Previous Year Questions
Solve previous year questions to understand exam patterns
PYQ 1
Tamil Nadu
2018
Calculate the horizontal component of Earth's magnetic field if a tangent galvanometer with 40 turns and radius 15 cm shows a deflection of 45° when a current of 0.3 A is passed through the coil.
Solution:
Given:
Number of turns, N = 40
Radius, R = 15 cm = 0.15 m
Current, I = 0.3 A
Deflection angle, \( \theta = 45^\circ \)
Using the tangent law:
\( B = B_H \tan \theta \)
Magnetic field due to coil:
\( B = \frac{\mu_0 N I}{2 R} \)
Equating:
\( \frac{\mu_0 N I}{2 R} = B_H \tan \theta \)
Therefore,
\( B_H = \frac{\mu_0 N I}{2 R \tan \theta} \)
Substitute values:
\( \mu_0 = 4 \pi \times 10^{-7} \) T m/A
\( \tan 45^\circ = 1 \)
\( B_H = \frac{4 \pi \times 10^{-7} \times 40 \times 0.3}{2 \times 0.15 \times 1} \)
\( = \frac{4 \pi \times 10^{-7} \times 12}{0.3} \)
\( = \frac{4 \pi \times 10^{-7} \times 12}{0.3} \)
\( = 5.0265 \times 10^{-5} \) T
Hence, the horizontal component of Earth's magnetic field is approximately \( 5.03 \times 10^{-5} \) Tesla.
Number of turns, N = 40
Radius, R = 15 cm = 0.15 m
Current, I = 0.3 A
Deflection angle, \( \theta = 45^\circ \)
Using the tangent law:
\( B = B_H \tan \theta \)
Magnetic field due to coil:
\( B = \frac{\mu_0 N I}{2 R} \)
Equating:
\( \frac{\mu_0 N I}{2 R} = B_H \tan \theta \)
Therefore,
\( B_H = \frac{\mu_0 N I}{2 R \tan \theta} \)
Substitute values:
\( \mu_0 = 4 \pi \times 10^{-7} \) T m/A
\( \tan 45^\circ = 1 \)
\( B_H = \frac{4 \pi \times 10^{-7} \times 40 \times 0.3}{2 \times 0.15 \times 1} \)
\( = \frac{4 \pi \times 10^{-7} \times 12}{0.3} \)
\( = \frac{4 \pi \times 10^{-7} \times 12}{0.3} \)
\( = 5.0265 \times 10^{-5} \) T
Hence, the horizontal component of Earth's magnetic field is approximately \( 5.03 \times 10^{-5} \) Tesla.
Detailed Explanation:
The tangent galvanometer works on the principle that the magnetic field produced by the coil and the horizontal component of Earth's magnetic field are perpendicular.
When current passes through the coil, the needle deflects at an angle \( \theta \) such that the tangent of this angle equals the ratio of the coil's magnetic field to Earth's horizontal magnetic field.
Using the formula for magnetic field at the center of a coil and the tangent law, we can rearrange to find the horizontal component of Earth's magnetic field.
This involves substituting known values and calculating carefully to get the final result.
This problem tests the application of the tangent law and understanding of magnetic fields in coils.
The tangent galvanometer works on the principle that the magnetic field produced by the coil and the horizontal component of Earth's magnetic field are perpendicular.
When current passes through the coil, the needle deflects at an angle \( \theta \) such that the tangent of this angle equals the ratio of the coil's magnetic field to Earth's horizontal magnetic field.
Using the formula for magnetic field at the center of a coil and the tangent law, we can rearrange to find the horizontal component of Earth's magnetic field.
This involves substituting known values and calculating carefully to get the final result.
This problem tests the application of the tangent law and understanding of magnetic fields in coils.
PYQ 2
Tamil Nadu
2023
Define strong nuclear force and mention its key characteristics.
Solution:
Strong nuclear force is the force that holds protons and neutrons together inside the nucleus.
Key characteristics:
1. It is a short-range force acting only up to a few femtometers.
2. It is attractive and stronger than the electrostatic repulsion between protons.
3. It acts equally between proton-proton, neutron-neutron, and proton-neutron pairs.
4. It does not act on electrons.
5. It is charge independent and saturates, meaning each nucleon interacts only with its nearest neighbors.
Key characteristics:
1. It is a short-range force acting only up to a few femtometers.
2. It is attractive and stronger than the electrostatic repulsion between protons.
3. It acts equally between proton-proton, neutron-neutron, and proton-neutron pairs.
4. It does not act on electrons.
5. It is charge independent and saturates, meaning each nucleon interacts only with its nearest neighbors.
Detailed Explanation:
The strong nuclear force is fundamental in maintaining the stability of the nucleus.
