Physics Cheatsheet

Cheatsheet Content

### Multiple Choice Questions (Answers Included) * A diver in a swimming pool bends his head before diving to: Decrease his moment of inertia. * The angular momentum of a system is strictly conserved when: No external torque acts upon the system. * The optimum banking angle of a road is independent of: The mass of the vehicle. * The SI unit of Coefficient of Viscosity is: Ns/m². * If the radius of a capillary tube is doubled, the height of the liquid column becomes: Half. * The surface tension of a liquid at its critical temperature is: Zero. * Free-falling raindrops assume a spherical shape due to: Surface tension. * The First Law of Thermodynamics is a manifestation of the law of conservation of: Energy. * In an ideal isothermal process, the change in internal energy is: Zero. * During an adiabatic process, the thermodynamic quantity that remains constant is: Heat. * The rms speed of gas molecules is directly proportional to: $\sqrt{T}$. * The value of the universal gas constant (R) is: 8.314 J/mol·K. * In a stationary wave, the distance between a node and an adjacent antinode is: $\lambda/4$. * The fundamental frequency of a vibrating pipe closed at one end is: v/4L. * The phenomenon of polarization definitively proves that light waves are: Transverse. * In Young's Double Slit Experiment, if the distance between the two slits is halved, the fringe width becomes: Doubled. * The phase difference between any two points situated on the same wavefront is: Zero. * Kirchhoff’s First Law (Current Law) is based on the conservation of: Charge. * The sensitivity of a potentiometer can be effectively increased by: Increasing the length of the potentiometer wire. * A capacitor stores its energy entirely within its: Electric field. * When a dielectric slab is inserted between the plates of a capacitor, its capacitance: Increases. * The working principle of a transformer is based on: Mutual Induction. * In an AC circuit containing a pure inductor, the alternating current: Lags the voltage by $90^\circ$. * Eddy currents are practically utilized in the construction of: Dead beat galvanometers. * The energy of a photon corresponding to a frequency f is given by: hf. * The de Broglie wavelength of a moving particle is inversely proportional to its: Momentum. * According to Bohr’s atomic theory, the angular momentum of an orbiting electron is an integral multiple of: $h/2\pi$. * For stable voltage regulation, a Zener diode is always operated in: Reverse bias. * A Solar Cell operates on the principle of the: Photovoltaic effect. * The logic gates that can be used to construct any other basic gate are: NAND and NOR gates. ### One Mark Questions * **Define Radius of Gyration.** The radius of gyration is the perpendicular distance from the axis of rotation to a point where the entire mass of the body is supposed to be concentrated to have the same moment of inertia as the body. * **State the Theorem of Parallel Axes.** The moment of inertia of a body about any axis is equal to its moment of inertia about a parallel axis passing through its center of mass plus the product of the mass of the body and the square of the perpendicular distance between the two axes. * **Define Surface Tension and state its dimensions.** Surface tension is the force per unit length acting perpendicularly on an imaginary line drawn on the liquid surface, tending to contract the surface. Dimensions: $[M^1 L^0 T^{-2}]$. * **State the effect of temperature on surface tension.** Surface tension of a liquid decreases with an increase in temperature and becomes zero at its critical temperature. * **State Mayer’s Relation.** Mayer's relation states that $C_p - C_v = R$, where $C_p$ is the molar specific heat at constant pressure, $C_v$ is the molar specific heat at constant volume, and R is the universal gas constant. * **Define an Isothermal Process.** An isothermal process is a thermodynamic process in which the temperature of the system remains constant throughout the process. * **What is a Perfectly Black Body?** A perfectly black body is a body that absorbs all the electromagnetic radiation incident on it, regardless of frequency or angle of incidence. It is also an ideal emitter of radiation. * **Define Angle of Contact.** The angle of contact is the angle subtended by the tangent to the liquid surface at the point of contact, inside the liquid, with the solid surface. * **State the formula for the period of a simple pendulum.** The period of a simple pendulum is given by $T = 2\pi\sqrt{L/g}$, where L is the length of the pendulum and g is the acceleration due to gravity. * **Define Overtones.** Overtones are the frequencies of sound produced by a vibrating body that are integer multiples of the fundamental frequency, excluding the fundamental frequency itself. * **Define a Wavefront.** A wavefront is the locus of all points in a medium that are in the same phase of oscillation at any given instant. * **State Brewster’s Law.** Brewster's law states that when unpolarized light is incident on a transparent medium at the polarizing angle ($i_p$), the reflected light is completely plane-polarized perpendicular to the plane of incidence, and the reflected and refracted rays are mutually perpendicular. Mathematically, $\tan i_p = \mu$. * **State the SI unit of Permittivity of free space.** The SI unit of permittivity of free space ($\epsilon_0$) is Farad per meter (F/m) or Coulomb squared per Newton per meter squared ($C^2 N^{-1} m^{-2}$). * **Define Potential Gradient.** Potential gradient is the rate of change of electric potential with respect to distance. It is given by $dV/dx$ and its unit is V/m. * **State Kirchhoff’s Voltage Law.** Kirchhoff's Voltage Law (KVL) states that the algebraic sum of all potential differences (voltages) around any closed loop in a circuit is zero. * **Define Magnetization.** Magnetization is the net magnetic dipole moment per unit volume of a material. It represents the extent to which a material is magnetized. * **What is a Cyclotron?** A cyclotron is a device used to accelerate charged particles (like protons, deuterons) to very high energies using a combination of a constant magnetic field and a high-frequency alternating electric field. * **Define Magnetic Susceptibility.** Magnetic susceptibility ($\chi_m$) is a dimensionless quantity that indicates the degree to which a material can be magnetized in an external magnetic field. It is the ratio of magnetization (M) to the magnetic intensity (H), i.e., $\chi_m = M/H$. * **State Lenz’s Law.** Lenz's Law states that the direction of the induced electromotive force (EMF) and hence the induced current is such that it opposes the cause producing it. * **Define Self-Induction.** Self-induction is the phenomenon where a changing current in a coil induces an EMF in the same coil, which opposes the change in current. * **State the condition for resonance in a series LCR circuit.** For resonance in a series LCR circuit, the inductive reactance ($X_L$) must be equal to the capacitive reactance ($X_C$), i.e., $X_L = X_C$. This results in the circuit impedance being purely resistive and minimum. * **Define Threshold Frequency.** Threshold frequency is the minimum frequency of incident light required to eject electrons from a metal surface in the photoelectric effect. If the incident light's frequency is below this value, no photoelectrons will be emitted, regardless of intensity. * **State Einstein’s Photoelectric Equation.** Einstein's Photoelectric Equation is $h\nu = \phi_0 + K_{max}$, where $h\nu$ is the energy of the incident photon, $\phi_0$ is the work function of the metal, and $K_{max}$ is the maximum kinetic energy of the emitted photoelectron. * **Define de-Broglie wavelength.** The de-Broglie wavelength ($\lambda$) is the wavelength associated with a particle, given by $\lambda = h/p$, where h is Planck's constant and p is the momentum of the particle. It signifies the wave-particle duality of matter. * **State Bohr’s second postulate.** Bohr's second postulate states that an electron can revolve only in those orbits for which its angular momentum is an integral multiple of $h/2\pi$, where h is Planck's constant. * **Define Half-life of a radioactive element.** The half-life of a radioactive element is the time taken for half of the radioactive nuclei in a sample to undergo radioactive decay. * **Name the majority carriers in n-type semiconductors.** The majority carriers in n-type semiconductors are electrons. * **Draw the circuit symbol for the NAND gate.** ```mermaid graph LR A -- Input A --> NAND B -- Input B --> NAND NAND -- Output --> Q ``` (Note: Actual symbol is a 'D' shape with a small circle at the output) * **State one use of a Solar Cell.** One use of a solar cell is to convert solar energy into electrical energy for power generation in satellites, calculators, or residential systems. * **Define Work Function.** Work function ($\phi_0$) is the minimum amount of energy required to remove an electron from the surface of a metal. ### Two Mark Questions * **Distinguish between Centripetal and Centrifugal force.