Previous Year Questions
Solve previous year questions to understand exam patterns
PYQ 1
CBSE
2017
How does Huygens' principle help in explaining the bending of light while passing from one medium to another?
Solution:
Huygens' principle explains bending of light (refraction) as follows:
1. Each point on the incident wavefront at the interface acts as a source of secondary wavelets.
2. The speed of light differs in the two media, so the wavelets in the second medium have different radii compared to those in the first.
3. The envelope of these wavelets forms the refracted wavefront.
4. Due to the difference in wave speeds, the refracted wavefront changes direction, causing the light to bend.
Thus, the bending of light is a natural consequence of the change in wave speed and the construction of the new wavefront from secondary wavelets.
1. Each point on the incident wavefront at the interface acts as a source of secondary wavelets.
2. The speed of light differs in the two media, so the wavelets in the second medium have different radii compared to those in the first.
3. The envelope of these wavelets forms the refracted wavefront.
4. Due to the difference in wave speeds, the refracted wavefront changes direction, causing the light to bend.
Thus, the bending of light is a natural consequence of the change in wave speed and the construction of the new wavefront from secondary wavelets.
Detailed Explanation:
The principle models wave propagation at the interface.
Different speeds in media cause wavelets to expand differently.
The new wavefront bends as the envelope of these wavelets.
This explains refraction as a wave phenomenon rather than just ray bending.
It provides a physical basis for Snell's law.
The principle models wave propagation at the interface.
Different speeds in media cause wavelets to expand differently.
The new wavefront bends as the envelope of these wavelets.
This explains refraction as a wave phenomenon rather than just ray bending.
It provides a physical basis for Snell's law.
PYQ 2
CBSE
2019
Arrange the following steps in the correct sequence to find the force on a charge due to multiple charges:\n1. Use vector addition\n2. Calculate individual forces using Coulomb’s law\n3. Apply principle of superposition
Solution:
2,3,1
Detailed Explanation:
First, calculate individual forces between the charge and each other charge using Coulomb's law.
Second, apply the principle of superposition which states that total force is the vector sum of individual forces.
Third, perform vector addition to find the resultant force.
This sequence logically follows the method to find net force.
First, calculate individual forces between the charge and each other charge using Coulomb's law.
Second, apply the principle of superposition which states that total force is the vector sum of individual forces.
Third, perform vector addition to find the resultant force.
This sequence logically follows the method to find net force.
PYQ 3
ICSE
2019
Calculate the equivalent capacitance of capacitors \(C_1 = 3 \mu F\), \(C_2 = 3 \mu F\), and \(C_3 = 6 \mu F\) connected first in parallel and then in series.
Solution:
First, parallel combination: \( C_p = C_1 + C_2 + C_3 = 3 + 3 + 6 = 12 \mu F \). Then, series combination of these three capacitors: \( \frac{1}{C_s} = \frac{1}{3} + \frac{1}{3} + \frac{1}{6} = 0.333 + 0.333 + 0.167 = 0.833 \). \( C_s = \frac{1}{0.833} = 1.2 \mu F \).
Detailed Explanation:
Calculate equivalent capacitance for parallel connection by summing capacitances. Calculate equivalent capacitance for series connection by summing reciprocals and then taking reciprocal. These steps help analyze complex capacitor arrangements.
Calculate equivalent capacitance for parallel connection by summing capacitances. Calculate equivalent capacitance for series connection by summing reciprocals and then taking reciprocal. These steps help analyze complex capacitor arrangements.
PYQ 4
Andhra Pradesh State Board
2019
Write down the expression for charge as a function of time in an LC circuit if initial charge is zero and initial current is maximum.
Solution:
Charge \( q(t) = q_m \sin \omega t \), where \( q_m \) is maximum charge and \( \omega = \frac{1}{\sqrt{LC}} \).
Detailed Explanation:
When initial charge is zero and initial current is maximum, the charge varies sinusoidally as a sine function. This is because current is the time derivative of charge, and maximum current at \( t=0 \) corresponds to zero charge at \( t=0 \). The angular frequency \( \omega \) is the natural frequency of the LC circuit.
When initial charge is zero and initial current is maximum, the charge varies sinusoidally as a sine function. This is because current is the time derivative of charge, and maximum current at \( t=0 \) corresponds to zero charge at \( t=0 \). The angular frequency \( \omega \) is the natural frequency of the LC circuit.
