Previous Year Questions
Solve previous year questions to understand exam patterns
PYQ 1
Tamil Nadu
2014
What is voltage magnification in resonance? How is it related to Q-factor?
Solution:
Voltage magnification in resonance is the phenomenon where the voltage across the inductor or capacitor in a series RLC circuit becomes much larger than the applied voltage at resonant frequency.
At resonance, the inductive reactance equals the capacitive reactance, and the circuit behaves like a pure resistor.
However, voltages across L and C can be significantly higher due to energy oscillation between them.
This magnification is quantified by the Q-factor.
Mathematically, voltage magnification = Q-factor = \( \frac{\text{Voltage across } L \text{ or } C}{\text{Applied voltage}} \).
At resonance, the inductive reactance equals the capacitive reactance, and the circuit behaves like a pure resistor.
However, voltages across L and C can be significantly higher due to energy oscillation between them.
This magnification is quantified by the Q-factor.
Mathematically, voltage magnification = Q-factor = \( \frac{\text{Voltage across } L \text{ or } C}{\text{Applied voltage}} \).
Detailed Explanation:
At resonance, the series RLC circuit's impedance is minimum, allowing maximum current.
The voltage across L and C can be much greater than the source voltage due to energy exchange.
This effect is called voltage magnification.
The Q-factor measures how many times the voltage across L or C is greater than the applied voltage.
Thus, voltage magnification and Q-factor are directly related and represent the same concept.
At resonance, the series RLC circuit's impedance is minimum, allowing maximum current.
The voltage across L and C can be much greater than the source voltage due to energy exchange.
This effect is called voltage magnification.
The Q-factor measures how many times the voltage across L or C is greater than the applied voltage.
Thus, voltage magnification and Q-factor are directly related and represent the same concept.
PYQ 2
Tamil Nadu
2020
An unregulated power supply has an input voltage varying between 12 V to 14 V. Design a voltage regulator using a Zener diode for 12 V output and explain its working.
Solution:
To design a voltage regulator for 12 V output:
1. Choose a Zener diode with breakdown voltage V_Z = 12 V.
2. Determine the series resistor R_s to limit the current.
Given:
- Input voltage varies from 12 V to 14 V.
- Output voltage V_o = 12 V.
Assuming load current I_L and Zener current I_Z(min) to maintain regulation are known or estimated.
Calculate R_s using:
R_s = (V_in(max) - V_Z) / (I_L + I_Z(min))
Working:
- When input voltage increases, the extra voltage is dropped across R_s.
- The Zener diode maintains 12 V across the load by shunting excess current.
- When input voltage decreases, the Zener diode current decreases but voltage remains constant.
This ensures a stable 12 V output despite input voltage variations.
1. Choose a Zener diode with breakdown voltage V_Z = 12 V.
2. Determine the series resistor R_s to limit the current.
Given:
- Input voltage varies from 12 V to 14 V.
- Output voltage V_o = 12 V.
Assuming load current I_L and Zener current I_Z(min) to maintain regulation are known or estimated.
Calculate R_s using:
R_s = (V_in(max) - V_Z) / (I_L + I_Z(min))
Working:
- When input voltage increases, the extra voltage is dropped across R_s.
- The Zener diode maintains 12 V across the load by shunting excess current.
- When input voltage decreases, the Zener diode current decreases but voltage remains constant.
This ensures a stable 12 V output despite input voltage variations.
Detailed Explanation:
Select a Zener diode with breakdown voltage equal to desired output voltage (12 V).
Calculate the series resistor to limit current and protect the diode.
The resistor drops the excess voltage when input voltage is higher than 12 V.
The Zener diode maintains constant voltage across the load by operating in breakdown region.
This design ensures output voltage remains at 12 V even if input varies between 12 V and 14 V.
Select a Zener diode with breakdown voltage equal to desired output voltage (12 V).
Calculate the series resistor to limit current and protect the diode.
The resistor drops the excess voltage when input voltage is higher than 12 V.
The Zener diode maintains constant voltage across the load by operating in breakdown region.
This design ensures output voltage remains at 12 V even if input varies between 12 V and 14 V.
PYQ 3
ICSE
2020
Problem: A spectral line in an emission spectrum has a frequency of \(4.5 \times 10^{14} \ \text{Hz}\). Calculate the wavelength of this light.
Solution:
Given frequency \(f = 4.5 \times 10^{14} \ \text{Hz}\)
Speed of light \(c = 3 \times 10^{8} \ \text{m/s}\)
Wavelength \(\lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{4.5 \times 10^{14}} = 6.67 \times 10^{-7} \ \text{m}\)
Therefore, the wavelength of the light is \(6.67 \times 10^{-7}\) meters or 667 nm.
