potential difference x charge. Since potential at infinity and origin are zero, work will be zero. (d is wrong). Ans.a
92. Flux density the centre = 0 2 r 4
Idl π
µ by Biot-Savart law. Since resistance is proportional to length, flux density = constant x I1R1 /r2 for one arc for the other arc constant x I2R2/r2. As the potential difference across the two paths are same I1R1 = I2R2, The fields will be same in magnitude. Since currents are in the opposite direction, the fields ancel. Ans.d
93. We have here two semi circular coils of radii R1 and R2. The field due to one µoi/4R1 ( half of a coil of one turn) and due to the other µoi/4R2. Since the currents are in the opposite directions field oppose. Ans.c
94. Since the proton is projected at an angle (not parallel or perpendicular) to the field its path will be a helix. Hence choices a and b are wrong. Time period (independent of velocity) = 2πm/qB = 2πx1.67x10-27 / 1.6 x 10-19 x 0.104 = 6.3 x 10-7 s or 2πx10-7s. Ans.c
95. The radius of a charged particle in a uniform magnetic field r = p/qB = 2mE/qB, where E is the kinetic energy and p momentum. Thus we find r α m/q Deflection is proportional to 1/r and hence q/m Deflection of H+ is proportional to q/√1. (choice a) Deflection of He+ is proportional to q/√4 = q/2. Deflection of O2+ is proportional to 2q/√16 = q/2. choice c).
Ans.a & c
96. The shunt resistance S to make a galvanometer an ammeter is given by S = Ig /I-Ig, where Ig is the current through galvanometer I is current through the main circuit, G galvanometer resistance. To convert into an ammeter we have to add a resistor in parallel. Since the answer carries 1 Ω we substitute S = 1Ω, Ig = 50 µA, G = 100 Ω, we get I = 5 mA. (choice c). To convert into voltmeter we have to connect a high resistor in series. If R is the value of the resist or in series R = (V/Ig) – G. Here substituting V = 10 volt, Ig = 50 µA, G = 100 Ω, we get R = 200000 Ω, i.e. 200 kΩ. (choice. b) Ans. b & c
97. Copper is a metal and its resistance decreases on cooling. Germanium is a semi conductor.
Its resistance increases on cooling. Ans.d
98. Redrawing thediagram we find the arrangement is a balanced Wheasone’s bridge. The effective value of resistance between A and B is RAB = 3R x 6R/9R = 2R. Maximum power is
transfered when internal resistance is equal to external resistance . That is 2R = 4, R = 2 Ω.
100. Use the information supplied in theory notes. Ratio of two equal resistors when connected in series and in parallel is n2. Here n = 2. Hence ratio of resistances in series and parallel will be 4. Heat produced = E2/R. The required ratio will be 1:4. Ans.d
101. Electric flux by Gauss’s theorem φ = E ds cosθ. Here the area intercepting the lines of forces which is parallel to the axis of the circular plane is the surface area of the hemisphere.
Therefore φ = 2πR2E. Ans.b
102. The current through the arrangement, which consists of n cells each of emf E and internal resistance r connected in series with m rows in parallel is
nR
103. Kinetic energy of a charged particle in an electric field = Charge x potential difference = charge x electric field x distance if the field is uniform. Here it is qEy. Ans.c
104. The torque on a dipole when it makes an angle θ = pEθ, when θ is small. This restoring torque should be equal to 1α. Thus 1α = -pEθ. α = -(pE/I)θ. The angular acceleration is proportional to angular displacement. The motion is S.H.M. The period is 2π I/pEThis answer can also be found from similarity with period of magnet in a uniform field 2π I/mB. In the place of a magnetic field B, electrif field E will come. Ans.a
105. Current through one bulb = 60/220 A. If n is maximum number of bulbs that can be used, then n x 60/220 = 9. This gives n = 33. Ans.d
106. Thermo electric power = dE/dT = emf/temperature difference. Ans b
107. Use the information supplied in indirect theory notes. When the wire is halved in length the resistance becomes 1/4. Ans.b