** | Feature | Centripetal Force | Centrifugal Force | | :-------------- | :------------------------------------------------ | :----------------------------------------------- | | **Definition** | Real force directed towards the center of a circular path. | Fictitious force directed away from the center. | | **Nature** | Real force, provides necessary acceleration for circular motion. | Pseudo force, experienced in a rotating frame of reference. | | **Origin** | Caused by external agents (tension, gravity, friction). | Arises due to inertia in a non-inertial frame. | | **Reference Frame** | Inertial frame of reference. | Non-inertial (rotating) frame of reference. | * **State the law of conservation of angular momentum with an example.** **Law:** The law of conservation of angular momentum states that if no external torque acts on a system, the total angular momentum of the system remains constant. **Example:** A ballet dancer spinning on ice. When she pulls her arms and legs inward, her moment of inertia decreases. To conserve angular momentum ($L = I\omega$), her angular velocity ($\omega$) increases, making her spin faster. * **Derive an expression for the time period of a conical pendulum.** Consider a bob of mass 'm' revolving in a horizontal circle of radius 'r' with uniform angular velocity '$\omega$'. Let 'L' be the length of the string and '$\theta$' be the angle the string makes with the vertical. The forces acting on the bob are: 1. Tension 'T' in the string. 2. Weight 'mg' acting vertically downwards. Resolving tension T: * $T\cos\theta = mg$ (vertical equilibrium) * $T\sin\theta = mr\omega^2$ (centripetal force) Dividing the two equations: $\tan\theta = \frac{mr\omega^2}{mg} = \frac{r\omega^2}{g}$ From geometry, $r = L\sin\theta$. So, $\tan\theta = \frac{L\sin\theta \omega^2}{g} \implies \frac{\sin\theta}{\cos\theta} = \frac{L\sin\theta \omega^2}{g}$ $\frac{1}{\cos\theta} = \frac{L\omega^2}{g} \implies \omega^2 = \frac{g}{L\cos\theta}$ $\omega = \sqrt{\frac{g}{L\cos\theta}}$ The time period $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L\cos\theta}{g}}$. * **Explain the concept of Mean Free Path.** In a gas, molecules are in constant random motion and frequently collide with each other. The mean free path ($\lambda$) is defined as the average distance travelled by a molecule between two successive collisions. It is inversely proportional to the number density of the gas and the square of the molecular diameter. * **State any four assumptions of Kinetic Theory of Gases.** 1. Gases consist of a large number of identical, tiny particles (molecules/atoms) that are in continuous, random motion. 2. The volume occupied by the gas molecules themselves is negligible compared to the total volume of the gas. 3. The collisions between molecules and with the container walls are perfectly elastic. 4. There are no attractive or repulsive forces between gas molecules, except during collisions. 5. The duration of a collision is negligible compared to the time between collisions. * **State and explain Bernoulli's Principle.** **Principle:** Bernoulli's principle states that for an incompressible, non-viscous fluid in streamline flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume (or per unit mass) remains constant at all points along a streamline. **Explanation:** It's essentially the conservation of energy for fluid flow. When a fluid flows through a narrower section, its velocity increases (kinetic energy increases), and consequently, its pressure must decrease to keep the total energy constant. Conversely, if the velocity decreases, the pressure increases. * **Define Free Expansion in thermodynamics.** Free expansion is an irreversible thermodynamic process in which an ideal gas expands into a vacuum without doing any work and without any heat exchange with the surroundings. Consequently, the internal energy and temperature of the gas remain constant ($W=0, Q=0, \Delta U=0$). * **Distinguish between Reversible and Irreversible processes.** | Feature | Reversible Process | Irreversible Process | | :-------------- | :------------------------------------------------ | :---------------------------------------------------- | | **Definition** | Can be reversed such that both system and surroundings return to their initial states without any net change. | Cannot be reversed without leaving some net change in the system or surroundings. | | **Conditions** | Occurs infinitesimally slowly (quasi-static), no dissipative forces (friction, viscosity). | Occurs rapidly, involves friction, viscosity, heat transfer across finite temperature difference. | | **Equilibrium** | System is always in thermodynamic equilibrium with surroundings. | System passes through non-equilibrium states. | | **Example** | Slow isothermal expansion/compression of an ideal gas. | Free expansion, combustion, heat flow from hot to cold. | * **State the laws of a simple pendulum.** 1. **Law of Isochronism:** For small angular displacements (up to about $5^\circ$), the period of oscillation is independent of the amplitude. 2. **Law of Length:** The period of oscillation is directly proportional to the square root of its effective length ($T \propto \sqrt{L}$). 3. **Law of Mass:** The period of oscillation is independent of the mass of the bob. 4. **Law of Gravity:** The period of oscillation is inversely proportional to the square root of the acceleration due to gravity ($T \propto 1/\sqrt{g}$). * **Define Linear S.H.M. and state its differential equation.** **Definition:** Linear Simple Harmonic Motion (SHM) is a type of oscillatory motion in which the restoring force acting on the oscillating particle is directly proportional to its displacement from the mean position and is always directed towards the mean position. **Differential Equation:** $m\frac{d^2x}{dt^2} = -kx$ or $\frac{d^2x}{dt^2} + \omega^2x = 0$, where $x$ is displacement, $m$ is mass, $k$ is force constant, and $\omega = \sqrt{k/m}$ is angular frequency. * **Distinguish between Free and Forced vibrations.** | Feature | Free Vibrations | Forced Vibrations | | :-------------- | :------------------------------------------------ | :---------------------------------------------------- | | **Cause** | Body oscillates with its natural frequency after an initial displacement/impulse. | Body oscillates under the influence of an external periodic force. | | **Frequency** | Natural frequency of the body. | Frequency of the external periodic force. | | **Amplitude** | Decreases over time due to damping (if not ideal). | Can be large if driving frequency is close to natural frequency (resonance). | | **Damping** | Damped by internal forces or air resistance. | Energy input from external force compensates for damping. | | **Example** | Simple pendulum swinging in a vacuum. | A bridge vibrating due to rhythmic marching of soldiers. | * **State any four properties of Stationary Waves.** 1. **Nodes and Antinodes:** There are fixed positions (nodes) where particle displacement is always zero and points (antinodes) where displacement is maximum. 2. **Energy Transfer:** Energy does not propagate across the medium; it is confined and oscillates within segments between nodes. 3. **Amplitude Variation:** The amplitude of oscillation varies from zero at nodes to maximum at antinodes. 4. **Phase Relationship:** All particles between two consecutive nodes oscillate in the same phase, but particles in adjacent segments (separated by a node) oscillate in opposite phases. * **State the conditions for constructive and destructive interference.** **Constructive Interference:** * **Path Difference:** $\Delta x = n\lambda$, where $n = 0, \pm 1, \pm 2, ...$ (an integer multiple of wavelength). * **Phase Difference:** $\Delta\phi = 2n\pi$, where $n = 0, \pm 1, \pm 2, ...$ (an even multiple of $\pi$). **Destructive Interference:** * **Path Difference:** $\Delta x = (2n+1)\lambda/2$, where $n = 0, \pm 1, \pm 2, ...$ (an odd multiple of half wavelength). * **Phase Difference:** $\Delta\phi = (2n+1)\pi$, where $n = 0, \pm 1, \pm 2, ...$ (an odd multiple of $\pi$). * **Distinguish between Fresnel and Fraunhofer diffraction.** | Feature | Fresnel Diffraction | Fraunhofer Diffraction | | :-------------- | :------------------------------------------------ | :---------------------------------------------------- | | **Source/Screen Distance** | Finite distance from the obstacle/aperture. | Effectively infinite distance from the obstacle/aperture. | | **Wavefront** | Spherical or cylindrical wavefronts. | Plane wavefronts. | | **Lenses** | No lenses are required to observe the pattern. | Lenses are required to make light parallel and focus the pattern. | | **Pattern** | Pattern changes significantly with distance, sharp central maximum. | Pattern is fixed, observed at focal plane of lens, broader central maximum. | * **State the advantages of a Potentiometer over a Voltmeter.** 1. **High Accuracy:** A potentiometer measures EMF accurately because it draws no current from the source at the null point, thus eliminating internal resistance effects. A voltmeter always draws some current. 2. **Measures EMF:** It can measure the EMF of a cell directly, whereas a voltmeter measures the terminal potential difference. 