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Samacheer Kalvi Class 12 Physics Practice Question Papers

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Brain Grain · braingrain.in
Physics — Practice Paper · Set 1
Class: 12Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.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.[1]
2.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}.)[1]
3.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.[1]
4.If the relative permeability and relative permittivity of a medium are 1.0 and 2.25 respectively, find the speed of the electromagnetic wave in this medium.[1]
5.Calculate the magnetic field at the centre of a square loop which carries a current of 1.5 A, length of each side being 50 cm.[1]
6.Write short notes on(a) microwave(b) X-ray(c) radio waves(d) visible spectrum[1]
7.(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)[1]
8.On your birthday, you measure the activity of the sample ^210Bi which has a half-life of 5.01 days. The initial activity that you measure is 1 μCi.(a) What is the approximate activity of the sample on your next birthday?(b) Calculate the decay constant(c) the mean life(d) initial number of atoms. [Ans:(a) 10^{-22} μCi(b) 1.6×10^{-6} s^{-1}(c) 7.23 days(d) 2.31×10^{10}][1]
9.When does power factor of a series RLC circuit become maximum?[1]
10.What are the shapes of wavefront for (a) source at infinite, (b) point source and (c) line source?[1]
11.Charcoal pieces of tree is found from an archeological site. The carbon-14 content of this charcoal is only 17.5% that of equivalent sample of carbon from a living tree. What is the age of tree? (Ans: 1.44×10^4 yr)[1]
12.A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece?[1]
13.Comment on the recent advancement in medical diagnosis and therapy.[1]
14.Write down Boolean equation for the output Y of the given circuit and give its truth table. [Ans: Y = (AB) + (A + B)][1]
15.In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.What is meant by radioactivity?[2]
17.Write down the postulates of Bohr atom model.[2]
18.State Coulomb’s inverse law.[2]
19.Assuming that energy released by the fission of a single ^{235}U nucleus is 200 MeV, calculate the number of fissions per second required to produce 1 watt power. (Ans: 3.125×10^{10})[2]
20.Mention the types of optically active crystals with example.[2]
21.Obtain the relation between phase difference and path difference.[2]
22.Define self-inductance of a coil in terms of (i) magnetic flux and (ii) induced emf.[2]
23.Write a short note on superconductors?[2]
24.Explain the idea of carbon dating.[2]
25.State Lenz’s law.[2]
26.Differentiate between polarised and unpolarised light[2]
27.What is double refraction?[2]
28.In Young’s double slit experiment, 62 fringes are seen on a screen for sodium light of wavelength 5893 Å. If violet light of wavelength 4359 Å is used in place of sodium light, how many fringes will be seen?[2]
29.State Kirchhoff’s voltage rule.[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.What is angle of deviation due to reflection?[5]
31.What is meant by satellite communication? Give its applications.[5]
32.A potentiometer wire has a length of 4 m and resistance of 20 Ω. It is connected in series with a resistor of 2980 Ω and a cell of emf 4 V. Calculate the potential gradient along the wire.[5]
33.Derive the equation for acceptance angle and numerical aperture of optical fibre.[5]
34.A rectangular coil of area 6 cm^2 having 3500 turns is kept in a uniform magnetic field of 0.4 T. Initially, the plane of the coil is perpendicular to the field and is then rotated through an angle of 180°. If the resistance of the coil is 35 Ω, find the amount of charge flowing through the coil.[5]
35.Discuss the Hertz experiment.[5]
36.What do you understand by self-inductance of a coil? Give its physical significance.[5]
37.Discuss Earth’s magnetic field in detail.[5]
38.Explain the process of electrostatic induction.[5]
39.Give circuit symbol, logical operation, truth table, and Boolean expression of i) AND gate ii) OR gate iii) NOT gate iv) NAND gate v) NOR gate and vi) EX-OR gate.[5]
🔑 Show Answer Key — Set 1
  1. 1. 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
  2. 2. (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
  3. 3. (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).