108. Using the equation F = q vxB with q = 1 C, we have the force as –3i. Ans.d
109. ML2T3 is the dimension of power. Hence the given quantity is power/current2 = resistance.
Ans.d
110. We start from Ohm’s law j = σE. Resistivity = 1/σ = E/j = electric field / current density.
Ans.a
111. Let the voltage across the source be E and resistance of each bulb be R. Then power across the bulbs will be E2/4R. This is given as P. When one of them is removed, the power consumed will be E2/3R. Since E2/3R = ( E2/4R) x (4/3), it will be equal to (4/3)P. Ans.d 112. A standard resistor should not change resistance. So it should have low temperature
coefficient. It should not melt quickly due to the heat produced in it. Hence it should have high melting point. Ans.d
114. A secondary cell has less internal resistance compared to a primary cell. Hence it gives more current. Ans.c
115. Using two resistors we can get 4 resistance values. They are each of their value, both of them in series and both of them in parallel. In a series connection the resistances add . So you should choose two values given in the question whose sum is the third value given there.
Hence the resistances should be 18 and 9. 18 and 9 in series gives 27 and in parallel gives 6 Ans.a
116. If E is emf of each cell and r its internal resistance, when connected in series current i = 4E/4r
= E/r. When one of the cells is reversed the emf is equal to E+E+E-E = 2E. However internal resistances still add to 4r. The new current i1 = 2E/4r = (1/2)i. The current reduces by 50%.
Ans.b
117. When n ampere enters the combination, 1 ampere should flow the ammeter while n-1 ampere should flow through the shunt. If S is the resistance of the shunt, then (n-1)S = 1xG. ⇒ S=
G/n-1. Ans.b
118. The increase in resistance when it is heated through ∆T is given by R1α1 ∆T. For the other resistor it should be R2α2 ∆T. These should be equal for all ∆T. This makes (b) correct and (a) wrong. If the combination in series should have same resistance at all temperatures, the resistance of one should increase by the same amount as that of the decrease of the other. So one should be a metal and the other a semi-conductor. This can also be answered from indirect theory notes. Ans. b & d
119. The network here is repeated network . Reduce it starting from left end. The value will be outer resistance 2 ohm. Current is equal to emf / resistance = 2/2= 1 A.
120. Use the information given in theory notes. Current through parallel network divides in the ratio of reciprocal of resistance. So divide 1.1 A in the ratio (1/1): (1/2): 1/3) = 6:3:2. Current through 2 ohm resistor equals 1.1 x (3/11) = 0.3 A
121. Use the information current divides in the reciprocal ratio of resistance. Take the current from 6V battery as 3i. Of these 2i passes through 1 ohm and i through 2 ohm. By Kirchoff’s law 1x2i+3x3i = 6. Solving, i = 6/11 A
122. The arrangement between P and Q (that is points connected to battery) is a balanced Wheatstone’s bridge. 3 ohm here is in the place of galvanometer, while 2 ohm resistors are 4
arms of the bridge. Since the current through galvanometer is zero in a balanced bridge, i =0.
123. The arrangement is three one ohm resistors in parallel. They are each half of the loop and diametrical wire AB. Take the current as 3i. Applying Kirchoff’s law to a closed loop containing i = 2A. Use the information supplied in theory notes. The number of electrons will be 2x6.25x1018 = 1.25x1019.
124. The balancing length L of a cell of emf E1 is given by E1 = irL, where r is the resistance per unit length of potentiometer wire and i current through it. When driving cell has more internal resistance, current i decreases and for a given E1, L increases. Ans.a
125. Use the information supplied in theory notes. We look for wire of maximum resistance.
Resistance R = ρL/πr2. Resistance is maximum for the wire which has L/r2 maximum. Their values are 1/1, √2/2, √3/3, 2/4. Hence it is maximum for the wire P. Ans.a
126. Usually in a circuit consisting of one cell the terminal voltage cannot exceed emf. However if the circuits has two or more cells connected in opposition, the terminal voltage of a cell of smaller emf can exceed its emf. Hence statements b,c are correct (a) and (d) are wrong.