Its short range ensures it acts only within the nucleus, overcoming repulsive forces.
Its equal action on all nucleon pairs keeps the nucleus bound.
Understanding these characteristics helps explain nuclear structure and reactions.
Teachers emphasize these points to clarify why nuclei are stable despite proton repulsion.
The strong nuclear force is fundamental in maintaining the stability of the nucleus.
Its short range ensures it acts only within the nucleus, overcoming repulsive forces.
Its equal action on all nucleon pairs keeps the nucleus bound.
Understanding these characteristics helps explain nuclear structure and reactions.
Teachers emphasize these points to clarify why nuclei are stable despite proton repulsion.
PYQ 3
CBSE
2018
Explain why a magnetic dipole experiences torque but zero net force when placed in a uniform magnetic field.
Solution:
A magnetic dipole in a uniform magnetic field experiences equal and opposite forces on its north and south poles, resulting in zero net force.
However, these forces form a couple that produces a torque, causing the dipole to rotate to align with the magnetic field.
Thus, the dipole undergoes rotational motion (torque) but no translational motion (net force).
However, these forces form a couple that produces a torque, causing the dipole to rotate to align with the magnetic field.
Thus, the dipole undergoes rotational motion (torque) but no translational motion (net force).
Detailed Explanation:
Each pole of the dipole experiences a force \( q_m \vec{B} \) but in opposite directions.
Since the forces are equal in magnitude and opposite in direction, they cancel out, producing no net force.
But because these forces act at different points separated by the length of the dipole, they create a torque.
This torque tends to rotate the dipole to align with the magnetic field.
Hence, the dipole experiences torque but no net force in a uniform magnetic field.
Each pole of the dipole experiences a force \( q_m \vec{B} \) but in opposite directions.
Since the forces are equal in magnitude and opposite in direction, they cancel out, producing no net force.
But because these forces act at different points separated by the length of the dipole, they create a torque.
This torque tends to rotate the dipole to align with the magnetic field.
Hence, the dipole experiences torque but no net force in a uniform magnetic field.
PYQ 4
Tamil Nadu Class 12
2023
In a transistor oscillator circuit, if the inductance \( L = 100 \mu H \) and capacitance \( C = 10 pF \), calculate the frequency of oscillations.
Solution:
Given:
\( L = 100 \mu H = 100 \times 10^{-6} \text{ H} \)
\( C = 10 pF = 10 \times 10^{-12} \text{ F} \)
Frequency of oscillations,
\[ f = \frac{1}{2 \pi \sqrt{LC}} \]
Substitute values:
\[ f = \frac{1}{2 \times 3.1416 \times \sqrt{100 \times 10^{-6} \times 10 \times 10^{-12}}} \]
\[ = \frac{1}{6.2832 \times \sqrt{1 \times 10^{-15}}} \]
\[ = \frac{1}{6.2832 \times 10^{-7.5}} \]
\[ = \frac{1}{6.2832 \times 3.1623 \times 10^{-8}} \approx \frac{1}{1.986 \times 10^{-7}} \]
\[ = 5.04 \times 10^{6} \text{ Hz} = 5.04 \text{ MHz} \]
Therefore, the frequency of oscillations is approximately 5.04 MHz.
\( L = 100 \mu H = 100 \times 10^{-6} \text{ H} \)
\( C = 10 pF = 10 \times 10^{-12} \text{ F} \)
Frequency of oscillations,
\[ f = \frac{1}{2 \pi \sqrt{LC}} \]
Substitute values:
\[ f = \frac{1}{2 \times 3.1416 \times \sqrt{100 \times 10^{-6} \times 10 \times 10^{-12}}} \]
\[ = \frac{1}{6.2832 \times \sqrt{1 \times 10^{-15}}} \]
\[ = \frac{1}{6.2832 \times 10^{-7.5}} \]
\[ = \frac{1}{6.2832 \times 3.1623 \times 10^{-8}} \approx \frac{1}{1.986 \times 10^{-7}} \]
\[ = 5.04 \times 10^{6} \text{ Hz} = 5.04 \text{ MHz} \]
Therefore, the frequency of oscillations is approximately 5.04 MHz.
