[PMC Practice 8]
If the peak voltage of an AC signal is \( 9\text{ V} \), the corresponding peak-to-peak voltage of the waveform is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:This covers the conversion between peak voltage (amplitude) and peak-to-peak voltage for a symmetric alternating wave.
Formula:$$V_{p-p} = 2 V_p$$
Solution:Given \( V_p = 9\text{ V} \):
$$V_{p-p} = 2 \times 9\text{ V} = 18\text{ V}$$
Why other options are incorrect:- 4.5 V is the result of dividing by 2 instead of multiplying.
- 9 V is simply the peak amplitude.
[PMC Practice 28]
The time duration required to complete one entire cycle of an alternating quantity is defined as the:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:This is a definition-based question covering basic parameters of alternating current theory.
Solution:The
Time Period (or simply period, \( T \)) is defined as the precise duration in seconds required for an alternating signal to complete one full cycle of positive and negative values.
Why other options are incorrect:- Cycle is the physical repeating unit itself, not its duration.
- Instantaneous value is the value of voltage or current at any specific split second of time.
[PMC Practice 28]
The duration of one complete wave cycle of alternating current is known as its:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:This is a duplicate test question with a modified option arrangement to ensure consistency.
Solution:The time taken for an alternating wave to complete one full cycle is called its
period.
Why other options are incorrect:- This matches the core definition of the waveform period, rendering other choices incorrect.
[PMC Practice 33]
The ozone layer in the upper atmosphere protects life on Earth by absorbing and scattering which type of radiation from the sun?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The ozone layer acts as a protective shield in the Earth's stratosphere.
Solution:The
ozone layer absorbs a significant portion (97% to 99%) of the high-energy, harmful
ultraviolet (UV) radiation emitted by the sun, particularly UV-B and UV-C rays. This protection prevents skin cancer and ecological damage.
Why other options are incorrect:- Ozone does not significantly absorb Infrared (heat) radiation or Alpha particles (which are absorbed by the upper atmosphere long before reaching the stratosphere).
[PMC Test 1]
Energy emitted in the form of visible radiation is called:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Visible radiation is a portion of the electromagnetic spectrum that can be detected by the human eye.
Solution:The electromagnetic energy emitted in the form of visible radiation is defined as
light energy. It allows us to see objects and travels through a vacuum in the form of photons at the speed of light.
Why other options are incorrect:- Sound energy is mechanical vibrational energy and is not electromagnetic or visible.
- Geothermal and Tidal energy are mechanical and thermal energy forms harvested from natural planetary forces, not visible electromagnetic emissions.
#6 of 54
Balochistan MDCAT 2025
[Balochistan MDCAT 2025]
If the capacitance in a purely capacitive AC circuit is doubled, the current will be:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:In an AC circuit containing a pure capacitor, alternating current is inversely proportional to capacitive reactance and directly proportional to capacitance.
Formula / Reaction:$$I = \frac{V}{X_C} = 2\pi f C V$$
Solution:- Capacitive reactance is defined as \(X_C = \frac{1}{2\pi f C}\).
- Substituting \(X_C\) into the current equation gives \(I = \frac{V}{X_C} = 2\pi f C V\).
- Because current is directly proportional to capacitance (\(I \propto C\)), doubling the capacitance (\(C' = 2C\)) results in doubling the current (\(I' = 2I\)).
Why other options are incorrect:Because current increases in direct proportion to capacitance (\(I \propto C\)), it cannot be halved, remain unchanged, or decrease to one-fourth.
#7 of 54
Federal MDCAT 2025
[Federal MDCAT 2025]
A capacitor is connected to an ac source. If the frequency of the AC source is doubled the current in a purely capacitive circuit will:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In a purely capacitive AC circuit, the alternating current is directly proportional to the supply frequency because capacitive reactance decreases as frequency rises.
Formula / Reaction:$$I = \frac{V}{X_C} = 2\pi f C V$$
Solution:- The reactance of a capacitor is given by \(X_C = \frac{1}{2\pi f C}\).
- The alternating current is \(I = \frac{V}{X_C} = V(2\pi f C) = 2\pi f C V\).
- Since \(I \propto f\), doubling the source frequency (\(f' = 2f\)) directly doubles the current in the circuit (\(I' = 2I\)).
Why other options are incorrect:Increasing frequency reduces capacitive opposition to current flow; hence, current cannot remain unchanged, halve, or drop to zero.
#8 of 54
Sindh MDCAT 2025
[Sindh MDCAT 2025]
If frequency of AC is doubled, the inductive reactance will:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Inductive reactance is the opposition offered by an inductor to the flow of alternating current, and it is directly proportional to the frequency of the AC source.
Formula / Reaction:$$X_L = 2\pi f L$$
Solution:- Inductive reactance is expressed by the relation \(X_L = 2\pi f L\).
- For a given inductor of fixed inductance \(L\), \(X_L \propto f\).
- When the AC frequency is doubled (\(f' = 2f\)), the inductive reactance doubles accordingly (\(X_L' = 2 X_L\)).
Why other options are incorrect:Since \(X_L\) is directly proportional to frequency, doubling \(f\) doubles \(X_L\); it cannot remain constant, reduce by half, or become zero.
[KPK MDCAT 2025]
In a pure capacitance AC circuit, the current:
A
Lags behind voltage by \(90^\circ\)
B
Is in phase with the voltage
C
Leads the voltage by \(45^\circ\)
D
Leads the voltage by \(90^\circ\)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:In a purely capacitive AC circuit, the alternating current leads the alternating voltage across the capacitor by a phase angle of \(90^\circ\) (\(\frac{\pi}{2}\text{ rad}\)).
Formula / Reaction:$$q = C V \implies I = \frac{\Delta q}{\Delta t}$$
Solution:- When an alternating voltage \(V = V_0 \sin(\omega t)\) is applied across a capacitor, the instantaneous current is \(I = I_0 \cos(\omega t) = I_0 \sin\left(\omega t + \frac{\pi}{2}\right)\).
- The rate of change of charge (and hence current) reaches its maximum when the voltage is zero.
- Therefore, current leads the alternating voltage by a phase difference of \(90^\circ\) (or \(\frac{\pi}{2}\text{ rad}\)).
