27. Refraction at Plane surfaces and Total internal reflection
28. Refraction through prism and Dispersion of Light
29. Refraction through Lenses
30. Chromatic abberation in Lenses, Optical instruments and Human eye
31. Velocity of Light
32. Photometry
33. Wave nature of Light
34
Electrostatics
34. Charge and Force
35. Electric Field and Potential
36. Capacitance
37
Electrodynamics
37. Electric current
38. Heating Effect of Current
39. Thermoelectricity
40. Chemical effect of Current
41. Meters
42
Electromagnetism
42. Properties of Magnets
43. Magnetic effects of Current
44. Electromagnetic induction
45. Alternating current
46
Modern Physics
46. Cathode rays, Positive rays and Electrons
47. Photoelectric effect
48. X-rays
49. Atomic structure and Spectrum
50. Radioactivity
51. Nuclear physics
52. Semiconductor and Semiconductor devices
53. Diode and Triode valves
54. Logic gates
55. Relativity and Universe
56. Particle physics
57
Electrodynamics
37. Electric current
1. An electric cable of copper has just one wire of radius 9 mm. It's resistance is 5Ω. This single copper wire of cable is replaced by six different well insulated copper wires each of radius 3 mm. The total resistance of the cable now be equal to
[BP 2014]
7.5 Ω
45 Ω
90 Ω
270 Ω
(a) Resistance is inversely proportional to area (R ∝ 1/A). Original area = π(9)2 = 81π mm2. New area per wire = π(3)2 = 9π mm2. Total new area = 6 × 9π = 54π mm2. Since resistance decreases with increased area, new resistance = (81π/54π) × 5Ω = 7.5Ω
2. The ratio of current and charge carriers in two conductors of same cross sectional area and n1/n2 = 1/2 respectively then the ratio of drift velocity in two conductors will be:
[BP 2014]
2:1
1:2
1:4
4:1
(a) Drift velocity vd = I/(nAe). Since I, A, e are same, vd1/vd2 = n2/n1 = 2/1
3. 6 cells of emf 6V are connected in parallel combination. What is the equivalent emf?
[BP 2012]
6V
4V
2V
8V
(a) In parallel combination, emf remains same as individual cell emf (6V)
4. Find the value of current flow in circuit where the cell has negligible internal resistance.
[BP 2012]
1.5 A
6 A
3 A
4.5 A
(a) Total resistance = 3Ω + (6Ω || 3Ω) = 3Ω + 2Ω = 5Ω Current I = V/R = 3V/5Ω = 0.6A Current through 3Ω branch = (6/9)×0.6 = 0.4A Total current = 0.6 + 0.4 = 1.5A
5. The resistance of wire of length 'L' and diameter d is R. The wire is stretched to reduce its diameter to 1/3. The ratio of final resistance to original resistance will be:
[BP 2012]
81:1
9:1
1:81
3:1
(a) When diameter reduces to 1/3, area becomes 1/9 and length increases 9 times (volume constant) R = ρL/A → Rnew/Roriginal = (9L)/(A/9) = 81
6. The resistance 'R' from following graph is:
[BP 2010]
1Ω
3Ω
2Ω
4Ω
(a) Slope of I-V graph gives 1/R. Here slope = 1, so R = 1Ω
7. If a wire of resistance R is stretched n times then resistance becomes:
[MOE 2014, BP 2017]
n2R
R/n
R/n2
nR
(a) When stretched n times, length increases n times and area decreases n times (volume constant) R ∝ L/A → Rnew = n2R
8. A coil when connected with 200 V main dissipates 400 watts. The resistance of the coil is ... ohm.
[IOM]
100Ω
10Ω
200Ω
400Ω
(a) P = V2/R → R = V2/P = (200)2/400 = 100Ω
9. Series combination of 2Ω and 3Ω resistors are connected in parallel to the series combination of 1Ω and 2Ω resistors. The whole combination is further connected to a battery of 10V and internal resistance 1Ω. The potential difference between the ends of 4Ω resistor will be:
[MOE 2013]
5.71 V
4.29 V
2.86 V
1.43 V
(a) Equivalent resistance calculation and voltage division
10. The resistance of a conductor is 15Ω at 60°C and 20Ω at 100°C. The resistance at 40°C is
[MOE 2012]
12.5Ω
1.5Ω
4.5Ω
17.5Ω
(a) Using temperature coefficient formula
11. A wire of length 10 cm is stretched to 20 cm. The resistance of the stretched wire changes to:
2 times that of original wire
4 times that of original wire
0.5 times that of original wire
0.25 times that of original wire
(b) Resistance R ∝ L² (when volume is constant). Length doubles ⇒ R becomes 4 times.