Detailed Explanation:
First, convert the given inductance and capacitance to standard units (Henries and Farads).
Use the formula for frequency of oscillation \( f = \frac{1}{2\pi\sqrt{LC}} \).
Calculate the product \( LC \) and then its square root.
Calculate the denominator \( 2\pi\sqrt{LC} \).
Take the reciprocal to find the frequency.
The final answer is approximately 5.04 MHz.
This problem helps students apply the formula and unit conversions to find oscillation frequency.
First, convert the given inductance and capacitance to standard units (Henries and Farads).
Use the formula for frequency of oscillation \( f = \frac{1}{2\pi\sqrt{LC}} \).
Calculate the product \( LC \) and then its square root.
Calculate the denominator \( 2\pi\sqrt{LC} \).
Take the reciprocal to find the frequency.
The final answer is approximately 5.04 MHz.
This problem helps students apply the formula and unit conversions to find oscillation frequency.
PYQ 5
Tamil Nadu Class 12
2021
What type of signal does an input transducer produce? Explain with an example.
Solution:
An input transducer produces an electrical signal called the baseband signal which is the electrical equivalent of the original information.
For example a microphone converts sound energy (acoustic signal) into an electrical signal.
This electrical signal carries the information in a form suitable for further processing and transmission.
For example a microphone converts sound energy (acoustic signal) into an electrical signal.
This electrical signal carries the information in a form suitable for further processing and transmission.
Detailed Explanation:
Input transducers convert physical signals such as sound, light, or pressure into electrical signals.
This conversion is essential because electronic communication systems operate using electrical signals.
The microphone is a common example where sound waves are converted into corresponding electrical signals.
Teachers can demonstrate this with simple examples to help students relate physical phenomena to electrical signals.
Input transducers convert physical signals such as sound, light, or pressure into electrical signals.
This conversion is essential because electronic communication systems operate using electrical signals.
The microphone is a common example where sound waves are converted into corresponding electrical signals.
Teachers can demonstrate this with simple examples to help students relate physical phenomena to electrical signals.
PYQ 6
Tamil Nadu
2019
Verify the associative law for the OR operation with three Boolean variables \( A, B, C \).
Solution:
Step 1: Write the associative law for OR operation: \( A + (B + C) = (A + B) + C \)
Step 2: Construct truth tables for both sides.
Step 3: For all combinations of \( A, B, \) and \( C \) (0 or 1), calculate \( B + C \), then \( A + (B + C) \).
Step 4: Calculate \( A + B \), then \( (A + B) + C \).
Step 5: Compare the outputs of both expressions for all input combinations.
Step 6: Since outputs match for all cases, associative law for OR is verified.
Step 2: Construct truth tables for both sides.
Step 3: For all combinations of \( A, B, \) and \( C \) (0 or 1), calculate \( B + C \), then \( A + (B + C) \).
Step 4: Calculate \( A + B \), then \( (A + B) + C \).
Step 5: Compare the outputs of both expressions for all input combinations.
Step 6: Since outputs match for all cases, associative law for OR is verified.
Detailed Explanation:
The associative law states that the way in which variables are grouped in OR operation does not affect the result.
By verifying the truth table for both expressions \( A + (B + C) \) and \( (A + B) + C \), we confirm that their outputs are identical for all input combinations.
This confirms the associative property for the OR operation in Boolean algebra.
The associative law states that the way in which variables are grouped in OR operation does not affect the result.
By verifying the truth table for both expressions \( A + (B + C) \) and \( (A + B) + C \), we confirm that their outputs are identical for all input combinations.
This confirms the associative property for the OR operation in Boolean algebra.
PYQ 7
Tamil Nadu Class 12
2023
Arrange the following hydrogen spectral series in increasing order of their wavelengths: Lyman, Balmer, Paschen.
Solution:
Lyman < Balmer < Paschen
Detailed Explanation:
The wavelength of spectral lines increases as the final energy level \( n \) increases.
Lyman series corresponds to transitions ending at \( n=1 \) and lies in the ultraviolet region with shortest wavelengths.
Balmer series corresponds to transitions ending at \( n=2 \) and lies in the visible region with intermediate wavelengths.
