All (28)Unattempted (28)Skipped (0)Correct (0)Wrong (0)
Q.1
WBCS Prelims 2004
What is stored in a storage cell?
A.Electric charge
B.Electric potential
C.Lead or some other metal
D.Chemical energy
Explanation
Why Correct: Storage cells (rechargeable batteries) fundamentally store chemical energy that converts to electrical energy during discharge. Distractor Analysis: Electric charge accumulates on electrodes during charging but represents energy transfer, not storage. Electric potential (voltage) is the energy per charge, not the stored quantity. Lead is an electrode material in lead-acid batteries, not what the cell stores. Takeaway: Primary cells (non-rechargeable) also store chemical energy but convert it irreversibly.
Answer or skip previous question to unlock.
Q.2
WBCS Prelims 2023
Combine three resistors 5Ω, 4.5Ω and 3Ω in such a way that the total resistance of this combination is maximum with value
A.12.5 Ω
B.13.5 Ω
C.14.5 Ω
D.16.5 Ω
Explanation
Core Formula/Logic: For maximum total resistance, connect all resistors in series: R_total = R1 + R2 + R3. For minimum resistance, connect in parallel: 1/R_total = 1/R1 + 1/R2 + 1/R3. Step-by-Step Solution: 1. To maximize resistance, use series combination.
2. Add all resistances: 5Ω + 4.5Ω + 3Ω = 12.5Ω. Common Pitfall: Adding incorrectly gives 5+4.5+3=12.5Ω (correct), but misadding as 5+4.5+3=12.5Ω matches option A. Option B (13.5Ω) comes from adding 5+4.5+4, option C (14.5Ω) from 5+4.5+5, option D (16.5Ω) from 5+4.5+7. Shortcut/Takeaway: For maximum resistance with given resistors, always connect in series and simply sum them. For minimum resistance, use parallel combination and calculate reciprocal sum.
Answer or skip previous question to unlock.
Q.3
WBCS Prelims 2021
A battery consists of 10 cells, each of emf 1V. If 2 cells are wrongly connected, the emf of the battery becomes
A.8V
B.10V
C.6V
D.12V
Explanation
Core Formula/Logic: Net emf = (Number of cells in correct polarity - Number of cells in reverse polarity) × emf per cell. Reverse-connected cells subtract their emf from the total. Step-by-Step Solution: 1. Total cells = 10, each emf = 1V.
2. 2 cells are reverse-connected, so they contribute -2V total.
3. 8 cells are correctly connected, contributing +8V total.
4. Net emf = 8V + (-2V) = 6V. Common Pitfall: Adding all cells ignoring reverse connection gives 10V (option B). Subtracting reverse cells only once (10-2=8) gives 8V (option A). Adding reverse cells instead of subtracting gives 12V (option D). Shortcut/Takeaway: For series-connected cells, net emf = (N_correct - N_reverse) × emf_per_cell. Each reverse cell reduces total by twice its individual emf relative to all being correct.
Answer or skip previous question to unlock.
Q.4
WBCS Prelims 2017
The electric appliances in a house are connected
A.in series
B.in parallel
C.either in series or in parallel
D.both in series and in parallel
Explanation
Why Correct: Household appliances connect in parallel to maintain independent operation at the same voltage (230V in India). Distractor Analysis: Series connection would cause all appliances to turn off if one fails, and voltage would divide among them. Appliances never use both series and parallel simultaneously in residential wiring. Some industrial or decorative lighting may use series, but not household appliances. Takeaway: Parallel circuits provide constant voltage across all branches, while series circuits provide constant current through all components.
Answer or skip previous question to unlock.
Q.5
WBCS Prelims 2014
Basically domestic electric wiring is a
A.parallel connection
B.series connection
C.combination of series and parallel connections
D.None of the above
Explanation
Why Correct: Domestic wiring uses parallel connections to maintain uniform voltage (230V in India) across all appliances, enabling independent operation. Distractor Analysis: Series connections cause cumulative voltage drops, preventing appliances from receiving full rated voltage. Combination circuits introduce unnecessary complexity and voltage regulation problems in standard home wiring. None of the above is wrong because parallel connection is the fundamental design. Takeaway: Parallel circuits keep voltage constant across branches while total current equals the sum of individual branch currents.