[There is a question which follows to illustrate this ]. Ans.b,c.
127. The current through the circuit is equal to net emf / total resistance = (12-8)/(1+2+9) = (1/3) A. The drop across E1 = current x internal resistance = (1/3)V. So terminal voltage is equal to 8+(1/3) = 8.33V. Ans.b (Note:-The drop is added with the emf because current flows from positive to negative inside the cell. This is due to the presence of a cell of higher emf 12V) 128. The drop across E2 = current x internal resistance = 2/3 V. Terminal voltage = 12-(2/3) =
11.34V. Ans.c
129. The potential difference across the points P and Q is the same as the potential difference across the resistor 9 ohm, because the point P is one end of the resistor and Q the other end.
The required voltage = currentxresistance = (1/3)x9 = 3 V. Ans.d
130. Whatever may be the value of R, the potential difference across the 3 resistors 1,2,3 ohm are same. Let this potential difference be V. It is given V2/3= H. We are asked V2/1. This is 3H. Ans.d
131. Since the lamp lights with maximum brightness, maximum power is transferred from the source to the lamp. This happens when resistance of the lamp is equal to internal resistance of the source (Maximum power transfer theorem)i.e. 4 ohm Ans.a
132. (Fig) .Looking carefully into the network, we can find that it has symmetry. The upper and lower parts are two balanced Wheatstone’s bridges. Since the current through galvanometer is zero, remove those 2 ohm resistors at centre and redraw the figure as shown in
figure above. This reduces as shown in adjacent figure. Here three resistors 4,2,4 are in parallel. The effective resistance R is given by 1/R=(1/4)+(1/2)+(1/4), which gives R = 1 ohm Ans.a
133. Since all the resistors are in parallel, use the equation for heat produced V2/R per second. Let V2/200 = H. When all the resisitors are connected in parallel, the effective resistance (adding them) is 40 ohm. The heat produced now will be V2/40 per second which is 5H. So time to boil same amount of water will be (1/5) of 5 minutes. Ans.b
134. Unless you read carefully, you will miss the correct answer. Since the rate of flow of charge is constant, current is constant. Ans c
135. The resistivity of a wire depends on its material. Note that the resistivity does not depend on the length or cross sectional area It is the resistance which depends on them. So (c) is correct while (a) and (b) are wrong. The resistivity also depends on the temperature, which makes (d) correct. Ans. c,d
136. The resistance of the bulb R = V2/P = 1002/500 = 20 ohm. When 150 V is supplied 100 V should be across the bulb and 50 V should be across a series resistor. In a series connection potential difference across a resistor is proportional to the resistance. Hence the value of the resistor should be half of that of the bulb = 10 ohm. Ans.d
137. Use information supplied in theory notes The resistance of the wire becomes 22 = 4 times.
The conductance of the wire which is reciprocal of resistance becomes 1/4. Ans.c 138. If E is the emf of each cell and r internal resistance i = 3
10 3 E
+ r . When in parallel i will be E
r
10+ ( / )3 . Equating the two and solving r = 10 ohm. Ans.c
139. If v is the drift speed of electron, current i = nev. Multiplying both sides by m, mi = mnev = e(nmv). Here nmv is a total momentum p of all electrons. This gives p= mi/e = i/(e/m)= i/K, where K is the specific charge e/m of the electron. Ans.c
140. Use the information supplied in theory notes. Both electrons will have same period and hence the same frequency. Ans.a
141. Use the information supplied in theory notes The speed of the particle = E/B = 1000/20 = 50.
Its momentum is equal to mass x velocity = 50m. Ans.b
142. When a particle is accelerated, its velocity increases. According to the theory of relativity, its mass also increases. The mass increase is more for lighter particle electron as it can be accelerated to close to the speed of light. The period T = 2πm/qB does not remain constant and hence the resonance condition will not be satisfied. Ans c.
