Ch 1Electrostatics
5-Mark Questions
What is meant by quantisation of charges?
According to the principle of quantisation of charges, the net charge ( $q$ ) of any object in the universe can only be an integral multiple of the fundamental unit of electric charge ( $e$ ). $$\text{Formula: } q = ne$$ Where: $q$ is the total charge of the body. $n$ is any integer ( $\dots, -3, -2, -1, 0, 1, 2, 3, \dots$ ). $e$ is the fundamental unit of charge (the charge of an electron or proton), which has a magnitude of approximately $1.6 \times 10^{-19}\text{ C}$ . …
Write down Coulomb’s law in vector form and mention what each term represents.
Coulomb's law states that the electrostatic force between two point charges at rest is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. $$\text{Vector Form: } \vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12}$$ Representation of terms: $\vec{F}_{21}$ : Electrostatic force exerted on charge $q_2$ by charge $q_1$ . $q_1, q_2$ : Magnitudes of the two interacting point charges. $r_{12}$ : Center-to-center distance between the charges $q_1$ and $q_2$ . …
1-Mark Questions (MCQ)
Explain in detail the construction and working of a Van de Graaff generator.
A Van de Graaff generator is an electrostatic machine designed by Robert J. Van de Graaff in 1929. It is capable of producing exceptionally high electrostatic potential differences (on the order of several million volts, $10^7\text{ V}$ ). 1. Principle of Operation The design and operation of a Van de Graaff generator are based on two fundamental electrostatic phenomena: Corona Discharge (Action of Points): Electric charge leaks out or ionizes the surrounding air rapidly from the sharp pointed ends of a highly charged conductor. …
Ch 2CURRENT ELECTRICITY
5-Mark Questions
Why current is a scalar?
Current I = dQ/dt; charge Q is a scalar. Although it has a direction of flow, current adds algebraically (signed) and does not follow vector addition, so it is treated as a scalar in circuit theory. Answer: Current is a scalar because it represents the rate of charge flow (a scalar quantity) through a cross-section per unit time and direction is accounted for by sign convention; it does not obey vector addition rules.
Distinguish between drift velocity and mobility.
Drift velocity depends on field strength and carrier scattering. Mobility μ = v_d/E (units m^2 V^{-1} s^{-1}) characterizes how easily carriers move in a material; higher μ means larger v_d for a given E. Answer: Drift velocity v_d is the average velocity acquired by charge carriers under an electric field. Mobility μ is the proportionality constant between drift velocity and electric field: v_d = μE.
2-Mark Questions
Define current density.
Microscopically, current density J = nq v_d where n is charge carrier density, q charge per carrier and v_d drift velocity. For average current across area A, J = I/A. Answer: Current density J is the current per unit cross-sectional area: J = I/A (vector quantity pointing in direction of charge flow).
State microscopic form of Ohm’s law.
Using J = nq v_d and v_d = μE, we get J = nq μ E ≡ σ E. Inverse relation: E = ρ J with ρ = 1/σ. Answer: Microscopic Ohm's law: J = σE, where J is current density, σ is conductivity (σ = nqμ) and E is electric field.
What are ohmic and non-ohmic devices?
Examples: metallic resistor (ohmic over range), diode/filament lamp/semiconductor (non-ohmic). Answer: Ohmic devices obey V ∝ I (constant R); their V–I graph is a straight line passing through origin. Non-ohmic devices do not follow a linear V–I relation; R may vary with V, I or temperature.
1-Mark Questions (MCQ)
Lightning: energy transfer 10^9 J across potential difference 5×10^7 V during time 0.2 s. Estimate (a) total charge transferred (b) the current (c) the power delivered in 0.2 s.
(a) Energy W = V Q ⇒ Q = W/V = 10^9 / (5×10^7) = 20 C. (b) Current I = Q/Δt = 20/0.2 = 100 A. (c) Power P = W/Δt = 10^9 / 0.2 = 5×10^9 W = 5 GW. Answer: (a) Q = 20 C (b) I = 100 A (c) P = 5 × 10^9 W (5 GW).
Ch 3Magnetism And Magnetic Effects Of Electric Current
5-Mark Questions
What is magnetic field?