Paschen series corresponds to transitions ending at \( n=3 \) and lies in the infrared region with longest wavelengths among these three.
Hence, the increasing order of wavelengths is Lyman, Balmer, Paschen.
The wavelength of spectral lines increases as the final energy level \( n \) increases.
Lyman series corresponds to transitions ending at \( n=1 \) and lies in the ultraviolet region with shortest wavelengths.
Balmer series corresponds to transitions ending at \( n=2 \) and lies in the visible region with intermediate wavelengths.
Paschen series corresponds to transitions ending at \( n=3 \) and lies in the infrared region with longest wavelengths among these three.
Hence, the increasing order of wavelengths is Lyman, Balmer, Paschen.
PYQ 8
CBSE
2019
Write the expression for the angular width of the central maximum in single slit diffraction and explain its significance.
Solution:
The angular width of the central maximum in single slit diffraction is given by \( \Delta \theta = 2 \theta = \frac{2 \lambda}{a} \), where \( \lambda \) is the wavelength of light and \( a \) is the width of the slit.
Explanation:
1. The first minimum on either side of the central maximum occurs at angles where \( a \sin \theta = \lambda \). For small angles, \( \sin \theta \approx \theta \), so \( \theta = \frac{\lambda}{a} \).
2. The central maximum extends from the first minimum on one side to the first minimum on the other side, so the total angular width is twice this angle, \( \Delta \theta = 2 \theta = \frac{2 \lambda}{a} \).
Significance:
- The angular width determines how broad the central bright fringe is in the diffraction pattern.
- It shows that the width of the central maximum is inversely proportional to the slit width; narrower slits produce wider central maxima.
- This concept is important in understanding the resolving power of optical instruments and the limits of image sharpness due to diffraction.
Explanation:
1. The first minimum on either side of the central maximum occurs at angles where \( a \sin \theta = \lambda \). For small angles, \( \sin \theta \approx \theta \), so \( \theta = \frac{\lambda}{a} \).
2. The central maximum extends from the first minimum on one side to the first minimum on the other side, so the total angular width is twice this angle, \( \Delta \theta = 2 \theta = \frac{2 \lambda}{a} \).
Significance:
- The angular width determines how broad the central bright fringe is in the diffraction pattern.
- It shows that the width of the central maximum is inversely proportional to the slit width; narrower slits produce wider central maxima.
- This concept is important in understanding the resolving power of optical instruments and the limits of image sharpness due to diffraction.
Detailed Explanation:
The angular width of the central maximum is derived from the condition of minima in single slit diffraction.
The first minima occur at angles where \( a \sin \theta = \lambda \). For small angles, this approximates to \( \theta = \frac{\lambda}{a} \).
The central maximum extends between these two minima on either side, so the angular width is twice this angle.
This means the central bright fringe becomes wider as the slit narrows or the wavelength increases.
This is significant because it limits the resolution of optical instruments; the diffraction causes spreading of light which affects image clarity.
Understanding this helps in designing instruments with better resolving power by choosing appropriate slit widths and wavelengths.
The angular width of the central maximum is derived from the condition of minima in single slit diffraction.
The first minima occur at angles where \( a \sin \theta = \lambda \). For small angles, this approximates to \( \theta = \frac{\lambda}{a} \).
The central maximum extends between these two minima on either side, so the angular width is twice this angle.
This means the central bright fringe becomes wider as the slit narrows or the wavelength increases.
This is significant because it limits the resolution of optical instruments; the diffraction causes spreading of light which affects image clarity.
Understanding this helps in designing instruments with better resolving power by choosing appropriate slit widths and wavelengths.
PYQ 9
Tamil Nadu Class 12
2021
Match the following components with their functions:
1. Transmitter a) Converts sound waves into electrical signals
2. Antenna b) Sends signals over long distances
3. Repeater c) Regenerates weak signals
Solution:
1 a, 2 b, 3 c
Detailed Explanation:
The transmitter converts the original information into electrical signals, often by modulating a carrier wave.
However, the conversion of sound waves into electrical signals is actually done by the input transducer, but since the option 'a' is given for transmitter, it is understood here as the initial stage of signal processing.
The antenna is responsible for sending signals over long distances by radiating electromagnetic waves.
The repeater regenerates weak signals received over long distances to maintain signal strength and quality.
Matching these components with their functions helps understand the roles of each in the communication system.
The transmitter converts the original information into electrical signals, often by modulating a carrier wave.