Answer or skip previous question to unlock.
Q.6
WBCS Prelims 2012
When a positively charged conductor is earthed, then –
A.Electrons flow from conductor to earth
B.Protoms flow from conductor to earth
C.Electrons flow from earth to conductor
D.Protoms flow from earth to conductor
Explanation
Why Correct: Earth provides an infinite reservoir of electrons that flow to the positively charged conductor to neutralize its charge deficiency. Distractor Analysis: Electrons flowing from conductor to earth would increase positive charge. Protons cannot flow in conductors as they're bound in atomic nuclei. Protons flowing from earth to conductor would increase positive charge. Takeaway: For negatively charged conductors, electrons flow from conductor to earth. Current direction convention (positive charge flow) is opposite to actual electron flow direction.
Answer or skip previous question to unlock.
Q.7
WBCS Prelims 2011
Two copper wires A and B have the same weight and the radius of B is half that of A. The ratio RA/RB of their resistances is
A.1/4
B.1/8
C.1/16
D.1/2
Explanation
Core Formula/Logic: Resistance R = ρl/A, where ρ is resistivity, l is length, A is cross-sectional area. For same material and weight, volume and thus A*l is constant. Step-by-Step Solution: 1. Let radius of A = r, radius of B = r/2. 2. Area A_A = πr^2, A_B = π(r/2)^2 = πr^2/4. 3. Same weight implies A_A*l_A = A_B*l_B. 4. So l_A/l_B = A_B/A_A = (πr^2/4)/(πr^2) = 1/4. 5. R_A = ρ*l_A/A_A, R_B = ρ*l_B/A_B. 6. Ratio R_A/R_B = (l_A/A_A)/(l_B/A_B) = (l_A/l_B)*(A_B/A_A) = (1/4)*(1/4) = 1/16. Common Pitfall: Using only area ratio (1/4) gives 1/4 (option A). Using only length ratio (1/4) gives 1/4 (option A). Multiplying incorrectly as 1/2*1/4 = 1/8 gives option B. Shortcut/Takeaway: For same material and weight, resistance is proportional to 1/(radius^4). Halving the radius increases resistance 16 times, so the ratio of the thicker wire's resistance to the thinner is 1/16.
Answer or skip previous question to unlock.
Q.8
WBCS Prelims 2010
In a house-hold wiring the appliances are connected in series.
A.True
B.False
C.In parallel
D.In mixed configuration
Explanation
Why Correct: Household appliances connect in parallel to maintain independent operation and constant voltage across each device. Distractor Analysis: Series wiring would cause all appliances to fail if one device breaks and would force current through each device sequentially. Parallel wiring allows each appliance to operate independently at full voltage. Mixed configurations exist in complex systems but not in basic household wiring. Takeaway: Parallel circuits dominate domestic wiring for reliability, while series circuits appear in simple strings like decorative lights.
Answer or skip previous question to unlock.
Q.9
WBCS Prelims 2010
Electrical energy is converted into mechanical energy by a
A.Thermostat
B.Motor
C.Dynamo
D.Rectifier
Explanation
Why Correct: An electric motor uses electromagnetic forces to convert electrical energy into rotational mechanical energy, powering devices from fans to industrial machinery. Distractor Analysis: A thermostat regulates temperature by switching circuits on/off. A dynamo (generator) converts mechanical energy into electrical energy, the reverse process. A rectifier converts alternating current to direct current electricity. Takeaway: Remember the energy conversion pairs: motor (electrical→mechanical), generator/dynamo (mechanical→electrical), microphone (sound→electrical), speaker (electrical→sound).
Answer or skip previous question to unlock.