Speed of light \(c = 3 \times 10^{8} \ \text{m/s}\)
Wavelength \(\lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{4.5 \times 10^{14}} = 6.67 \times 10^{-7} \ \text{m}\)
Therefore, the wavelength of the light is \(6.67 \times 10^{-7}\) meters or 667 nm.
Detailed Explanation:
To find the wavelength from the frequency, use the relation \(\lambda = \frac{c}{f}\).
Substitute the given frequency and speed of light values.
Perform the division carefully to get the wavelength in meters.
This problem reinforces the direct relationship between frequency and wavelength of electromagnetic waves.
To find the wavelength from the frequency, use the relation \(\lambda = \frac{c}{f}\).
Substitute the given frequency and speed of light values.
Perform the division carefully to get the wavelength in meters.
This problem reinforces the direct relationship between frequency and wavelength of electromagnetic waves.
PYQ 4
CBSE Class 12
2020
True or False: The primary health hazard when humans only monitor robots is physical injury from robot malfunction.
Solution:
False
Detailed Explanation:
Although robot malfunctions can cause physical injuries, the primary health hazard when humans only monitor robots is the increase in health problems due to inactivity and lack of physical work.
Monitoring robots often leads to sedentary behavior, which can cause various health issues.
Therefore, the main concern is not physical injury but health hazards related to inactivity.
Teachers should emphasize the difference between direct physical risks and indirect health consequences of robotic workplaces.
Although robot malfunctions can cause physical injuries, the primary health hazard when humans only monitor robots is the increase in health problems due to inactivity and lack of physical work.
Monitoring robots often leads to sedentary behavior, which can cause various health issues.
Therefore, the main concern is not physical injury but health hazards related to inactivity.
Teachers should emphasize the difference between direct physical risks and indirect health consequences of robotic workplaces.
PYQ 5
CBSE
2018
Explain the role of the multiplier band in the colour coding of carbon resistors.
Solution:
The multiplier band indicates the power of ten by which the two significant figures should be multiplied to get the resistance value. It scales the base number formed by the first two bands to the actual resistance value in ohms.
Detailed Explanation:
In the colour coding system for carbon resistors, the first two bands represent the significant digits of the resistance value. The multiplier band determines how many zeros to add or the factor by which to multiply these digits. For example, a multiplier band of Red means multiply by \(10^2\), Orange means multiply by \(10^3\), and so on. This allows a compact representation of a wide range of resistance values using just a few colour bands.
In the colour coding system for carbon resistors, the first two bands represent the significant digits of the resistance value. The multiplier band determines how many zeros to add or the factor by which to multiply these digits. For example, a multiplier band of Red means multiply by \(10^2\), Orange means multiply by \(10^3\), and so on. This allows a compact representation of a wide range of resistance values using just a few colour bands.
PYQ 6
ICSE
2020
Calculate the image distance formed by refraction at a convex spherical surface of radius \(20 \text{ cm}\) when the object distance is \(25 \text{ cm}\) and refractive indices are \(n_1=1.0\) and \(n_2=1.5\).
Solution:
Given: Object distance \(u = -25 \text{ cm}\) (object in front of surface), Radius of curvature \(R = +20 \text{ cm}\) (convex surface), Refractive indices \(n_1 = 1.0\), \(n_2 = 1.5\).
Using the refraction formula:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
Substitute values:
\[ \frac{1.5}{v} - \frac{1.0}{-25} = \frac{1.5 - 1.0}{20} \]
\[ \frac{1.5}{v} + 0.04 = 0.025 \]
\[ \frac{1.5}{v} = 0.025 - 0.04 = -0.015 \]
\[ v = \frac{1.5}{-0.015} = -100 \text{ cm} \]
The image is formed at 100 cm on the same side as the object (virtual image).
Using the refraction formula:
\[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \]
Substitute values:
\[ \frac{1.5}{v} - \frac{1.0}{-25} = \frac{1.5 - 1.0}{20} \]
\[ \frac{1.5}{v} + 0.04 = 0.025 \]
\[ \frac{1.5}{v} = 0.025 - 0.04 = -0.015 \]
\[ v = \frac{1.5}{-0.015} = -100 \text{ cm} \]
The image is formed at 100 cm on the same side as the object (virtual image).
Detailed Explanation:
This problem is similar to the previous one but with different parameters.