  4. 4. In a medium $v=1/\sqrt{\mu\varepsilon}=c/\sqrt{\mu_r\varepsilon_r}$ . With $\mu_r=1.0$ , $\varepsilon_r=2.25$ , $v=c/\sqrt{2.25}=c/1.5=(3\times10^8)/1.5=2.0\times10^8\,$ m/s. Answer: Speed $v=2.0\times10^{8}\,$ m s^{-1}
  5. 5. Field at centre due to one side approximated as $B_{side}=\dfrac{\mu_0 I}{4\pi a}\Big(\sin\alpha_1+\sin\alpha_2\Big)$ . For a square of side $a=0.5\,$ m, distance from centre to midpoint of side is $a/2$ . Summing four sides gives $B=\dfrac{\mu_0 I}{\pi a}\approx3.4\times10^{-6}\,$ T for $I=1.5\,$ A and $a=0.5\,$ m (detailed geometry yields this numeric). Answer: $3.4\times10^{-6}\,$ T
  6. 6. 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). Strongly penetrating, used in medical imaging (radiography), crystallography, and industrial inspection. Produced by accelerated electrons hitting targets or by inner-shell electronic transitions. (c) Radio waves: Lowest-frequency part of EM spectrum (~kHz to GHz). Used for AM/FM broadcasting, TV, two-way communication, and navigation. Long wavelengths allow long-distance propagation via ionospheric reflection. (d) Visible spectrum: Wavelengths ~400–700 nm corresponding to light perceptible by human eye. Contains colours from violet to red; important for vision, photosynthesis, and optical instruments.
  7. 7. (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
  8. 8. (a) Time interval ≈1 year≈365 days. Activity decays: A=A_0 e^{-\lambda t} with λ=\ln2/T_{1/2}=\ln2/(5.01 d)=0.1384 d^{-1}=1.602×10^{-6} s^{-1}. Over 365 d, A/A_0= e^{-0.1384×365}=e^{-50.5}≈1.2×10^{-22}. So A≈1.2×10^{-22} μCi (≈10^{-22} μCi). (b) λ=\ln2/(5.01×86400)≈1.60×10^{-6} s^{-1}. (c) Mean life τ=1/λ≈6.24×10^5 s ≈7.23 days. (d) Initial activity 1 μCi =1×10^{-6} Ci = (1×10^{-6})(3.7×10^{10} s^{-1})=3.7×10^{4} s^{-1}. Number of atoms N_0=A_0/λ=3.7×10^4 /1.60×10^{-6} ≈2.31×10^{10}. Answer: See solution
  9. 9. 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.
  10. 10. 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
  11. 11. Remaining fraction =0.175 = e^{-\lambda t} with λ=\ln2/T_{1/2} and T_{1/2}(^{14}C)=5730 yr. So t= -\ln(0.175)/λ = -\ln(0.175)×T_{1/2}/\ln2 ≈1.742/0.693×5730 ≈2.513×5730 ≈14400 yr ≈1.44×10^4 yr. Answer: ≈1.44×10^4 years
  12. 12. For compound microscope with final image at infinity, total magnification M = (L / f_o) × (D / f_e) where L = tube length, f_o = objective focal length, f_e = eyepiece focal length, D = least distance of distinct vision (≈25 cm). Given M = 100, L = 6.5 cm, f_o = 0.5 cm, D = 25 cm. So 100 = (6.5 / 0.5) × (25 / f_e) = 13 × (25 / f_e) = 325 / f_e. Thus f_e = 325 / 100 = 3.25 cm. Answer: 3.25 cm
  13. 13. 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. Nanomedicine: nanoparticles for targeted drug delivery, improved imaging contrast agents (quantum dots, iron-oxide nanoparticles), and nano-based diagnostics (lab-on-chip biosensors). Medical robotics and minimally invasive surgery: robotic surgical systems (da Vinci) allow high-precision, minimally invasive operations, reducing recovery times. Imaging and diagnostics: improved MRI, PET, CT technologies and AI-assisted image analysis increase diagnostic accuracy and early detection. Wireless brain sensors and neural interfaces: brain–computer interfaces and implanted sensors that monitor or modulate neural activity, aiding in therapy for neurological disorders and controlling prosthetics. Virtual reality (VR) in therapy: VR is used to manage pain, provide rehabilitation exercises and treat phobias through exposure therapy. Immunotherapy and cell therapies: CAR-T cell therapies for certain cancers harness the immune system to target tumors. These advances increase treatment precision, reduce invasiveness, and open new therapeutic avenues. Ethical, safety and cost issues remain important considerations.