PYQ 5
Maharashtra SSC
2018
Calculate the number of electrons that must be removed from a neutral conductor to give it a charge of +1.0 μC.
Solution:
Charge to be given \( q = +1.0 \, \mu\mathrm{C} = 1.0 \times 10^{-6} \ \mathrm{C} \). Charge of one electron \( e = 1.6 \times 10^{-19} \ \mathrm{C} \). Number of electrons removed \( n = \frac{q}{e} = \frac{1.0 \times 10^{-6}}{1.6 \times 10^{-19}} = 6.25 \times 10^{12} \) electrons.
Detailed Explanation:
Removing electrons from a neutral conductor makes it positively charged. The number of electrons removed is total charge divided by charge per electron.
Removing electrons from a neutral conductor makes it positively charged. The number of electrons removed is total charge divided by charge per electron.
PYQ 6
Andhra Pradesh State Board
2015
Calculate the radius of radio horizon for an antenna 50 m high. Use Earth's radius as 6400 km.
Solution:
Given height of antenna, h = 50 m Radius of Earth, R = 6400 km = 6.4 \times 10^{6} m Formula for radius of radio horizon, d = \( \sqrt{2Rh} \) Calculate: d = \( \sqrt{2 \times 6.4 \times 10^{6} \times 50} \) = \( \sqrt{6.4 \times 10^{8}} \) = 2.53 \times 10^{4} m = 25.3 km
Detailed Explanation:
The radio horizon distance depends on the height of the antenna and the Earth's radius. Using the formula \( d = \sqrt{2Rh} \), we substitute the values to find the horizon distance. This distance represents the maximum line-of-sight distance from the antenna to the horizon.
The radio horizon distance depends on the height of the antenna and the Earth's radius. Using the formula \( d = \sqrt{2Rh} \), we substitute the values to find the horizon distance. This distance represents the maximum line-of-sight distance from the antenna to the horizon.
PYQ 7
Andhra Pradesh State Board
2023
Explain Young’s double slit experiment and derive the expression for fringe width.
Solution:
Detailed Explanation:
Young’s experiment demonstrates interference of light waves from two coherent sources. The path difference between waves reaching a point on the screen determines whether constructive or destructive interference occurs. Using geometry and the condition for maxima, the fringe positions are derived. The fringe width formula shows how wavelength, slit separation, and screen distance affect the pattern. This experiment provides strong evidence for the wave nature of light.
Young’s experiment demonstrates interference of light waves from two coherent sources. The path difference between waves reaching a point on the screen determines whether constructive or destructive interference occurs. Using geometry and the condition for maxima, the fringe positions are derived. The fringe width formula shows how wavelength, slit separation, and screen distance affect the pattern. This experiment provides strong evidence for the wave nature of light.
PYQ 8
Andhra Pradesh State Board
2017
A capacitor is connected to an AC source with voltage \( v = 50 \sin 1000 t \) volts. If the capacitor value is 20 \( \mu F \), calculate the instantaneous current at \( t = 1 \text{ ms} \).
Solution:
Given: \( v = 50 \sin 1000 t \), \( C = 20 \times 10^{-6} \) F, Time \( t = 1 \times 10^{-3} \) s. Current \( i = C \frac{dv}{dt} = C \times 50 \times 1000 \cos 1000 t = 20 \times 10^{-6} \times 50 \times 1000 \cos(1000 \times 0.001) \). Calculate \( \cos 1 \) radian \( \approx 0.5403 \). \( i = 20 \times 10^{-6} \times 50 \times 1000 \times 0.5403 = 0.5403 \) A.
Detailed Explanation:
Current is derivative of voltage times capacitance. Differentiate voltage function with respect to time. Substitute given values and evaluate cosine at specified time. Calculate instantaneous current value.
Current is derivative of voltage times capacitance. Differentiate voltage function with respect to time. Substitute given values and evaluate cosine at specified time. Calculate instantaneous current value.
PYQ 9
Karnataka SSLC
2019
Fill in the blank: The purpose of the bandpass filter in AM transmitter is to __________.
Solution:
remove unwanted frequencies
Detailed Explanation:
The bandpass filter removes frequencies other than the carrier and sidebands produced by the square law device.
This ensures only the desired AM signal is transmitted.