However, the conversion of sound waves into electrical signals is actually done by the input transducer, but since the option 'a' is given for transmitter, it is understood here as the initial stage of signal processing.
The antenna is responsible for sending signals over long distances by radiating electromagnetic waves.
The repeater regenerates weak signals received over long distances to maintain signal strength and quality.
Matching these components with their functions helps understand the roles of each in the communication system.
PYQ 10
CBSE
2022
Write the nuclear equation for beta plus decay of sodium-22 and identify all emitted particles.
Solution:
The beta plus decay of sodium-22 is represented as: \( _{11}^{22}\mathrm{Na} \rightarrow _{10}^{22}\mathrm{Ne} + e^{+} + \nu \). The emitted particles are: 1. Positron \( e^{+} \) 2. Neutrino \( \nu \).
Detailed Explanation:
In beta plus decay, a proton in the nucleus converts into a neutron, emitting a positron and a neutrino. For sodium-22, the atomic number decreases by one while the mass number remains the same. The nuclear equation shows the parent nucleus transforming into the daughter nucleus with the emission of a positron and neutrino. This process is important in nuclear transmutations and positron emission tomography.
In beta plus decay, a proton in the nucleus converts into a neutron, emitting a positron and a neutrino. For sodium-22, the atomic number decreases by one while the mass number remains the same. The nuclear equation shows the parent nucleus transforming into the daughter nucleus with the emission of a positron and neutrino. This process is important in nuclear transmutations and positron emission tomography.
PYQ 11
CBSE
2019
In a circuit applying Kirchhoff’s laws, if the sum of voltages in one loop is nonzero, what is the implication?
Solution:
The circuit is not in a steady state or Kirchhoff’s voltage rule is violated indicating an error in analysis or presence of time-varying magnetic fields.
Detailed Explanation:
Kirchhoff’s voltage rule states that the algebraic sum of all voltages around any closed loop in a circuit must be zero.
If the sum of voltages in one loop is nonzero, it implies that either the circuit is not in a steady state or there is an error in the application of the rule.
Another possibility is the presence of changing magnetic fields inducing emf, which violates the assumptions of Kirchhoff’s voltage law.
This indicates that the circuit analysis needs to be reviewed or the physical conditions reconsidered.
Kirchhoff’s voltage rule states that the algebraic sum of all voltages around any closed loop in a circuit must be zero.
If the sum of voltages in one loop is nonzero, it implies that either the circuit is not in a steady state or there is an error in the application of the rule.
Another possibility is the presence of changing magnetic fields inducing emf, which violates the assumptions of Kirchhoff’s voltage law.
This indicates that the circuit analysis needs to be reviewed or the physical conditions reconsidered.
PYQ 12
Tamil Nadu
2019
If the current in a circular loop is doubled and the area is halved, how does the magnetic dipole moment change?
Solution:
Original magnetic dipole moment \( p_m = I \times A \) New current \( I' = 2I \) New area \( A' = \frac{A}{2} \) New magnetic dipole moment \( p'_m = I' \times A' = 2I \times \frac{A}{2} = I \times A = p_m \) Therefore, the magnetic dipole moment remains the same.
Detailed Explanation:
Magnetic dipole moment depends on the product of current and area. Doubling the current doubles the magnetic moment, but halving the area halves it. These two changes cancel each other out, so the magnetic dipole moment remains unchanged.
Magnetic dipole moment depends on the product of current and area. Doubling the current doubles the magnetic moment, but halving the area halves it. These two changes cancel each other out, so the magnetic dipole moment remains unchanged.
PYQ 13
CBSE Class 12
2019
Calculate the surface area to volume ratio of a spherical nanoparticle of radius 5 nm. Compare this with a bulk particle of radius 5 \u03bc m.