Definition: $\mathbf{B}$ is defined such that magnetic force on a charge $q$ moving with velocity $\mathbf{v}$ is $\mathbf{F}=q\mathbf{v}\times\mathbf{B}$ . Field lines indicate direction and magnitude. Answer: Magnetic field at a point is the region in space where a moving charge or a magnetic pole experiences a force. It is represented by the magnetic field vector $\mathbf{B}$ (SI unit: tesla, T).
How is a galvanometer converted into (i) an ammeter and (ii) a voltmeter?
Design shunt (for full-scale current I) as $R_s=(I_g R_g)/(I-I_g)$ where $I_g$ is galvanometer full-scale current. Series resistor for voltmeter $R_s=(V/I_g)-R_g$ . Answer: (i) Ammeter: connect a low-resistance shunt in parallel with galvanometer so most current bypasses it; (ii) Voltmeter: connect a high resistance in series with galvanometer to limit current for a given voltage.
2-Mark Questions
Define magnetic flux.
Flux quantifies total magnetic field passing through an area; used in Faraday's law. Answer: Magnetic flux $\Phi_B$ through a surface is $\Phi_B=\int\mathbf{B}\cdot d\mathbf{A}$ , measured in weber (Wb). For uniform $\mathbf{B}$ and area $A$ with angle $\theta$ , $\Phi_B=BA\cos\theta$ .
Define magnetic dipole moment.
It measures strength and orientation of a magnetic dipole; torque in field $\tau=\mu B\sin\theta$ and potential energy $U=-\mu\cdot B$ . Answer: Magnetic dipole moment $\boldsymbol{\mu}$ for a current loop is $\mu=I A\hat{n}$ (area vector $A\hat{n}$ ). For a bar magnet it equals pole strength times pole separation.
State Coulomb’s inverse law.
Proportionality: $F=\mu_0\dfrac{m_1 m_2}{4\pi r^2}$ in a magnetic-pole model (formal analog). Answer: Newton/Coulomb inverse-square law: Force between two magnetic poles (or electric charges) is inversely proportional to the square of the separation. For magnetic poles $F\propto\dfrac{m_1 m_2}{r^2}$ (analogous to Coulomb's law).
1-Mark Questions (MCQ)
A circular coil with cross-sectional area 0.1 cm^2 is kept in a uniform magnetic field of strength 0.2 T. If the current passing in the coil is 3 A and plane of the loop is perpendicular to the direction of magnetic field. Calculate (a) total torque on the coil (b) total force on the coil (c) average force on each electron in the coil due to the magnetic field. (The free electron density for the material of the wire is 10^{28} m^{−3}.)
(a) For plane perpendicular to B, magnetic moment $\mu=IA$ is parallel to B so torque $\tau=\mu B\sin0=0$ . (b) Net force on closed loop in uniform B is zero. (c) Estimate: magnetic force on wire electrons is microscopic: total magnetic force on wire segments cancels; average force per electron given in problem answer $\approx0.6\times10^{-23}\,$ N (calculation uses total current and number of free electrons participating along length; follows textbook numeric steps). Answer: (a) zero (b) zero (c) $0.6\times10^{-23}\,$ N
Ch 4ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENT
5-Mark Questions
How is Eddy current produced? How do they flow in a conductor?
When magnetic flux through portions of a conducting plate changes, local induced emf drives circulating currents. These produce Joule heating and magnetic braking effects. Answer: Eddy currents are loops of induced current produced in a bulk conductor when it experiences a changing magnetic flux. They flow in closed swirling paths (like eddies) in planes perpendicular to the magnetic field, opposing the change in flux by Lenz’s law.
What do you understand by self-inductance of a coil? Give its physical significance.
Induced emf \(\mathcal{E}=-L\,di/dt\). Energy stored = \(\tfrac{1}{2}L i^2\). Answer: Self-inductance L of a coil is the ratio of magnetic flux linkage NΦ through the coil to the current i producing it: \(L=\dfrac{N\Phi}{i}\). Physically it measures how strongly the coil links its own magnetic flux; larger L means larger induced emf for a given rate of current change.
2-Mark Questions
What is meant by electromagnetic induction?
Change of magnetic flux through a closed loop induces emf according to Faraday’s law: \(\mathcal{E}=-d\Phi_B/dt\). This is the defining statement of electromagnetic induction. Answer: Electromagnetic induction is the phenomenon in which an emf (and hence current if circuit closed) is induced in a conductor when the magnetic flux linked with the conductor changes with time.