Applying the refraction formula at a spherical surface allows calculation of image distance.
Sign conventions are important: object distance is negative because the object is in front of the surface; radius of curvature is positive for convex surfaces.
The negative image distance indicates a virtual image formed on the same side as the object.
This exercise reinforces understanding of refraction at curved surfaces and the use of sign conventions.
This problem is similar to the previous one but with different parameters.
Applying the refraction formula at a spherical surface allows calculation of image distance.
Sign conventions are important: object distance is negative because the object is in front of the surface; radius of curvature is positive for convex surfaces.
The negative image distance indicates a virtual image formed on the same side as the object.
This exercise reinforces understanding of refraction at curved surfaces and the use of sign conventions.
PYQ 7
CBSE
2016
Explain why the deflection needle in a tangent galvanometer aligns along the bisector of the angle between Earth's horizontal magnetic field and the coil's magnetic field.
Solution:
The deflection needle aligns along the resultant magnetic field which is the vector sum of Earth's horizontal magnetic field \( B_H \) and the magnetic field \( B \) produced by the coil.
Since these two fields are perpendicular to each other, the resultant field lies along the diagonal of the rectangle formed by \( B_H \) and \( B \).
The needle aligns along this resultant field to minimize potential energy.
This resultant field bisects the angle between the two perpendicular fields, so the needle points along the bisector of the angle between Earth's horizontal magnetic field and the coil's magnetic field.
This explains the observed deflection in the tangent galvanometer experiment.
Since these two fields are perpendicular to each other, the resultant field lies along the diagonal of the rectangle formed by \( B_H \) and \( B \).
The needle aligns along this resultant field to minimize potential energy.
This resultant field bisects the angle between the two perpendicular fields, so the needle points along the bisector of the angle between Earth's horizontal magnetic field and the coil's magnetic field.
This explains the observed deflection in the tangent galvanometer experiment.
Detailed Explanation:
The magnetic needle experiences torque due to the combined effect of Earth's horizontal magnetic field and the magnetic field from the coil.
Because these two fields are perpendicular, their vector sum forms the resultant field.
The needle aligns along this resultant field to achieve equilibrium.
Since the resultant is the diagonal of the rectangle formed by the two perpendicular fields, it lies along the bisector of the angle between them.
This physical reasoning helps students visualize magnetic field vectors and understand the behavior of magnetic needles in combined fields.
The magnetic needle experiences torque due to the combined effect of Earth's horizontal magnetic field and the magnetic field from the coil.
Because these two fields are perpendicular, their vector sum forms the resultant field.
The needle aligns along this resultant field to achieve equilibrium.
Since the resultant is the diagonal of the rectangle formed by the two perpendicular fields, it lies along the bisector of the angle between them.
This physical reasoning helps students visualize magnetic field vectors and understand the behavior of magnetic needles in combined fields.
PYQ 8
Tamil Nadu Class 12
2023
The deflection of cathode rays decreases when the strength of the magnetic field is increased. Explain why.
Solution:
When the magnetic field strength is increased, the magnetic force on the cathode rays increases. Since the magnetic force acts perpendicular to the velocity and electric force, if the magnetic force becomes stronger than the electric force, the net force causes the electrons to curve less in the direction of the electric field. If the magnetic force balances the electric force exactly, the deflection becomes zero. Therefore, increasing magnetic field strength reduces the net deflection of cathode rays.
Detailed Explanation:
Cathode rays experience forces due to both electric and magnetic fields.
The electric force causes deflection in one direction, while the magnetic force causes deflection in the opposite direction.
Increasing the magnetic field increases the magnetic force, which opposes the electric force.
As a result, the net deflection decreases.
This explains why increasing magnetic field strength reduces the deflection of cathode rays.
Cathode rays experience forces due to both electric and magnetic fields.
The electric force causes deflection in one direction, while the magnetic force causes deflection in the opposite direction.
Increasing the magnetic field increases the magnetic force, which opposes the electric force.
As a result, the net deflection decreases.
This explains why increasing magnetic field strength reduces the deflection of cathode rays.
PYQ 9
CBSE
2019
Explain how the wattless current arises in an AC circuit containing only inductance or capacitance.
Solution:
In a purely inductive or capacitive AC circuit, the current and voltage are out of phase by 90 degrees.
This means the current either leads or lags the voltage by 90°.
Because of this phase difference, the average power consumed over a cycle is zero.
The current component that is 90° out of phase with voltage does not contribute to real power consumption and is called wattless current.
It represents energy that is alternately stored and released by the inductor or capacitor without being dissipated.