3. **Comparison of EMFs:** It can be used to compare the EMFs of two cells. 4. **Measurement of Internal Resistance:** It can determine the internal resistance of a cell. * **Distinguish between Paramagnetic and Ferromagnetic substances.** | Feature | Paramagnetic Substances | Ferromagnetic Substances | | :-------------- | :------------------------------------------------ | :--------------------------------------------------- | | **Magnetic Moment** | Permanent atomic magnetic moments, randomly oriented. | Permanent atomic magnetic moments, aligned in domains. | | **External Field** | Weakly attracted, moments align parallel to field. | Strongly attracted, domains align with field. | | **Removal of Field** | Lose magnetism. | Retain magnetism (hysteresis). | | **Susceptibility** | Small, positive, temperature dependent ($\chi_m \propto 1/T$). | Large, positive, temperature dependent (Curie-Weiss law). | | **Example** | Aluminum, Platinum, Oxygen. | Iron, Nickel, Cobalt. | * **State the formula for magnetic force on a straight current-carrying wire.** The magnetic force ($\vec{F}$) on a straight current-carrying wire of length $\vec{L}$ (vector in the direction of current) carrying current I, placed in a uniform magnetic field $\vec{B}$, is given by: $\vec{F} = I(\vec{L} \times \vec{B})$ The magnitude of the force is $F = I L B \sin\theta$, where $\theta$ is the angle between $\vec{L}$ and $\vec{B}$. * **Define Magnetic Dipole Moment and state its SI unit.** **Definition:** Magnetic dipole moment ($\vec{\mu}$ or $\vec{M}$) is a measure of the strength and orientation of a magnetic source. For a current loop, it is the product of the current (I) and the area (A) enclosed by the loop, directed perpendicular to the plane of the loop. **SI Unit:** Ampere-meter squared ($A m^2$). * **State Faraday's laws of Electromagnetic Induction.** **First Law:** Whenever the magnetic flux linked with a coil changes, an electromotive force (EMF) is induced in the coil. **Second Law:** The magnitude of the induced EMF is directly proportional to the rate of change of magnetic flux linked with the coil. Mathematically, $\mathcal{E} = -N \frac{d\Phi_B}{dt}$, where N is the number of turns and the negative sign is due to Lenz's Law. * **State any two applications of Eddy Currents.** 1. **Magnetic Braking in Trains:** Powerful electromagnets create eddy currents in the metal rails, generating a braking force that opposes the train's motion. 2. **Induction Furnaces:** High-frequency alternating currents induce large eddy currents in metals, producing enough heat to melt them. 3. **Dead Beat Galvanometers:** Eddy currents are used to damp the oscillations of the coil quickly, bringing the pointer to rest without overshooting. * **Explain Inductive Reactance and Capacitive Reactance.** **Inductive Reactance ($X_L$):** It is the opposition offered by an inductor to the flow of alternating current. It arises due to the back EMF induced in the inductor opposing the change in current. It is given by $X_L = \omega L = 2\pi f L$, where $\omega$ is angular frequency, f is frequency, and L is inductance. $X_L$ increases with frequency. **Capacitive Reactance ($X_C$):** It is the opposition offered by a capacitor to the flow of alternating current. It arises because the capacitor stores charge and opposes a sudden change in voltage across it. It is given by $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$, where C is capacitance. $X_C$ decreases with frequency. * **Draw a neat labelled circuit diagram of an LCR series resonator.** ```mermaid graph LR V("AC Voltage Source (V)") -- Current (I) --> R("Resistor (R)") R -- Current (I) --> L("Inductor (L)") L -- Current (I) --> C("Capacitor (C)") C -- Current (I) --> V ``` (Note: This is a simplified text-based representation. A proper diagram would show the components in series with an AC voltage source.) * **State any four characteristics of the Photoelectric Effect.** 1. **Threshold Frequency:** For a given metal, there exists a minimum frequency (threshold frequency) below which no photoelectrons are emitted, regardless of the intensity of incident light. 2. **Instantaneous Emission:** Photoelectric emission is an instantaneous process. Electrons are emitted almost immediately (within $10^{-9}$ s) after light strikes the surface, provided the frequency is above the threshold. 3. **Intensity Dependence:** The number of photoelectrons emitted (and thus the photocurrent) is directly proportional to the intensity of incident light, provided the frequency is above the threshold. 4. **Kinetic Energy Dependence:** The maximum kinetic energy of the emitted photoelectrons is independent of the intensity of incident light but increases linearly with the frequency of incident light. * **State the dual nature of matter (de Broglie’s hypothesis).** De Broglie's hypothesis states that just as light exhibits both wave-like and particle-like properties (wave-particle duality), matter particles (like electrons, protons, atoms) also exhibit wave-like properties. Every moving particle has a wave associated with it, called a matter wave or de Broglie wave, whose wavelength is inversely proportional to its momentum. * **State Bohr's first and third postulates of the hydrogen atom.** **First Postulate (Stationary Orbits):** An electron in an atom can revolve in certain stable, non-radiating orbits (called stationary orbits) without emitting any energy. Each such orbit has a definite energy. **Third Postulate (Frequency Condition):** An electron can make a transition from a higher energy orbit (initial state $E_i$) to a lower energy orbit (final state $E_f$), emitting a photon whose energy is equal to the energy difference between the two states. The frequency of the emitted photon is given by $h\nu = E_i - E_f$. * **Define Binding Energy per Nucleon.** Binding energy per nucleon is the average energy required to remove a single nucleon (proton or neutron) from a nucleus. It is calculated by dividing the total binding energy of a nucleus by its mass number (total number of nucleons). It is a measure of the stability of a nucleus. * **Distinguish between p-type and n-type semiconductors.** | Feature | p-type Semiconductor | n-type Semiconductor | | :-------------- | :---------------------------------------------- | :----------------------------------------------- | | **Doping** | Doped with trivalent impurities (e.g., Boron, Aluminum). | Doped with pentavalent impurities (e.g., Phosphorus, Arsenic). | | **Majority Carriers** | Holes | Electrons | | **Minority Carriers** | Electrons | Holes | | **Charge** | Electrically neutral (overall). | Electrically neutral (overall). | | **Mechanism** | Impurities create "holes" (electron vacancies) for conduction. | Impurities provide "extra" electrons for conduction. | * **Explain the working of a Photodiode.** A photodiode is a p-n junction diode operated in reverse bias. When light of appropriate energy (greater than the band gap of the semiconductor) strikes the depletion region, electron-hole pairs are generated. The electric field in the depletion region sweeps these carriers (electrons to n-side, holes to p-side) across the junction, creating a reverse current. The magnitude of this current is proportional to the intensity of the incident light, allowing it to detect light signals. * **State the advantages of a Full Wave Rectifier.** 1. **Higher Efficiency:** A full-wave rectifier has higher rectification efficiency compared to a half-wave rectifier (typically around 81.2% versus 40.6%). 2. **Less Ripple:** It produces a smoother DC output with less ripple because both halves of the AC input cycle are utilized, resulting in a higher ripple frequency (twice the input frequency). This makes filtering easier. 3. **Higher DC Output Voltage:** The average DC output voltage is higher than that of a half-wave rectifier for the same input AC voltage. 4. **Better Transformer Utilization:** In a center-tap full-wave rectifier, the transformer secondary is fully utilized. * **Define Current Amplification Factors ($\alpha$ and $\beta$).** **Current Amplification Factor ($\alpha$) in Common Base (CB) configuration:** It is the ratio of the change in collector current ($\Delta I_C$) to the change in emitter current ($\Delta I_E$) at constant collector-base voltage ($V_{CB}$). $\alpha = \frac{\Delta I_C}{\Delta I_E} \Big|_{V_{CB}=\text{constant}}$ (typically 0.95 to 0.99) **Current Amplification Factor ($\beta$) in Common Emitter (CE) configuration:** It is the ratio of the change in collector current ($\Delta I_C$) to the change in base current ($\Delta I_B$) at constant collector-emitter voltage ($V_{CE}$). $\beta = \frac{\Delta I_C}{\Delta I_B} \Big|_{V_{CE}=\text{constant}}$ (typically 50 to 500) **Relation:** $\beta = \frac{\alpha}{1-\alpha}$ and $\alpha = \frac{\beta}{1+\beta}$. ### Three Mark Questions * **Derive an expression for the maximum safety speed of a vehicle on a banked road.** Consider a vehicle of mass 'm' moving on a banked road with banking angle '$\theta$'. Forces acting on the vehicle: 1. Weight 'mg' acting vertically downwards. 