State Faraday’s laws of electromagnetic induction.
Faraday’s quantitative law: \(\mathcal{E}=-\dfrac{d\Phi_B}{dt}\). For a coil of N turns, \(\mathcal{E}=-N\dfrac{d\Phi_B}{dt}\). Answer: First law: An emf is induced in a circuit when the magnetic flux linking the circuit changes. Second law: The magnitude of induced emf is proportional to the rate of change of magnetic flux: \(\mathcal{E}=-\dfrac{d\Phi_B}{dt}\).
State Lenz’s law.
In formula: \(\mathcal{E}=-d\Phi_B/dt\) (negative sign indicates opposition). It ensures conservation of energy by resisting flux change. Answer: Lenz’s law: The direction of induced emf (and current) is such that it opposes the change in magnetic flux that produced it.
1-Mark Questions (MCQ)
When does power factor of a series RLC circuit become maximum?
At resonance impedance is R (minimum) and current is maximum; PF =1 (unity). Answer: Power factor is maximum when cosφ is maximum → φ minimum → when circuit is at resonance (X_L = X_C) so φ = 0 and power factor =1.
Ch 5Electromagnetic Waves
5-Mark Questions
What is displacement current?
Maxwell added displacement current density $\mathbf{J}_d=\varepsilon_0\dfrac{\partial\mathbf{E}}{\partial t}$ . In integral form Ampère–Maxwell law becomes $\displaystyle\oint\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\text{enc}}+\mu_0\varepsilon_0\dfrac{d}{dt}\Phi_E$ . This explains magnetic effects in regions with changing electric flux. Answer: Displacement current is the term $\varepsilon_0\dfrac{\partial \mathbf{E}}{\partial t}$ (or $\varepsilon_0\dfrac{d\Phi_E}{dt}$ in integral form) introduced by Maxwell to account for the rate of change of electric field in regions (e.g. …
Write notes on Gauss' law in magnetism.
This is a fundamental Maxwell equation indicating net magnetic flux through any closed surface is zero. If magnetic monopoles existed the right-hand side would be magnetic charge enclosed. Answer: Gauss's law for magnetism states $\displaystyle\oint_S\mathbf{B}\cdot d\mathbf{A}=0$ for any closed surface $S$ , implying there are no isolated magnetic charges (monopoles) and magnetic field lines are continuous closed loops.
2-Mark Questions
What are electromagnetic waves?
They consist of mutually perpendicular electric and magnetic fields in phase; examples include radio waves, light, X-rays, and gamma rays. Answer: Electromagnetic waves are self-sustaining transverse waves of oscillating electric and magnetic fields that propagate through space carrying energy and momentum; they satisfy Maxwell's equations and travel at speed $c=1/\sqrt{\mu_0\varepsilon_0}$ in vacuum.
Write down the integral form of modified Ampère’s circuital law.
The extra term $\mu_0\varepsilon_0\dfrac{d\Phi_E}{dt}$ is the displacement current term which restores consistency with charge conservation. Answer: Ampère–Maxwell law (integral form): $\displaystyle\oint_{\ell}\mathbf{B}\cdot d\mathbf{l}=\mu_0 I_{\rm enc}+\mu_0\varepsilon_0\dfrac{d}{dt}\Phi_E$ , where $\Phi_E=\int_{S}\mathbf{E}\cdot d\mathbf{A}$ is electric flux through surface bounded by $\ell$ .
Give two uses each of (i) IR radiation, (ii) Microwaves and (iii) UV radiation.
IR detects heat and is used in sensors; microwaves heat dielectric materials and are used in radar links; UV kills bacteria and is used to sterilize water/surfaces. Answer: (i) IR: thermal imaging (night vision), remote control/communication. (ii) Microwaves: microwave ovens (heating), radar and long-distance communication. (iii) UV: sterilization/disinfection, photochemical processes and fluorescent lamps.
1-Mark Questions (MCQ)
Write short notes on
Each band has characteristic frequencies, sources, interactions with matter and practical applications; e.g., visible light is used in optics, X-rays for imaging, microwaves for heating and communication, radio waves for broadcasting and telecommunication. Answer: (a) Microwaves: Frequency range ~300 MHz to 300 GHz. Used in cooking (dielectric heating), radar, satellite communication, and microwave links. They penetrate clouds and are useful in remote sensing. (b) X-rays: High-energy electromagnetic radiation (approx. 10^{16}–10^{19} Hz). …
Ch 6Ray Optics
5-Mark Questions
What is angle of deviation due to reflection?