Q.10
WBCS Prelims 2008
The resistance of which of the following decreases with the rise of temperature ?
A.Copper
B.Iron
C.Silicon
D.Mercury
Explanation
Why Correct: Silicon is a semiconductor whose resistance decreases with temperature rise because increased thermal energy excites more electrons into the conduction band, increasing conductivity. Distractor Analysis: Copper, iron, and mercury are metals whose resistance increases with temperature due to enhanced lattice vibrations that scatter conduction electrons more effectively. Takeaway: Germanium, carbon, and other semiconductors also exhibit negative temperature coefficients of resistance, while metals like aluminum, silver, and gold show positive coefficients.
Answer or skip previous question to unlock.
Q.11
WBCS Prelims 2008
The resistance of a 200 V-100 W bulb is
A.400 Ω
B.400 Ω only when it is connected to 200 volt mains
C.400 Ω when it is not glowing
D.2 Ω
Explanation
Core Formula/Logic: Electrical power P = V2/R, where P is power in watts, V is voltage in volts, and R is resistance in ohms. Step-by-Step Solution: 1. Given: P = 100 W, V = 200 V.
2. Rearrange formula: R = V2/P.
3. Calculate: R = (200*200)/100 = 40000/100 = 400 Ω.
4. This resistance value is the nominal resistance at rated voltage, regardless of connection status. Common Pitfall: Confusing resistance with power rating leads to R = P/V = 100/200 = 0.5 Ω, which is not among options. Misunderstanding that resistance changes with temperature might lead to selecting option B or C, but the question asks for the resistance of the bulb, which is its rated resistance. Shortcut/Takeaway: For any bulb rated V volts and P watts, resistance R = V2/P. Memorize: 200V-100W bulb always has R = 400 Ω.
Answer or skip previous question to unlock.
Q.12
WBCS Prelims 2007
The motion of which particle through a metallic wire is called electric current?
A.Electron
B.Positron
C.Neutron
D.Photon
Explanation
Why Correct: Electric current in metallic conductors consists of electron flow—negatively charged particles moving from lower to higher potential. Distractor Analysis: Positrons are antimatter particles with positive charge, not present in normal metallic conduction. Neutrons are neutral particles within atomic nuclei, not involved in charge transport. Photons are massless particles of electromagnetic radiation, not charge carriers in wires. Takeaway: Conventional current direction (positive to negative) opposes actual electron flow direction (negative to positive) in metallic conductors.
Answer or skip previous question to unlock.
Q.13
WBCS Prelims 2004
If the length and cross-section of a wire are both doubled the resistance will
A.increased 8 times
B.decreases 4 times
C.increase twice
D.remain unchanged
Explanation
Core Formula/Logic: Resistance R = ρL/A, where ρ is resistivity, L is length, A is cross-sectional area. Step-by-Step Solution: 1. Initial resistance: R1 = ρL/A. 2. New length L' = 2L. 3. New area A' = 2A (since cross-section doubled). 4. New resistance R2 = ρ(2L)/(2A) = ρL/A = R1. 5. Therefore resistance remains unchanged. Common Pitfall: Multiplying only length gives R2 = ρ(2L)/A = 2R1, which produces option C. Multiplying only area gives R2 = ρL/(2A) = R1/2, which produces option B. Multiplying both incorrectly as 2L × 2A = 4 gives R2 = ρ(4L)/A = 4R1, which produces option A. Shortcut/Takeaway: When both length and area change by factor k, resistance changes by factor k/k = 1. So doubling both cancels out exactly.
Answer or skip previous question to unlock.
Q.14
WBCS Prelims 2004
The specific resistance of a conductor depends on its
A.length
B.width
C.temperature
D.shape of the cross-section
Explanation
Why Correct: Specific resistance (resistivity) is an intrinsic property of the material, defined as ρ = R × A / L, where R is resistance, A is cross-sectional area, and L is length. It depends on temperature because atomic vibrations increase with temperature, scattering electrons more and increasing resistivity. Distractor Analysis: Length and width affect the conductor's resistance but not its specific resistance. Shape of the cross-section affects area but not the material's intrinsic resistivity. Takeaway: For most metals, resistivity increases linearly with temperature: ρ = ρ0[1 + α(T - T0)], where α is the temperature coefficient of resistivity.