The bandpass filter removes frequencies other than the carrier and sidebands produced by the square law device.
This ensures only the desired AM signal is transmitted.
PYQ 10
ICSE
2019
What is the shape of the magnetic field lines around a long straight current-carrying conductor? Also, explain how the direction of the field is decided.
Solution:
The magnetic field lines around a long straight current-carrying conductor form concentric circles centered on the wire.
The direction of the magnetic field is given by the right-hand thumb rule: if the thumb of the right hand points in the direction of the current, the curled fingers show the direction of the magnetic field lines around the wire.
The direction of the magnetic field is given by the right-hand thumb rule: if the thumb of the right hand points in the direction of the current, the curled fingers show the direction of the magnetic field lines around the wire.
Detailed Explanation:
Magnetic field lines form closed loops around current-carrying conductors.
The right-hand thumb rule is a simple mnemonic to determine the direction of these field lines.
Teachers often demonstrate this with a wire and compass or by using hand gestures.
This concept helps students visualize magnetic fields in three dimensions.
Magnetic field lines form closed loops around current-carrying conductors.
The right-hand thumb rule is a simple mnemonic to determine the direction of these field lines.
Teachers often demonstrate this with a wire and compass or by using hand gestures.
This concept helps students visualize magnetic fields in three dimensions.
PYQ 11
CBSE
2019
Explain the concept of angular resolution and how diffraction limits the resolving power of optical instruments.
Solution:
Step 1 Angular resolution is the smallest angular separation between two objects that an optical instrument can distinguish.
Step 2 Due to diffraction, light from a point source spreads out forming a diffraction pattern with a central bright spot.
Step 3 When two sources are close, their diffraction patterns overlap.
Step 4 The Rayleigh criterion states that two sources are just resolved when the central maximum of one coincides with the first minimum of the other.
Step 5 Diffraction thus limits the resolving power by setting a minimum angular separation.
Step 6 Increasing aperture size or using shorter wavelengths improves resolution by reducing diffraction effects.
Step 2 Due to diffraction, light from a point source spreads out forming a diffraction pattern with a central bright spot.
Step 3 When two sources are close, their diffraction patterns overlap.
Step 4 The Rayleigh criterion states that two sources are just resolved when the central maximum of one coincides with the first minimum of the other.
Step 5 Diffraction thus limits the resolving power by setting a minimum angular separation.
Step 6 Increasing aperture size or using shorter wavelengths improves resolution by reducing diffraction effects.
Detailed Explanation:
Angular resolution defines the clarity of images formed by optical instruments.
Diffraction causes spreading of light, limiting how close two points can appear.
The Rayleigh criterion provides a practical limit for resolution.
Understanding diffraction effects is essential for designing better telescopes and microscopes.
This explanation connects wave optics with practical imaging limits.
Angular resolution defines the clarity of images formed by optical instruments.
Diffraction causes spreading of light, limiting how close two points can appear.
The Rayleigh criterion provides a practical limit for resolution.
Understanding diffraction effects is essential for designing better telescopes and microscopes.
This explanation connects wave optics with practical imaging limits.
PYQ 12
Andhra Pradesh State Board
2019
MCQ: When two waves of the same frequency and amplitude interfere, the resultant amplitude is maximum when the phase difference is\n(a) \(0^\circ\)\n(b) \(90^\circ\)\n(c) \(180^\circ\)\n(d) \(270^\circ\)
Solution:
a
Detailed Explanation:
Maximum resultant amplitude occurs when waves are in phase.
This happens when phase difference is zero degrees.
Other phase differences lead to partial or complete cancellation.
Thus, option (a) is correct.
Maximum resultant amplitude occurs when waves are in phase.
This happens when phase difference is zero degrees.
Other phase differences lead to partial or complete cancellation.
Thus, option (a) is correct.
PYQ 13
A parallel plate capacitor is partially filled with a dielectric. Draw the electric field lines and explain the effect on the electric field distribution.
Solution:
Detailed Explanation:
When a dielectric partially fills the capacitor, it polarizes and reduces the electric field inside it. The electric field in the dielectric is less than in the air gap because the dielectric's induced charges oppose the external field. The field lines concentrate more in the air region and spread less in the dielectric region. This results in a non-uniform electric field distribution inside the capacitor. The diagram above shows the field lines denser in air and sparser in the dielectric region.