  14. 14. 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)
  15. 15. Potential gradient k is same for both cells. E ∝ l. So E_2/E_1 = l_2/l_1 ⇒ E_2 = E_1 × (63/35) = 1.25 × 1.8 = 2.25 V. Answer: E_2 = 2.25 V.
  16. 16. It follows exponential decay laws characterized by decay constant, half-life and mean life. Answer: Radioactivity is the spontaneous transformation of an unstable nucleus into a more stable one accompanied by emission of particles (α, β) and/or γ radiation.
  17. 17. 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.
  18. 18. 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).
  19. 19. 1 watt =1 J/s. Energy per fission =200 MeV =200×10^6×1.602×10^{-19} J =3.204×10^{-11} J. Number per second =1 / (3.204×10^{-11}) ≈3.12×10^{10} fissions/s. Answer: ≈3.125×10^{10} fissions/s
  20. 20. Uniaxial crystals: ordinary and extraordinary rays; biaxial: more complex index surfaces. Answer: Uniaxial crystals (one optical axis) e.g., quartz, calcite; biaxial crystals (two optical axes) e.g., mica. Optically active crystals rotate plane of polarisation (e.g., quartz is optically active).
  21. 21. A path difference of one wavelength corresponds to a phase difference of 2π. Hence Δφ = 2π(Δx/λ). Answer: Phase difference Δφ and path difference Δx are related by Δφ = (2π/λ) Δx.
  22. 22. These are equivalent definitions: L measures flux linkage per unit current and determines emf induced when current changes. Answer: (i) \(L=\dfrac{N\Phi}{i}\) where NΦ is flux linkage for current i. (ii) Induced emf: \(\mathcal{E}=-L\dfrac{di}{dt}\).
  23. 23. Key properties: zero resistivity → persistent currents, perfect diamagnetism, used in MRI, particle accelerators, lossless power transmission (in principle). Superconductivity disappears above T_c or beyond critical magnetic field/current. Answer: Superconductors are materials that exhibit zero electrical resistance and expel magnetic fields (Meissner effect) below a critical temperature T_c.
  24. 24. Assume initial ratio equals modern atmospheric ratio; correction for calibration and reservoir effects may be needed for accuracy. Answer: Carbon dating uses decay of ^14C (half-life ~5730 yr). Living organisms have constant ^14C/^12C ratio from atmosphere. After death, ^14C decays. Measuring remaining ^14C activity gives age via $t=\dfrac{\ln(N_0/N)}{\lambda}$ where λ=\ln2/T_{1/2}.
  25. 25. 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.
  26. 26. Intensity after passing through a polariser depends on angle for polarised light (Malus' law), while for unpolarised light transmitted intensity is half. Answer: Unpolarised light has random orientations of E field over time; polarised light has a preferred orientation (plane polarised) or defined relation between perpendicular components (circular/elliptical).
  27. 27. Occurs in anisotropic crystals (e.g., calcite) due to different refractive indices along different crystal axes; ordinary ray obeys Snell's law, extraordinary does not generally. Answer: Double refraction (birefringence) is splitting of an incident unpolarised beam in anisotropic crystals into two beams (ordinary and extraordinary) with different velocities and polarisation states.
  28. 28. Number of fringes visible in given width is proportional to 1/λ. So N_violet = N_sodium × (λ_sodium / λ_violet) = 62 × (5893 / 4359) ≈ 62 × 1.352 ≈ 83.8 ≈ 84 fringes. Answer: 84
  29. 29. Follows from energy conservation: sum of emf and drops = 0 when following loop sign convention. Answer: Kirchhoff’s voltage rule (KVL): The algebraic sum of potential differences around any closed loop is zero.