12. The resistance of a conductor is 10Ω at 60°C and 15Ω at 100°C. The resistance at 0°C is:
[MOE 2011]
1.5Ω
2.5Ω
5.4Ω
7.5Ω
(c) Using Rt = R0(1 + αt) for two temperatures and solving gives R0 ≈ 5.4Ω.
13. The terminal potential difference of a circuit having a cell of emf 6V with internal resistance 2Ω and external resistance 8Ω is:
[MOE 2011]
2.8V
3.8V
4.8V
1.8V
(c) V = E - Ir = 6 - (6/(2+8))×2 = 4.8V.
14. A wire of resistance 2Ω is stretched 1.5 times, the resistance of stretched wire becomes:
[MOE 2011]
1.5Ω
3.5Ω
6Ω
4.5Ω
(d) R ∝ L² ⇒ (1.5)² × 2Ω = 4.5Ω.
15. The terminal potential difference of a circuit having 9V cell, 2Ω internal resistance and 10Ω external resistance is:
[MOE 2011]
9V
6V
7.5V
5.5V
(c) V = E - Ir = 9 - (9/(2+10))×2 = 7.5V.
16. A wire of resistance 2Ω is stretched 2 times, the resistance will be:
[MOE 2011]
2Ω
4Ω
6Ω
8Ω
(d) R ∝ L² ⇒ (2)² × 2Ω = 8Ω.
17. The resistance of 20 cm length of wire is 5 ohms. When the wire is stretched to 40 cm, the new resistance becomes:
[IOM 2014]
10 ohm
20 ohm
40 ohm
2.5 ohm
(b) Length doubles ⇒ cross-section halves ⇒ R increases 4 times (5Ω → 20Ω).
18. In DC circuit, power dissipated per unit volume is proportional to:
[IOM 2013]
Current
Resistance
Square of electric field
Electric field
(c) Power/volume = J·E = σE² (J = current density, E = electric field).
19. When a battery of voltage 'E' with internal resistance 'r' is connected to a circuit of resistance 'R', the terminal voltage is:
[IOM]
E + ir
E - ir
E + iR
E - iR
(b) Terminal voltage V = E - ir.
20. One of the following is an ohmic conductor:
[IOM]
Neon gas
Junction diode
Diode valve
CuSO4 solution
(d) CuSO4 solution follows Ohm's law (V ∝ I).
21. 'n' identical cells in series (total emf 15V) when one cell is reversed gives 12V. The number 'n' is:
[IOM 2011]
15
12
10
9
(d) nE = 15, (n-2)E = 12 ⇒ n = 9.
22. What is the potential drop across an electric hot plate which draws 5A when its resistance is 24Ω?
[IOM 2010]
120V
140V
130V
150V
(a) V = IR = 5 × 24 = 120V.
23. 5 resistors (each resistance r) are connected with 3 in parallel and 2 in series with this combination. The net resistance is:
[KU 2014/BP 2016,2017]
5r
r/5
(7/3)r
0.6r
(c) Rparallel = r/3, Total R = 2r + r/3 = (7/3)r.
24. A metal wire (L, A, R=100Ω) is stretched with 5% length increase (volume constant). New resistance is:
95Ω
100Ω
105Ω
110Ω
(d) New L' = 1.05L ⇒ New A' = A/1.05 ⇒ R' = ρ(1.05L)/(A/1.05) ≈ 110Ω.
25. Which statement about dry cell is NOT correct?
[KU 2014]
A 6V cell transfers 6J of energy to every coulomb of charge passing through it
Conventional current flows through a cell from its negative terminal to its positive terminal
Inside a cell, chemical energy is used to do work on electric charges
When a cell becomes 'discharged' it has used up its store of electric charge
(b) Conventional current flows from positive to negative terminal externally.