Why other options are incorrect:Current lagging behind voltage by \(90^\circ\) occurs in purely inductive circuits; current in-phase with voltage occurs in purely resistive circuits; a \(45^\circ\) phase angle occurs only in series \(RC\) circuits where \(R = X_C\).
[KPK MDCAT 2025]
The wave that has the highest frequency and penetrating power is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:In the electromagnetic spectrum, photon energy and penetrating power increase directly with frequency and inversely with wavelength.
Formula / Reaction:$$E = h f = \frac{h c}{\lambda}$$
Solution:- The electromagnetic spectrum ordered by increasing frequency is: Radio waves < Microwaves < Infrared < Visible light < Ultraviolet < X-rays < Gamma rays.
- Gamma rays have the shortest wavelengths (\(\lambda < 10^{-11}\text{ m}\)) and highest frequencies (\(f > 10^{19}\text{ Hz}\)).
- Because photon energy is directly proportional to frequency (\(E = hf\)), Gamma rays carry the highest energy per quantum and have the greatest penetrating power.
Why other options are incorrect:X-rays, Ultraviolet rays, and Microwaves possess longer wavelengths, lower frequencies, and lower photon energies than Gamma rays, giving them significantly lower penetrating power.
Which electromagnetic wave is used to kill bacteria and insects in food?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Food irradiation utilizes highly penetrating ionizing electromagnetic radiation to neutralize pathogens, bacteria, and pests without raising the temperature of food.
Formula / Reaction:Qualitative concept / No formula required.
Solution:- Gamma rays are high-energy electromagnetic waves emitted by radioisotopes (such as Cobalt-60).
- Due to their high penetrating capability, gamma rays pass completely through packaged food items and disrupt the cellular DNA of bacteria, parasites, and insects.
- This sterilizes the food, retards spoilage, and extends shelf life safely.
Why other options are incorrect:Alpha rays (helium nuclei) and Beta rays (high-speed electrons) are particulate radiations with extremely low penetration depth; UV rays are non-ionizing electromagnetic radiation that only disinfect surface layers.
#12 of 54
2022 PMC Test 1
[2022 PMC Test 1]
A simple AC generator consists of how many magnetic poles?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:An elementary AC generator requires a magnetic dipole to provide a uniform magnetic field across the rotating armature coil.
Formula / Reaction:$$\Phi = B A \cos(\omega t), \quad \varepsilon = -N \frac{d\Phi}{dt} = N B A \omega \sin(\omega t)$$
Solution:- A standard basic generator consists of at least one pair of magnetic poles: a North pole and a South pole (2 poles).
- Magnetic field lines pass from the North to South pole across the armature.
Why other options are incorrect:- Opt_A: Magnetic fields cannot exist without poles.
- Opt_B: Isolated magnetic monopoles do not exist in nature.
- Opt_D: Magnetic poles always occur in conjugate pairs (even numbers: 2, 4, 6...).
#13 of 54
PMC 2021 Tested
[PMC 2021 Tested]
Which of the following types of radiation is an electromagnetic wave and therefore requires NO material medium to propagate?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Electromagnetic waves consist of self-propagating electric and magnetic fields, allowing them to travel through a vacuum without a supporting physical medium.
Solution:- Gamma rays are high-energy photons located at the short-wavelength end of the electromagnetic spectrum. Because they are electromagnetic waves, they require no physical medium to propagate.
- Alpha rays (helium nuclei) and Beta rays (high-speed electrons or positrons) are streams of physical particles with mass and charge. They are matter streams, not electromagnetic waves.
Why other options are incorrect:- Alpha and Beta rays are matter waves/streams that carry mass, so they are not electromagnetic waves and do not propagate in the same manner.
#14 of 54
PMC 2021 Tested
[PMC 2021 Tested]
The narrow spectral region that lies between the Infrared (IR) and Ultraviolet (UV) regions of the electromagnetic spectrum is called:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The electromagnetic spectrum is classified into bands based on wavelength and frequency.
Solution:The
visible light band occupies the narrow range of wavelengths from approximately \( 400\text{ nm} \) to \( 700\text{ nm} \). This band is positioned directly between the longer-wavelength Infrared (IR) band and the shorter-wavelength Ultraviolet (UV) band.
Why other options are incorrect:- X-rays lie at shorter wavelengths beyond the UV band.
- Radio waves lie at longer wavelengths beyond the IR band.
#15 of 54
PMC 2021 Tested
[PMC 2021 Tested]
Microwaves are best classified as ____ waves.
A
Longitudinal and mechanical
B
Longitudinal and electromagnetic
C
Transverse and electromagnetic
D
Transverse and mechanical
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Electromagnetic waves consist of electric and magnetic fields that vibrate perpendicular to the direction of wave travel.
Solution:Microwaves belong to the electromagnetic spectrum, making them
electromagnetic waves.
Because their oscillating electric and magnetic field vectors are oriented perpendicular to each other and perpendicular to the direction of propagation, they are also classified as
transverse waves. Therefore, they are both
transverse and electromagnetic.
Why other options are incorrect:- Options containing longitudinal or mechanical are incorrect because microwaves do not vibrate parallel to their direction of propagation and do not require a physical material medium to travel.
[ETEA 2019]
Which of the following is NOT an electromagnetic wave?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:This question distinguishes between electromagnetic waves and mechanical waves.
Solution:Sound waves are mechanical waves. They consist of pressure fluctuations (compressions and rarefactions) in a physical medium and require a material medium (like air, water, or solids) to travel. They cannot travel through a vacuum.
Radio waves, X-rays, and visible light waves are all electromagnetic waves made of oscillating electric and magnetic fields, so they can travel through a vacuum.
Why other options are incorrect:- Radio waves, X-rays, and light waves are all classified as electromagnetic waves because they consist of fluctuating electric and magnetic fields.
In an AC circuit containing only a pure inductor, what is the phase relationship between voltage and current?
A
Current leads voltage by \(90^\circ\)
B
Voltage leads current by \(90^\circ\)
C
Voltage and current are in phase
D
Voltage lags current by \(180^\circ\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In an inductor, back-EMF opposes rapid current changes. The voltage across the inductor peaks a quarter-cycle before the current.
Formula / Reaction:$$v_L(t) = V_0 \sin(\omega t), \quad i(t) = I_0 \sin\left(\omega t - \frac{\pi}{2}\right)$$
Solution:- Self-induction causes current to lag voltage by \(90^\circ\).