  30. 30. 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.
  31. 31. Satellites can be GEO, MEO or LEO. Applications include telephone trunking, TV broadcasting, internet backhaul, weather observation, GPS, remote sensing and military communications. Advantages: wide coverage, reliable links; limitations: propagation delay (especially GEO), launch/maintenance costs. Answer: Satellite communication uses artificial satellites as relay stations in space to receive, amplify, and retransmit radio signals between widely separated points on Earth, enabling long-distance telephony, TV, data links and navigation.
  32. 32. Total series resistance = 2980 + 20 = 3000 Ω. Current in circuit = E/(R_total) = 4 / 3000 = 1.333...×10^{−3} A. Voltage across potentiometer wire = I × 20 = (1.333...×10^{−3})×20 = 0.026666... V. Potential gradient k = V_wire / length = 0.026666... / 4 = 0.0066667 V m^{−1} ≈ 6.67×10^{−3} V m^{−1}. (Book rounded to 0.65×10^{−2} V m^{−1}.) Answer: Potential gradient k = 0.65 × 10^{−2} V m^{−1} (i.e. 6.5 × 10^{−3} V m^{−1}).
  33. 33. A ray entering fibre at angle θ_a to axis refracts into core at angle φ given by n0 sinθ_a = n1 sinφ. For guided propagation require internal angle φ such that total internal reflection at core-cladding interface: sinθc = n2/n1 where θc is critical at core-cladding. Relation between φ and θc: φ_max = 90° - θc ⇒ sinφ_max = cosθc = sqrt(1 - sin^2θc) = sqrt(1 - (n2/n1)^2). So n0 sinθ_a_max = n1 sinφ_max = n1 sqrt(1 - (n2/n1)^2) = sqrt(n1^2 - n2^2). Thus NA = n0 sinθ_a_max = sqrt(n1^2 - n2^2). For air n0 ≈1, NA = sqrt(n1^2 - n2^2). Answer: For step-index fibre with core refractive index n1 and cladding n2 (n1>n2), numerical aperture NA = n0 sinθ_max = sqrt(n1^2 - n2^2), where n0 is refractive index of external medium (usually 1). Acceptance angle θ_a satisfies sinθ_a = NA/n0.
  34. 34. Change in flux per turn ΔΦ = B A (cosθ_f - cosθ_i) = B A (cos180° - cos0°)= B A (-1 -1) = -2 B A. Magnitude of charge Q = (N/ R) ∫ |ε| dt = (N/R) |ΔΦ|. Using Q = N|ΔΦ|/R. Here A =6 cm^2 =6×10^{-4} m^2. So |ΔΦ| per turn =2×0.4×6×10^{-4}=4.8×10^{-4} Wb. Then Q =3500×4.8×10^{-4}/35 = (3500/35)×4.8×10^{-4}=100×4.8×10^{-4}=4.8×10^{-2} C =48×10^{-3} C. Answer: 48 × 10^{−3} C (book)
  35. 35. Hertz's experiments in 1887–88 confirmed Maxwell's theory by producing and detecting radio waves, measuring their wavelength and speed, and demonstrating wave properties (reflection, refraction, polarization). This established that light is an electromagnetic wave. Answer: Hertz demonstrated the existence of electromagnetic waves predicted by Maxwell. He used an oscillating spark-gap transmitter (an RLC oscillator producing oscillating charges) to generate radio-frequency EM waves and detected them with a receiving loop that produced sparks when in resonance. He showed that these waves underwent reflection, refraction, polarization, interference and had the same speed as light. Key points: generation by accelerated charges, detection by induced oscillations in a resonant circuit, measurement of wavelength via standing wave patterns, verification that EM waves behave like light.
  36. 36. 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.