2. Normal reaction 'N' perpendicular to the banked road surface. Resolve N into components: * $N\cos\theta$ (vertical, balances weight) * $N\sin\theta$ (horizontal, provides centripetal force) From vertical equilibrium: $N\cos\theta = mg \quad (1)$ From horizontal motion (centripetal force): $N\sin\theta = \frac{mv^2}{r} \quad (2)$ Dividing (2) by (1): $\frac{N\sin\theta}{N\cos\theta} = \frac{mv^2/r}{mg}$ $\tan\theta = \frac{v^2}{rg}$ Therefore, the maximum safety speed (or optimum speed) is $v = \sqrt{rg\tan\theta}$. This is the speed at which the vehicle can take the turn without relying on friction. * **Derive an expression for the excess pressure inside a liquid drop.** Consider a spherical liquid drop of radius 'R' and surface tension 'T'. If the radius of the drop increases by a small amount 'dR', the increase in surface area is: $\Delta A = 4\pi(R+dR)^2 - 4\pi R^2 = 4\pi(R^2 + 2RdR + dR^2) - 4\pi R^2$ Neglecting $dR^2$ (as it is very small), $\Delta A = 8\pi R dR$. Work done by surface tension to increase the surface area: $dW = T \times \Delta A = T (8\pi R dR) \quad (1)$ Let 'P' be the excess pressure inside the drop. This excess pressure exerts an outward force on the surface. Force = $P \times \text{Area} = P \times 4\pi R^2$. Work done by the excess pressure in expanding the drop by 'dR': $dW = \text{Force} \times \text{distance} = (P \times 4\pi R^2) \times dR \quad (2)$ Equating (1) and (2) based on energy conservation: $T (8\pi R dR) = P (4\pi R^2) dR$ $8\pi R T = P (4\pi R^2)$ $P = \frac{8\pi R T}{4\pi R^2} = \frac{2T}{R}$ Thus, the excess pressure inside a liquid drop is $P = \frac{2T}{R}$. * **Derive Torricelli's law using Bernoulli's equation.** Torricelli's law describes the speed of efflux of a fluid from an orifice under gravity. Consider a tank containing a liquid of density $\rho$ up to height H. An orifice is at a depth 'h' below the free surface. Apply Bernoulli's equation at two points: * Point 1: On the free surface of the liquid. * Point 2: At the orifice. Bernoulli's equation: $P + \frac{1}{2}\rho v^2 + \rho g y = \text{constant}$ At Point 1: * Pressure $P_1 = P_{atm}$ (atmospheric pressure) * Velocity $v_1 \approx 0$ (if the tank is large, the free surface drops very slowly) * Height $y_1 = H$ (taking the orifice level as $y=0$) At Point 2: * Pressure $P_2 = P_{atm}$ (liquid emerges into atmosphere) * Velocity $v_2 = v$ (speed of efflux) * Height $y_2 = 0$ Applying Bernoulli's equation: $P_{atm} + \frac{1}{2}\rho (0)^2 + \rho g H = P_{atm} + \frac{1}{2}\rho v^2 + \rho g (0)$ $\rho g H = \frac{1}{2}\rho v^2$ $g H = \frac{1}{2} v^2$ $v^2 = 2gH$ $v = \sqrt{2gH}$ Here, H is the height of the free surface above the orifice, which is the depth of the orifice below the free surface. So, the speed of efflux is $v = \sqrt{2gh}$. This is Torricelli's Law. * **Derive an expression for capillary rise using the forces method.** Consider a capillary tube of radius 'r' dipped in a liquid of surface tension 'T', density '$\rho$', and angle of contact '$\theta$'. The liquid rises to a height 'h' in the tube. The forces involved are: 1. **Upward force due to surface tension:** The surface tension acts along the circumference of the liquid meniscus. The component of surface tension acting vertically upwards is $T\cos\theta$. The total upward force ($F_u$) is the product of surface tension component and the circumference of contact: $F_u = (T\cos\theta) \times (2\pi r)$ 2. **Downward force due to the weight of the liquid column:** This is the weight of the liquid raised in the capillary tube. Volume of liquid column = $\pi r^2 h$ Mass of liquid column = Volume $\times$ Density $= \pi r^2 h \rho$ Downward force ($F_d$) = Weight = Mass $\times$ g $= \pi r^2 h \rho g$ At equilibrium, the upward force balances the downward force: $F_u = F_d$ $2\pi r T\cos\theta = \pi r^2 h \rho g$ Solving for h: $h = \frac{2\pi r T\cos\theta}{\pi r^2 \rho g} = \frac{2T\cos\theta}{r\rho g}$ This is the expression for capillary rise (or fall, if $\cos\theta$ is negative). * **State the law of equipartition of energy and find the specific heat of a monatomic gas.** **Law of Equipartition of Energy:** This law states that for a system in thermal equilibrium, the total energy is equally distributed among all degrees of freedom, and the average energy associated with each degree of freedom is $\frac{1}{2}k_B T$, where $k_B$ is Boltzmann's constant and T is the absolute temperature. **Specific Heat of a Monatomic Gas:** A monatomic gas (like He, Ne, Ar) has only translational degrees of freedom. It has 3 translational degrees of freedom ($f=3$). According to the law of equipartition of energy, the average energy per molecule is: $E_{avg} = f \times \frac{1}{2}k_B T = 3 \times \frac{1}{2}k_B T = \frac{3}{2}k_B T$ For one mole of gas, the total internal energy (U) is: $U = N_A E_{avg} = N_A (\frac{3}{2}k_B T) = \frac{3}{2} (N_A k_B) T = \frac{3}{2} R T$, where $R = N_A k_B$ is the universal gas constant. Molar specific heat at constant volume ($C_v$) is given by: $C_v = \frac{dU}{dT} = \frac{d}{dT}(\frac{3}{2} RT) = \frac{3}{2} R$ Using Mayer's relation, $C_p - C_v = R$, so $C_p = C_v + R = \frac{3}{2} R + R = \frac{5}{2} R$. The ratio of specific heats, $\gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3} \approx 1.67$. * **Derive the expression for work done in an Isothermal process.** An isothermal process is one where the temperature (T) remains constant. For an ideal gas, internal energy ($\Delta U$) is also constant, so $\Delta U = 0$. According to the first law of thermodynamics, $\Delta Q = \Delta U + \Delta W$. Since $\Delta U = 0$, $\Delta Q = \Delta W$. The work done by the gas during expansion from initial volume $V_1$ to final volume $V_2$ is given by: $W = \int_{V_1}^{V_2} P dV$ For an ideal gas, $PV = nRT \implies P = \frac{nRT}{V}$. Since T is constant, we can take n, R, and T out of the integral: $W = \int_{V_1}^{V_2} \frac{nRT}{V} dV = nRT \int_{V_1}^{V_2} \frac{1}{V} dV$ $W = nRT [\ln V]_{V_1}^{V_2}$ $W = nRT (\ln V_2 - \ln V_1)$ $W = nRT \ln \left(\frac{V_2}{V_1}\right)$ Since $T$ is constant, $P_1V_1 = P_2V_2 \implies \frac{V_2}{V_1} = \frac{P_1}{P_2}$. So, the work done can also be expressed as: $W = nRT \ln \left(\frac{P_1}{P_2}\right)$ * **Show that linear S.H.M. is the projection of U.C.M. on any diameter.** Consider a particle P performing Uniform Circular Motion (UCM) with angular velocity $\omega$ and radius 'A' in the XY-plane. Let the projection of P on the X-axis be M and on the Y-axis be N. We will show that the motion of M (or N) is SHM. Assume at time t=0, the particle P is at $(A,0)$. At time 't', the angle made by OP with the X-axis is $\theta = \omega t$. The coordinates of P are $(x, y)$, where: $x = A\cos(\omega t)$ $y = A\sin(\omega t)$ The motion of point M along the X-axis is described by $x = A\cos(\omega t)$. To prove this is SHM, we need to find its acceleration and show it's proportional to displacement and directed towards the mean position (origin). Velocity of M: $v_x = \frac{dx}{dt} = \frac{d}{dt}(A\cos(\omega t)) = -A\omega\sin(\omega t)$ Acceleration of M: $a_x = \frac{dv_x}{dt} = \frac{d}{dt}(-A\omega\sin(\omega t)) = -A\omega^2\cos(\omega t)$ Substitute $x = A\cos(\omega t)$: $a_x = -\omega^2 x$ This equation shows that the acceleration ($a_x$) is directly proportional to the displacement ($x$) from the mean position (origin) and is always directed towards the mean position (indicated by the negative sign). This is the defining characteristic of Simple Harmonic Motion. Similarly, the motion of N along the Y-axis ($y = A\sin(\omega t)$) can be shown to be SHM. Therefore, linear SHM is indeed the projection of UCM on any of its diameters. * **Derive the expression for the time period of a simple pendulum.** Consider a simple pendulum consisting of a bob of mass 'm' suspended by a light inextensible string of length 'L' from a rigid support. Let the pendulum be displaced by a small angle '$\theta$' from its equilibrium position. Forces acting on the bob: 1. Tension 'T' in the string, directed towards the support. 2. Weight 'mg' acting vertically downwards. Resolve 'mg' into two components: * $mg\cos\theta$ along the string, balanced by tension 'T'. * $mg\sin\theta$ tangential to the arc, directed towards the mean position. This is the restoring force ($F_{restoring}$). So, $F_{restoring} = -mg\sin\theta$ (negative sign indicates it opposes displacement). According to Newton's second law, $F = ma$. $ma = -mg\sin\theta$ $a = -g\sin\theta$ For small angles, $\sin\theta \approx \theta$ (in radians). Also, for a circular arc, displacement $x = L\theta$, so $\theta = x/L$. Substituting these into the acceleration equation: $a = -g(\frac{x}{L}) = -(\frac{g}{L})x$ This is the differential equation for SHM: $a = -\omega^2 x$, where $\omega^2 = g/L$. The angular frequency is $\omega = \sqrt{g/L}$. The time period ($T$) of SHM is related to angular frequency by $T = \frac{2\pi}{\omega}$. Therefore, $T = \frac{2\pi}{\sqrt{g/L}} = 2\pi\sqrt{\frac{L}{g}}$. * **Explain the formation of stationary waves using the analytical method.