- Equivalently, the voltage across the inductor leads current by \(90^\circ\) (\(\pi/2\text{ rad}\)).
Why other options are incorrect:- Opt_A: Current leads voltage by \(90^\circ\) in a pure capacitor.
- Opt_C: Voltage and current are in phase in a pure resistor.
- Opt_D: A \(180^\circ\) lag represents complete phase opposition.
[ETEA 2017]
Which of the following arrangements correctly orders the regions of the electromagnetic spectrum in terms of decreasing wavelength (longest to shortest)?
A
Infrared \( > \) Ultraviolet \( > \) Visible \( > \) Microwave \( > \) Radio frequency
B
Radio frequency \( > \) Microwave \( > \) Infrared \( > \) Visible \( > \) Ultraviolet
C
Microwave \( > \) Infrared \( > \) Visible \( > \) Ultraviolet \( > \) Radio frequency
D
Visible \( > \) Infrared \( > \) Ultraviolet \( > \) Microwave \( > \) Radio frequency
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:This question requires organizing the electromagnetic spectrum by wavelength from longest to shortest.
Solution:Wavelength (\( \lambda \)) is inversely proportional to frequency. The correct order from longest wavelength (lowest frequency) to shortest wavelength (highest frequency) is:
$$\text{Radio frequency} > \text{Microwave} > \text{Infrared} > \text{Visible Light} > \text{Ultraviolet}$$
Why other options are incorrect:- Other options place shorter wavelength bands (like ultraviolet or visible light) before longer wavelength bands (like microwave or infrared), violating the order of decreasing wavelength.
[ETEA 2016]
A.C and D.C have the same:
A
Effect in charging battery
B
Effect in charging capacitor
C
Heating effect through a resistance
D
Effect passing through an inductance
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The root mean square (RMS) value of alternating current is defined strictly by the thermal Joule heating effect in a pure resistor, which is independent of current direction.
Formula / Reaction:$$H = I_{\text{rms}}^2 R t = I_{\text{DC}}^2 R t$$
Solution:- Joule's heating law states \(P = I^2 R\). Because current is squared, heat dissipation is always positive regardless of instantaneous polarity.
- An AC of root-mean-square value \(I_{\text{rms}}\) produces the exact same average thermal power in a resistor as a steady DC of magnitude \(I_{\text{DC}} = I_{\text{rms}}\).
Why other options are incorrect:- Opt_A: Battery charging relies on chemical electrolysis, which requires unidirectional DC. AC produces zero net chemical deposition over a cycle.
- Opt_B: DC charges a capacitor to a constant state and then stops, whereas AC continuously charges and discharges it.
- Opt_D: An inductor opposes AC via inductive reactance \(X_L = 2\pi f L\), but offers zero reactance to steady DC.
[ETEA 2016]
In purely resistive circuit the current _____.
A
Leads voltage by one half of a cycle
B
Leads voltage by one fourth of a cycle
C
Leads voltage by one and a half of a cycle
D
Is in phase with voltage
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:In an AC circuit containing only a pure ohmic resistor, the opposition to current flow is constant and does not depend on time or rate of change of current.
Formula / Reaction:$$v = V_0 \sin(\omega t), \quad i = \frac{v}{R} = \frac{V_0}{R} \sin(\omega t) = I_0 \sin(\omega t)$$
Solution:- Both voltage and current waveforms cross zero, increase, and reach their positive and negative maximums at the exact same instants.
- The phase difference \(\phi = 0^\circ\), meaning current and voltage are strictly in-phase.
Why other options are incorrect:- Opt_A: Leading by one-half cycle corresponds to \(\phi = 180^\circ\).
- Opt_B: Leading by one-fourth cycle corresponds to \(\phi = 90^\circ\), which occurs in a pure capacitor.
- Opt_C: Leading by \(1.5\) cycles corresponds to \(540^\circ\).
[ETEA 2016]
Which of the following electromagnetic rays has the smallest wavelength?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Across the electromagnetic spectrum, wavelength decreases as photon energy and frequency increase.
Formula / Reaction:$$\lambda_{\gamma} < \lambda_{\text{X-ray}} < \lambda_{\text{UV}} < \lambda_{\text{Visible}} < \lambda_{\text{IR}} < \lambda_{\text{Micro}} < \lambda_{\text{Radio}}$$
Solution:- Gamma rays have the highest frequencies (\(> 10^{19}\text{ Hz}\)) and the shortest wavelengths (\(\lambda < 10^{-11}\text{ m}\) or \(< 0.01\text{ nm}\)).
- Therefore, gamma rays have the smallest wavelength among all electromagnetic waves.
Why other options are incorrect:- Opt_A: X-rays have wavelengths around \(0.01\text{ nm to }10\text{ nm}\), which are longer than gamma rays.
- Opt_C: Microwaves have long wavelengths (\(1\text{ mm to }1\text{ m}\)).
- Opt_D: Ultraviolet rays have wavelengths between \(10\text{ nm}\) and \(400\text{ nm}\).
[ETEA 2016]
The potential difference and the current flowing through a component in an A.C. circuit are given by:
\(V = 5 \cos \omega t\text{ volts}\)
\(i = 2 \sin \omega t\text{ amperes}\)
The power dissipated in the instrument is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Average real power in an AC circuit depends on the cosine of the phase angle difference \(\phi\) between voltage and current.
Formula / Reaction:$$P = V_{\text{rms}} I_{\text{rms}} \cos\phi$$
Solution:- Using \(\cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right)\), the voltage equation becomes \(V = 5 \sin\left(\omega t + \frac{\pi}{2}\right)\).
- Current is \(i = 2 \sin(\omega t)\).
- The phase difference is \(\phi = 90^\circ\) (\(\pi/2\text{ rad}\)).
- \(P = V_{\text{rms}} I_{\text{rms}} \cos(90^\circ) = 0\text{ W}\).
Why other options are incorrect:- Opt_B: \(10\text{ W}\) is the product of peak values \(V_0 I_0\) assuming \(\cos\phi = 1\).
- Opt_C: \(5\text{ W}\) is \(\frac{1}{2} V_0 I_0\), valid only for in-phase waveforms.
- Opt_D: \(2.5\text{ W}\) is an incorrect fraction.