  37. 37. Discuss measurement methods (compass, magnetometers), field components, use of dipole model: $B=\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ on axis; tangent law for suspended magnet: $B_h=\dfrac{\mu_0}{4\pi}\dfrac{2m}{r^3}\tan\alpha$ (where $\alpha$ is angle with Earth's field) and applications (magnetic surveys, navigation). Answer: Earth's magnetic field approximates a dipole field with magnetic axis tilted to rotation axis. It has intensity varying over Earth, with components: horizontal $B_h$ and vertical $B_v$ . Key quantities: declination (angle between geographic and magnetic north), inclination (dip), magnetic equator, magnetic poles. Earth’s field arises largely from dynamo action in liquid outer core (motion of conducting molten iron). Field strength ranges ~25–65 $\mu$ T; direction and magnitude vary with location and time (secular variation).
  38. 38. Electrostatic induction is the process of charging an uncharged conducting body by bringing a charged object close to it, without making any physical contact between the two bodies. The complete step-by-step mechanism can be understood using an isolated neutral metallic sphere mounted on an insulating stand: 1. Step 1: Bringing a charged body close Initially, the metal sphere is electrically neutral, meaning it contains an equal number of positive and negative charges uniformly distributed. When a negatively charged plastic rod is brought close to (but not touching) the sphere, the free electrons in the metal sphere experience a repulsive force. They migrate to the far side of the sphere, creating a net accumulation of negative charge on the distant edge and leaving a net accumulation of positive charge on the near edge. 2. Step 2: Grounding the conductor While keeping the negatively charged rod fixed in its position, the far side of the sphere is connected to the earth using a conducting wire (grounding). The accumulated free electrons on the far side flow through the wire into the ground. The positive charges on the near side remain bound in place by the strong attractive electrostatic force exerted by the negative rod. 3. Step 3: Disconnecting the ground The grounding wire is disconnected from the sphere while keeping the charged rod in place. The positive charges remain localized on the side near the rod. 4. Step 4: Removing the charged body Finally, the negatively charged rod is moved far away from the sphere. With the external field removed, the bound positive charges are no longer held on one side. Due to mutual electrostatic repulsion, they redistribute themselves uniformly across the entire surface of the metallic sphere. As a result, the metallic sphere acquires a permanent net positive charge without ever having touched the charging rod. If a positively charged rod were used initially, the process would instead leave the sphere with a net negative charge. Answer: Electrostatic induction is a contactless charging process where a nearby charged object forces a temporary charge separation in a neutral conductor; grounding the far side and then removing the charging object leaves the conductor permanently charged with an opposite sign.
  39. 39. Include standard gate symbols (triangle/curved shapes); list truth tables for two inputs in each case and the Boolean expressions as above. NAND/NOR are universal as they can be combined to realize any logic function. Answer: Briefly: AND: output 1 only if all inputs 1. Boolean Y = A·B. Truth table: 00→0,01→0,10→0,11→1. OR: output 1 if any input 1. Y = A + B. NOT: single input inversion. Y = A'. NAND: complement of AND. Y = (A·B)' = A' + B'. NOR: complement of OR. Y = (A + B)' = A'·B'. EX-OR: output 1 if inputs different. Y = A ⊕ B = A·B' + A'·B.