This means the current either leads or lags the voltage by 90°.
Because of this phase difference, the average power consumed over a cycle is zero.
The current component that is 90° out of phase with voltage does not contribute to real power consumption and is called wattless current.
It represents energy that is alternately stored and released by the inductor or capacitor without being dissipated.
Detailed Explanation:
In inductors and capacitors, energy is stored temporarily in magnetic and electric fields respectively.
The current associated with this energy storage leads or lags the voltage by 90 degrees.
Because power is the product of voltage and current, and these are out of phase, the net power over a cycle is zero.
This current is called wattless because it does not perform any real work but affects the circuit's reactive properties.
Understanding wattless current is important for power factor correction and efficient circuit design.
In inductors and capacitors, energy is stored temporarily in magnetic and electric fields respectively.
The current associated with this energy storage leads or lags the voltage by 90 degrees.
Because power is the product of voltage and current, and these are out of phase, the net power over a cycle is zero.
This current is called wattless because it does not perform any real work but affects the circuit's reactive properties.
Understanding wattless current is important for power factor correction and efficient circuit design.
PYQ 10
Tamil Nadu
2022
Two coherent sources of light having wavelength 600 nm interfere to produce bright and dark fringes on a screen 1 m away. If distance between the sources is 0.2 mm, calculate the fringe width.
Solution:
Given wavelength \( \lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m} \), distance between sources \( d = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m} \), distance to screen \( D = 1 \text{ m} \).
Fringe width \( \beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 3 \times 10^{-3} \text{ m} = 3 \text{ mm} \).
Fringe width is 3 mm.
Fringe width \( \beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1}{0.2 \times 10^{-3}} = 3 \times 10^{-3} \text{ m} = 3 \text{ mm} \).
Fringe width is 3 mm.
Detailed Explanation:
Fringe width in interference is the distance between two consecutive bright or dark fringes.
It depends on the wavelength of light, distance between the sources, and distance to the screen.
The formula \( \beta = \frac{\lambda D}{d} \) is used to calculate fringe width.
Substituting the given values yields the fringe width.
This problem illustrates the practical application of interference in optics.
Fringe width in interference is the distance between two consecutive bright or dark fringes.
It depends on the wavelength of light, distance between the sources, and distance to the screen.
The formula \( \beta = \frac{\lambda D}{d} \) is used to calculate fringe width.
Substituting the given values yields the fringe width.
This problem illustrates the practical application of interference in optics.
PYQ 11
Tamil Nadu Class 12
2023
Explain the role of applied mathematics in nanotechnology research with one suitable example.
Solution:
Applied mathematics plays a crucial role in nanotechnology research by providing the mathematical models and computational techniques needed to understand and predict the behavior of nanoscale materials and devices.
For example, differential equations and linear algebra are used to model quantum mechanical effects in nanoparticles, helping researchers design materials with desired optical or electronic properties.
For example, differential equations and linear algebra are used to model quantum mechanical effects in nanoparticles, helping researchers design materials with desired optical or electronic properties.
Detailed Explanation:
Applied mathematics is fundamental in nanotechnology because it allows scientists to create models that describe how nanoscale systems behave.
These models help in simulating physical phenomena like electron transport, heat transfer, and mechanical properties at the nanoscale.
By solving these mathematical models, researchers can predict how nanomaterials will perform before physically creating them, saving time and resources.
For example, using mathematical modeling, one can simulate how quantum dots absorb and emit light, which is essential for designing new optical devices.
Applied mathematics is fundamental in nanotechnology because it allows scientists to create models that describe how nanoscale systems behave.
These models help in simulating physical phenomena like electron transport, heat transfer, and mechanical properties at the nanoscale.
By solving these mathematical models, researchers can predict how nanomaterials will perform before physically creating them, saving time and resources.
For example, using mathematical modeling, one can simulate how quantum dots absorb and emit light, which is essential for designing new optical devices.
PYQ 12
Explain how the proton-proton cycle maintains energy generation in stars like the Sun.
Solution:
The proton-proton cycle involves three main steps:
1. Two protons fuse to form a deuterium nucleus, a positron, and a neutrino.
2. The deuterium nucleus fuses with another proton to form helium-3 and emits a gamma ray.
3. Two helium-3 nuclei fuse to form helium-4 and release two protons.
These reactions convert hydrogen into helium, releasing energy in the form of gamma rays, positrons, and neutrinos.
The energy released balances gravitational collapse and sustains the star's radiation output.