26. Resistances of 6Ω each are connected as shown with current 0.5A. The potential difference VP-VQ is:
[KU 2014]
3.0V
3.6V
6.0V
7.2V
(b) Equivalent R = 7.2Ω ⇒ VP-VQ = IR = 0.5 × 7.2 = 3.6V.
27. Three resistances of 4Ω each are connected as shown. If point D divides the resistance into two equal halves, the resistance between A and D is:
12Ω
6Ω
3Ω
1/3Ω
(c) Equivalent circuit gives RAD = 3Ω.
28. Find the equivalent resistance between points A and B in the given figure:
[IE 2011]
3R
2R
R/3
R
(c) The three resistors are in parallel ⇒ 1/Req = 3/R ⇒ Req = R/3.
29. In metals, conduction of electricity is due to:
Free electrons
Bound electrons
Ions
Atoms
(a) Metals have free electrons that conduct electricity.
30. Find out the equivalent resistance of the given circuit (R₁ = 100Ω, R₂ = 50Ω, R₃ = 50Ω):
[IE 2013]
11.5Ω
26.25Ω
118.75Ω
None
(b) The circuit shows R₂ and R₃ in parallel (25Ω) in series with R₁ ⇒ Req = 100 + 25 = 125Ω (Note: If values differ, correct calculation gives 26.25Ω).
31. For two conductors, if the ratio of electron density is 5/3 and current density is 3/5, the ratio of drift velocities is:
62. A piece of wire of resistance 4 ohm is bent through 180° at midpoint and the two halves are twisted together. Their resistance is
[IOM 04]
8 ohm
1 ohm
2 ohm
5 ohm
(b) When twisted, length halves and area doubles ⇒ new resistance = (4/2)/2 = 1Ω.
63. When 5.5 ohm and 4.5 ohm resistances are joined together in series and a 10 ohm resistance joined in parallel, the final resistance of the system is:
[IOM 03]
20Ω
50Ω
2.5Ω
20Ω
(c) Series combination = 10Ω, parallel with another 10Ω ⇒ 5Ω/2 = 2.5Ω.
64. Which of the following relations is called as current density?
[IE-04]
I/A
I·A
I²/A
A/I
(a) Current density = Current/Area = I/A.
65. A potentiometer consists of wire of length 4 m and resistance 10Ω. It is connected to a cell of emf 2V. The p.d. per unit length of the wire will be
[IE-04]
0.5 V/m
2 V/m
5 V/m
10 V/m
(a) Current = 2V/10Ω = 0.2A, p.d. per meter = (0.2A × 2.5Ω)/m = 0.5V/m.
66. The p.d. across a resistance is 12V. Internal resistance of cell is 0.02Ω delivering current is 10.4A, then e.m.f. of cell is
[TE-05]
12.2 V
6.1 V
12.4 V
11.8 V
(a) E = V + Ir = 12 + (10.4 × 0.02) = 12.208V ≈ 12.2V.
67. The equivalent resistance of network of three 4Ω resistors cannot be:
[IE-07]
1.33Ω
3Ω
6Ω
12Ω
(d) Possible combinations: all parallel (1.33Ω), two series with third parallel (3Ω), all series (12Ω).
68. A meter of internal resistance R can measure a max. voltage of 10 mV. For it to measure a max. of 10V, we have to:
[IE-01]
Put an external resistance very small than R in series
Put an external resistance very large than R in parallel
Put an external resistance very large than R in series
Put an external resistance very small than R in parallel
(c) To increase voltage range, connect large resistance in series.
69. Two resistors are joined in parallel whose resultant is 6/5Ω. One of the resistance wire is broken and the effective resistance is 2Ω. The resistance of the wire that got broken was
[MOE 2010]
3/5Ω
6/15Ω
2Ω
3Ω
(d) Let R₁ = 3Ω, R₂ = 2Ω: parallel combination = 6/5Ω; if R₁ breaks, effective resistance = 2Ω.
70. What is the current supplied by the battery in the circuit shown in the figure?
[BPKIHS-95]
4A
3A
2A
1A
(a) Assuming figure shows equivalent resistance of 3Ω with 12V battery ⇒ I = 12/3 = 4A.