Solution:
Step 1 Calculate surface area of a sphere using formula \(4\pi r^{2}\) For nanoparticle radius \(r = 5 \text{ nm} = 5 \times 10^{-9} \text{ m}\) Surface area nanoparticle = \(4 \times 3.1416 \times (5 \times 10^{-9})^{2} = 4 \times 3.1416 \times 25 \times 10^{-18} = 314.16 \times 10^{-18} = 3.1416 \times 10^{-16} \text{ m}^{2}\) Step 2 Calculate volume of a sphere using formula \(\frac{4}{3}\pi r^{3}\) Volume nanoparticle = \(\frac{4}{3} \times 3.1416 \times (5 \times 10^{-9})^{3} = \frac{4}{3} \times 3.1416 \times 125 \times 10^{-27} = 523.6 \times 10^{-27} = 5.236 \times 10^{-25} \text{ m}^{3}\) Step 3 Calculate surface area to volume ratio (SA/V) for nanoparticle \(\text{SA/V nanoparticle} = \frac{3.1416 \times 10^{-16}}{5.236 \times 10^{-25}} = 0.6 \times 10^{9} = 6 \times 10^{8} \text{ m}^{-1}\) Step 4 Repeat calculations for bulk particle radius \(r = 5 \mu \text{m} = 5 \times 10^{-6} \text{ m}\) Surface area bulk = \(4 \times 3.1416 \times (5 \times 10^{-6})^{2} = 4 \times 3.1416 \times 25 \times 10^{-12} = 314.16 \times 10^{-12} = 3.1416 \times 10^{-10} \text{ m}^{2}\) Volume bulk = \(\frac{4}{3} \times 3.1416 \times (5 \times 10^{-6})^{3} = \frac{4}{3} \times 3.1416 \times 125 \times 10^{-18} = 523.6 \times 10^{-18} = 5.236 \times 10^{-16} \text{ m}^{3}\) Step 5 Calculate SA/V for bulk particle \(\text{SA/V bulk} = \frac{3.1416 \times 10^{-10}}{5.236 \times 10^{-16}} = 0.6 \times 10^{5} = 6 \times 10^{4} \text{ m}^{-1}\) Step 6 Compare ratios \(\text{SA/V nanoparticle} = 6 \times 10^{8} \text{ m}^{-1}\) is much larger than \(\text{SA/V bulk} = 6 \times 10^{4} \text{ m}^{-1}\) This shows nanoparticles have a much higher surface area to volume ratio than bulk particles.
Detailed Explanation:
To find the surface area to volume ratio of spherical particles, we use the formulas for surface area and volume of a sphere. First, calculate surface area and volume for both nanoparticle and bulk particle using their respective radii. Then, divide surface area by volume to get the ratio. Because the nanoparticle radius is much smaller, its surface area to volume ratio is significantly higher. This means nanoparticles have more surface area available per unit volume, which affects their physical and chemical properties. This concept is important in nanoscience as it explains why nanoparticles behave differently from bulk materials.
To find the surface area to volume ratio of spherical particles, we use the formulas for surface area and volume of a sphere. First, calculate surface area and volume for both nanoparticle and bulk particle using their respective radii. Then, divide surface area by volume to get the ratio. Because the nanoparticle radius is much smaller, its surface area to volume ratio is significantly higher. This means nanoparticles have more surface area available per unit volume, which affects their physical and chemical properties. This concept is important in nanoscience as it explains why nanoparticles behave differently from bulk materials.
PYQ 14
ICSE
2018
A magnetic dipole of moment \(1.5 \, \text{A\cdot m}^2\) is placed in a uniform magnetic field of \(0.2 \, \text{T}\). Find the potential energy when the dipole is perpendicular to the field.
Solution:
Potential energy \(U = - p_m B \cos \theta\)
Given:
\(p_m = 1.5 \, \text{A\cdot m}^2\), \(B = 0.2 \, \text{T}\), \(\theta = 90^\circ\)
Since \(\cos 90^\circ = 0\),
\(U = -1.5 \times 0.2 \times 0 = 0\)
Therefore, the potential energy is 0 joules when the dipole is perpendicular to the magnetic field.
Given:
\(p_m = 1.5 \, \text{A\cdot m}^2\), \(B = 0.2 \, \text{T}\), \(\theta = 90^\circ\)
Since \(\cos 90^\circ = 0\),
\(U = -1.5 \times 0.2 \times 0 = 0\)
Therefore, the potential energy is 0 joules when the dipole is perpendicular to the magnetic field.
Detailed Explanation:
Potential energy depends on the angle between the magnetic moment and the magnetic field.
When the dipole is perpendicular to the field, the angle \(\theta = 90^\circ\) and \(\cos 90^\circ = 0\).
Substituting these values into the formula \(U = - p_m B \cos \theta\) gives zero potential energy.
This means the dipole neither gains nor loses potential energy in this orientation.
Potential energy depends on the angle between the magnetic moment and the magnetic field.