** A stationary wave is formed when two progressive waves of the same amplitude, frequency, and wavelength, travelling in opposite directions, superimpose. Let the two progressive waves be: Wave 1 (travelling in +x direction): $y_1 = A\sin(kx - \omega t)$ Wave 2 (travelling in -x direction): $y_2 = A\sin(kx + \omega t)$ According to the principle of superposition, the resultant displacement $y = y_1 + y_2$: $y = A\sin(kx - \omega t) + A\sin(kx + \omega t)$ Using the trigonometric identity $\sin C + \sin D = 2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$: $y = 2A\sin\left(\frac{kx - \omega t + kx + \omega t}{2}\right)\cos\left(\frac{kx - \omega t - (kx + \omega t)}{2}\right)$ $y = 2A\sin(kx)\cos(-\omega t)$ Since $\cos(-\theta) = \cos\theta$: $y = (2A\sin kx)\cos\omega t$ This is the equation of a stationary wave. **Analysis:** * The term $(2A\sin kx)$ represents the amplitude of the resultant wave. This amplitude is not constant but varies with position 'x'. * The term $\cos\omega t$ indicates that all particles oscillate with the same angular frequency $\omega$. * **Nodes:** These are points where the amplitude is zero. This occurs when $\sin kx = 0$. $kx = n\pi \implies \frac{2\pi}{\lambda}x = n\pi \implies x = \frac{n\lambda}{2}$, where $n=0, 1, 2, ...$ Nodes occur at $x=0, \lambda/2, \lambda, 3\lambda/2, ...$ * **Antinodes:** These are points where the amplitude is maximum ($2A$). This occurs when $\sin kx = \pm 1$. $kx = (n + \frac{1}{2})\pi \implies \frac{2\pi}{\lambda}x = (n + \frac{1}{2})\pi \implies x = (n+\frac{1}{2})\frac{\lambda}{2} = (2n+1)\frac{\lambda}{4}$, where $n=0, 1, 2, ...$ Antinodes occur at $x=\lambda/4, 3\lambda/4, 5\lambda/4, ...$ This analysis clearly shows the characteristic features of stationary waves: position-dependent amplitude and fixed points of zero displacement (nodes) and maximum displacement (antinodes). * **Show that for a pipe open at both ends, all harmonics are present.** For a pipe open at both ends, antinodes are formed at both ends. **Fundamental Mode (First Harmonic):** The simplest standing wave pattern has an antinode at each end and a node in the middle. The length of the pipe L corresponds to half a wavelength: $L = \frac{\lambda_1}{2}$. So, $\lambda_1 = 2L$. The fundamental frequency $f_1 = \frac{v}{\lambda_1} = \frac{v}{2L}$. **Second Harmonic (First Overtone):** The next possible mode has an antinode at each end and two nodes inside. The length of the pipe L corresponds to one full wavelength: $L = \lambda_2$. So, $\lambda_2 = L$. The frequency $f_2 = \frac{v}{\lambda_2} = \frac{v}{L} = 2\left(\frac{v}{2L}\right) = 2f_1$. **Third Harmonic (Second Overtone):** The next mode has an antinode at each end and three nodes inside. The length of the pipe L corresponds to three half-wavelengths: $L = \frac{3\lambda_3}{2}$. So, $\lambda_3 = \frac{2L}{3}$. The frequency $f_3 = \frac{v}{\lambda_3} = \frac{v}{2L/3} = 3\left(\frac{v}{2L}\right) = 3f_1$. In general, for the $n^{th}$ harmonic, the length of the pipe is $L = n\frac{\lambda_n}{2}$. So, $\lambda_n = \frac{2L}{n}$. The frequency $f_n = \frac{v}{\lambda_n} = \frac{v}{2L/n} = n\left(\frac{v}{2L}\right) = nf_1$. Since 'n' can be any integer (1, 2, 3, ...), it means that all integer multiples of the fundamental frequency (i.e., all harmonics) are present in an open organ pipe. * **Derive an expression for the fringe width in Young’s Double Slit Experiment.** Consider two coherent monochromatic light sources $S_1$ and $S_2$ separated by a distance 'd'. Let a screen be placed at a distance 'D' from the slits. Let P be a point on the screen at a distance 'x' from the central bright fringe O. The path difference between the waves reaching P from $S_1$ and $S_2$ is $\Delta x = S_2P - S_1P$. From the geometry, we can use Pythagoras theorem: $S_2P^2 = D^2 + (x + d/2)^2$ $S_1P^2 = D^2 + (x - d/2)^2$ Subtracting the two equations: $S_2P^2 - S_1P^2 = (D^2 + (x + d/2)^2) - (D^2 + (x - d/2)^2)$ $(S_2P - S_1P)(S_2P + S_1P) = (x + d/2)^2 - (x - d/2)^2$ $(S_2P - S_1P)(S_2P + S_1P) = (x^2 + xd + d^2/4) - (x^2 - xd + d^2/4)$ $(S_2P - S_1P)(S_2P + S_1P) = 2xd$ Since D >> d and D >> x, $S_1P \approx S_2P \approx D$. So, $S_2P + S_1P \approx 2D$. Therefore, $\Delta x (2D) = 2xd \implies \Delta x = \frac{xd}{D}$. **For Bright Fringes (Constructive Interference):** Path difference $\Delta x = n\lambda$, where $n = 0, 1, 2, ...$ $\frac{xd}{D} = n\lambda \implies x_n = \frac{n\lambda D}{d}$ This gives the position of the $n^{th}$ bright fringe. **For Dark Fringes (Destructive Interference):** Path difference $\Delta x = (2n+1)\frac{\lambda}{2}$, where $n = 0, 1, 2, ...$ $\frac{xd}{D} = (2n+1)\frac{\lambda}{2} \implies x_n' = (2n+1)\frac{\lambda D}{2d}$ This gives the position of the $n^{th}$ dark fringe. **Fringe Width ($\beta$):** Fringe width is the distance between two consecutive bright fringes or two consecutive dark fringes. $\beta = x_{n+1} - x_n = \frac{(n+1)\lambda D}{d} - \frac{n\lambda D}{d} = \frac{\lambda D}{d}$ Or, $\beta = x_{n+1}' - x_n' = (2(n+1)+1)\frac{\lambda D}{2d} - (2n+1)\frac{\lambda D}{2d} = (2n+3 - 2n - 1)\frac{\lambda D}{2d} = 2\frac{\lambda D}{2d} = \frac{\lambda D}{d}$ Thus, the fringe width is $\beta = \frac{\lambda D}{d}$. * **Explain diffraction at a single slit and find the condition for secondary minima.** **Diffraction at a Single Slit:** When light passes through a narrow single slit (width 'a') whose size is comparable to the wavelength of light, it bends around the edges and spreads into the region of geometrical shadow. This phenomenon is called diffraction. The diffraction pattern observed on a screen consists of a broad central bright maximum, flanked by alternating dark and progressively fainter bright fringes (secondary maxima and minima). The pattern is explained by considering the wavefront incident on the slit as being divided into a large number of Huygens' secondary wavelets. These wavelets interfere to produce the observed pattern. **Condition for Secondary Minima:** Consider a single slit of width 'a'. Let light of wavelength '$\lambda$' be incident on it. For a minimum (dark fringe) at an angle $\theta$ with respect to the central axis, we consider the slit to be divided into an even number of equal parts. If we divide the slit into two halves, the path difference between wavelets from corresponding points in the two halves is $(a/2)\sin\theta$. For destructive interference, this path difference must be $\lambda/2$. So, $\frac{a}{2}\sin\theta = \frac{\lambda}{2} \implies a\sin\theta = \lambda$. This gives the first minimum. More generally, for the $n^{th}$ secondary minimum, the path difference between the wavelets from the top and bottom edges of the slit must be an integer multiple of $\lambda$. The condition for secondary minima is $a\sin\theta = n\lambda$, where $n = \pm 1, \pm 2, \pm 3, ...$. (Note: $n=0$ corresponds to the central maximum). This means that for the first minimum ($n=1$), $a\sin\theta = \lambda$. For the second minimum ($n=2$), $a\sin\theta = 2\lambda$, and so on. * **Show how Brewster’s law is used to produce plane-polarized light.** Brewster's Law states that when unpolarized light is incident on a transparent medium at a specific angle (Brewster's angle, $i_p$), the reflected light is completely plane-polarized, and the reflected and refracted rays are perpendicular to each other. The relationship is $\tan i_p = \mu$, where $\mu$ is the refractive index of the medium. **Production of Plane-Polarized Light:** 1. **Selection of Material:** Choose a transparent medium (e.g., glass, water) with a known refractive index. 2. **Angle of Incidence:** Calculate Brewster's angle $i_p = \tan^{-1}(\mu)$ for the chosen material. 3. **Setup:** Arrange a light source (emitting unpolarized light), the transparent medium (e.g., a glass plate), and a detector (e.g., a polaroid analyzer). 4. **Incidence:** Allow the unpolarized light to be incident on the surface of the transparent medium at precisely Brewster's angle ($i_p$). 5. **Reflection:** Observe the reflected light. According to Brewster's Law, this reflected light will be completely plane-polarized. The vibrations of the electric field vector in the reflected light will be perpendicular to the plane of incidence (i.e., parallel to the surface of the medium). 6. **Refraction:** The refracted light will be partially polarized, with the component perpendicular to the plane of incidence being dominant. By reflecting unpolarized light from a transparent surface at Brewster's angle, we can obtain plane-polarized light. This method is often used in situations where a large beam of polarized light is required, for example, in lasers. * **Derive an expression for the electric potential due to an electric dipole.