[ETEA 2016]
The potential difference and the current flowing through a component in an A.C. circuit are given by:
\(V = 3 \cos \omega t\text{ V}\)
\(i = 4 \sin \omega t\text{ A}\)
The circuit is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Rewriting both quantities as sine functions reveals their relative phase angle. If voltage leads current by \(90^\circ\), the circuit is purely inductive.
Formula / Reaction:$$V = 3 \cos(\omega t) = 3 \sin\left(\omega t + \frac{\pi}{2}\right), \quad i = 4 \sin(\omega t)$$
Solution:- Comparing phases: Voltage phase is \(\omega t + \pi/2\) and current phase is \(\omega t\).
- Voltage leads current by \(\frac{\pi}{2}\) (\(90^\circ\)).
- A \(90^\circ\) voltage lead is the characteristic signature of an inductive circuit.
Why other options are incorrect:- Opt_A: In a capacitive circuit, current leads voltage by \(90^\circ\).
- Opt_C: In a resistive circuit, voltage and current are in-phase.
- Opt_D: The phase relationship uniquely identifies an inductive circuit.
[ETEA 2016]
In pure inductance, the average power dissipated is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:In an ideal inductor, the current lags the applied voltage by \(90^\circ\). No electrical energy is converted into heat.
Formula / Reaction:$$P = V_{\text{rms}} I_{\text{rms}} \cos(90^\circ) = 0$$
Solution:- Energy is stored in the magnetic field during the first quarter-cycle and completely returned to the source during the next quarter-cycle.
- Therefore, the net average power dissipated over any complete cycle is identically zero.
Why other options are incorrect:- Opt_A, Option B, Option C: Non-zero values are incorrect because ideal inductors do not dissipate real power.
[ETEA 2016]
A \(35\ \mu\text{F}\) capacitor is connected to a source of sinusoidal e.m.f with a frequency of \(400\text{ Hz}\) and a maximum e.m.f of \(20\text{ V}\). The maximum current is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Peak current through a capacitor is determined by peak voltage divided by capacitive reactance: \(I_0 = \omega C V_0 = 2\pi f C V_0\).
Formula / Reaction:$$I_0 = 2\pi f C V_0$$
Solution:- Given: \(f = 400\text{ Hz}\), \(C = 35 \times 10^{-6}\text{ F}\), \(V_0 = 20\text{ V}\).
- \(I_0 = 2 \times 3.1416 \times 400 \times (35 \times 10^{-6}) \times 20\).
- \(I_0 = 2 \times 3.1416 \times 400 \times 700 \times 10^{-6} = 2 \times 3.1416 \times 0.28 \approx 1.76\text{ A} \approx 1.8\text{ A}\).
Why other options are incorrect:- Opt_A: Current is non-zero in an active AC circuit.
- Opt_B: \(0.28\text{ A}\) misses multiplying by \(2\pi\).
- Opt_D: Calculation error.
[ETEA 2016]
Which of the following are also called heat radiations?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Infrared radiation is readily absorbed by molecular vibrational modes in matter, causing an increase in thermal kinetic energy and temperature.
Formula / Reaction:$$\lambda_{\text{IR}} \approx 700\text{ nm to } 1\text{ mm}$$
Solution:- Warm objects emit electromagnetic radiation predominantly in the infrared band.
- Because its absorption produces noticeable heating in objects and biological tissue, infrared radiation is commonly termed heat radiation.
Why other options are incorrect:- Opt_A: Visible light stimulates human vision.
- Opt_B: Radio waves are used for communication.
- Opt_C: Microwaves are used in radar and dielectric heating.
[ETEA 2016]
The changing electric flux in a certain region of space produces:
C
Both electric and magnetic fields
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:According to the Maxwell-Ampere law, a time-varying electric field acts as a source of a magnetic field.
Formula / Reaction:$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_C + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$
Solution:- A changing electric flux \(\frac{d\Phi_E}{dt}\) induces a magnetic field \(\vec{B}\) in the surrounding space, forming the foundation of electromagnetic wave propagation.
Why other options are incorrect:- Opt_A: A changing magnetic flux produces an electric field (Faraday's law), not a changing electric flux.
- Opt_C: The direct entity generated by changing electric flux is a magnetic field.
- Opt_D: Option B is correct.
[NUMS 2015]
An A.C. emf of \( E = 100 \sin(100 \pi t) \) volt is connected across a choke of negligible resistance. In order to produce a current of amplitude \( 1\text{ A} \), the inductance of the choke should be:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:A choke is an inductor designed to limit AC current. For a purely inductive circuit, the opposition to the alternating current is called inductive reactance.
Formula:The inductive reactance is defined as:
$$X_L = \omega L$$
and the current amplitude (peak current) is:
$$I_0 = \frac{E_0}{X_L}$$
Solution:Compare the given voltage equation \( E = 100 \sin(100 \pi t) \) to the standard equation \( E = E_0 \sin(\omega t) \):
- Peak EMF, \( E_0 = 100\text{ V} \)
- Angular frequency, \( \omega = 100 \pi\text{ rad/s} \)
- Peak current, \( I_0 = 1\text{ A} \)
Substitute these values to solve for inductance \( L \):
$$I_0 = \frac{E_0}{\omega L} \implies 1 = \frac{100}{100 \pi \times L}$$
$$L = \frac{100}{100 \pi} = \frac{1}{\pi}\text{ H}$$
Why other options are incorrect:- Other options like \( 200\text{ H} \), \( 2\pi\text{ H} \), or \( 2/\pi\text{ H} \) are incorrect due to standard errors in algebraic rearrangement, such as multiplying by \( 2 \) instead of dividing or incorrect cancellation of \( 100 \).
[NUMS 2015]
In a series RLC circuit at resonance, the AC voltages across the Resistance \( R \), Inductance \( L \), and Capacitance \( C \) are \( 5\text{ V} \), \( 10\text{ V} \), and \( 10\text{ V} \) respectively. The AC voltage applied to the circuit will be:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:In a series RLC circuit, AC voltages must be added vectorially (as phasors) rather than algebraically because of the phase differences between them.