When the dipole is perpendicular to the field, the angle \(\theta = 90^\circ\) and \(\cos 90^\circ = 0\).
Substituting these values into the formula \(U = - p_m B \cos \theta\) gives zero potential energy.
This means the dipole neither gains nor loses potential energy in this orientation.
PYQ 15
CBSE
2021
What fundamental force mediates the interaction responsible for beta decay?
Solution:
The fundamental force responsible for beta decay is the weak nuclear force.
Detailed Explanation:
Beta decay involves the transformation of a neutron into a proton or vice versa.
This process is mediated by the weak nuclear force, one of the four fundamental forces of nature.
The weak force allows quarks to change flavor, enabling beta decay.
It is characterized by its very short range and its role in processes like nuclear decay and fusion in stars.
Beta decay involves the transformation of a neutron into a proton or vice versa.
This process is mediated by the weak nuclear force, one of the four fundamental forces of nature.
The weak force allows quarks to change flavor, enabling beta decay.
It is characterized by its very short range and its role in processes like nuclear decay and fusion in stars.
PYQ 16
CBSE Class 12
2021
Calculate the period of circular motion of a proton in a magnetic field of 0.8 T, given the charge and mass of the proton.
Solution:
Given:
Magnetic field, \( B = 0.8 \) T
Charge of proton, \( q = 1.60 \times 10^{-19} \) C
Mass of proton, \( m = 1.67 \times 10^{-27} \) kg
The period of circular motion (cyclotron period) is given by:
\( T = \frac{2 \pi m}{q B} \)
Substitute values:
\( T = \frac{2 \pi \times 1.67 \times 10^{-27}}{1.60 \times 10^{-19} \times 0.8} \)
\( T = \frac{10.5 \times 10^{-27}}{1.28 \times 10^{-19}} = 8.20 \times 10^{-8} \) seconds
Therefore, the period of the proton's circular motion is approximately \( 8.20 \times 10^{-8} \) seconds.
Magnetic field, \( B = 0.8 \) T
Charge of proton, \( q = 1.60 \times 10^{-19} \) C
Mass of proton, \( m = 1.67 \times 10^{-27} \) kg
The period of circular motion (cyclotron period) is given by:
\( T = \frac{2 \pi m}{q B} \)
Substitute values:
\( T = \frac{2 \pi \times 1.67 \times 10^{-27}}{1.60 \times 10^{-19} \times 0.8} \)
\( T = \frac{10.5 \times 10^{-27}}{1.28 \times 10^{-19}} = 8.20 \times 10^{-8} \) seconds
Therefore, the period of the proton's circular motion is approximately \( 8.20 \times 10^{-8} \) seconds.
Detailed Explanation:
The cyclotron period depends only on the charge-to-mass ratio and magnetic field strength.
It is independent of the particle's velocity or radius of the path.
By substituting the known constants for the proton and magnetic field strength, the period can be calculated.
This concept is used in cyclotrons to accelerate charged particles.
The cyclotron period depends only on the charge-to-mass ratio and magnetic field strength.
It is independent of the particle's velocity or radius of the path.
By substituting the known constants for the proton and magnetic field strength, the period can be calculated.
This concept is used in cyclotrons to accelerate charged particles.
PYQ 17
Karnataka SSLC
2020
Explain the applications of X-rays in industry and medical diagnostics.
Solution:
In medical diagnostics, X-rays are used to create radiographs that show bones and tissues, helping detect fractures, foreign bodies, and diseased organs.
In industry, X-rays are employed to check for flaws in welded joints, motor tyres, tennis balls, and wood.
They are also used at customs for detection of contraband goods.
In industry, X-rays are employed to check for flaws in welded joints, motor tyres, tennis balls, and wood.
They are also used at customs for detection of contraband goods.
Detailed Explanation:
X-rays have the ability to penetrate materials differently depending on their density.
In medical diagnostics, this property allows visualization of bones and soft tissues, aiding in diagnosis.
In industry, X-rays help ensure quality control by detecting internal defects without damaging the objects.
These applications demonstrate the practical importance of X-rays in everyday life and safety.
X-rays have the ability to penetrate materials differently depending on their density.
In medical diagnostics, this property allows visualization of bones and soft tissues, aiding in diagnosis.
In industry, X-rays help ensure quality control by detecting internal defects without damaging the objects.