** Consider an electric dipole consisting of two charges, +q and -q, separated by a small distance 2l. Let the midpoint of the dipole be the origin O. We want to find the electric potential at a point P at a distance 'r' from O, making an angle '$\theta$' with the dipole axis. Let $r_1$ be the distance from +q to P, and $r_2$ be the distance from -q to P. The potential at P due to +q is $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_1}$. The potential at P due to -q is $V_2 = \frac{1}{4\pi\epsilon_0} \frac{-q}{r_2}$. The total potential at P is $V = V_1 + V_2 = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$. From geometry, using the cosine rule or approximations for $r >> l$: Draw a perpendicular from +q to OP, and from -q to the extension of OP. $r_1 \approx r - l\cos\theta$ $r_2 \approx r + l\cos\theta$ Substituting these approximations: $V = \frac{q}{4\pi\epsilon_0} \left(\frac{1}{r - l\cos\theta} - \frac{1}{r + l\cos\theta}\right)$ $V = \frac{q}{4\pi\epsilon_0} \left(\frac{(r + l\cos\theta) - (r - l\cos\theta)}{(r - l\cos\theta)(r + l\cos\theta)}\right)$ $V = \frac{q}{4\pi\epsilon_0} \left(\frac{2l\cos\theta}{r^2 - l^2\cos^2\theta}\right)$ Since $r >> l$, we can neglect $l^2\cos^2\theta$ in the denominator. $V = \frac{q(2l\cos\theta)}{4\pi\epsilon_0 r^2}$ The electric dipole moment is $p = q(2l)$. Therefore, $V = \frac{p\cos\theta}{4\pi\epsilon_0 r^2}$. **Special Cases:** * **On the axis ($\theta = 0^\circ$ or $180^\circ$):** $\cos\theta = \pm 1 \implies V = \pm \frac{p}{4\pi\epsilon_0 r^2}$. * **On the equatorial plane ($\theta = 90^\circ$):** $\cos\theta = 0 \implies V = 0$. * **Using Gauss’s law, find the electric field intensity due to an infinite plane sheet.** Consider a thin, infinite plane sheet of charge with uniform surface charge density $\sigma$. To find the electric field intensity $\vec{E}$ at a point P, we construct a Gaussian surface. **Gaussian Surface:** Choose a cylindrical Gaussian surface with its axis perpendicular to the plane sheet. The cylinder has cross-sectional area 'A' and extends symmetrically on both sides of the sheet, with its flat ends parallel to the sheet and passing through point P. **Applying Gauss's Law:** Gauss's Law states that $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$. The electric field lines are perpendicular to the plane sheet and directed away from positive charge (or towards negative charge). * **Flat ends of the cylinder:** The electric field $\vec{E}$ is perpendicular to the ends, and parallel to $d\vec{A}$. So $\vec{E} \cdot d\vec{A} = E dA$. Since there are two ends, the flux through them is $2EA$. * **Curved surface of the cylinder:** The electric field $\vec{E}$ is parallel to the plane sheet, and perpendicular to the area vector $d\vec{A}$ of the curved surface. So $\vec{E} \cdot d\vec{A} = 0$. Total electric flux $\Phi_E = \int_{ends} E dA + \int_{curved} \vec{E} \cdot d\vec{A} = 2EA + 0 = 2EA$. **Charge Enclosed ($Q_{enc}$):** The charge enclosed by the Gaussian cylinder is the charge on the area 'A' of the sheet within the cylinder. $Q_{enc} = \sigma A$. **Substituting into Gauss's Law:** $2EA = \frac{\sigma A}{\epsilon_0}$ $E = \frac{\sigma}{2\epsilon_0}$ **Conclusion:** The electric field intensity due to an infinite plane sheet of charge is uniform and independent of the distance from the sheet. It is directed perpendicular to the sheet, away from positive charge. * **Derive the expression for energy stored in a capacitor.** Consider a capacitor being charged by a battery. As charge is transferred from one plate to another against the electric field, work is done, and this work is stored as potential energy in the capacitor's electric field. Let 'q' be the charge on the capacitor plates at some instant, and 'V' be the potential difference across them. We know that $V = q/C$, where C is the capacitance. If an additional small charge 'dq' is transferred, the small amount of work done ($dW$) is: $dW = V dq = \frac{q}{C} dq$ To find the total work done (W) in charging the capacitor from 0 to a final charge Q, we integrate dW: $W = \int_0^Q \frac{q}{C} dq = \frac{1}{C} \int_0^Q q dq$ $W = \frac{1}{C} \left[\frac{q^2}{2}\right]_0^Q$ $W = \frac{1}{C} \left(\frac{Q^2}{2} - 0\right) = \frac{Q^2}{2C}$ This work done is stored as electrical potential energy (U) in the capacitor. So, $U = \frac{Q^2}{2C}$. We can express this in other forms using $Q = CV$: $U = \frac{(CV)^2}{2C} = \frac{C^2V^2}{2C} = \frac{1}{2}CV^2$ And $U = \frac{Q^2}{2(Q/V)} = \frac{1}{2}QV$ Thus, the energy stored in a capacitor can be expressed as $U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV$. * **Explain the comparison of EMF of two cells using the individual method.** The individual method using a potentiometer allows comparing the EMFs of two cells without drawing current from them at the balance point, thus providing accurate EMF values. **Circuit Diagram:** A potentiometer wire AB (uniform material and cross-section) is connected in series with a battery (driver cell) $E_D$, a rheostat (Rh), and a key (K). This forms the primary circuit. In the secondary circuit, the positive terminal of the first cell ($E_1$) is connected to point A. Its negative terminal is connected to a galvanometer (G) through a two-way key. The other terminal of the galvanometer is connected to a jockey (J). Similarly, the second cell ($E_2$) is connected to the other terminal of the two-way key, with its positive terminal also connected to A. **Procedure:** 1. **Setup:** Ensure all connections are tight and the driver cell has a higher EMF than $E_1$ and $E_2$. 2. **Cell $E_1$:** Close the key to connect cell $E_1$ to the galvanometer. Slide the jockey along the potentiometer wire (AB) until the galvanometer shows zero deflection (null point). Let the balance length from A be $l_1$. At the null point, no current flows through $E_1$. The potential difference across length $l_1$ of the potentiometer wire is equal to the EMF of $E_1$. $E_1 = \phi l_1 \quad (1)$ (where $\phi$ is the potential gradient of the wire). 3. **Cell $E_2$:** Now, open the key for $E_1$ and close the key to connect cell $E_2$ to the galvanometer. Find the new balance length $l_2$. Similarly, $E_2 = \phi l_2 \quad (2)$ 4. **Comparison:** Divide equation (1) by equation (2): $\frac{E_1}{E_2} = \frac{\phi l_1}{\phi l_2} = \frac{l_1}{l_2}$ This allows for the comparison of the EMFs of the two cells. If the EMF of one cell is known, the EMF of the other can be determined. * **Explain Kelvin’s method to find the resistance of a galvanometer.** Kelvin's method uses a Wheatstone bridge principle to accurately determine the resistance of a galvanometer ($R_g$). This method is particularly suitable because it avoids passing a large current through the galvanometer, protecting its delicate coil. **Circuit Diagram:** A Wheatstone bridge is set up with four arms. * Arm P: A known resistance. * Arm Q: A known resistance. * Arm R: A resistance box (variable known resistance). * Arm S: The galvanometer whose resistance $R_g$ is to be determined. A battery (E) and a key (K) are connected across two opposite corners of the bridge. The galvanometer and a tapping key ($K_g$) are connected across the other two opposite corners. **Procedure:** 1. **Initial Balance:** Close key K. Adjust the resistance R from the resistance box until the galvanometer shows a null deflection when key $K_g$ is pressed. This means the bridge is balanced. 2. **Bridge Balance Condition:** At balance, the ratio of resistances in adjacent arms is equal: $\frac{P}{Q} = \frac{R}{R_g}$ 3. **Calculate $R_g$:** From the balance condition, the resistance of the galvanometer can be calculated as: $R_g = R \times \frac{Q}{P}$ By choosing suitable values for P and Q (e.g., in a ratio of 1:1, 1:10, 10:1), and adjusting R, the galvanometer resistance can be determined with good precision. The key advantage is that at the null point, no current flows through the galvanometer, so its internal resistance does not affect the measurement, and it is protected from damage. * **Derive the magnetic induction at a point on the axis of a circular loop.** Consider a circular loop of radius 'R' carrying a steady current 'I'. We want to find the magnetic induction (magnetic field) at a point P on its axis, at a distance 'x' from the center O of the loop. Consider a small current element $d\vec{l}$ at the top of the loop. According to Biot-Savart Law, the magnetic field $d\vec{B}$ due to this element at P is: $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3}$ where $\vec{r}$ is the vector from $d\vec{l}$ to P, and its magnitude is $r = \sqrt{R^2 + x^2}$. The angle between $d\vec{l}$ and $\vec{r}$ is $90^\circ$, so $|d\vec{l} \times \vec{r}| = dl \cdot r \sin 90^\circ = dl \cdot r$. Magnitude of $d\vec{B}$: $dB = \frac{\mu_0}{4\pi} \frac{I dl}{r^2} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2}$. The direction of $d\vec{B}$ is perpendicular to both $d\vec{l}$ and $\vec{r}$. It makes an angle $\phi$ with the axis. Resolve $d\vec{B}$ into two components: * $dB\cos\phi$ perpendicular to the axis. * $dB\sin\phi$ along the axis. Due to symmetry, for any element $d\vec{l}$ at the top, there's a corresponding element at the bottom. The perpendicular components ($dB\cos\phi$) cancel each other out. Only the axial components ($dB\sin\phi$) add up. From geometry, $\sin\phi = R/r = R/\sqrt{R^2 + x^2}$. Total magnetic field B at P is the integral of the axial components: $B = \int dB\sin\phi = \int \frac{\mu_0}{4\pi} \frac{I dl}{R^2 + x^2} \frac{R}{\sqrt{R^2 + x^2}}$ $B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \int dl$ The integral $\int dl$ is the circumference of the loop, $2\pi R$. $B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} (2\pi R)$ $B = \frac{\mu_0 I R^2}{2 (R^2 + x^2)^{3/2}}$ **At the center of the loop (x=0):** $B_{center} = \frac{\mu_0 I R^2}{2 (R^2)^{3/2}} = \frac{\mu_0 I R^2}{2 R^3} = \frac{\mu_0 I}{2R}$. * **Find the magnetic induction inside an ideal solenoid using Ampere's Law.