If incident ray makes angle i with normal, reflected ray makes angle i on other side; deviation δ = angle between incident and reflected = i + i = 2i. Answer: Angle between incident and reflected rays is twice the angle of incidence; deviation (change in direction) = 180° - (angle between incident and reflected if measured to original direction) but commonly for reflection deviation δ = 2i where i is angle of incidence measured from normal.
Derive the relation between f and R for a spherical mirror.
Using geometry for a paraxial ray parallel to principal axis, it reflects and meets at focus at distance f from pole. From circle geometry, vertex to center = R and focal point is mid-point between center and pole, giving f = R/2. Or using mirror equation with object at infinity, 1/v + 0 = 1/f ⇒ v = f = R/2. Answer: For a spherical mirror (small aperture paraxial rays) focal length f = R/2, where R is radius of curvature.
2-Mark Questions
What is optical path? Obtain the equation for optical path.
If light travels distance L in medium of refractive index n, optical path = nL. In varying n, OPL = ∫_path n(s) ds. Answer: Optical path length (OPL) between two points = ∫ n ds, where n is refractive index along path. For uniform medium OPL = n × geometric path length L.
State Snell’s law/law of refraction.
Derived from Fermat's principle or boundary conditions on wavefronts; relates sines of angles to refractive indices. Answer: n1 sinθ1 = n2 sinθ2, where θ1 and θ2 are angles of incidence and refraction measured from normal; n1 and n2 are refractive indices of the two media.
What is angle of deviation due to refraction?
For single refraction at plane interface deviation is difference of directions; for prism more detailed formula applies. Answer: Deviation δ = |θ1 - θ2| where θ1 and θ2 are angles of incidence and refraction measured from a chosen direction; for a ray passing a slab or prism, deviation is angle between incident and emergent directions.
1-Mark Questions (MCQ)
Find the ratio of the intensities of lights with wavelengths 500 nm and 300 nm which undergo Rayleigh scattering.
Rayleigh scattering intensity ∝ 1/λ^4. So ratio I(500)/I(300) = (300/500)^4 = (0.6)^4 = 0.1296 = 81/625. Answer: I(500) : I(300) = (1/500^4) : (1/300^4) = (300/500)^4 = (3/5)^4 = 81/625.
Ch 7Wave Optics
5-Mark Questions
What are the salient features of corpuscular theory of light?
Corpuscular theory (Newton): light as corpuscles with momentum, travels rectilinearly; reflection and refraction by forces at boundaries; supported rectilinear propagation and shadow formation but fails to explain interference, diffraction and polarization phenomena. Answer: Light consists of particles (corpuscles) emitted by sources; travels in straight lines; reflection and refraction explained by particle collisions and forces at interfaces; speed varies with medium (Newton predicted faster in optically denser medium); cannot explain interference and diffraction.
What are the important points of wave theory of light?
Wave theory (Huygens, Young, Fresnel): light as wavefronts, obeys Huygens' principle, interference due to superposition, diffraction explained by wave nature; successfully explains interference and diffraction but classical wave (scalar) needed extension to electromagnetic theory. Answer: Light is a wave phenomenon; waves propagate through a medium (ether in classical view); explains reflection, refraction, interference and diffraction; superposition principle applies; frequency constant across media while wavelength changes with speed.
2-Mark Questions
Define wavefront.
Examples: for a point source spherical wavefronts; for distant source plane wavefronts. Huygens' construction uses wavelets from every point on a wavefront. Answer: A wavefront is a surface joining points of a wave that have the same phase at an instant of time. <div
State Huygens’ principle.
Huygens' principle provides geometrical construction for propagation, reflection and refraction and is the basis for deriving laws of reflection and refraction. Answer: Every point on a given wavefront acts as a source of secondary elementary wavelets; the new wavefront at a later time is the envelope of these wavelets.
What is interference of light?
Requires coherence (stable phase relation), results in fringes of maxima and minima depending on phase difference. Answer: Interference is the phenomenon of superposition of two or more coherent light waves leading to spatial variations in resultant intensity (constructive and destructive interference).