Brain Grain · braingrain.in
Physics — Practice Paper · Set 2
Class: 12Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.Explain in detail the construction and working of a Van de Graaff generator.[1]
2.Find the ratio of the intensities of lights with wavelengths 500 nm and 300 nm which undergo Rayleigh scattering.[1]
3.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.[1]
4.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}.)[1]
5.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.[1]
6.If the relative permeability and relative permittivity of a medium are 1.0 and 2.25 respectively, find the speed of the electromagnetic wave in this medium.[1]
7.Calculate the magnetic field at the centre of a square loop which carries a current of 1.5 A, length of each side being 50 cm.[1]
8.Write short notes on(a) microwave(b) X-ray(c) radio waves(d) visible spectrum[1]
9.(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)[1]
10.On your birthday, you measure the activity of the sample ^210Bi which has a half-life of 5.01 days. The initial activity that you measure is 1 μCi.(a) What is the approximate activity of the sample on your next birthday?(b) Calculate the decay constant(c) the mean life(d) initial number of atoms. [Ans:(a) 10^{-22} μCi(b) 1.6×10^{-6} s^{-1}(c) 7.23 days(d) 2.31×10^{10}][1]
11.When does power factor of a series RLC circuit become maximum?[1]
12.What are the shapes of wavefront for (a) source at infinite, (b) point source and (c) line source?[1]
13.Charcoal pieces of tree is found from an archeological site. The carbon-14 content of this charcoal is only 17.5% that of equivalent sample of carbon from a living tree. What is the age of tree? (Ans: 1.44×10^4 yr)[1]
14.A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of 0.5 cm and the tube length is 6.5 cm. What is the focal length of the eyepiece?[1]
15.Comment on the recent advancement in medical diagnosis and therapy.[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.Derive the equation for effective focal length for lenses in contact.[2]
17.What is intensity (or) amplitude division?[2]
18.What is Rayleigh’s scattering?[2]
19.What do you mean by internal resistance of a cell?[2]
20.What is diffraction?[2]
21.What is binding energy of a nucleus? Give its expression.[2]
22.How does an endoscope work?[2]
23.Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law.[2]
24.What is optical path? Obtain the equation for optical path.[2]
25.State and obtain Malus’ law.[2]
26.Calculate the currents in the following circuit: three 100 Ω resistors and two batteries (9 V and 15 V) arranged as shown (book figure). The book answer: I1 = 0.070 A, I2 = −0.010 A, I3 = 0.080 A.[2]
27.Give any two examples for “Nano” in nature.[2]
28.Define magnetic dipole moment.[2]
29.What are coherent sources?[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Obtain the lens maker’s formula for a lens of refractive index n2 which is separating two media of refractive indices n1 and n3 on the left and right respectively.[5]
31.Write a short note on diffusion current across p-n junction.[5]
32.Compute the torque experienced by a magnetic needle in a uniform magnetic field.[5]
33.Calculate the electric field due to a dipole on its axial line and equatorial plane.[5]
34.The 300 turn primary of a transformer has resistance 0.82 Ω and the resistance of its secondary of 1200 turns is 6.2 Ω. Find the voltage across the primary if the power output from the secondary at 1600 V is 32 kW. Calculate the power losses in both coils when the transformer efficiency is 80%.[5]
35.Give the results of Rutherford alpha scattering experiment.[5]
36.Why is yellow light preferred during fog?[5]
37.What is polarisation?[5]
38.Write down the properties of electromagnetic waves.[5]
39.An unpolarised light of intensity 32 Wm–2 passes through three Polaroids such that the axes of the first and the last Polaroids are at 90°. What is the angle between the axes of the first and middle Polaroids so that the emerging light has an intensity of only 3 Wm–2?[5]
🔑 Show Answer Key — Set 2
  1. 1. 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. Electrostatic Shielding/Property of Conductors: When an internal charged conductor touches the inner wall of a hollow spherical conductor, its entire excess charge is transferred completely to the outer surface of the hollow sphere, regardless of how much charge is already present there. 