Answer or skip previous question to unlock.
Q.15
WBPSC Miscellaneous Prelims 2009WBCS Prelims 2004
The only vector quantity among the following is
A.electric charge
B.electric potential
C.electric field intensity
D.electric resistance
Asked 2 times in WBCS. High priority question.
Explanation
Why Correct: Electric field intensity (E = F/q) is a vector quantity with both magnitude and direction, measured in volts/meter. Distractor Analysis: Electric charge is a scalar quantity measured in coulombs. Electric potential is a scalar quantity measured in volts. Electric resistance is a scalar quantity measured in ohms. Takeaway: In electromagnetism, other vector quantities include magnetic field intensity, current density, and Poynting vector, while scalar quantities include electric flux, capacitance, and inductance.
Answer or skip previous question to unlock.
Q.16
WBCS Prelims 2004
The current on a 100W, 220V electric bulb is
A.2.2 amp.
B.1.1 amp.
C.5/11 amp.
D.22000 amp.
Explanation
Core Formula/Logic: Power formula: I = P ÷ V, where I is current in amperes, P is power in watts, and V is voltage in volts. Step-by-Step Solution: 1. Given P = 100 W, V = 220 V.
2. Apply formula: I = 100 ÷ 220.
3. Simplify: I = 10 ÷ 22 = 5 ÷ 11 A. Common Pitfall: Dividing 220 by 100 gives 2.2 A (option A). Forgetting to divide gives 22000 A (option D). Incorrect simplification yields 1.1 A (option B). Shortcut/Takeaway: For quick mental calculation, note that 100/220 = 10/22 = 5/11 ≈ 0.4545 A. Memorize that I = P/V always gives current in amperes when power is in watts and voltage in volts.
Answer or skip previous question to unlock.
Q.17
WBCS Prelims 2001
The emf of a cell does not depend on
A.the size of the cell
B.the material of cathode
C.the material of anode
D.electrolyte used
Explanation
Why Correct: Electromotive force (emf) depends on the electrode materials and electrolyte composition, which determine the standard electrode potentials and concentration differences, but not on the physical dimensions of the cell. Distractor Analysis: The material of cathode and anode directly determines the standard reduction potentials that contribute to the cell potential. The electrolyte used affects the concentration and activity of ions, influencing the Nernst equation calculation of emf. The size of the cell only affects current capacity and internal resistance, not the thermodynamic potential difference. Takeaway: For a galvanic cell, Ecell = Ecathode - Eanode under standard conditions, and E = E0 - (RT/nF)lnQ under non-standard conditions, where Q depends on electrolyte concentrations.
Answer or skip previous question to unlock.
Q.18
WBPSC Miscellaneous Prelims 2023
What kind of electricity is associated with Nikola Tesla’s invention?
A.Alternating current
B.Direct current
C.Static electricity
D.Ball lightning
Explanation
Why Correct: Nikola Tesla developed and promoted alternating current (AC) electrical systems, which became the standard for power transmission. Distractor Analysis: Direct current (DC) was associated with Thomas Edison. Static electricity involves stationary charges, not Tesla's main work. Ball lightning is a rare atmospheric phenomenon, not Tesla's invention. Takeaway: Tesla's key contributions include the AC motor, Tesla coil, and wireless communication. The AC vs. DC conflict is known as the 'War of Currents'.
Answer or skip previous question to unlock.