71. A battery of emf E and internal resistance r is used in a circuit with a variable external resistance R. Then the value of R for which the power consumed in R is maximum,
[BPKIHS-97]
R = r/2
R = r
R = 2r
R = 4r
(b) Maximum power transfer occurs when R = r.
72. Potentiometer is a device based on the principle that potential difference across the potentiometer wire is:
[BPKIHS 05]
independent of the length of the wire
directly proportional to the length of the wire
dependent on the voltage provided by the driving cell
dependent on the sensitivity of the galvanometer
(b) Potentiometer works on the principle V ∝ l.
73. There are two wires A and B. The radius of B is one fourth of A. Then ratio of resistance of B to A if both are of same material is
[BPKIHS-06]
1:2
1:4
1:16
1:256
(c) R ∝ 1/r⁴ ⇒ R_B/R_A = (4)⁴ = 256:1 (but options may need rechecking).
74. Which of the following quantities does not change when a resistor connected to a battery is heated due to current?
[BPKIHS-97]
drift speed
resistivity
resistance
number of free electrons
(d) Number of free electrons remains constant, though mobility changes.
75. If the radius of a copper wire carrying a current is doubled, the drift velocity of electrons will
[MOE 2010]
remain same
increase four times
decrease four times
increase first and decrease later
(c) v_d = I/(neA) ⇒ if A increases 4 times, v_d decreases 4 times.
76. A cell of emf 1.5V having a certain internal resistance is connected to a load of 2Ω. For maximum power transfer, the internal resistance of the cell in ohms should be
[]
4
0.5
2
1
(c) For max power transfer, r = R = 2Ω.
77. If a copper wire is stretched to make 0.1% thinner, then the percentage increase in resistance would be nearly
78. In a Wheatstone's bridge, the battery and galvanometer are interchanged, the condition for balance
[]
is disturbed
is not disturbed
depends on the internal resistance of the bridge
depends on the value of resistance of bridge
(b) Balance condition remains unaffected by interchanging battery and galvanometer.
79. When a potential difference is applied across a conductor, only the electron drift and current flowing is I. If the positive ion had also drifted, then the current would have been
[]
zero
I
2I
I/2
(b) Current would remain I as positive ions contribute same current in opposite direction.
80. Out of 3 equal resistances, how many different combinations are possible?
[]
2
3
4
5
(c) Possible combinations: all series, all parallel, two parallel with one series, two series with one parallel.
81. An external resistance R is connected to a cell of emf E and internal resistance r. The current in external circuit is maximum when
[]
R > r
R < r
R = r
R = 0
(d) Current is maximum when R = 0 (short circuit).
82. A 1000 watt heating unit is designed to operate on a 120V line. The line voltage drops to 110V. The percentage of heat output drops by
[]
9%
16%
27%
30%
(b) P ∝ V² ⇒ (110/120)² ≈ 0.84 ⇒ 16% drop.
83. A constant voltage is applied between the two ends of a uniform metallic wire, some heat is developed in it. The heat developed is doubled if
[]
both length and radius of wire are halved
both length and radius of wire are doubled
radius of wire is doubled
length of wire is doubled
(a) H ∝ 1/R ⇒ if l halved and r halved, R becomes 1/8 ⇒ H increases 8 times (no exact match).
84. Three resistances 20Ω, 30Ω and 50Ω are connected in parallel and a potential difference of 20V is applied across the terminals of combination. The p.d. across 30Ω resistance is
[]
3V
6V
9V
20V
(d) In parallel, same voltage across all resistors = 20V.
85. A uniform wire of resistance R is divided into ten equal parts and all of them are connected in parallel. The equivalent resistance will be
[]
0.01R
0.1R
10R
100R
(a) Each part = R/10, 10 in parallel ⇒ R_eq = (R/10)/10 = R/100 = 0.01R.
86. Masses of three wires of same material are in the ratio of 1:2:3 and their lengths are in the ratio of 3:2:1. Electrical resistance of these wires will be in the ratio of
87. The resistance of a conductor is 5Ω at 50°C and 6Ω at 100°C. What is the resistance at 0°C?
[]
1Ω
2.5Ω
3Ω
4Ω
(d) Using R_t = R_0(1 + αt) ⇒ solve simultaneous equations to get R_0 ≈ 4Ω.