** An ideal solenoid is a long cylindrical coil of wire with a large number of closely wound turns. Inside an ideal solenoid, the magnetic field is uniform and parallel to the axis, while outside it is very weak and can be approximated as zero. Consider a long solenoid with 'n' turns per unit length carrying current 'I'. **Ampere's Law:** $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}$ **Amperian Loop:** Choose a rectangular Amperian loop ABCD. * Side AB of length 'L' is taken inside the solenoid, parallel to its axis. * Side CD of length 'L' is taken outside the solenoid (where $B \approx 0$). * Sides BC and DA are perpendicular to the axis. **Applying Ampere's Law:** $\oint \vec{B} \cdot d\vec{l} = \int_A^B \vec{B} \cdot d\vec{l} + \int_B^C \vec{B} \cdot d\vec{l} + \int_C^D \vec{B} \cdot d\vec{l} + \int_D^A \vec{B} \cdot d\vec{l}$ * **Along AB:** $\vec{B}$ is parallel to $d\vec{l}$, so $\vec{B} \cdot d\vec{l} = B dl$. $\int_A^B B dl = BL$. * **Along BC and DA:** $\vec{B}$ is perpendicular to $d\vec{l}$, so $\vec{B} \cdot d\vec{l} = 0$. $\int_B^C \vec{B} \cdot d\vec{l} = 0$ and $\int_D^A \vec{B} \cdot d\vec{l} = 0$. * **Along CD:** Outside the solenoid, $B \approx 0$. So $\int_C^D \vec{B} \cdot d\vec{l} = 0$. Thus, $\oint \vec{B} \cdot d\vec{l} = BL$. **Current Enclosed ($I_{enc}$):** The current enclosed by the loop is the current passing through the 'L' length of the solenoid. Number of turns in length L = $nL$. Total current enclosed = (number of turns) $\times$ (current per turn) $= nLI$. **Equating:** $BL = \mu_0 (nLI)$ $B = \mu_0 n I$ This is the magnetic induction inside a long ideal solenoid. * **Explain the working of a Moving Coil Galvanometer and define sensitivity.** **Working Principle:** A moving coil galvanometer works on the principle that a current-carrying coil placed in a uniform magnetic field experiences a torque. This torque causes the coil to rotate, and the rotation is proportional to the current flowing through it. **Construction:** It consists of a rectangular coil (many turns of insulated copper wire) wound on a non-magnetic frame, suspended by a phosphor bronze strip between the pole pieces of a strong permanent horse-shoe magnet. A soft iron core is placed inside the coil to make the field radial and uniform, and to increase magnetic flux. A pointer is attached to the coil to indicate deflection on a scale. **Working:** When current flows through the coil, it produces a magnetic field. This field interacts with the external radial magnetic field of the permanent magnet, producing a torque ($\tau = NIAB$, where N is turns, I is current, A is area, B is magnetic field). This torque causes the coil to rotate. As the coil rotates, the suspension wire gets twisted, producing a restoring torque ($\tau_r = k\alpha$, where k is torsional constant, $\alpha$ is deflection angle). At equilibrium, deflecting torque = restoring torque: $NIAB = k\alpha$ $\alpha = \left(\frac{NAB}{k}\right)I$ Since N, A, B, and k are constants for a given galvanometer, $\alpha \propto I$. This means the deflection is directly proportional to the current. **Sensitivity:** **Current Sensitivity ($S_I$):** It is defined as the deflection produced per unit current flowing through the galvanometer. $S_I = \frac{\alpha}{I} = \frac{NAB}{k}$ **Voltage Sensitivity ($S_V$):** It is defined as the deflection produced per unit voltage applied across the galvanometer. $S_V = \frac{\alpha}{V} = \frac{\alpha}{IR_g} = \frac{NAB}{kR_g}$ (where $R_g$ is the resistance of the galvanometer). A galvanometer is said to be sensitive if it produces a large deflection for a small current or voltage. * **Explain Retentivity and Coercivity using a B-H curve.** The B-H curve (or hysteresis loop) illustrates the relationship between magnetic flux density (B) and magnetizing field intensity (H) for a ferromagnetic material as it is taken through a cycle of magnetization. **Retentivity (or Remanence):** When a ferromagnetic material is fully magnetized (point C on a typical B-H curve, where H is maximum), and then the magnetizing field H is reduced to zero, the magnetic flux density B in the material does not drop to zero. The value of B that remains in the material when H = 0 is called **retentivity** (or remanence). This is represented by point D on the B-H curve. * **Significance:** It indicates the ability of a material to retain magnetism after the external field is removed. Materials with high retentivity are suitable for making permanent magnets. **Coercivity (or Coercive Force):** To demagnetize the material completely (bring B to zero) after it has been magnetized, a reverse magnetizing field needs to be applied. The magnitude of this reverse magnetizing field (H) required to reduce the residual magnetism (B) to zero is called **coercivity** (or coercive force). This is represented by point E on the B-H curve. * **Significance:** It indicates the resistance of a material to demagnetization. Materials with high coercivity are also suitable for permanent magnets (hard magnetic materials), while those with low coercivity are used for electromagnets and transformer cores (soft magnetic materials). (A diagram of a B-H curve would visually represent these points.) * **Define Self-Induction and find the inductance of a long solenoid.** **Self-Induction:** Self-induction is the phenomenon by which a changing current in a coil induces an electromotive force (EMF) in the same coil. This induced EMF opposes the change in current that produced it (Lenz's Law). The magnetic flux ($\Phi_B$) linked with a coil is directly proportional to the current (I) flowing through it: $\Phi_B \propto I$. The constant of proportionality is the self-inductance (L) of the coil: $\Phi_B = LI$. The induced EMF is $\mathcal{E} = -\frac{d\Phi_B}{dt} = -L\frac{dI}{dt}$. **Inductance of a Long Solenoid:** Consider a long solenoid of length 'l', cross-sectional area 'A', and 'N' total turns. Let 'n' be the number of turns per unit length ($n = N/l$). When a current 'I' flows through the solenoid, the magnetic field inside it is approximately uniform and given by $B = \mu_0 n I$. The magnetic flux ($\Phi_B$) through a single turn of the solenoid is: $\Phi_{single} = B \times A = (\mu_0 n I) A$ The total magnetic flux linked with the entire solenoid (all N turns) is: $\Phi_{total} = N \times \Phi_{single} = N (\mu_0 n I A)$ Substitute $N = nl$: $\Phi_{total} = (nl) (\mu_0 n I A) = \mu_0 n^2 A l I$ By definition of self-inductance, $\Phi_{total} = LI$. Comparing the two expressions for $\Phi_{total}$: $LI = \mu_0 n^2 A l I$ $L = \mu_0 n^2 A l$ This is the expression for the self-inductance of a long solenoid. * **Describe the construction and working of a Transformer.** **Construction:** A transformer consists of two coils, the primary coil ($N_P$ turns) and the secondary coil ($N_S$ turns), wound on a common soft iron laminated core. * **Coils:** Made of insulated copper wire. The primary coil is connected to the AC input, and the secondary coil is connected to the load. * **Core:** The soft iron core is laminated (made of thin sheets insulated from each other) to reduce energy losses due to eddy currents. The core provides a continuous path for magnetic flux, ensuring maximum flux linkage between the primary and secondary coils. **Working:** 1. **AC Input to Primary:** When an alternating voltage ($V_P$) is applied across the primary coil, an alternating current flows through it. 2. **Changing Magnetic Flux:** This alternating current produces a continuously changing magnetic flux in the primary coil. 3. **Flux Linkage:** The soft iron core is designed to channel almost all of this changing magnetic flux through the secondary coil. 4. **Induced EMF in Secondary:** According to Faraday's law of electromagnetic induction, this changing magnetic flux linked with the secondary coil induces an alternating electromotive force (EMF) or voltage ($V_S$) across its terminals. 5. **Mutual Induction:** This phenomenon is called mutual induction, as a changing current in one coil induces an EMF in an adjacent coil. **Transformer Equation:** From Faraday's law: $V_P = -N_P \frac{d\Phi_B}{dt}$ $V_S = -N_S \frac{d\Phi_B}{dt}$ Assuming perfect flux linkage, $\frac{d\Phi_B}{dt}$ is the same for both coils. Dividing the equations: $\frac{V_S}{V_P} = \frac{N_S}{N_P}$ Also, for an ideal transformer, input power = output power ($P_P = P_S$). $V_P I_P = V_S I_S \implies \frac{V_S}{V_P} = \frac{I_P}{I_S}$. Combining these, $\frac{V_S}{V_P} = \frac{N_S}{N_P} = \frac{I_P}{I_S} = k$ (transformer ratio). * **Step-up transformer:** $N_S > N_P \implies V_S > V_P$ (k > 1). * **Step-down transformer:** $N_S X_C$: $\phi$ is positive, voltage leads current (circuit is inductive). * If $X_C > X_L$: $\phi$ is negative, voltage lags current (circuit is capacitive). * If $X_L = X_C$: $\phi = 0$, voltage and current are in phase (resonance, circuit is purely resistive). * **Describe the Davisson-Germer experiment.