Q.19
WBPSC Miscellaneous Prelims 2019
The most important safety method used for protecting home appliances from short circuiting or overloading is
A.earthing
B.use of fuse
C.use of stabilizer
D.use of electric meter
Explanation
Why Correct: A fuse is a thin wire that melts and breaks the circuit when current exceeds a safe value, directly protecting appliances from damage due to short circuit or overload. Distractor Analysis: Earthing provides a path for leakage current to ground, preventing shock but not overloading. A stabilizer regulates voltage fluctuations, not overcurrent. An electric meter measures energy consumption; it does not provide protection. Takeaway: Fuse and circuit breaker are the primary overcurrent protection devices. Earthing protects against electric shock, not overloading.
Answer or skip previous question to unlock.
Q.20
WBPSC Miscellaneous Prelims 2019
Which quantity remains constant in parallel connection of resistance?
A.Electric current flow rate
B.Potential Difference
C.Amount of electricity
D.Both the potential difference and amount of electricity
Explanation
Why Correct: In a parallel circuit, each resistor is connected directly across the voltage source, so the potential difference (voltage) across each branch is equal and remains constant. Current divides inversely with resistance, and the amount of electricity (charge) depends on current and time, which can vary. Distractor Analysis: Electric current flow rate (current) is not constant; it splits among branches. Amount of electricity (charge) = current × time, so it also varies with current. Option D incorrectly claims both voltage and charge remain constant. Takeaway: In parallel circuits, voltage is same across all components; in series circuits, current is same. This is a fundamental rule for solving circuit problems.
Answer or skip previous question to unlock.
Q.21
WBPSC Miscellaneous Prelims 2012
For measuring the flow of current, the device is used
A.Voltmeter
B.Galvanometer
C.Ammeter
D.Potentiometer
Explanation
Why Correct: An ammeter measures electric current (in amperes) and is connected in series with the circuit.Distractor Analysis: A voltmeter measures potential difference (voltage) in parallel. A galvanometer detects small currents but does not directly measure magnitude; it is a sensitive meter used in bridges. A potentiometer measures emf or potential difference by null deflection.Takeaway: Connect ammeter in series, voltmeter in parallel. Amperes = current, Volts = voltage.
Answer or skip previous question to unlock.
Q.22
WBPSC Miscellaneous Prelims 2010
A positively charged glass rod first attracts and then repels a suspended object. The suspended object is
A.Negatively charged
B.Positively charged
C.Uncharged and insulated
D.Earthed
Explanation
Why Correct: When a positively charged rod is brought near an uncharged insulated object, it induces opposite charge on the near side, causing attraction. After contact, the object acquires the same positive charge as the rod, leading to repulsion. Distractor Analysis: A negatively charged object would be attracted initially but upon contact charge neutralization may occur, preventing repulsion. A positively charged object would repel from the start. An earthed object cannot retain charge after contact, so no repulsion occurs. Takeaway: A charged object attracting then repelling a suspended object is classic evidence that the object was initially uncharged and insulated.
Answer or skip previous question to unlock.
Q.23
WBPSC Miscellaneous Prelims 2009
The current in a 100 W, 220V electric light is:
A.5/11 amp
B.1.1 amp
C.2.2 amp
D.2 amp
Explanation
Core Formula/Logic: I = P/V (Current = Power / Voltage) Step-by-Step Solution:
1. Given power P = 100 W, voltage V = 220 V.
2. Use formula I = P/V = 100 / 220.
3. Simplify fraction: divide numerator and denominator by 20 -> 5/11 A. Common Pitfall: Common mistake: using I = V/P gives 220/100 = 2.2 A, option C. Also using I = V/P incorrectly gives 2.2 A. Another error: forgetting to divide and multiplying gives 22000, not an option. Shortcut/Takeaway: For resistive loads, I = P/V directly. For quick fraction, 100/220 = 10/22 = 5/11. Alternatively, compute approximate decimal 100/220 ≈ 0.4545 A, matching 5/11 ≈ 0.4545.
Answer or skip previous question to unlock.