143. A current loop is like a magnet. In a uniform magnetic field a magnet experiences only a torque. No force Torque G G G
τ = i A x B . When plane of the loop is parallel to the field, the area vector is perpendicular to the field. Hence τ = iAB sin 90o. Ans.a
144. Here angle between area vector and field vector is 0. Hence τ = iAB sin0o = 0. No torque, no force. Ans.d
145. Since the lengths are in the ratio 1:2, the radii of loops will be also in the ratio 1:2. Areas will be in the ratio 1:4. Magnetic moment = current x area of loop. ⇒ Ratio of moments 1:4.
Ans.c
146. Use the information supplied in theory notes. The charged particle here has uniform cir- ular motion. In a uniform circular motion speed is constant. Hence kinetic energy = (1/2) mv2 = constant. But the velocity varies. Hence momentum also varies. Ans.b
147. F(axis) = µonR i
149. Use information supplied in theory notes The final momentum p = qBR. The final kinetic energy = p2/2m = q2B2R2/2m which is independent of the speed. Ans.d
150. Use the information supplied already in theory notes Since E is same, radius of the tracks will
be proportional to the ratio (√m)/q. 2
e
151. Since the velocity components are equal the electron should be entering at an angle 45o with the field. Use the information supplied in theory notes. The path will be a helix. Ans.c 152. If B is the magnetic field, then by Ampere’s circuital theorem, G
B dl. oi
∫
= µ , where i is the current enclosed by the path along which integral is taken. If the point is outside , path will enclose a current, and hence the field B will not be zero. If the point is inside, the integralpath will not enclose a current. The current i and hence the field B will be zero. Ans.a
153. Use the information supplied in theory notes. The energy E = qV, where V is the accelerating potential. Since r is proportional to √E, it will be proportional to √V as well.
Ans.d
154. The equation for force between the parallel wires is a line. The graph between F and a will be a rectangular hyperbola while the graph between F and 1/a will be a straight line. Ans.a
155. Since the electron is moving from north to south, it is equal to a current flowing from south to north. Thus the electron current and the current through the conductor are in opposite direction. Hence they repel. The electron beam will be pulled up. Ans.c
156. The magnetic field due to each conductor X has same magnitude. Its direction is given by the vector i dl x r and will be pointing into the plane of the paper . All the fields add in magnitude and direction Ans.b
157. Current i = charge / time = 2e/T, where 2e is charge of helium nucleus. The magnetic field at
centre B = o 19
158. Let p be the pole strength. Then the magnetic moment M is equal to px L. When the needle is bent the poles are at A and B as shown in the adjacent figure. The length of the magnet is equal to the distance between poles = L/√2. So the new magnetic moment = pL/√2 = M√2.
Ans.d
159. The magnetic field at the centre is µoiLsin90o/4πr2. = constant x iL, where L is the length of the arc. Since the loop is uniform its length is proportional to the resistance. So the field becomes constant x iR. For a parallel connection i1R1=i2R2 i.e. iR =constant. So the magnetic field will be same due to the two arcs. But the field will be the opposite direction as the currents are in opposite directions. They cancel. Ans.d
160. For a tangent galvanometer i = K tanθ, where K is 2rBH/µon. When the coil is arranged at an angle 10o with the meridian, the horizontal component of earth’s field BH will only be BH cos 10o. For a constant current i the deflection θ will increase. Ans.a
161. The period of oscillation of a magnet is given by 2π I MB/ , where I is moment of inertia of the magnet. For two identical magnets, one placed over the other as given, I becomes 2I and M becomes 2M. Thus the period remains the same. Ans.b
162. The field at one end point along the perpendicular bisector of a straight conductor is got from the equation µ 163. Diamagnetism is produced due to induced effect when the
atom is placed in a magnetic field Hence it is present in all substances. But it shows only in substances where para and ferro effects are absent. (a) is correct. Since it is due to electromagnetic induction it can be explained by Lenz’s law.
atom is placed in a magnetic field Hence it is present in all substances. But it shows only in substances where para and ferro effects are absent. (a) is correct. Since it is due to electromagnetic induction it can be explained by Lenz’s law.