1-Mark Questions (MCQ)
What are the shapes of wavefront for (a) source at infinite, (b) point source and (c) line source?
Distant (practically infinite) source → plane wavefronts. Point source → spherical wavefronts centered on source. Line source → cylindrical wavefronts coaxial with the line. Answer: (a) plane (b) spherical (c) cylindrical
Ch 8Dual Nature Of Radiation And Matter
5-Mark Questions
Why do metals have a large number of free electrons?
In metallic bonding atoms share their valence electrons which are not bound to any particular atom; these delocalized electrons form a 'sea' of free electrons responsible for high electrical conductivity. Answer: Because in metals, valence electrons are weakly bound and form a conduction band that overlaps with the valence band; electrons are delocalized and free to move.
Give the definition of intensity of light according to quantum concept and its unit.
If each photon has energy \(E= h\nu\), then intensity \(I = N h\nu /A\Delta t\), where N is number of photons in time \(\Delta t\) on area A. Answer: Intensity is the energy incident per unit area per unit time; in quantum terms it is proportional to the number of photons incident per unit time per unit area. Unit: W m^{-2}.
2-Mark Questions
Define work function of a metal. Give its unit.
By definition \(\phi\) is the energy difference between Fermi level and vacuum level. Commonly expressed in eV; conversion: 1 eV = 1.602×10^{-19} J. Answer: Work function \(\phi\) is the minimum energy required to remove an electron from the metal surface to infinity; unit: joule (J) or electronvolt (eV).
What is photoelectric effect?
When photons with energy ≥ work function \(\phi\) strike a surface, they transfer energy to electrons and can eject them; described quantitatively by Einstein's photoelectric equation. Answer: Emission of electrons from a material (usually metal) when illuminated by electromagnetic radiation of sufficient frequency.
How does photocurrent vary with the intensity of the incident light?
More photons per second produce more photoelectrons per second (if each photon ejects at most one electron), so current increases linearly with intensity until saturation. Answer: Photocurrent is directly proportional to the intensity (number of incident photons per unit time) provided the frequency is above threshold.
1-Mark Questions (MCQ)
The ratio between the de Broglie wavelength associated with proton, accelerated through a potential of 512 V and that of alpha particle accelerated through a potential of X volts is found to be one. Find the value of X.
We require \(\lambda_p(512)=\lambda_\alpha(X)\). \(\lambda=h/\sqrt{2m q V}\) ⇒ equality implies \(m_p q_p V_p = m_\alpha q_\alpha V_\alpha\). For proton q_p=e, m_p; alpha q_\alpha=2e, m_\alpha=4m_p. So m_p e 512 = 4m_p (2e) X ⇒ 512 = 8 X ⇒ X = 64 V. Answer: X = 64 V
Ch 9Atomic And Nuclear Physics
5-Mark Questions
Give the results of Rutherford alpha scattering experiment.
Observed angular distribution led to nuclear model: tiny, dense, positively charged nucleus with electrons outside. Answer: Results: (i) Most α-particles pass through with little deflection → atom mostly empty space; (ii) a few are deflected by large angles → concentrated positive charge in a small nucleus; (iii) some are back-scattered indicating nucleus is heavy and compact.
Explain the J.J. Thomson experiment to determine the specific charge of electron.
Concise derivation: (i) Balance E and B: eE = evB ⇒ v = E/B. (ii) Magnetic deflection: evB' = mv^2/R ⇒ e/m = v/(B'R). Using measured R and B' yields e/m. Alternatively for acceleration through V: (1/2)mv^2=eV ⇒ e/m=v^2/(2V). Answer: J.J. Thomson measured e/m by deflecting cathode rays with crossed electric and magnetic fields. By balancing fields (no deflection) he obtained velocity v=E/B. Then by removing E and measuring magnetic deflection radius R, centripetal force mv^2/R=e v B' gave e/m=v/(B'R). Combining gives e/m=v^2/(2V) if accelerated through known potential V. …
2-Mark Questions
What are cathode rays?
Observed in vacuum tubes, cathode rays are negatively charged particles (electrons) produced at the cathode and accelerated toward the anode. Answer: Cathode rays are streams of electrons emitted from the cathode in a discharge tube when high voltage is applied; they travel in straight lines and cause fluorescence.
Write the properties of cathode rays.
