Important Questions
Practice these important questions to strengthen your understanding
Question 1
Arrange the following modulation types in order of increasing complexity.
Answer:
Amplitude Modulation (AM), Frequency Modulation (FM), Phase Modulation (PM), Quadrature Amplitude Modulation (QAM)
Explanation:
Amplitude Modulation is the simplest form.
Frequency Modulation is more complex than AM.
Phase Modulation is similar in complexity to FM but slightly more advanced.
Quadrature Amplitude Modulation combines amplitude and phase modulation and is the most complex.
Amplitude Modulation is the simplest form.
Frequency Modulation is more complex than AM.
Phase Modulation is similar in complexity to FM but slightly more advanced.
Quadrature Amplitude Modulation combines amplitude and phase modulation and is the most complex.
Question 2
Consider two point charges, +3 µC and -2 µC, separated by 5 m. What is the expression for the electric potential at a point P located 2 m from the +3 µC charge and 3 m from the -2 µC charge?
Answer:
A
Explanation:
The electric potential at point P due to multiple point charges is the algebraic sum of potentials due to each charge. Using \(k = \frac{1}{4\pi\epsilon_0}\), the potential due to +3 µC charge at 2 m is \(\frac{k \times 3 \times 10^{-6}}{2}\), and due to -2 µC charge at 3 m is \(\frac{k \times (-2) \times 10^{-6}}{3}\). Adding these gives option A. Options B, C, and D have incorrect distances or signs.
The electric potential at point P due to multiple point charges is the algebraic sum of potentials due to each charge. Using \(k = \frac{1}{4\pi\epsilon_0}\), the potential due to +3 µC charge at 2 m is \(\frac{k \times 3 \times 10^{-6}}{2}\), and due to -2 µC charge at 3 m is \(\frac{k \times (-2) \times 10^{-6}}{3}\). Adding these gives option A. Options B, C, and D have incorrect distances or signs.
Question 3
Derive the expression for the instantaneous current \( i(t) \) in an inductor connected to an AC voltage source given by \( v(t) = V_m \sin \omega t \). Show that the current lags the voltage by \( 90^\circ \).
Answer:
Step 1 The voltage across an inductor is \( v(t) = L \frac{di}{dt} \)
Step 2 Substitute \( v(t) = V_m \sin \omega t \), so \( L \frac{di}{dt} = V_m \sin \omega t \)
Step 3 Rearrange to \( \frac{di}{dt} = \frac{V_m}{L} \sin \omega t \)
Step 4 Integrate both sides:
\( i(t) = -\frac{V_m}{\omega L} \cos \omega t + C \)
Step 5 For zero initial current, constant \( C = 0 \), so
\( i(t) = \frac{V_m}{\omega L} \sin \left( \omega t - \frac{\pi}{2} \right) \)
Step 6 This shows current lags voltage by \( 90^\circ \).
Step 2 Substitute \( v(t) = V_m \sin \omega t \), so \( L \frac{di}{dt} = V_m \sin \omega t \)
Step 3 Rearrange to \( \frac{di}{dt} = \frac{V_m}{L} \sin \omega t \)
Step 4 Integrate both sides:
\( i(t) = -\frac{V_m}{\omega L} \cos \omega t + C \)
Step 5 For zero initial current, constant \( C = 0 \), so
\( i(t) = \frac{V_m}{\omega L} \sin \left( \omega t - \frac{\pi}{2} \right) \)
Step 6 This shows current lags voltage by \( 90^\circ \).
Explanation:
Use the inductor voltage-current relation \( v = L \frac{di}{dt} \).
Substitute the sinusoidal voltage and solve the differential equation by integration.
The resulting current expression is a sine wave shifted by \( -90^\circ \) relative to voltage.
This phase lag means current reaches its peak a quarter cycle after voltage.
Use the inductor voltage-current relation \( v = L \frac{di}{dt} \).
Substitute the sinusoidal voltage and solve the differential equation by integration.
The resulting current expression is a sine wave shifted by \( -90^\circ \) relative to voltage.
This phase lag means current reaches its peak a quarter cycle after voltage.