Formula:The total source voltage \( V \) is:
$$V = \sqrt{V_R^2 + (V_L - V_C)^2}$$
Solution:Given information:
- \( V_R = 5\text{ V} \)
- \( V_L = 10\text{ V} \)
- \( V_C = 10\text{ V} \)
At the resonance condition, the inductive reactance matches the capacitive reactance, which means the voltages across them are equal and in exact phase opposition (\( 180^\circ \) out of phase):
$$V = \sqrt{5^2 + (10 - 10)^2} = \sqrt{5^2 + 0} = 5\text{ V}$$
The total applied voltage is simply equal to the voltage drop across the resistor.
Why other options are incorrect:- 20 V and 25 V result from simple algebraic addition (e.g., \( 5 + 10 + 10 = 25\text{ V} \)), which completely ignores phase vector alignment.
- 10 V is the value of the reactive components alone.
[ETEA 2015]
The instantaneous value of e.m.f at time t is \(V = 20 \sin (100\pi t)\). The frequency of the alternating quantity is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The time-dependent sinusoidal voltage equation relates the argument of the sine function to angular frequency \(\omega = 2\pi f\).
Formula / Reaction:$$V = V_0 \sin(2\pi f t)$$
Solution:- Given: \(V = 20 \sin(100\pi t)\).
- Comparing coefficients: \(\omega = 2\pi f = 100\pi\).
- Solving for frequency: \(f = \frac{100\pi}{2\pi} = 50\text{ Hz}\).
Why other options are incorrect:- Opt_A: \(100\text{ Hz}\) incorrectly takes \(\omega / \pi\) instead of \(\omega / 2\pi\).
- Opt_C: \(25\text{ Hz}\) results from dividing by \(4\pi\).
- Opt_D: \(20\text{ V}\) is the peak voltage \(V_0\), not the frequency.
[ETEA 2015]
An A.C current varies with time (t) sec as \(i = 4 \sin (200\pi t)\), the r.m.s value of current in ampere is:
B
\(4 \sqrt{2}\text{ A}\)
C
\(\frac{2}{\sqrt{2}}\text{ A}\)
D
\(2 \sqrt{2}\text{ A}\)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The root mean square (RMS) current of a sinusoidal wave is related to the peak current by \(I_{\text{rms}} = \frac{I_0}{\sqrt{2}}\).
Formula / Reaction:$$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{I_0 \sqrt{2}}{2}$$
Solution:- From the equation \(i = 4 \sin(200\pi t)\), peak current \(I_0 = 4\text{ A}\).
- \(I_{\text{rms}} = \frac{4}{\sqrt{2}} = \frac{2 \times 2}{\sqrt{2}} = 2\sqrt{2}\text{ A}\).
Why other options are incorrect:- Opt_A: \(2\text{ A}\) incorrectly divides \(I_0\) by \(2\) instead of \(\sqrt{2}\).
- Opt_B: Multiplies \(I_0\) by \(\sqrt{2}\) instead of dividing.
- Opt_C: While \(4/\sqrt{2}\) is mathematically equivalent to \(2\sqrt{2}\), the simplified textbook form \(2\sqrt{2}\) is the standard final answer.
[ETEA 2015]
Phase angle between voltage and current in A.C through a pure inductor is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In a pure inductor, Faraday's law of electromagnetic induction generates a self-induced back emf \(e = -L \frac{di}{dt}\) that opposes changes in current.
Formula / Reaction:$$v_L(t) = L \frac{di}{dt} \implies v_L = V_0 \sin\left(\omega t + \frac{\pi}{2}\right)$$
Solution:- Voltage reaches its peak when the rate of change of current is maximum (at \(i = 0\)).
- Thus, alternating voltage leads the alternating current by \(90^\circ\) (\(\pi/2\text{ rad}\)), or current lags voltage by \(90^\circ\).
Why other options are incorrect:- Opt_A: \(0^\circ\) phase angle occurs in a pure resistor.
- Opt_C: \(60^\circ\) occurs in combination circuits containing both resistance and reactance.
- Opt_D: \(180^\circ\) represents complete phase inversion.
[ETEA 2015]
Peak value of alternating current is \(5\sqrt{2}\text{ A}\). The mean square value of current will be:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The mean square value of an alternating current waveform is half of the square of its peak value.
Formula / Reaction:$$\langle I^2 \rangle = \frac{I_0^2}{2}$$
Solution:- Given peak current: \(I_0 = 5\sqrt{2}\text{ A}\).
- \(I_0^2 = (5\sqrt{2})^2 = 25 \times 2 = 50\text{ A}^2\).
- \(\langle I^2 \rangle = \frac{50}{2} = 25\text{ A}^2\).
Why other options are incorrect:- Opt_A: \(5\text{ A}\) is the RMS value (\(I_{\text{rms}} = \sqrt{25} = 5\text{ A}\)), not the mean square value.
- Opt_C: \(5\sqrt{2}\text{ A}\) is the peak current \(I_0\).
- Opt_D: \(50\text{ A}^2\) is the peak square \(I_0^2\), without dividing by 2.
[ETEA 2015]
Which of the following is the smoothest waveform?
C
Saw-tooth shape waveform
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:A pure sinusoidal wave consists of a single, continuous harmonic frequency without higher-order harmonic distortions or sharp discontinuities.
Formula / Reaction:$$y(t) = A \sin(\omega t)$$
Solution:- Square, triangular, and saw-tooth waves contain sharp corners and infinite series of higher-frequency odd/even harmonics (Fourier components).
- A sine wave has a perfectly smooth, infinitely differentiable mathematical profile containing zero harmonic distortion, making it the smoothest AC waveform.
Why other options are incorrect:- Opt_B: Triangular waveforms have sharp periodic slope reversals.
- Opt_C: Saw-tooth waveforms have discontinuous vertical drop-offs.
- Opt_D: Rectangular waveforms have instantaneous step discontinuities.
[ETEA 2015]
At what frequency will a \(1\text{ H}\) inductor have a reactance of \(500\ \Omega\)?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Inductive reactance is related to frequency and inductance by \(X_L = 2\pi f L\).
Formula / Reaction:$$f = \frac{X_L}{2\pi L}$$
Solution:- Given: \(X_L = 500\ \Omega\) and \(L = 1\text{ H}\).
- \(f = \frac{500}{2\pi(1)} = \frac{250}{\pi} \approx \frac{250}{3.1416} \approx 79.6\text{ Hz} \approx 80\text{ Hz}\).
Why other options are incorrect:- Opt_B: \(60\text{ Hz}\) gives \(X_L \approx 377\ \Omega\).