When a dielectric partially fills the capacitor, it polarizes and reduces the electric field inside it. The electric field in the dielectric is less than in the air gap because the dielectric's induced charges oppose the external field. The field lines concentrate more in the air region and spread less in the dielectric region. This results in a non-uniform electric field distribution inside the capacitor. The diagram above shows the field lines denser in air and sparser in the dielectric region.
PYQ 14
Telangana Board
2019
Two waves \(y_1 = a \sin(kx - \omega t)\) and \(y_2 = a \sin(kx - \omega t + \pi)\) interfere. Find the resultant amplitude and explain the type of interference.
Solution:
The phase difference \(\phi = \pi\). Using \( A = 2a \cos(\phi/2) = 2a \cos(\pi/2) = 2a \times 0 = 0 \). The resultant amplitude is zero. This is complete destructive interference where the waves cancel each other out at all points.
Detailed Explanation:
A phase difference of \(\pi\) means the waves are exactly out of phase. Their displacements are equal in magnitude but opposite in direction. This causes total cancellation resulting in zero resultant amplitude.
A phase difference of \(\pi\) means the waves are exactly out of phase. Their displacements are equal in magnitude but opposite in direction. This causes total cancellation resulting in zero resultant amplitude.
PYQ 15
CBSE
2018
Calculate the binding energy per nucleon if the total binding energy of a nucleus is 160 MeV and it contains 40 nucleons.
Solution:
Step 1 Given total binding energy E_b = 160 MeV and number of nucleons A = 40
Step 2 Binding energy per nucleon E_bn = E_b / A = 160 / 40 = 4 MeV
Step 3 Binding energy per nucleon is 4 MeV
Step 2 Binding energy per nucleon E_bn = E_b / A = 160 / 40 = 4 MeV
Step 3 Binding energy per nucleon is 4 MeV
Detailed Explanation:
Binding energy per nucleon is found by dividing total binding energy by the number of nucleons.
This gives the average energy that binds each nucleon in the nucleus.
This measure helps compare nuclear stability across different nuclei.
Binding energy per nucleon is found by dividing total binding energy by the number of nucleons.
This gives the average energy that binds each nucleon in the nucleus.
This measure helps compare nuclear stability across different nuclei.
PYQ 16
Andhra Pradesh State Board
2018
Derive the expression for the net force on a charge \(q\) placed at the centroid of an equilateral triangle with charges \(q\) at each vertex.
Solution:
Consider an equilateral triangle with side length \(a\). The centroid is at distance \(r = \frac{a}{\sqrt{3}}\) from each vertex. Each vertex has charge \(q\); the charge at centroid is \(q\). Force due to one vertex charge on centroid charge is \(F = \frac{k q^2}{r^2} = \frac{k q^2}{\frac{a^2}{3}} = \frac{3 k q^2}{a^2}\). There are three such forces, each directed from centroid to vertex. These three forces are symmetrically spaced at 120\degree. Vector sum of three equal magnitude forces at 120\degree angles is zero. Therefore, net force on charge at centroid is zero.
Detailed Explanation:
Calculate distance from centroid to vertices. Use Coulomb’s law for force magnitude. Recognize symmetry and equal spacing of forces. Use vector addition of forces at 120\degree to find resultant. Resultant is zero due to symmetry. This derivation shows importance of symmetry in electrostatics.
Calculate distance from centroid to vertices. Use Coulomb’s law for force magnitude. Recognize symmetry and equal spacing of forces. Use vector addition of forces at 120\degree to find resultant. Resultant is zero due to symmetry. This derivation shows importance of symmetry in electrostatics.
PYQ 17
CBSE
2017
Two coherent waves have intensities \(25 \ \mathrm{W/m^2}\) and \(16 \ \mathrm{W/m^2}\). Calculate the maximum and minimum resultant intensities.
Solution:
Given \(I_1 = 25 \ \mathrm{W/m^2}\), \(I_2 = 16 \ \mathrm{W/m^2}\). Amplitudes: \(a_1 = \sqrt{25} = 5\), \(a_2 = \sqrt{16} = 4\). Maximum intensity (constructive interference): \(I_{\max} = (a_1 + a_2)^2 = (5 + 4)^2 = 81 \ \mathrm{W/m^2}\). Minimum intensity (destructive interference): \(I_{\min} = (a_1 - a_2)^2 = (5 - 4)^2 = 1 \ \mathrm{W/m^2}\).