These applications demonstrate the practical importance of X-rays in everyday life and safety.
PYQ 18
Tamil Nadu
2022
The number of nuclei of a radioactive substance decreases to one-eighth of the original in 33 hours. Calculate the half-life.
Solution:
Given that the number of nuclei decreases to one-eighth, so \( N = \frac{N_0}{8} \) after time \( t = 33 \) hours.
We know that \( N = N_0 \left( \frac{1}{2} \right)^n \), where \( n = \frac{t}{T_{1/2}} \) is the number of half-lives.
Since \( N = \frac{N_0}{8} = N_0 \left( \frac{1}{2} \right)^3 \), it means \( n = 3 \).
Therefore, \( n = \frac{t}{T_{1/2}} = 3 \Rightarrow T_{1/2} = \frac{t}{3} = \frac{33}{3} = 11 \) hours.
Hence, the half-life of the substance is 11 hours.
We know that \( N = N_0 \left( \frac{1}{2} \right)^n \), where \( n = \frac{t}{T_{1/2}} \) is the number of half-lives.
Since \( N = \frac{N_0}{8} = N_0 \left( \frac{1}{2} \right)^3 \), it means \( n = 3 \).
Therefore, \( n = \frac{t}{T_{1/2}} = 3 \Rightarrow T_{1/2} = \frac{t}{3} = \frac{33}{3} = 11 \) hours.
Hence, the half-life of the substance is 11 hours.
Detailed Explanation:
The problem involves understanding the concept of half-life and how the number of undecayed nuclei decreases exponentially.
Since the number of nuclei reduces to one-eighth, which is \( \left( \frac{1}{2} \right)^3 \), it means three half-lives have passed.
By dividing the total time by the number of half-lives, we find the duration of one half-life.
This approach is straightforward and relies on the definition of half-life and exponential decay.
The problem involves understanding the concept of half-life and how the number of undecayed nuclei decreases exponentially.
Since the number of nuclei reduces to one-eighth, which is \( \left( \frac{1}{2} \right)^3 \), it means three half-lives have passed.
By dividing the total time by the number of half-lives, we find the duration of one half-life.
This approach is straightforward and relies on the definition of half-life and exponential decay.
PYQ 19
Tamil Nadu
2021
Explain the Peltier effect and state how heat is evolved or absorbed at the junctions when current flows in a thermocouple circuit.
Solution:
The Peltier effect is the phenomenon where heat is either absorbed or evolved at the junctions of two different metals when an electric current flows through a thermocouple circuit. When current passes through the junction of two dissimilar metals, one junction absorbs heat and becomes cold, while the other junction evolves heat and becomes hot. The direction of heat flow depends on the direction of current. Reversing the current reverses the heat absorption and evolution at the junctions.
Detailed Explanation:
The Peltier effect is the reverse of the Seebeck effect. Instead of generating emf from temperature difference, it involves heat transfer due to electric current. When current flows through a thermocouple, one junction absorbs heat (cooling effect) and the other junction releases heat (heating effect). This happens because the energy carried by electrons changes as they cross the junctions of different metals. The effect is reversible, meaning changing the direction of current reverses which junction heats or cools. This effect is used in thermoelectric cooling devices.
The Peltier effect is the reverse of the Seebeck effect. Instead of generating emf from temperature difference, it involves heat transfer due to electric current. When current flows through a thermocouple, one junction absorbs heat (cooling effect) and the other junction releases heat (heating effect). This happens because the energy carried by electrons changes as they cross the junctions of different metals. The effect is reversible, meaning changing the direction of current reverses which junction heats or cools. This effect is used in thermoelectric cooling devices.
PYQ 20
Tamil Nadu
2022
Arrange the following robot parts in the sequence starting from control to actuation: Controller, Sensors, Manipulators, Power supply.
Solution:
The correct sequence starting from control to actuation is: Power supply, Sensors, Controller, Manipulators
Detailed Explanation:
The power supply provides energy to the robot system. Sensors detect information from the environment and send it to the controller. The controller processes this information and sends commands to the manipulators which perform the mechanical actions. Hence, the sequence is Power supply → Sensors → Controller → Manipulators.
The power supply provides energy to the robot system. Sensors detect information from the environment and send it to the controller. The controller processes this information and sends commands to the manipulators which perform the mechanical actions. Hence, the sequence is Power supply → Sensors → Controller → Manipulators.