1. Two protons fuse to form a deuterium nucleus, a positron, and a neutrino.
2. The deuterium nucleus fuses with another proton to form helium-3 and emits a gamma ray.
3. Two helium-3 nuclei fuse to form helium-4 and release two protons.
These reactions convert hydrogen into helium, releasing energy in the form of gamma rays, positrons, and neutrinos.
The energy released balances gravitational collapse and sustains the star's radiation output.
Detailed Explanation:
The proton-proton cycle is the primary fusion process in stars like the Sun.
Each step releases energy by converting mass into energy.
The cycle ensures continuous energy production as hydrogen nuclei fuse.
Understanding this cycle helps students appreciate how stars generate energy over billions of years.
The proton-proton cycle is the primary fusion process in stars like the Sun.
Each step releases energy by converting mass into energy.
The cycle ensures continuous energy production as hydrogen nuclei fuse.
Understanding this cycle helps students appreciate how stars generate energy over billions of years.
PYQ 13
ICSE
2019
State and explain the Biot-Savart law for the magnetic field due to a current element.
Solution:
Biot-Savart law states that the magnetic field \( d\vec{B} \) at a point P due to a small current element \( I d\vec{l} \) is directly proportional to the current \( I \), the length of the element \( dl \), and the sine of the angle \( \theta \) between the current element and the line joining the element to the point P. It is inversely proportional to the square of the distance \( r \) between the element and point P. Mathematically, \[ d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} \], where \( \hat{r} \) is the unit vector from the current element to point P and \( \mu_0 \) is the permeability of free space. The direction of \( d\vec{B} \) is perpendicular to both the current element and the line joining the element to the point, determined by the right-hand rule.
Detailed Explanation:
The Biot-Savart law helps us calculate the magnetic field created by a small segment of current-carrying wire.
It shows how the magnetic field depends on the current, the size and orientation of the wire segment, and the distance to the observation point.
The vector nature of the formula means the magnetic field direction is perpendicular to both the current element and the line to the point.
This law is fundamental for understanding magnetic fields from any shaped conductor by integrating over the entire wire.
Teachers emphasize the cross product and the inverse square dependence to help students visualize the magnetic field generation.
The Biot-Savart law helps us calculate the magnetic field created by a small segment of current-carrying wire.
It shows how the magnetic field depends on the current, the size and orientation of the wire segment, and the distance to the observation point.
The vector nature of the formula means the magnetic field direction is perpendicular to both the current element and the line to the point.
This law is fundamental for understanding magnetic fields from any shaped conductor by integrating over the entire wire.
Teachers emphasize the cross product and the inverse square dependence to help students visualize the magnetic field generation.
PYQ 14
Tamil Nadu
2023
What happens to the mass number and atomic number during an alpha decay? Write the corresponding nuclear equation for the alpha decay of Uranium-238.
Solution:
In alpha decay, the atomic number decreases by 2 and the mass number decreases by 4.
For Uranium-238, the alpha decay is:
\[ _{92}^{238}U \rightarrow _{90}^{234}Th + _{2}^{4}He \]
For Uranium-238, the alpha decay is:
\[ _{92}^{238}U \rightarrow _{90}^{234}Th + _{2}^{4}He \]
Detailed Explanation:
Alpha decay involves emission of an alpha particle which is a helium nucleus containing 2 protons and 2 neutrons.
Thus, the parent nucleus loses 2 protons and 2 neutrons.
This reduces the atomic number by 2 and mass number by 4.
The nuclear equation represents this transformation clearly.
Alpha decay involves emission of an alpha particle which is a helium nucleus containing 2 protons and 2 neutrons.
Thus, the parent nucleus loses 2 protons and 2 neutrons.
This reduces the atomic number by 2 and mass number by 4.
The nuclear equation represents this transformation clearly.
PYQ 15
Tamil Nadu Class 12
2022
Arrange the following devices according to their application of heating effect from highest to lowest heat generation: Electric furnace, Electric lamp filament, Electric heater, Fuse wire.
Solution:
The order of devices according to their heat generation from highest to lowest is Electric furnace Electric heater Electric lamp filament Fuse wire
Detailed Explanation:
The electric furnace is designed to generate very high temperatures for industrial processes, so it produces the highest heat. Electric heaters are used for domestic heating and produce significant heat but less than furnaces. Electric lamp filaments produce heat as a byproduct of light generation but less than heaters. Fuse wires produce heat only to the extent necessary to melt and break the circuit, so they generate the least heat among these devices.