Detailed Explanation:
Calculate amplitudes from given intensities. Add amplitudes for maximum intensity and square the sum. Subtract amplitudes for minimum intensity and square the difference. These calculations give the range of intensities observed in interference.
Calculate amplitudes from given intensities. Add amplitudes for maximum intensity and square the sum. Subtract amplitudes for minimum intensity and square the difference. These calculations give the range of intensities observed in interference.
PYQ 18
CBSE
2019
Why is a bandpass filter necessary in the AM transmitter? What happens if it is not used?
Solution:
A bandpass filter is necessary to remove unwanted frequency components produced by the square law device such as dc, ω_m, 2ω_m, and 2ω_c.
It allows only the carrier frequency ω_c and the sideband frequencies ω_c ± ω_m to pass.
If the bandpass filter is not used, unwanted frequencies will be transmitted causing interference and distortion.
This will degrade the quality of the transmitted AM wave and reduce communication efficiency.
It allows only the carrier frequency ω_c and the sideband frequencies ω_c ± ω_m to pass.
If the bandpass filter is not used, unwanted frequencies will be transmitted causing interference and distortion.
This will degrade the quality of the transmitted AM wave and reduce communication efficiency.
Detailed Explanation:
The square law device generates many frequency components beyond the desired AM frequencies.
The bandpass filter selects only the carrier and sidebands for transmission.
Without filtering, the transmitter would emit spurious signals causing interference.
Thus, the bandpass filter ensures signal clarity and efficient use of bandwidth.
The square law device generates many frequency components beyond the desired AM frequencies.
The bandpass filter selects only the carrier and sidebands for transmission.
Without filtering, the transmitter would emit spurious signals causing interference.
Thus, the bandpass filter ensures signal clarity and efficient use of bandwidth.
PYQ 19
Andhra Pradesh State Board
2020
What is the significance of the direction assumed for currents when applying Kirchhoff’s laws? How does one correct if the assumed direction is wrong?
Solution:
The assumed direction of current is arbitrary but must be consistent.
If the calculated current is positive, the assumption is correct.
If the current is negative, it means the actual current flows opposite to the assumed direction.
No correction in equations is needed; the sign indicates the true direction.
If the calculated current is positive, the assumption is correct.
If the current is negative, it means the actual current flows opposite to the assumed direction.
No correction in equations is needed; the sign indicates the true direction.
Detailed Explanation:
When applying Kirchhoff’s laws, current directions are guessed to set up equations.
This assumption simplifies writing equations but may not be correct.
After solving, a negative value means the current flows opposite to the assumed direction.
This approach avoids guesswork and ensures correct results.
When applying Kirchhoff’s laws, current directions are guessed to set up equations.
This assumption simplifies writing equations but may not be correct.
After solving, a negative value means the current flows opposite to the assumed direction.
This approach avoids guesswork and ensures correct results.
PYQ 20
Andhra Pradesh State Board
2023
Calculate the amplitude of current in a series LCR circuit if the maximum applied voltage is \( 120 \, V \), resistance is \( 40 \Omega \), inductive reactance is \( 30 \Omega \) and capacitive reactance is \( 20 \Omega \).
Solution:
Given: Maximum voltage, \( v_m = 120 \, V \) Resistance, \( R = 40 \Omega \) Inductive reactance, \( X_L = 30 \Omega \) Capacitive reactance, \( X_C = 20 \Omega \) Calculate impedance \( Z \): \( Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{40^2 + (30 - 20)^2} = \sqrt{1600 + 100} = \sqrt{1700} = 41.23 \Omega \) Amplitude of current: \( i_m = \frac{v_m}{Z} = \frac{120}{41.23} = 2.91 \, A \)
Detailed Explanation:
The amplitude of current in a series LCR circuit is given by the voltage amplitude divided by the impedance. Impedance combines resistance and net reactance. Calculate impedance first, then divide voltage by impedance. This question tests application of impedance formula and current calculation in AC circuits.
The amplitude of current in a series LCR circuit is given by the voltage amplitude divided by the impedance. Impedance combines resistance and net reactance. Calculate impedance first, then divide voltage by impedance. This question tests application of impedance formula and current calculation in AC circuits.