- Opt_C: \(40\text{ Hz}\) gives \(X_L \approx 251\ \Omega\).
- Opt_D: \(20\text{ Hz}\) gives \(X_L \approx 126\ \Omega\).
[ETEA 2015]
An alternating voltage \(\varepsilon = 200 \sin (100 t)\) is connected to a \(1\ \mu\text{F}\) capacitor through an A.C. ammeter. The peak reading of current shall be:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The peak current through a pure capacitor is given by \(I_0 = \frac{V_0}{X_C} = \omega C V_0\).
Formula / Reaction:$$I_0 = \omega C V_0$$
Solution:- From \(\varepsilon = 200 \sin(100t)\): \(V_0 = 200\text{ V}\) and \(\omega = 100\text{ rad/s}\).
- Given: \(C = 1\ \mu\text{F} = 1 \times 10^{-6}\text{ F}\).
- \(I_0 = (100)(1 \times 10^{-6})(200) = 2 \times 10^{-2}\text{ A} = 20\text{ mA}\).
Why other options are incorrect:- Opt_A: \(10\text{ mA}\) uses \(V_0 = 100\text{ V}\).
- Opt_B: \(80\text{ mA}\) is an incorrect arithmetic scaling.
- Opt_D: \(40\text{ mA}\) results from doubling the voltage.
[ETEA 2015]
The slope of the graph between inductive reactance and the frequency of an A.C source is related to:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Inductive reactance is a linear function of frequency: \(X_L = (2\pi L) f\). Plotting \(X_L\) versus \(f\) produces a straight line through the origin with slope proportional to inductance.
Formula / Reaction:$$X_L = (2\pi L) f \implies \text{Slope} = \frac{\Delta X_L}{\Delta f} = 2\pi L$$
Solution:- The gradient of the \(X_L - f\) graph is \(2\pi L\).
- Dividing the slope by the constant \(2\pi\) gives the self-inductance \(L\) of the coil.
Why other options are incorrect:- Opt_A: Impedance is total opposition in combination circuits.
- Opt_B: Resistance is independent of frequency.
- Opt_C: Power factor is a dimensionless ratio.
[ETEA 2015]
Which of the following is not an electromagnetic radiation?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Electromagnetic radiations are massless photons propagating at the speed of light. Cathode rays are streams of material particles (electrons).
Formula / Reaction:$$\text{Cathode rays} = \text{beam of electrons } (m_e = 9.11 \times 10^{-31}\text{ kg}, q = -e)$$
Solution:- Cathode rays possess mass, carry negative electric charge, and are deflected by electrostatic and magnetic fields.
- X-rays, UV rays, and infrared rays are uncharged, massless electromagnetic waves.
Why other options are incorrect:- Opt_A, Option B, Option D: All are standard regions of the electromagnetic spectrum.
[ETEA 2015]
For the production of electromagnetic waves, the electric charges must be:
B
Charges moving with uniform velocity
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Stationary charges produce static electric fields, charges in uniform motion produce steady magnetic fields, and accelerated charges radiate electromagnetic energy.
Formula / Reaction:$$a = \frac{dv}{dt} \ne 0 \implies \text{Electromagnetic radiation}$$
Solution:- An accelerating charge creates time-varying electric and magnetic fields that propagate outward as electromagnetic waves.
Why other options are incorrect:- Opt_A: Stationary charges produce only electrostatic fields.
- Opt_B: Uniform motion produces constant magnetic fields without radiation.
- Opt_D: Only accelerated charges radiate.
[ETEA 2015]
The current which is produced due to changing electric flux is called:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Maxwell introduced displacement current to account for time-varying electric fields producing magnetic fields, ensuring Ampere's Law satisfies charge conservation.
Formula / Reaction:$$I_D = \varepsilon_0 \frac{d\Phi_E}{dt}$$
Solution:- The rate of change of electric flux between capacitor plates produces a displacement current \(I_D\) that generates a magnetic field just like a conduction current.
Why other options are incorrect:- Opt_B: Conduction current is the actual drift of physical charges through a conductor (\(I = dq/dt\)).
- Opt_C: Eddy currents are closed loops of induced conduction current circulating inside bulk conductors.
- Opt_D: Option A is correct.
[ETEA 2015]
For electromagnetic waves, Maxwell generalized which of the following laws?
A
Gauss's law for magnetism
B
Gauss's law for electricity
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Ampere's original circuital law was incomplete for non-steady currents. Maxwell added the displacement current term \(\varepsilon_0 \frac{d\Phi_E}{dt}\).
Formula / Reaction:$$\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( I + \varepsilon_0 \frac{d\Phi_E}{dt} \right)$$
Solution:- Maxwell generalized Ampere's Law by incorporating the displacement current, making the equations symmetric and predicting the existence of electromagnetic waves.
Why other options are incorrect:- Opt_A, Option B: Gauss's laws describe static field divergences.
- Opt_C: Faraday formulated electromagnetic induction, which Maxwell adopted directly.
[ETEA 2012]
The sinusoidal AC current in a circuit is \(i = 50 \sin (20 t)\). The peak value of current is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The instantaneous alternating current is modeled by the standard sinusoidal equation \(i = I_0 \sin(\omega t)\), where \(I_0\) represents the peak amplitude.
Formula / Reaction:$$i(t) = I_0 \sin(\omega t)$$
Solution:- Comparing \(i = 50 \sin(20t)\) directly with the standard equation \(i = I_0 \sin(\omega t)\):
- Peak current \(I_0 = 50\text{ A}\).
- Angular frequency \(\omega = 20\text{ rad/s}\).
Why other options are incorrect:- Opt_A: \(100\text{ A}\) represents the peak-to-peak value \(2I_0\), not the peak value.
- Opt_B: \(25\text{ A}\) is half of the peak current.
- Opt_D: \(20\text{ rad/s}\) is the angular frequency \(\omega\), not the current amplitude.
[ETEA 2012]
An alternating current is represented by the equation \(i = i_0 \sin \omega t\). Which one of the following equations represents an alternating current that has half the amplitude and double the frequency:
A
\(i = 2 i_0 \sin \omega t\)
B
\(2 i = i_0 \sin^{-1} \omega t\)
C
\(i = \frac{1}{2} i_0 \sin 2\omega t\)
D
\(2 i = i_0 \sin \omega t\)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The general equation for a sinusoidal AC is \(i(t) = I_{\text{new}} \sin(\omega_{\text{new}} t)\), where \(I_{\text{new}}\) is the amplitude and \(\omega_{\text{new}}\) is the angular frequency.