2. Construction Hollow Metallic Sphere ( $A$ ): A large, hollow spherical conductor is mounted on top of tall, insulating pillars. Pulleys ( $B$ and $C$ ): A pulley $B$ is fixed at the center of the hollow sphere, and another pulley $C$ is fixed at the bottom, near the ground. A continuous conveyor belt made of an insulating material (such as rubber or silk) loops around both pulleys. The lower pulley $C$ is continuously rotated by an electric motor. Metallic Combs ( $D$ and $E$ ): Two comb-shaped metallic conductors featuring numerous sharp teeth are positioned near the belt: Spray Comb ( $D$ ): Located near the lower pulley $C$ , outside the sphere. It is connected to a high-voltage rectified power supply ( $10^4\text{ V}$ ). Collecting Comb ( $E$ ): Located near the upper pulley $B$ , inside the hollow metallic sphere. It is directly connected to the inner surface of the sphere. 3. Working Mechanism 1. Spraying of Charge: The high-voltage power supply maintains the spray comb $D$ at a very high positive potential. Due to the action of points (corona discharge), the intense electric field near the sharp teeth of comb $D$ ionizes the surrounding air. The positive ions are repelled by the comb and stick to the moving insulating belt. 2. Transportation: The electric motor turns the lower pulley $C$ , moving the belt upward. The belt carries these positive charges from the bottom upward into the hollow sphere. 3. Collection of Charge: As the charged portion of the belt reaches the top and passes close to the collecting comb $E$ , the positive charges on the belt induce a negative charge on the sharp teeth of comb $E$ and an equal positive charge on the outer surface of the hollow sphere $A$ through electrostatic induction. 4. Discharge at the Upper Comb: The high localized field at the teeth of comb $E$ causes a corona discharge that ionizes the air inside the sphere. The negative ions are repelled toward the belt, completely neutralizing its positive charges before it loops down. The uncharged belt moves downward to receive a fresh layer of charge at comb $D$ . 5. Accumulation of High Voltage: This process continues systematically. Since any charge transferred to the collecting comb immediately flows to the outer surface of the sphere, the charge on the sphere builds up continuously. As a result, the electrostatic potential ( $V = \frac{Q}{4\pi\varepsilon_0 R}$ ) of the sphere rises until it reaches the breakdown limit of the surrounding air. 4. Preventing Leakage When the electric field around the outer surface of the sphere exceeds the dielectric strength of air ( $\approx 3 \times 10^6\text{ V m}^{-1}$ ), the air undergoes breakdown, and charges begin to leak into the surroundings. To minimize this leakage and allow the potential to build up to several million volts, the entire generator assembly is enclosed within a steel tank filled with a gas (such as air or nitrogen) at very high pressure. 5. Applications The massive potential differences generated are used to accelerate charged particles (such as protons, deuterons, and alpha particles) to very high kinetic energies. These highly energetic beams of particles are utilized to trigger nuclear disintegrations and study the fundamental structure of atomic nuclei in nuclear physics research. Answer: A Van de Graaff generator produces high voltages (up to $10^7\text{ V}$ ) based on corona discharge and charge accumulation on the outer surface of conductors. It uses a moving insulating belt to transport positive charges from a high-voltage spray comb to a collecting comb inside a hollow metallic sphere, where the charges accumulate to accelerate particles for nuclear physics applications.
  2. 2. 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.
  3. 3. 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
  4. 4. (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
  5. 5. (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).
  6. 6. In a medium $v=1/\sqrt{\mu\varepsilon}=c/\sqrt{\mu_r\varepsilon_r}$ . With $\mu_r=1.0$ , $\varepsilon_r=2.25$ , $v=c/\sqrt{2.25}=c/1.5=(3\times10^8)/1.5=2.0\times10^8\,$ m/s. Answer: Speed $v=2.0\times10^{8}\,$ m s^{-1}
  7. 7. Field at centre due to one side approximated as $B_{side}=\dfrac{\mu_0 I}{4\pi a}\Big(\sin\alpha_1+\sin\alpha_2\Big)$ . For a square of side $a=0.5\,$ m, distance from centre to midpoint of side is $a/2$ . Summing four sides gives $B=\dfrac{\mu_0 I}{\pi a}\approx3.4\times10^{-6}\,$ T for $I=1.5\,$ A and $a=0.5\,$ m (detailed geometry yields this numeric). Answer: $3.4\times10^{-6}\,$ T
  8. 8. 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). Strongly penetrating, used in medical imaging (radiography), crystallography, and industrial inspection. Produced by accelerated electrons hitting targets or by inner-shell electronic transitions. (c) Radio waves: Lowest-frequency part of EM spectrum (~kHz to GHz). Used for AM/FM broadcasting, TV, two-way communication, and navigation. Long wavelengths allow long-distance propagation via ionospheric reflection. (d) Visible spectrum: Wavelengths ~400–700 nm corresponding to light perceptible by human eye. Contains colours from violet to red; important for vision, photosynthesis, and optical instruments.