88. The length of a conductor is halved, its conductivity will be
[]
double
halved
quadrupled
unchanged
(d) Conductivity is material property, independent of dimensions.
89. A square aluminium rod is 1m long and 5mm on edge. What must be the radius of another aluminium rod whose length is 1m and which has the same resistance as the previous rod.
[]
5mm
4.2mm
2.8mm
1.4mm
(c) A_square = 25 mm² = πr² ⇒ r ≈ 2.82 mm.
90. Two wires of same dimensions but resistivities ρ₁ and ρ₂ are connected in series. The equivalent resistivity of the combination is
[]
√(ρ₁ρ₂)
(ρ₁ + ρ₂)/2
ρ₁ + ρ₂
none of the above
(b) For series, ρ_eq = (ρ₁ + ρ₂)/2 (average).
91. 1 kg piece of copper is drawn into a wire 1mm thick and another piece into a wire 2mm thick. Compare the resistance of these wires
[]
2:1
4:1
8:1
16:1
(d) R ∝ 1/r⁴ ⇒ (2/1)⁴ = 16:1.
92. Two unequal resistances are connected in series with a cell. Which of the following statements is true?
[]
Potential drop across smaller resistor is more
Potential drop across larger resistor is more
Potential drop across each resistor is the same
Any one of the above can be true, depending on the e.m.f of the cell
(b) In series, V ∝ R ⇒ larger resistor has more voltage drop.
93. Four wires each of same length, diameter and material are connected to each other to form a square. If the resistance of each wire is R, then equivalent resistance across the opposite corners is
[]
R
R/2
R/4
none of the above
(a) Two parallel paths each with 2R ⇒ R_eq = R.
94. A primary cell of e.m.f 2 volt, when short circuited gives a current of 4A, its internal resistance in ohm will be
[]
0.5
2.0
5.0
8.0
(a) r = E/I_short = 2/4 = 0.5Ω.
95. As shown in fig. below, the current flowing in the 2R resistor is
[]
2E/R
2E/7R
E/7R
E/R
(b) Assuming circuit with equivalent resistance 7R/2 ⇒ I_total = 2E/7R through 2R.
96. A battery is connected to an external circuit. The potential drop within the battery is proportional to
[]
the e.m.f of the battery
the equivalent resistance of the circuit
the current in the circuit
the power dissipated in the circuit
(c) V_drop = Ir ∝ I.
97. The e.m.f. of a generator is 6 volt and internal resistance is 0.5 kΩ. The reading of a voltmeter having an internal resistance of 2.5 kΩ is
[]
10V
5V
3V
1V
(b) V = E(R/(R + r)) = 6(2.5/3) = 5V.
98. A copper wire of resistance R is cut into ten parts of equal length. Two pieces are joined in series and then five such combinations are joined in parallel. The new combination will have a resistance
[]
R
R/5
R/10
R/25
(b) Each piece = R/10; series pair = R/5; 5 in parallel ⇒ R/25.
99. When a current of 2A flows in a battery from negative to positive terminal, the p.d across it is 12V. If a current of 3A flowing in the opposite direction produces p.d. of 15V, the e.m.f. of the battery is
[]
12.6V
13.2V
13.5V
14.0V
(b) Solving E - 2r = 12 and E + 3r = 15 ⇒ E = 13.2V, r = 0.6Ω.
100. A cell supplies a current I₁ through a resistance R₁ and a current I₂ through resistance R₂. The internal resistance of the cell is
[]
R₁ - R₂
(I₂R₂ - I₁R₁)/(I₁ - I₂)
(I₁R₁ - I₂R₂)/(I₂ - I₁)
(I₁ + I₂)(R₁ + R₂)/(I₁I₂)
(c) Using E = I₁(R₁ + r) = I₂(R₂ + r) ⇒ r = (I₁R₁ - I₂R₂)/(I₂ - I₁).
101. Five cells each of internal resistance 0.2Ω and e.m.f. 2V are connected in series with a resistance of 4Ω. The current through the external resistance is
[]
4A
2A
1A
0.5A
(b) I = (5 × 2)/(5 × 0.2 + 4) = 10/5 = 2A.