The electric furnace is designed to generate very high temperatures for industrial processes, so it produces the highest heat. Electric heaters are used for domestic heating and produce significant heat but less than furnaces. Electric lamp filaments produce heat as a byproduct of light generation but less than heaters. Fuse wires produce heat only to the extent necessary to melt and break the circuit, so they generate the least heat among these devices.
PYQ 16
CBSE
2022
An oil drop of density \( \rho = 900 \) kg/m\(^3\) and radius \( r = 1 \times 10^{-6} \) m falls with terminal velocity \( v = 3 \times 10^{-5} \) m/s in air of density \( 1.2 \) kg/m\(^3\) and viscosity \( \eta = 1.8 \times 10^{-5} \) Ns/m\(^2\). Calculate the radius of the drop and the viscous force acting on it.
Solution:
Given data:
Density of oil drop, \( \rho = 900 \) kg/m\(^3\)
Radius, \( r = 1 \times 10^{-6} \) m
Terminal velocity, \( v = 3 \times 10^{-5} \) m/s
Density of air, \( \sigma = 1.2 \) kg/m\(^3\)
Viscosity of air, \( \eta = 1.8 \times 10^{-5} \) Ns/m\(^2\)
Step 1: Calculate the radius using formula
\[ r = \sqrt{ \frac{9 \eta v}{2 (\rho - \sigma) g} } \]
Substitute values:
\( g = 9.8 \) m/s\(^2\)
\[ r = \sqrt{ \frac{9 \times 1.8 \times 10^{-5} \times 3 \times 10^{-5}}{2 \times (900 - 1.2) \times 9.8} } \]
\[ r = \sqrt{ \frac{4.86 \times 10^{-9}}{17642.5} } = \sqrt{2.756 \times 10^{-13}} = 1.66 \times 10^{-6} \text{ m} \]
Step 2: Calculate viscous force using Stokes' law
\[ F_v = 6 \pi r \eta v \]
\[ F_v = 6 \times 3.1416 \times 1.66 \times 10^{-6} \times 1.8 \times 10^{-5} \times 3 \times 10^{-5} \]
\[ F_v = 1.68 \times 10^{-14} \text{ N} \]
Therefore, the radius of the drop is approximately \(1.66 \times 10^{-6}\) m and the viscous force acting on it is \(1.68 \times 10^{-14}\) N.
Density of oil drop, \( \rho = 900 \) kg/m\(^3\)
Radius, \( r = 1 \times 10^{-6} \) m
Terminal velocity, \( v = 3 \times 10^{-5} \) m/s
Density of air, \( \sigma = 1.2 \) kg/m\(^3\)
Viscosity of air, \( \eta = 1.8 \times 10^{-5} \) Ns/m\(^2\)
Step 1: Calculate the radius using formula
\[ r = \sqrt{ \frac{9 \eta v}{2 (\rho - \sigma) g} } \]
Substitute values:
\( g = 9.8 \) m/s\(^2\)
\[ r = \sqrt{ \frac{9 \times 1.8 \times 10^{-5} \times 3 \times 10^{-5}}{2 \times (900 - 1.2) \times 9.8} } \]
\[ r = \sqrt{ \frac{4.86 \times 10^{-9}}{17642.5} } = \sqrt{2.756 \times 10^{-13}} = 1.66 \times 10^{-6} \text{ m} \]
Step 2: Calculate viscous force using Stokes' law
\[ F_v = 6 \pi r \eta v \]
\[ F_v = 6 \times 3.1416 \times 1.66 \times 10^{-6} \times 1.8 \times 10^{-5} \times 3 \times 10^{-5} \]
\[ F_v = 1.68 \times 10^{-14} \text{ N} \]
Therefore, the radius of the drop is approximately \(1.66 \times 10^{-6}\) m and the viscous force acting on it is \(1.68 \times 10^{-14}\) N.
Detailed Explanation:
First, we use the relation derived from force balance at terminal velocity to find the radius of the oil drop.
This formula relates viscosity, terminal velocity, density difference, and gravity to the radius.
Substituting the given values carefully gives the radius.
Next, using Stokes' law, the viscous force is calculated as proportional to radius, viscosity, and velocity.
This stepwise calculation helps understand how physical quantities relate in the experiment.
First, we use the relation derived from force balance at terminal velocity to find the radius of the oil drop.
This formula relates viscosity, terminal velocity, density difference, and gravity to the radius.
Substituting the given values carefully gives the radius.
Next, using Stokes' law, the viscous force is calculated as proportional to radius, viscosity, and velocity.
This stepwise calculation helps understand how physical quantities relate in the experiment.