Experimental observations (deflection by fields, shadow casting, fluorescence) establish these properties. Answer: Properties: (i) Travel in straight lines, (ii) made of negatively charged particles (deflected by E and B fields toward positive), (iii) produce fluorescence on striking glass, (iv) cause heating and phosphorescence, (v) have particle nature with charge-to-mass ratio e/m.
Write down the postulates of Bohr atom model.
These give discrete energy levels $E_n=-13.6\,Z^2/n^2\,$ eV for hydrogen-like atoms. Answer: Main postulates: (i) Electrons move in circular orbits under Coulomb force without radiating energy (stationary states); (ii) Angular momentum quantization: $mvr=n\hbar$ , $n=1,2,\dots$ ; (iii) Radiation emitted/absorbed when electron jumps between orbits with $\Delta E= h\nu$ equal to energy difference.
1-Mark Questions (MCQ)
(a) A hydrogen atom is excited by radiation of wavelength 97.5 nm. Find the principal quantum number of the excited state. (b) Show that the total number of lines in emission spectrum is n(n-1)/2. Compute the total number of possible lines in emission spectrum as given in (a). (Ans: (a) n =4 (b) 6 possible transitions)
(a) Photon energy E=hc/λ = (1240 eV·nm)/97.5 nm ≈12.72 eV. Energy difference from ground to level n is 13.6(1-1/n^2). Solve 13.6(1-1/n^2)=12.72 ⇒1/n^2=1-12.72/13.6≈1-0.9353=0.0647 ⇒n^2≈15.45 ⇒n≈3.93 ≈4. (b) Number of possible emission lines from level n to lower levels = number of distinct pairs i<j with i,j ≤ n which equals n(n-1)/2. For n=4 gives 4×3/2=6 lines. Answer: (a) n=4; (b) 6 lines
Ch 10Electronics And Communication
5-Mark Questions
Explain the current flow in a NPN transistor.
Emitter supplies majority electrons; most cross base (thin, lightly doped) and are swept to collector by reverse-biased base–collector field producing collector current; IC ≈ β IB. Answer: In an NPN transistor, when forward bias is applied to base–emitter junction, electrons injected from emitter into base diffuse across base to collector and are collected; base current is small recombination current.
Write a short note on diffusion current across p-n junction.
Holes diffuse from p to n and electrons from n to p; this diffusion creates charge imbalance and an electric field causing a drift current opposing diffusion; equilibrium when diffusion = drift. Answer: Diffusion current arises from majority carriers moving from regions of high concentration to low concentration across the junction, producing a net current until equilibrium (balanced by drift) is reached.
2-Mark Questions
Define forbidden energy gap.
In semiconductors, the band gap Eg separates filled valence band and empty conduction band; Eg determines electrical and optical properties. Answer: The forbidden energy gap (band gap) is the energy difference between the conduction band minimum and the valence band maximum in a solid; electrons cannot have energies in this range.
Why is temperature co-efficient of resistance negative for semiconductor?
In semiconductors conductivity σ increases rapidly with temperature (n ∝ e^{-Eg/2kT}), so resistance falls; hence temperature coefficient is negative. Answer: Because increasing temperature produces more electron–hole pairs (increases carrier concentration) which decreases the resistance.
What do you mean by doping?
Donor impurities (pentavalent) provide extra electrons (n-type); acceptor impurities (trivalent) create holes (p-type). Answer: Doping is the intentional introduction of impurity atoms into an intrinsic semiconductor to change its electrical properties (create n-type or p-type material).
1-Mark Questions (MCQ)
Write down Boolean equation for the output Y of the given circuit and give its truth table. [Ans: Y = (AB) + (A + B)]
Simplify the expression: (AB) + (A + B) = A + B (since A + B already covers AB). Truth table for A,B: A B | AB | A+B | Y 0 0 | 0 | 0 | 0 0 1 | 0 | 1 | 1 1 0 | 0 | 1 | 1 1 1 | 1 | 1 | 1 Thus Y equals A + B (OR operation). The derived expression matches the book answer; the truth table is above. Answer: Y = (AB) + (A + B)
Ch 11Recent Developments In Physics
5-Mark Questions
Distinguish between Nanoscience and Nanotechnology.