Question 4
Calculate the frequency and energy of the photon emitted when an electron in a hydrogen atom transitions from the \( n=3 \) level to the \( n=2 \) level.
Answer:
Step 1 Calculate energies: \( E_3 = -13.6 / 3^2 = -1.51 \) eV, \( E_2 = -13.6 / 2^2 = -3.4 \) eV.
Step 2 Energy difference \( \Delta E = E_2 - E_3 = -3.4 - (-1.51) = -1.89 \) eV.
Step 3 Take absolute value: \( 1.89 \) eV.
Step 4 Convert eV to joules: \( 1.89 \times 1.6 \times 10^{-19} = 3.02 \times 10^{-19} \) J.
Step 5 Frequency \( f = \frac{\Delta E}{h} = \frac{3.02 \times 10^{-19}}{6.626 \times 10^{-34}} = 4.56 \times 10^{14} \) Hz.
Step 6 Energy of photon is \( 3.02 \times 10^{-19} \) J or 1.89 eV.
Step 2 Energy difference \( \Delta E = E_2 - E_3 = -3.4 - (-1.51) = -1.89 \) eV.
Step 3 Take absolute value: \( 1.89 \) eV.
Step 4 Convert eV to joules: \( 1.89 \times 1.6 \times 10^{-19} = 3.02 \times 10^{-19} \) J.
Step 5 Frequency \( f = \frac{\Delta E}{h} = \frac{3.02 \times 10^{-19}}{6.626 \times 10^{-34}} = 4.56 \times 10^{14} \) Hz.
Step 6 Energy of photon is \( 3.02 \times 10^{-19} \) J or 1.89 eV.
Explanation:
Energy difference between levels determines photon energy.
Frequency is energy divided by Planck's constant.
This calculation shows how spectral lines correspond to specific photon energies.
It also illustrates the quantised nature of atomic transitions.
The frequency lies in the visible range corresponding to Balmer series.
Energy difference between levels determines photon energy.
Frequency is energy divided by Planck's constant.
This calculation shows how spectral lines correspond to specific photon energies.
It also illustrates the quantised nature of atomic transitions.
The frequency lies in the visible range corresponding to Balmer series.
Question 5
Describe how the terminal voltage of a cell changes as the current drawn from it increases.
Answer:
As the current drawn increases, the voltage drop across the internal resistance increases.
This causes the terminal voltage to decrease below the emf.
Hence, terminal voltage falls with increasing current.
This causes the terminal voltage to decrease below the emf.
Hence, terminal voltage falls with increasing current.
Explanation:
Terminal voltage is emf minus the voltage drop inside the cell.
More current means more voltage drop inside due to internal resistance.
This explains why devices may receive lower voltage under heavy load.
Terminal voltage is emf minus the voltage drop inside the cell.
More current means more voltage drop inside due to internal resistance.
This explains why devices may receive lower voltage under heavy load.
Question 6
Select the correct formula for displacement current density \(J_d\):
Answer:
b
Explanation:
The displacement current density is defined as epsilon naught times the time derivative of the electric flux.
This is Maxwell's correction to Ampere's law to include time-varying electric fields.
Option B correctly states this formula.
The displacement current density is defined as epsilon naught times the time derivative of the electric flux.
This is Maxwell's correction to Ampere's law to include time-varying electric fields.
Option B correctly states this formula.
Question 7
What is the main function of the secondary mirror in a Cassegrain reflecting telescope?
Answer:
The main function of the secondary mirror in a Cassegrain reflecting telescope is to reflect light to the eyepiece through a hole in the primary mirror.
Explanation:
In a Cassegrain telescope, the primary concave mirror collects light and reflects it to a convex secondary mirror. The secondary mirror then reflects the light back through a hole in the primary mirror to the eyepiece, allowing a compact design and convenient viewing position.
In a Cassegrain telescope, the primary concave mirror collects light and reflects it to a convex secondary mirror. The secondary mirror then reflects the light back through a hole in the primary mirror to the eyepiece, allowing a compact design and convenient viewing position.
Question 8
Derive the formula for the effective focal length when two thin lenses are placed in contact and explain its practical significance.