** The Davisson-Germer experiment (1927) provided the first experimental confirmation of the wave nature of electrons, a prediction made by Louis de Broglie. **Experimental Setup:** 1. **Electron Gun:** A filament heated by a low-voltage power supply emits electrons via thermionic emission. These electrons are then accelerated by an adjustable potential difference (V) to a specific velocity. 2. **Collimator:** The accelerated electrons are passed through a small aperture to produce a fine beam. 3. **Nickel Crystal Target:** This electron beam is directed onto the surface of a nickel crystal. The crystal has a regular atomic arrangement, acting as a diffraction grating. 4. **Detector:** A movable detector (Faraday cup) measures the intensity of the scattered electrons at different angles ($\phi$) with the incident beam. **Procedure and Observations:** * The electron beam was directed at the nickel crystal. * The detector was rotated around the crystal to measure the intensity of the scattered electrons at various angles. * The accelerating voltage was varied. * Davisson and Germer observed a strong peak in the intensity of scattered electrons at a particular angle ($\phi = 50^\circ$) for a specific accelerating voltage (V = 54 V). This indicated a constructive interference phenomenon, characteristic of wave diffraction. **Explanation and Conclusion:** * The observed diffraction pattern (peak intensity at specific angles) was analogous to X-ray diffraction from crystals, confirming the wave nature of electrons. * Using Bragg's law ($2d\sin\theta = n\lambda$), where 'd' is the interatomic spacing in the nickel crystal and $\theta$ is the glancing angle, they calculated the wavelength of the diffracted electrons. For nickel, $d=0.091$ nm. For $V=54$ V and $\phi=50^\circ$, the glancing angle $\theta = (180^\circ - 50^\circ)/2 = 65^\circ$. * The calculated wavelength matched very closely with the de Broglie wavelength predicted by $\lambda = h/\sqrt{2meV}$, where h is Planck's constant, m is electron mass, and e is electron charge. For V = 54 V, the de Broglie wavelength is $\approx 0.167$ nm. This experiment provided crucial evidence for de Broglie's hypothesis of matter waves and established the wave-particle duality of matter. * **Derive an expression for the total energy of an electron in the n^{th} Bohr orbit.** According to Bohr's model, an electron of mass 'm' revolves in a circular orbit of radius 'r' around a nucleus with charge Ze (for hydrogen-like atoms, Z=1 for hydrogen). **Kinetic Energy (K.E.):** The electrostatic force of attraction between the electron and nucleus provides the necessary centripetal force: $\frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2} = \frac{mv^2}{r}$ $mv^2 = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}$ K.E. $= \frac{1}{2}mv^2 = \frac{1}{2} \left(\frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}\right) = \frac{Ze^2}{8\pi\epsilon_0 r}$ **Potential Energy (P.E.):** The potential energy of an electron at a distance 'r' from a nucleus with charge Ze is: P.E. $= \frac{1}{4\pi\epsilon_0} \frac{(Ze)(-e)}{r} = -\frac{Ze^2}{4\pi\epsilon_0 r}$ **Total Energy (E):** Total Energy E = K.E. + P.E. $E = \frac{Ze^2}{8\pi\epsilon_0 r} - \frac{Ze^2}{4\pi\epsilon_0 r} = -\frac{Ze^2}{8\pi\epsilon_0 r}$ Now, we need to substitute the expression for the radius of the $n^{th}$ Bohr orbit, $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$. $E_n = -\frac{Ze^2}{8\pi\epsilon_0} \left(\frac{\pi m Z e^2}{n^2 h^2 \epsilon_0}\right)$ $E_n = -\frac{Z^2 m e^4}{8 \epsilon_0^2 n^2 h^2}$ For a hydrogen atom, Z = 1. $E_n = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$ This expression shows that the energy of an electron in a Bohr orbit is quantized and inversely proportional to $n^2$. The negative sign indicates that the electron is bound to the nucleus. * **Explain the spectral series of the hydrogen atom with an energy level diagram.** When a hydrogen atom absorbs energy, its electron jumps from a lower energy level to a higher one. When it de-excites, it falls back to a lower energy level, emitting a photon of specific wavelength, resulting in a line spectrum. The collection of these spectral lines forms various series. **Energy Level Diagram:** The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2}$ eV. * $n=1$: Ground state ($E_1 = -13.6$ eV) * $n=2$: First excited state ($E_2 = -3.4$ eV) * $n=3$: Second excited state ($E_3 = -1.51$ eV) * ... * $n=\infty$: Ionization state ($E_\infty = 0$ eV) (A vertical diagram with energy levels decreasing downwards and different series transitions marked would be ideal here.) **Spectral Series:** 1. **Lyman Series:** * Transitions: Electrons jump from higher energy levels ($n_i = 2, 3, 4, ...$) to the ground state ($n_f = 1$). * Region: Ultraviolet (UV) region. * Formula: $\frac{1}{\lambda} = R\left(\frac{1}{1^2} - \frac{1}{n_i^2}\right)$, where R is Rydberg constant. 2. **Balmer Series:** * Transitions: Electrons jump from higher energy levels ($n_i = 3, 4, 5, ...$) to the second energy level ($n_f = 2$). * Region: Visible region (four lines: $H_\alpha, H_\beta, H_\gamma, H_\delta$). * Formula: $\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{n_i^2}\right)$. 3. **Paschen Series:** * Transitions: Electrons jump from higher energy levels ($n_i = 4, 5, 6, ...$) to the third energy level ($n_f = 3$). * Region: Infrared (IR) region. * Formula: $\frac{1}{\lambda} = R\left(\frac{1}{3^2} - \frac{1}{n_i^2}\right)$. 4. **Brackett Series:** * Transitions: Electrons jump from higher energy levels ($n_i = 5, 6, 7, ...$) to the fourth energy level ($n_f = 4$). * Region: Infrared (IR) region. * Formula: $\frac{1}{\lambda} = R\left(\frac{1}{4^2} - \frac{1}{n_i^2}\right)$. 5. **Pfund Series:** * Transitions: Electrons jump from higher energy levels ($n_i = 6, 7, 8, ...$) to the fifth energy level ($n_f = 5$). * Region: Far Infrared (IR) region. * Formula: $\frac{1}{\lambda} = R\left(\frac{1}{5^2} - \frac{1}{n_i^2}\right)$. * **Derive the relationship between Half-life and Mean life.** **Law of Radioactive Decay:** The rate of decay of radioactive nuclei is proportional to the number of nuclei present at that instant. $-\frac{dN}{dt} = \lambda N$ Integrating this gives $N = N_0 e^{-\lambda t}$, where $N_0$ is initial nuclei, N is nuclei at time t, and $\lambda$ is the decay constant. **Half-life ($T_{1/2}$):** Half-life is the time required for the number of radioactive nuclei to reduce to half of its initial value. At $t = T_{1/2}$, $N = N_0/2$. $\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}}$ $\frac{1}{2} = e^{-\lambda T_{1/2}}$ Taking natural logarithm on both sides: $\ln(1/2) = -\lambda T_{1/2}$ $-\ln 2 = -\lambda T_{1/2}$ $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}$ **Mean Life ($\tau$):** Mean life (or average life) is the average lifetime of all the radioactive nuclei in a sample. It can be shown by integration that $\tau = \frac{1}{\lambda}$. **Relationship:** Substitute $\lambda = 1/\tau$ into the half-life equation: $T_{1/2} = \frac{0.693}{1/\tau} = 0.693 \tau$ So, Half-life = 0.693 $\times$ Mean life. * **Show how NAND and NOR gates act as universal gates.** A logic gate is considered universal if it can be used to construct any other basic logic gate (AND, OR, NOT) using only that type of gate. Both NAND and NOR gates are universal gates. **NAND Gate as a Universal Gate:** 1. **NOT Gate from NAND:** * Connect both inputs of a NAND gate together. * If input A is 0, output is NAND(0,0) = 1. * If input A is 1, output is NAND(1,1) = 0. * This is a NOT gate: $\overline{A}$. 2. **AND Gate from NAND:** * Connect the output of a NAND gate (A NAND B) to the input of another NAND gate configured as a NOT gate. * Output = NOT(A NAND B) = NOT($\overline{A \cdot B}$) = $A \cdot B$. * This is an AND gate. 3. **OR Gate from NAND:** * Connect two NOT gates (made from NAND gates) to the inputs of a third NAND gate. * Input to third NAND: $\overline{A}$ and $\overline{B}$. * Output = $\overline{A}$ NAND $\overline{B}$ = $\overline{\overline{A} \cdot \overline{B}}$. * Using De Morgan's theorem: $\overline{\overline{A} \cdot \overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B$. * This is an OR gate. **NOR Gate as a Universal Gate:** 1. **NOT Gate from NOR:** * Connect both inputs of a NOR gate together. * If input A is 0, output is NOR(0,0) = 1. * If input A is 1, output is NOR(1,1) = 0. * This is a NOT gate: $\overline{A}$. 2. **OR Gate from NOR:** * Connect the output of a NOR gate (A NOR B) to the input of another NOR gate configured as a NOT gate. * Output = NOT(A NOR B) = NOT($\overline{A + B}$) = $A + B$. * This is an OR gate. 3. **AND Gate from NOR:** * Connect two NOT gates (made from NOR gates) to the inputs of a third NOR gate. * Input to third NOR: $\overline{A}$ and $\overline{B}$. * Output = $\overline{A}$ NOR $\overline{B}$ = $\overline{\overline{A} + \overline{B}}$. * Using De Morgan's theorem: $\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$. * This is an AND gate. Since both NAND and NOR gates can implement NOT, AND, and OR gates, they are considered universal gates.