Formula / Reaction:$$I_{\text{new}} = \frac{i_0}{2}, \quad \omega_{\text{new}} = 2\omega \implies i = \frac{1}{2}i_0 \sin(2\omega t)$$
Solution:- Original waveform: \(i = i_0 \sin(\omega t)\).
- Half amplitude \(\implies I_{\text{new}} = \frac{1}{2} i_0\).
- Double frequency \(\implies \omega_{\text{new}} = 2\omega\).
- Substituting these parameters gives \(i = \frac{1}{2} i_0 \sin(2\omega t)\).
Why other options are incorrect:- Opt_A: Represents doubled amplitude (\(2i_0\)) with unchanged frequency.
- Opt_B: Contains an inverse trigonometric function, which is mathematically invalid for an AC wave.
- Opt_D: \(i = \frac{1}{2} i_0 \sin(\omega t)\) has half amplitude but retains the original frequency \(\omega\).
[ETEA 2012]
An alternating current of r.m.s \(20\text{ mA}\) passes through a \(4\text{ k}\Omega\) resistor. What is the average power dissipated?
C
\(8 \times 10^8\text{ W}\)
D
\(1.6 \times 10^8\text{ W}\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In a pure resistor, the average electrical power dissipated as heat is calculated directly from the RMS current and resistance.
Formula / Reaction:$$P = I_{\text{rms}}^2 R$$
Solution:- Given: \(I_{\text{rms}} = 20\text{ mA} = 20 \times 10^{-3}\text{ A} = 2 \times 10^{-2}\text{ A}\).
- Resistance: \(R = 4\text{ k}\Omega = 4000\ \Omega = 4 \times 10^3\ \Omega\).
- \(P = (2 \times 10^{-2})^2 \times (4 \times 10^3) = (4 \times 10^{-4}) \times (4 \times 10^3) = 1.6\text{ W}\).
Why other options are incorrect:- Opt_A: \(0.8\text{ W}\) incorrectly divides the power by 2.
- Opt_C: Result of using incorrect metric exponents (\(+3\) instead of \(-3\)).
- Opt_D: Metric power calculation error.
[ETEA 2011]
The sinusoidal alternating voltage can be expressed by the equation \(V = V_m \cos \omega t\). The instantaneous value of alternating voltage at \(t = 3T/4\) is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The instantaneous value is found by substituting the time \(t\) and \(\omega = \frac{2\pi}{T}\) into the given cosine waveform equation.
Formula / Reaction:$$V(t) = V_m \cos(\omega t) = V_m \cos\left(\frac{2\pi}{T} \cdot t\right)$$
Solution:- At \(t = \frac{3T}{4}\):
- \(\theta = \omega t = \left(\frac{2\pi}{T}\right) \left(\frac{3T}{4}\right) = \frac{3\pi}{2} = 270^\circ\).
- \(V = V_m \cos(270^\circ) = V_m (0) = 0\).
Why other options are incorrect:- Opt_A: \(+V_m\) occurs at \(t = 0\) and \(t = T\).
- Opt_B: \(-V_m\) occurs at \(t = T/2\) (\(\theta = \pi\)).
- Opt_D: Option C is the exact mathematical result.
[ETEA 2010-14]
In alternating current the average value of current in a cycle is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:A symmetrical sinusoidal alternating current completes one full cycle consisting of equal and opposite positive and negative half-cycles.
Formula / Reaction:$$I_{\text{avg}} = \frac{1}{T} \int_{0}^{T} I_0 \sin(\omega t)\, dt = 0$$
Solution:- During the first half-cycle \((0 \to T/2)\), the area under the current waveform is positive: \(+I_{\text{avg(half)}} = +\frac{2I_0}{\pi}\).
- During the second half-cycle \((T/2 \to T)\), the area under the current waveform is negative: \(-I_{\text{avg(half)}} = -\frac{2I_0}{\pi}\).
- Summing these values over a full period gives an exact net sum of zero.
Why other options are incorrect:- Opt_B: The current is continuously varying with time, not constant.
- Opt_C: The average over a full cycle cannot be positive because the negative half-cycle completely cancels the positive half-cycle.
- Opt_D: Maximum value corresponds to peak current \(I_0\), not the cycle average.
[ETEA 2010]
The r.m.s value of alternating voltage for unit peak amplitude (\(V_0 = 1\text{ V}\)) is:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:For a unit peak sinusoidal voltage (\(V_0 = 1\text{ V}\)), the root mean square value is given by \(\frac{1}{\sqrt{2}} V_0\).
Formula / Reaction:$$V_{\text{rms}} = \frac{1}{\sqrt{2}} V_0 \approx 0.707 V_0$$
Solution:- Evaluating the numerical fraction: \(\frac{1}{\sqrt{2}} = \frac{1}{1.4142} \approx 0.707\).
- Thus, for a peak voltage of \(1\text{ V}\), the RMS voltage is \(0.707\text{ V}\).
Why other options are incorrect:- Opt_A: \(1.77\) is an incorrect arithmetic factor.
- Opt_B: \(17.7\) represents a scaling error by a factor of 25.
- Opt_D: \(0.0177\) represents an incorrect decimal shift.
[ETEA 2010]
The capacitive reactance of the AC circuit increases:
A
By increasing the frequency of AC
B
By decreasing the frequency of AC
C
Does not depend upon the frequency of AC voltage
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Capacitive reactance \(X_C\) is the opposition offered by a capacitor to the flow of alternating current and is inversely proportional to frequency.
Formula / Reaction:$$X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$$
Solution:- Because \(X_C \propto \frac{1}{f}\), decreasing the frequency \(f\) of the AC source causes the capacitive reactance \(X_C\) to increase.
- In the limit of direct current (\(f = 0\)), \(X_C \to \infty\), completely blocking steady DC.
Why other options are incorrect:- Opt_A: Increasing the frequency decreases \(X_C\).
- Opt_C: Reactance is explicitly frequency-dependent.
- Opt_D: Option B is correct.