Answer:
Let the focal lengths of two thin lenses be \( f_1 \) and \( f_2 \).
When placed in contact, the combined focal length \( f \) is given by:
\( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \).
This is derived by considering the powers of lenses add:
Power \( P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} \).
The combined lens behaves like a single lens with focal length f.
This principle is used in designing compound lenses and correcting vision.
When placed in contact, the combined focal length \( f \) is given by:
\( \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \).
This is derived by considering the powers of lenses add:
Power \( P = P_1 + P_2 = \frac{1}{f_1} + \frac{1}{f_2} \).
The combined lens behaves like a single lens with focal length f.
This principle is used in designing compound lenses and correcting vision.
Explanation:
Start with the definition of focal length and power of lenses.
Explain that powers add when lenses are in contact.
Derive the formula for combined focal length.
Discuss practical applications such as compound microscopes and eyeglasses.
This derivation is fundamental for understanding lens combinations.
Start with the definition of focal length and power of lenses.
Explain that powers add when lenses are in contact.
Derive the formula for combined focal length.
Discuss practical applications such as compound microscopes and eyeglasses.
This derivation is fundamental for understanding lens combinations.
Question 9
Identify the odd one out that is NOT a unit related to electromagnetism
Answer:
3
Explanation:
Weber, Tesla, and Volt are units related to electromagnetism.
Newton is a unit of force and not related to electromagnetism.
Hence Newton is the odd one out.
Weber, Tesla, and Volt are units related to electromagnetism.
Newton is a unit of force and not related to electromagnetism.
Hence Newton is the odd one out.
Question 10
Calculate the potential difference between the inner and outer spheres of a Van de Graaff generator given their radii and charge.
Answer:
Let the inner sphere radius be \( r_2 \), outer sphere radius be \( r_1 \), and charge be \( Q \).
The capacitance of the spherical capacitor is \( C = \frac{4 \pi \epsilon_0 r_1 r_2}{r_1 - r_2} \).
The potential difference \( V = \frac{Q}{C} = \frac{Q (r_1 - r_2)}{4 \pi \epsilon_0 r_1 r_2} \).
The capacitance of the spherical capacitor is \( C = \frac{4 \pi \epsilon_0 r_1 r_2}{r_1 - r_2} \).
The potential difference \( V = \frac{Q}{C} = \frac{Q (r_1 - r_2)}{4 \pi \epsilon_0 r_1 r_2} \).
Explanation:
The spherical capacitor consists of two concentric spheres.
The capacitance formula relates the geometry and permittivity.
Potential difference is charge divided by capacitance.
By substituting the capacitance expression, we find the potential difference in terms of charge and radii.
This formula allows calculation of voltage given charge and dimensions.
The spherical capacitor consists of two concentric spheres.
The capacitance formula relates the geometry and permittivity.
Potential difference is charge divided by capacitance.
By substituting the capacitance expression, we find the potential difference in terms of charge and radii.
This formula allows calculation of voltage given charge and dimensions.
Question 11
Using the right hand thumb rule, if the thumb points along the axis of a solenoid in the direction of the magnetic field, how do the fingers curl?
Answer:
A
Explanation:
According to the right-hand thumb rule for solenoids:
If you point your thumb along the axis of the solenoid in the direction of the magnetic field inside,
your fingers curl in the direction of the current flowing through the solenoid windings.
This rule helps visualize the relationship between current direction and magnetic field.
According to the right-hand thumb rule for solenoids:
If you point your thumb along the axis of the solenoid in the direction of the magnetic field inside,
your fingers curl in the direction of the current flowing through the solenoid windings.
This rule helps visualize the relationship between current direction and magnetic field.
Question 12
Which of the following best describes the wave-particle duality of electrons?
Answer:
Electrons exhibit both wave and particle properties depending on the experiment
Explanation:
Wave-particle duality means electrons sometimes behave like particles and sometimes like waves depending on the experimental setup.
Experiments such as electron diffraction show wave nature, while others show particle nature.
This duality is fundamental to quantum mechanics.
Wave-particle duality means electrons sometimes behave like particles and sometimes like waves depending on the experimental setup.