Nanoscience: investigates size-dependent physical, chemical and biological behaviours at nanoscale (e.g., quantum confinement, high surface-to-volume ratio). Nanotechnology: uses this knowledge to design and fabricate devices (e.g., nanoelectronics, drug-delivery nanoparticles). Answer: Nanoscience is the study of phenomena and properties of materials at the nanoscale (approximately 1–100 nm); it is focused on fundamental understanding. Nanotechnology is the practical application and engineering of nanoscale materials and devices to create useful products and systems.
What is the difference between Nano materials and Bulk materials?
Differences: (1) Surface area: nanomaterials have much larger surface-area-to-volume ratio leading to higher reactivity; (2) Electronic/optical properties: quantum confinement alters band structure in nanomaterials; (3) Mechanical/thermal properties can differ at nanoscale; (4) Bulk materials do not show these size-specific effects. Answer: Nanomaterials have at least one dimension in the nanoscale (1–100 nm) and show size-dependent properties (quantum effects, large surface-to-volume ratio). …
2-Mark Questions
Give any two examples for “Nano” in nature.
Other examples include butterfly wings (structural colour), gecko foot pads (nanoscale setae for adhesion) and diatom silica shells (nanoscale architectures). Answer: Examples: (1) Peacock feather — nanoscale photonic structures produce brilliant colours. (2) Lotus leaf — nanoscale roughness causes superhydrophobic, self-cleaning behaviour.
Why steel is preferred in making Robots?
Steel's mechanical properties (high yield strength, toughness) and availability make it ideal where structural rigidity and longevity are required; aluminium may be used where lower weight is important. Answer: Steel provides high strength, stiffness and durability at reasonable cost; it is easily machined and welded, making it suitable for robot frames and load-bearing parts.
1-Mark Questions (MCQ)
Comment on the recent advancement in medical diagnosis and therapy.
Example: targeted nanoparticle delivery can concentrate chemotherapeutic drugs at the tumor site, reducing systemic side effects; AI algorithms help radiologists detect abnormalities faster and more accurately. Answer: Recent advances include: Precision medicine and genomics: treatments tailored to a patient's genetic profile (e.g., targeted cancer therapies) and gene-editing tools like CRISPR for potential therapies. …
Frequently asked questions
- What is meant by quantisation of charges?
- According to the principle of quantisation of charges, the net charge ( $q$ ) of any object in the universe can only be an integral multiple of the fundamental unit of electric charge ( $e$ ). $$\text{Formula: } q = ne$$ Where: $q$ is the total charge of the body. $n$ is any integer ( $\dots, -3, -2, -1, 0, 1, 2, 3, \dots$ ). $e$ is the fundamental unit of charge (the charge of an electron or proton), which has a magnitude of approximately $1.6 \times 10^{-19}\text{ C}$ . …
- Write down Coulomb’s law in vector form and mention what each term represents.
- Coulomb's law states that the electrostatic force between two point charges at rest is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. $$\text{Vector Form: } \vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_{12}^2} \hat{r}_{12}$$ Representation of terms: $\vec{F}_{21}$ : Electrostatic force exerted on charge $q_2$ by charge $q_1$ . $q_1, q_2$ : Magnitudes of the two interacting point charges. $r_{12}$ : Center-to-center distance between the charges $q_1$ and $q_2$ . …
- Why current is a scalar?
- Current I = dQ/dt; charge Q is a scalar. Although it has a direction of flow, current adds algebraically (signed) and does not follow vector addition, so it is treated as a scalar in circuit theory. Answer: Current is a scalar because it represents the rate of charge flow (a scalar quantity) through a cross-section per unit time and direction is accounted for by sign convention; it does not obey vector addition rules.
- Distinguish between drift velocity and mobility.
- Drift velocity depends on field strength and carrier scattering. Mobility μ = v_d/E (units m^2 V^{-1} s^{-1}) characterizes how easily carriers move in a material; higher μ means larger v_d for a given E. Answer: Drift velocity v_d is the average velocity acquired by charge carriers under an electric field. Mobility μ is the proportionality constant between drift velocity and electric field: v_d = μE.
These important questions are selected from the Samacheer Kalvi Class 12 Physics textbook book-back exercises to help you revise the most useful questions. Mark weightage (5/2/1) follows the usual exam pattern and may vary by exam — always check your latest syllabus and question pattern. Open each chapter for the complete set of questions and answers.