[ETEA 2009]
Which of the following has least wavelength?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In the high-energy radiation spectrum, wavelength is inversely proportional to photon energy and frequency according to the Planck-Einstein relation \(E = \frac{hc}{\lambda}\).
Formula / Reaction:$$\lambda = \frac{c}{f}$$
Solution:- Cosmic rays consist of ultra-high-energy astronomical particles and photons with frequencies exceeding \(10^{22}\text{ Hz}\) and wavelengths shorter than \(10^{-14}\text{ m}\).
- X-rays have wavelengths typically ranging between \(10^{-8}\text{ m}\) and \(10^{-11}\text{ m}\).
- Therefore, cosmic ray photons have the shortest wavelength among the choices.
Why other options are incorrect:- Opt_A: \(\alpha\)-rays are helium nuclei (matter particles), not electromagnetic waves.
- Opt_C: \(\beta\)-rays are fast electrons/positrons (matter particles), not electromagnetic waves.
- Opt_D: X-rays have longer wavelengths than cosmic ray photons.
[ETEA 2009]
The average power loss in a capacitor in an AC circuit is:
A
\(\langle P \rangle = V_0 I_0\)
B
\(\langle P \rangle = V_0 I_0 \sin\omega t\)
C
\(\langle P \rangle = V_0 I_0 \cos\omega t\)
D
\(\langle P \rangle = \text{Zero}\)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Because current leads voltage by \(90^\circ\) in a pure capacitor, the power factor \(\cos\phi = \cos(90^\circ) = 0\).
Formula / Reaction:$$\langle P \rangle = V_{\text{rms}} I_{\text{rms}} \cos(90^\circ) = 0$$
Solution:- The capacitor acts as an energy storage element, storing energy electrostatically and releasing it back to the circuit without net thermal dissipation.
- Thus, average power loss over a complete cycle is zero.
Why other options are incorrect:- Opt_A: Apparent peak power.
- Opt_B, Option C: Time-dependent functions do not represent cycle-averaged power.
[ETEA 2007]
The alternating current can be measured from its:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Because the cycle average of AC is zero, standard moving-coil meters based on linear magnetic deflection or chemical electrolysis average to zero. Thermal heating depends on \(I^2\) and is unidirectional.
Formula / Reaction:$$P_{\text{avg}} = I_{\text{rms}}^2 R$$
Solution:- Hot-wire ammeters utilize the expansion of a metal wire heated by the current \(I^2 R\).
- Because heating \(H \propto I^2\), deflection is always in one direction regardless of AC alternating polarity, making it the standard method to measure RMS AC.
Why other options are incorrect:- Opt_A: Standard magnetic torque reverses every half cycle; a permanent-magnet moving-coil meter oscillates rapidly around zero and reads zero.
- Opt_C: Chemical electrolytic effects reverse every half cycle, resulting in zero net chemical deposition.
- Opt_D: Only thermal effect meters can directly measure AC without rectification.
[ETEA 2007]
Which of the following has the highest energy photon?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Photon energy is directly proportional to frequency and inversely proportional to wavelength across the electromagnetic spectrum.
Formula / Reaction:$$E = hf = \frac{hc}{\lambda}$$
Solution:- The electromagnetic spectrum in order of increasing frequency and photon energy is: Radio \(\to\) Microwaves \(\to\) Infrared \(\to\) Visible \(\to\) UV \(\to\) X-rays \(\to\) Gamma rays.
- Gamma rays possess the highest frequencies (\(f > 10^{19}\text{ Hz}\)) and therefore the most energetic photons in the electromagnetic spectrum.
Why other options are incorrect:- Opt_A: Visible light photons have low energies (\(1.8\text{ to }3.1\text{ eV}\)).
- Opt_B: X-rays have high energy (\(\text{keV}\)), but are less energetic than nuclear gamma rays.
- Opt_C: Ultraviolet photons are lower in energy than X-rays and gamma rays.
[ETEA 2006-08]
In a capacitive circuit, current and voltage phase relation is:
B
Current leads voltage by \(90^\circ\)
C
Voltage leads current by \(90^\circ\)
D
Current leads voltage by \(180^\circ\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In a pure capacitor, alternating current is proportional to the rate of change of voltage across its plates: \(i = C \frac{dv}{dt}\).
Formula / Reaction:$$\text{If } v(t) = V_0 \sin(\omega t), \quad i(t) = I_0 \cos(\omega t) = I_0 \sin\left(\omega t + \frac{\pi}{2}\right)$$
Solution:- When alternating voltage is applied to a capacitor, charge flows onto the plates immediately, so current reaches its peak before voltage builds up across the plates.
- Therefore, current leads the applied voltage by a phase angle of \(90^\circ\) (or \(\pi/2\text{ radians}\)).
Why other options are incorrect:- Opt_A: Current and voltage are in-phase only in a pure resistor (\(\phi = 0^\circ\)).
- Opt_C: Voltage leads current by \(90^\circ\) in a pure inductor, not a capacitor.
- Opt_D: A \(180^\circ\) phase lead represents complete phase opposition, which does not occur in a pure capacitor.
[ETEA 2006]
The mean square value of the alternating current is:
A
\(i_{\text{ms}} = i_0^2 / 2\)
B
\(i_{\text{ms}} = i_0^2 / \sqrt{2}\)
C
\(i_{\text{ms}} = 2 i_0^2\)
D
\(i_{\text{ms}} = \sqrt{2} i_0^2\)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The mean square value of a sinusoidal alternating current is the average of the squared instantaneous current over a complete cycle.
Formula / Reaction:$$\langle i^2 \rangle = \frac{1}{T} \int_{0}^{T} I_0^2 \sin^2(\omega t)\, dt = \frac{I_0^2}{2}$$
Solution:- The average value of \(\sin^2(\omega t)\) over one full cycle is \(\frac{1}{2}\).
- Therefore, the mean square current is \(i_{\text{ms}} = \langle i^2 \rangle = \frac{i_0^2}{2}\).
- Taking the square root gives the root-mean-square value: \(I_{\text{rms}} = \sqrt{\frac{i_0^2}{2}} = \frac{i_0}{\sqrt{2}}\).
Why other options are incorrect:- Opt_B: Divides by \(\sqrt{2}\), which confuses the mean square with the root mean square.
- Opt_C: Multiplies by 2 instead of dividing by 2.
- Opt_D: Multiplies by \(\sqrt{2}\).
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