Experiments such as electron diffraction show wave nature, while others show particle nature.
This duality is fundamental to quantum mechanics.
Question 13
321.
Assertion (A): Increasing transmission voltage reduces power loss.
Reason (R): Power loss is proportional to the square of the current.
- Both A and R are true, and R explains A
- Both A and R are true, but R does not explain A
- A is true, R is false
- A is false, R is true
Answer:
a
Explanation:
The assertion states that increasing transmission voltage reduces power loss, which is true because power loss in transmission lines is given by P = I^2 R. The reason states that power loss is proportional to the square of the current, which is also true. Since power loss depends on current squared, increasing voltage allows the same power to be transmitted at lower current, thus reducing power loss. Therefore, both A and R are true and R explains A.
The assertion states that increasing transmission voltage reduces power loss, which is true because power loss in transmission lines is given by P = I^2 R. The reason states that power loss is proportional to the square of the current, which is also true. Since power loss depends on current squared, increasing voltage allows the same power to be transmitted at lower current, thus reducing power loss. Therefore, both A and R are true and R explains A.
Question 14
Derive the expression for the electric potential at a point due to a system of \( n \) point charges located at different positions in space.
Answer:
Consider \( n \) point charges \( q_1, q_2, ..., q_n \) located at distances \( r_1, r_2, ..., r_n \) from the point P.
The potential due to charge \( q_i \) at P is \( V_i = \frac{1}{4 \pi \epsilon_0} \frac{q_i}{r_i} \).
By the superposition principle, total potential at P is \( V = \sum_{i=1}^n V_i = \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^n \frac{q_i}{r_i} \).
This formula accounts for all charges and their distances from the point, providing the net potential.
The potential due to charge \( q_i \) at P is \( V_i = \frac{1}{4 \pi \epsilon_0} \frac{q_i}{r_i} \).
By the superposition principle, total potential at P is \( V = \sum_{i=1}^n V_i = \frac{1}{4 \pi \epsilon_0} \sum_{i=1}^n \frac{q_i}{r_i} \).
This formula accounts for all charges and their distances from the point, providing the net potential.
Explanation:
This derivation uses the principle that potentials add algebraically.
Each charge contributes independently to the potential at the point.
Summing these contributions yields the total potential.
This expression is foundational for electrostatics involving multiple charges.
This derivation uses the principle that potentials add algebraically.
Each charge contributes independently to the potential at the point.
Summing these contributions yields the total potential.
This expression is foundational for electrostatics involving multiple charges.
Question 15
Which of the following best describes Faraday's and Henry's experiments?
Answer:
Explanation:
Faraday and Henry showed that a changing magnetic flux induces an emf and current.
Static fields do not induce current.
Electric current producing magnetic fields was shown by others.
Electron discovery is unrelated.
Faraday and Henry showed that a changing magnetic flux induces an emf and current.
Static fields do not induce current.
Electric current producing magnetic fields was shown by others.
Electron discovery is unrelated.
Question 16
Derive the expression relating the electric and magnetic field amplitudes in an electromagnetic wave.
Answer:
In an electromagnetic wave, electric and magnetic fields are perpendicular and related by the speed of light.
The magnitude of electric field amplitude E_0 and magnetic field amplitude B_0 satisfy E_0 = c B_0.
This follows from Maxwell's equations and the wave nature of the fields.
Thus, E_0 / B_0 = c.
The magnitude of electric field amplitude E_0 and magnetic field amplitude B_0 satisfy E_0 = c B_0.
This follows from Maxwell's equations and the wave nature of the fields.
Thus, E_0 / B_0 = c.
Explanation:
Maxwell's equations predict that changing electric fields produce magnetic fields and vice versa.
In free space, the ratio of electric to magnetic field amplitudes equals the speed of light.
This relation is fundamental to electromagnetic wave theory.
The derivation involves solving Maxwell's equations for plane waves.
Maxwell's equations predict that changing electric fields produce magnetic fields and vice versa.
In free space, the ratio of electric to magnetic field amplitudes equals the speed of light.
This relation is fundamental to electromagnetic wave theory.
The derivation involves solving Maxwell's equations for plane waves.