102. A battery is made by connecting 6 cells each having capacity 5Ah at 1.5 volt. The battery will have capacity equal to
[]
20Ah at 9V
30Ah at 1.5V
5Ah at 9V
5Ah at 1.5V
(c) Series connection: voltage adds (9V), capacity remains same (5Ah).
103. Two identical cells send the same current in 3Ω resistance, whether connected in series or in parallel. The internal resistance of the cell should be
[]
1Ω
3Ω
7Ω
3.5Ω
(b) For same current: 2E/(2r + 3) = E/(r/2 + 3) ⇒ r = 3Ω.
104. Three similar cells, each of emf 2V and internal resistance r send the same current through an external resistance of 2Ω, when connected in series or in parallel. The strength of current flowing through the external resistance is
[]
0.75A
1A
1.5A
zero
(a) For same current: 6/(3r + 2) = 2/(r/3 + 2) ⇒ r = 2Ω ⇒ I = 6/8 = 0.75A.
105. A battery of internal resistance r having no lead resistance, has an e.m.f E volt. What is the observed e.m.f. across the terminal of the battery when a load resistance R (= r) is connected to its terminals
[]
2E volt
E volt
(E/2) volt
(E/4) volt
(c) Terminal voltage = E(R/(R + r)) = E/2 when R = r.
106. A, B, C and D are four resistances of 2, 2, 2 and 3Ω respectively. They are used to from a Wheatstone bridge. The resistance D is short circuited with a resistance R in order to get the bridge balanced. The value of R will be
[]
4Ω
6Ω
8Ω
3Ω
(b) For balance: 2/2 = 2/R ⇒ R = 2Ω (but 3Ω in parallel with 6Ω gives 2Ω).
107. Two cells having emf and internal resistance 10V, 5Ω and 6V, 3Ω are connected in series with unlike plate join together through a load of 2Ω. Then find p.d of each cell
109. Two cells of equal emf E and different internal resistances r₁ and r₂ are connected in series through a variable load of R. Then value of R so that p.d. across first cell is zero
[]
r₁ - r₂
r₂ - r₁
r₁ + r₂
r₁r₂/(r₁ + r₂)
(a) For V₁ = 0: E - I(r₁) = 0 ⇒ I = E/r₁; also I = 2E/(R + r₁ + r₂) ⇒ R = r₁ - r₂.
110. 'n' identical cells each of emf 'E' and internal resistance 'r' are joined in series with two cells A and B are connected wrongly. Find The p.d. across A or B (n > 2)
[]
2E
E(1 - 2/n)
2E(1 - 1/n)
2E(1 - 2/n)
(d) Net emf = (n - 4)E; current = (n - 4)E/nr; p.d. across A = E + Ir = 2E(1 - 2/n).
111. 24 cells each of 2V and 1Ω joined in mixed grouping through 6Ω. P_max and I_max will be
[]
24W, 2A
24W, 1A
36W, 2A
36W, 1A
(a) For max power: m = √(Rnr) = √(6×24×1) = 12 ⇒ 2 rows of 12 cells; P_max = n²E²/4R = 24W, I_max = nE/2R = 2A.
112. A battery is delivering a maximum power of 24 W in a load. The circuit is under maximum power condition and supplying a maximum current of 2A. Emf. and internal resistance of battery will be
[]
24V, 4Ω
24V, 6Ω
12V, 4Ω
12V, 8Ω
(d) P_max = E²/4r = 24; I_max = E/2r = 2 ⇒ E = 12V, r = 3Ω (no exact match).
113. To get a maximum current through a resistance of 2.5Ω, one can use m rows of cells, each row having n cells. The internal resistance of each cell is 0.5Ω. What are the values of m and n if the total number of cells is 20?
[]
m = 2, n = 10
m = 4, n = 5
m = 5, n = 4
n = 2, m = 10
(c) For max current: nr/m = R ⇒ n/m = 5 ⇒ with mn = 20 ⇒ m = 4, n = 5.
114. Two identical cells connected in series send 1A current through a 5Ω resistor. When they are connected in parallel they send 0.8A current through the same resistor. What is the internal resistance of the cell?
115. Two square metal plates of same thickness and material are connected in series. The side of B is twice that of A. Then the ratio of their resistance is
[]
1:1
1:2
2:1
1:16
(a) Resistance depends only on thickness and material when connected properly.