Q.24
WBPSC Miscellaneous Prelims 2008
When the temperature of a metal rises, its electrical resistance:
A.Increases
B.Decreases
C.Does not change
D.None of the above
Explanation
Why Correct: In metals, as temperature increases, lattice vibrations intensify, scattering conduction electrons more, thereby increasing electrical resistance. Distractor Analysis: For semiconductors, resistance decreases with temperature. 'Does not change' is false; 'None of the above' is incorrect because 'Increases' is correct. Takeaway: For pure metals, the temperature coefficient of resistance is positive. Resistance increases approximately linearly with temperature.
Answer or skip previous question to unlock.
Q.25
WBPSC Miscellaneous Prelims 2008
Production of heat due to flow of electric current through a conductor is given by:
A.Joule effect
B.Joule-Thomson effect
C.Seebeck effect
D.Peltier effect
Explanation
Why Correct: Joule effect (Joule heating) states that heat produced in a conductor is proportional to the square of current times resistance (H = I^2 R t). Distractor Analysis: Joule-Thomson effect involves temperature change during adiabatic expansion of a gas; Seebeck effect generates voltage from a temperature difference; Peltier effect absorbs or releases heat at the junction of two dissimilar materials when current flows. None describe resistive heating. Takeaway: Resistive heating is always Joule heating. Formula: H = I^2 R t. Remember: 'Joule' alone for current heating; 'Joule-Thomson' for gas cooling.
Answer or skip previous question to unlock.
Q.26
WBPSC Miscellaneous Prelims 2008
A.C. is used in homes because:
A.It is safe
B.It is easily reproducible
C.It is cheap
D.It is economical in transmission
Explanation
Why Correct: Alternating current (AC) is used in homes primarily because it can be stepped up to high voltage for efficient long-distance transmission, reducing power loss (I2R loss), and then stepped down for safe domestic use. Distractor Analysis: AC is not inherently safer than DC; both can be dangerous. AC is not more easily reproducible or cheaper to generate than DC; the key advantage is economical transmission via transformers. Takeaway: The transformer is the key: AC can be transformed in voltage easily, making transmission economical. DC cannot be efficiently transformed.
Answer or skip previous question to unlock.
Q.27
WBPSC Miscellaneous Prelims 2007
A copper wire of circular cross section has a resistance of 2 Ω. If the length and radius of cross section of the wire both become half of its original value, then its resistance will be
A.2 Ω
B.1 Ω
C.8 Ω
D.4 Ω
Explanation
Why Correct: Resistance is directly proportional to length and inversely proportional to cross-sectional area. Halving the length halves the resistance, while halving the radius reduces the area to one-fourth, increasing the resistance fourfold. The net effect gives a new resistance of 4 Ω. Distractor Analysis: 2 Ω is the original resistance before any change. 1 Ω would be the resistance if only the length were halved without changing the radius. 8 Ω would result if only the radius were halved without changing the length. Takeaway: When both length and radius are halved, the resistance changes by a factor of (1/2) × (4) = 2, so it doubles from 2 Ω to 4 Ω.
Answer or skip previous question to unlock.
Q.28
WBCS prelims 2024
Resistance of a uniform metal wire of length 1 m and cross-sectional area 1 cm² is 10 Ω. Resistivity of the wire is
A.1/10 Ωcm
B.1 Ωcm
C.10 Ωcm
D.100 Ωcm
Explanation
Why Correct: Resistivity ρ = R × A / L. Convert length to cm: 1 m = 100 cm, area = 1 cm², R = 10 Ω. Thus ρ = 10 × 1 / 100 = 0.1 Ωcm, which equals 1/10 Ωcm. Distractor Analysis: 1 Ωcm would result from using length as 10 cm instead of 100 cm. 10 Ωcm equals the resistance value without any unit conversion. 100 Ωcm would result from multiplying resistance by length instead of dividing. Takeaway: Resistivity is a material property independent of the wire's dimensions; for a given material, changing length or cross-sectional area alters resistance but not resistivity. The SI unit of resistivity is ohm-meter (Ωm), and 1 Ωm equals 100 Ωcm.
Sign in to save progress
Sign in to Papersetters
Save your progress, unlock Smart Review, and track your performance.