Question 17
Describe the setup of a meter bridge and explain its working principle in detail. Include how the balance point is found and how unknown resistance is calculated.
Answer:
The meter bridge consists of a 1-meter long uniform wire stretched on a wooden or metallic base with terminals A and C at the ends.
Two gaps are provided to connect the unknown resistance R and a known standard resistance S.
A galvanometer and a jockey are connected across a sliding contact on the wire.
A battery and key are connected to provide current through the wire and resistors.
When the jockey is moved along the wire, a point is found where the galvanometer shows zero deflection; this is the balance point at length l from A.
At balance, the ratio of resistances is equal to the ratio of lengths: R/S = l/(100 - l).
From this, the unknown resistance R is calculated as R = S × l/(100 - l).
Two gaps are provided to connect the unknown resistance R and a known standard resistance S.
A galvanometer and a jockey are connected across a sliding contact on the wire.
A battery and key are connected to provide current through the wire and resistors.
When the jockey is moved along the wire, a point is found where the galvanometer shows zero deflection; this is the balance point at length l from A.
At balance, the ratio of resistances is equal to the ratio of lengths: R/S = l/(100 - l).
From this, the unknown resistance R is calculated as R = S × l/(100 - l).
Explanation:
The meter bridge is a practical application of the Wheatstone bridge principle.
The uniform wire ensures resistance is proportional to length.
At the balance point, no current flows through the galvanometer, indicating equal potential difference on both sides.
By measuring the balance length, the unknown resistance can be accurately determined.
Teachers should emphasize the importance of uniform wire and careful measurement of balance length for accuracy.
The meter bridge is a practical application of the Wheatstone bridge principle.
The uniform wire ensures resistance is proportional to length.
At the balance point, no current flows through the galvanometer, indicating equal potential difference on both sides.
By measuring the balance length, the unknown resistance can be accurately determined.
Teachers should emphasize the importance of uniform wire and careful measurement of balance length for accuracy.
Question 18
At resonance in a series LCR circuit, what is the phase relationship between voltage and current phasors?
Answer:
At resonance, the voltage and current phasors are in phase.
This means the phase difference \(\phi = 0\) degrees.
Therefore, the power factor is unity and the circuit behaves like a pure resistor.
This means the phase difference \(\phi = 0\) degrees.
Therefore, the power factor is unity and the circuit behaves like a pure resistor.
Explanation:
At resonance, inductive and capacitive reactances cancel each other.
Hence, the circuit's impedance is purely resistive.
Voltage and current reach their maximum values simultaneously.
This results in zero phase difference and unity power factor.
At resonance, inductive and capacitive reactances cancel each other.
Hence, the circuit's impedance is purely resistive.
Voltage and current reach their maximum values simultaneously.
This results in zero phase difference and unity power factor.
Question 19
Describe how noise affects signal quality in communication systems.
Answer:
Noise refers to unwanted signals that disturb the transmission and processing of message signals. It can originate from sources inside or outside the communication system. Noise degrades the quality of the received signal by causing errors and distortion, making it difficult to accurately extract the original message at the receiver.
Explanation:
Noise introduces random fluctuations that interfere with the desired signal, reducing the signal-to-noise ratio. This leads to errors in data transmission and loss of information. Understanding noise and its effects helps in designing systems with better noise immunity and error correction techniques.
Noise introduces random fluctuations that interfere with the desired signal, reducing the signal-to-noise ratio. This leads to errors in data transmission and loss of information. Understanding noise and its effects helps in designing systems with better noise immunity and error correction techniques.
Question 20
A magnetic material with high coercivity is best described as one that:
Answer:
B) Requires a strong reverse magnetic field to demagnetize
Explanation:
Coercivity is the measure of the reverse magnetic field strength needed to reduce the magnetization to zero.
High coercivity means the material is hard to demagnetize and is suitable for permanent magnets.
Low coercivity materials lose magnetization easily.
Coercivity is the measure of the reverse magnetic field strength needed to reduce the magnetization to zero.
High coercivity means the material is hard to demagnetize and is suitable for permanent magnets.
Low coercivity materials lose magnetization easily.