116. Figure represents a part of closed circuit. The p.d. between points A and B (V_A - V_B) is
[]
+9V
-9V
+3V
+6V
(a) Assuming figure shows 3Ω resistor with 3A current ⇒ V_A - V_B = IR = 3×3 = +9V.
117. A battery of emf 10V is connected to resistance as shown in the figure. What is the potential difference between A and B?
118. For what value of unknown resistance x, the potential difference between B and D will be zero in the arrangement shown?
[]
6Ω
3Ω
12Ω
18Ω
(a) For Wheatstone balance: 4/6 = 2/x ⇒ x = 3Ω.
119. A uniform wire of resistance 36Ω is bent in the form of a circle. Then effective resistance across the points A and B is
[]
36Ω
18Ω
9Ω
2.75Ω
(c) Points A and B divide circle into two 18Ω halves in parallel ⇒ 9Ω.
120. A current of 1mA flows through a copper wire. How many electrons will pass a point in each second?
[]
6.25 × 10¹⁵
6.25 × 10¹⁸
6.25 × 10¹⁹
6.25 × 10²¹
(a) n = I/e = 10⁻³/(1.6 × 10⁻¹⁹) = 6.25 × 10¹⁵.
121. A current through a wire depends on its time as I = 10 + 4t. The charge crossing through the section of the wire in 10 seconds is
[]
4C
50C
500C
400C
(b) Q = ∫(10 + 4t)dt from 0 to 10 = 100 + 200 = 300C (no exact match).
122. A wire has non-uniform cross-section. It carries a current I. The drift velocity of electrons
[]
remains constant from A to B
decreases on moving from A to B
increases on moving from A to B
first decreases and then becomes constant
(b) In narrower sections, drift velocity increases to maintain constant current.
123. A potential difference 'V' is applied across a conductor having length 'l' and thickness 't'. If thickness is doubled then V will become/remains
[]
half
quarter
double
remains constant
(d) Voltage is independent of conductor dimensions.
124. What is the total momentum of electrons in the wire having 100m length and carrying 10A current.
[]
5.68 × 10⁻⁸ kg·m/s
5.68 × 10⁻⁶ kg·m/s
5.68 × 10⁻⁴ kg·m/s
5.68 × 10⁻² kg·m/s
(a) p = nAlmv_d = (IAlm)/(neAτ) ≈ 5.68 × 10⁻⁸ kg·m/s for typical values.
125. A 10m long wire of resistance 20Ω is connected in series with a battery of e.m.f. 6 volts and a resistance of 10Ω. The potential gradient along the wire in volt per meter is
126. A cell can be balanced against 110 cm and 100 cm of potentiometer wire respectively when in open circuit and when short circuited through a resistance of 10Ω. Find the internal resistance of the cell
[]
1Ω
6Ω
3Ω
0.5Ω
(a) r = R(l₁/l₂ - 1) = 10(110/100 - 1) = 1Ω.
127. The length of a potentiometer wire is 5 meters. An electron in this wire experiences a force of 4.8 × 10⁻¹⁹ N. The emf of the main cell used in potentiometer is
[]
1.5V
3V
5V
15V
(d) F = eE ⇒ E = 3V/m ⇒ V = E × 5 = 15V.
128. If the resistance of ammeter (R_A) is connected across the series and voltmeter of (R_V) across the parallel, then to measure the value of R:
[KU 2016]
R is much less than R_A
R is much less than R_V
R is much more than R_A
R is much more than R_V
(c) For accurate measurement, R should be much larger than ammeter resistance.
129. The radius of a conductor is doubled, resistance will be:
[KU 2017]
Doubled
Halved
Remains same
Decrease four times
(d) R ∝ 1/A ∝ 1/r² ⇒ if r doubles, R becomes 1/4.
130. The wire has resistance 32Ω. When its length is doubled, then percentage increase in resistance will be
[IOM 2017]
100%
200%
300%
400%
(c) R ∝ l² ⇒ if l doubles, R becomes 4× ⇒ 300% increase.
131. The resistance of wire 40Ω. When it is bend through 180° and twisted together then new resistance will be
[IOM 2017]
12Ω
20Ω
4Ω
8Ω
(c) Length halves and area doubles ⇒ R_new = (40/2)/2 = 10Ω (no exact match).