PYQ 17
CBSE
2020
Problem: If the Zener voltage of a diode is \(5.6 \, V\), calculate the electric field strength across a depletion layer width of \(100 \, nm\).
Solution:
Given:
Zener voltage, \(V_Z = 5.6 \, V\)
Depletion layer width, \(d = 100 \, nm = 100 \times 10^{-9} \, m = 1 \times 10^{-7} \, m\)
Electric field strength, \(E = \frac{V_Z}{d} = \frac{5.6}{1 \times 10^{-7}} = 5.6 \times 10^{7} \, V/m\)
Therefore, the electric field strength across the depletion layer is \(5.6 \times 10^{7} \, V/m\).
Zener voltage, \(V_Z = 5.6 \, V\)
Depletion layer width, \(d = 100 \, nm = 100 \times 10^{-9} \, m = 1 \times 10^{-7} \, m\)
Electric field strength, \(E = \frac{V_Z}{d} = \frac{5.6}{1 \times 10^{-7}} = 5.6 \times 10^{7} \, V/m\)
Therefore, the electric field strength across the depletion layer is \(5.6 \times 10^{7} \, V/m\).
Detailed Explanation:
The electric field strength is the voltage divided by the distance over which the voltage is applied.
Here, the voltage is the Zener voltage and the distance is the depletion layer width.
Convert the width from nanometers to meters for consistent units.
Divide the voltage by the width to get the electric field strength in volts per meter.
This calculation shows the very strong electric field present in the depletion region during Zener breakdown.
The electric field strength is the voltage divided by the distance over which the voltage is applied.
Here, the voltage is the Zener voltage and the distance is the depletion layer width.
Convert the width from nanometers to meters for consistent units.
Divide the voltage by the width to get the electric field strength in volts per meter.
This calculation shows the very strong electric field present in the depletion region during Zener breakdown.
PYQ 18
Tamil Nadu
2021
Arrange the following steps in the correct sequence to solve a circuit using Kirchhoff’s rules: (i) Assign current directions, (ii) Apply junction rule, (iii) Write loop equations, (iv) Solve simultaneous equations.
Solution:
i, ii, iii, iv
Detailed Explanation:
To solve a circuit using Kirchhoff’s rules, the first step is to assign directions to the currents in each branch.
Next, apply Kirchhoff’s current rule (junction rule) at the junctions to write current equations.
Then, apply Kirchhoff’s voltage rule (loop rule) to write voltage equations for independent loops.
Finally, solve the simultaneous equations obtained from the junction and loop equations to find the unknown currents.
This sequence ensures systematic and correct analysis of the circuit.
To solve a circuit using Kirchhoff’s rules, the first step is to assign directions to the currents in each branch.
Next, apply Kirchhoff’s current rule (junction rule) at the junctions to write current equations.
Then, apply Kirchhoff’s voltage rule (loop rule) to write voltage equations for independent loops.
Finally, solve the simultaneous equations obtained from the junction and loop equations to find the unknown currents.
This sequence ensures systematic and correct analysis of the circuit.
PYQ 19
Tamil Nadu Class 12
2019
Two point charges are separated by distance \( r \). The force between them is \( F \). If the distance is doubled and both charges are halved, find the new force in terms of \( F \).
Solution:
Given:
Initial force: \( F = k \frac{q_1 q_2}{r^2} \)
New charges: \( q_1' = \frac{q_1}{2} \), \( q_2' = \frac{q_2}{2} \)
New distance: \( r' = 2r \)
New force:
\[ F' = k \frac{q_1' q_2'}{r'^2} = k \frac{(q_1/2)(q_2/2)}{(2r)^2} = k \frac{q_1 q_2}{4} \times \frac{1}{4 r^2} = \frac{k q_1 q_2}{16 r^2} = \frac{F}{16} \]
Therefore, the new force is \( \frac{F}{16} \).
Initial force: \( F = k \frac{q_1 q_2}{r^2} \)
New charges: \( q_1' = \frac{q_1}{2} \), \( q_2' = \frac{q_2}{2} \)
New distance: \( r' = 2r \)
New force:
\[ F' = k \frac{q_1' q_2'}{r'^2} = k \frac{(q_1/2)(q_2/2)}{(2r)^2} = k \frac{q_1 q_2}{4} \times \frac{1}{4 r^2} = \frac{k q_1 q_2}{16 r^2} = \frac{F}{16} \]
Therefore, the new force is \( \frac{F}{16} \).
Detailed Explanation:
The electrostatic force depends on the product of the charges and inversely on the square of the distance. When both charges are halved, their product becomes one-fourth. When the distance is doubled, the denominator becomes four times larger (since distance squared). Multiplying these effects, the force reduces by a factor of 16. This problem helps students apply the formula and understand how changes in charge and distance affect the force.
The electrostatic force depends on the product of the charges and inversely on the square of the distance. When both charges are halved, their product becomes one-fourth. When the distance is doubled, the denominator becomes four times larger (since distance squared). Multiplying these effects, the force reduces by a factor of 16. This problem helps students apply the formula and understand how changes in charge and distance affect the force.
PYQ 20
Karnataka SSLC
2022
Calculate the current through the galvanometer when the Wheatstone bridge has the following resistances: \( P = 10 \ \Omega \), \( Q = 20 \ \Omega \), \( R = 15 \ \Omega \), \( S = 18 \ \Omega \), and a supply voltage of \( 12 \ \text{V} \).
Solution:
First, check if the bridge is balanced using the condition \( \frac{P}{Q} = \frac{R}{S} \). Calculate \( \frac{P}{Q} = \frac{10}{20} = 0.5 \) and \( \frac{R}{S} = \frac{15}{18} \approx 0.833 \). Since \( 0.5 \neq 0.833 \), the bridge is not balanced, so current flows through the galvanometer. Using Kirchhoff’s rules, denote currents \( I_1 \) through \( P \), \( I_2 \) through \( R \), \( I_3 \) through \( Q \), \( I_4 \) through \( S \), and \( I_G \) through the galvanometer. Apply Kirchhoff’s current rule at junction B: \( I_1 - I_G - I_3 = 0 \). Apply Kirchhoff’s current rule at junction D: \( I_2 + I_G - I_4 = 0 \). Apply Kirchhoff’s voltage rule to loop ABDA: \( I_1 P + I_G G - I_2 R = 0 \) (assuming galvanometer resistance \( G \) is negligible, \( I_G G \approx 0 \)). Apply Kirchhoff’s voltage rule to loop ABCDA: \( I_1 P + I_3 Q - I_4 S - I_2 R = 0 \). Assuming galvanometer resistance \( G \) is very small and neglecting it, solve the system: From current rules: \( I_3 = I_1 - I_G \) and \( I_4 = I_2 + I_G \). Substitute into voltage loop: \( I_1 P + (I_1 - I_G) Q - (I_2 + I_G) S - I_2 R = 0 \). Since the problem is complex, approximate \( I_G \) by: \( I_G = \frac{V}{P + Q + R + S} \times \text{(difference in ratios)} \). Alternatively, use the formula for galvanometer current: \( I_G = \frac{V}{P + Q + R + S} \times \frac{P S - Q R}{P + Q + R + S} \). Calculate numerator: \( P S - Q R = 10 \times 18 - 20 \times 15 = 180 - 300 = -120 \). Sum resistances: \( 10 + 20 + 15 + 18 = 63 \ \Omega \). \( I_G = \left( \frac{12}{63} \right) \times \left( \frac{-120}{63} \right) = (0.1905) \times (-1.9048) = -0.362 \ \text{A} \). The negative sign indicates direction opposite to assumed. Therefore, current through galvanometer is approximately \( 0.36 \ \text{A} \).
Detailed Explanation:
The Wheatstone bridge is balanced when \( \frac{P}{Q} = \frac{R}{S} \), and no current flows through the galvanometer. Here, since the ratios are not equal, the bridge is unbalanced, and current flows through the galvanometer. To find this current, Kirchhoff’s current and voltage laws are applied to the circuit. By writing equations for currents at junctions and voltages around loops, a system of equations is formed. Solving these equations gives the galvanometer current. The calculation involves algebraic manipulation and understanding of circuit laws. Approximations such as neglecting galvanometer resistance simplify the problem. This example illustrates how Kirchhoff’s laws help analyze complex circuits with multiple branches.
The Wheatstone bridge is balanced when \( \frac{P}{Q} = \frac{R}{S} \), and no current flows through the galvanometer. Here, since the ratios are not equal, the bridge is unbalanced, and current flows through the galvanometer. To find this current, Kirchhoff’s current and voltage laws are applied to the circuit. By writing equations for currents at junctions and voltages around loops, a system of equations is formed. Solving these equations gives the galvanometer current. The calculation involves algebraic manipulation and understanding of circuit laws. Approximations such as neglecting galvanometer resistance simplify the problem. This example illustrates how Kirchhoff’s laws help analyze complex circuits with multiple branches.