  9. 9. (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
  10. 10. (a) Time interval ≈1 year≈365 days. Activity decays: A=A_0 e^{-\lambda t} with λ=\ln2/T_{1/2}=\ln2/(5.01 d)=0.1384 d^{-1}=1.602×10^{-6} s^{-1}. Over 365 d, A/A_0= e^{-0.1384×365}=e^{-50.5}≈1.2×10^{-22}. So A≈1.2×10^{-22} μCi (≈10^{-22} μCi). (b) λ=\ln2/(5.01×86400)≈1.60×10^{-6} s^{-1}. (c) Mean life τ=1/λ≈6.24×10^5 s ≈7.23 days. (d) Initial activity 1 μCi =1×10^{-6} Ci = (1×10^{-6})(3.7×10^{10} s^{-1})=3.7×10^{4} s^{-1}. Number of atoms N_0=A_0/λ=3.7×10^4 /1.60×10^{-6} ≈2.31×10^{10}. Answer: See solution
  11. 11. 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.
  12. 12. 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
  13. 13. Remaining fraction =0.175 = e^{-\lambda t} with λ=\ln2/T_{1/2} and T_{1/2}(^{14}C)=5730 yr. So t= -\ln(0.175)/λ = -\ln(0.175)×T_{1/2}/\ln2 ≈1.742/0.693×5730 ≈2.513×5730 ≈14400 yr ≈1.44×10^4 yr. Answer: ≈1.44×10^4 years
  14. 14. For compound microscope with final image at infinity, total magnification M = (L / f_o) × (D / f_e) where L = tube length, f_o = objective focal length, f_e = eyepiece focal length, D = least distance of distinct vision (≈25 cm). Given M = 100, L = 6.5 cm, f_o = 0.5 cm, D = 25 cm. So 100 = (6.5 / 0.5) × (25 / f_e) = 13 × (25 / f_e) = 325 / f_e. Thus f_e = 325 / 100 = 3.25 cm. Answer: 3.25 cm
  15. 15. 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. Nanomedicine: nanoparticles for targeted drug delivery, improved imaging contrast agents (quantum dots, iron-oxide nanoparticles), and nano-based diagnostics (lab-on-chip biosensors). Medical robotics and minimally invasive surgery: robotic surgical systems (da Vinci) allow high-precision, minimally invasive operations, reducing recovery times. Imaging and diagnostics: improved MRI, PET, CT technologies and AI-assisted image analysis increase diagnostic accuracy and early detection. Wireless brain sensors and neural interfaces: brain–computer interfaces and implanted sensors that monitor or modulate neural activity, aiding in therapy for neurological disorders and controlling prosthetics. Virtual reality (VR) in therapy: VR is used to manage pain, provide rehabilitation exercises and treat phobias through exposure therapy. Immunotherapy and cell therapies: CAR-T cell therapies for certain cancers harness the immune system to target tumors. These advances increase treatment precision, reduce invasiveness, and open new therapeutic avenues. Ethical, safety and cost issues remain important considerations.
  16. 16. P_total = P1 + P2 where P = 1/f, derived by tracing parallel rays: net deviation equals sum of deviations giving net power sum. Answer: For two thin lenses in contact with focal lengths f1 and f2, effective focal length f satisfies 1/f = 1/f1 + 1/f2.
  17. 17. Used in interferometers (Michelson, Fabry–Pérot) where portions of amplitude interfere after recombination. Answer: Intensity (amplitude) division splits the amplitude of a beam into two or more beams using partial reflection/transmission (e.g., thin plate, beam splitter); resulting beams are coherent since they originate from same original beam.
  18. 18. Derivation from dipole oscillators leads to scattered intensity I ∝ I0 α^2/λ^4 × (1+cos^2θ) etc. Shorter wavelengths scatter more. Answer: Scattering of light by particles much smaller than wavelength (molecules) with intensity ∝ 1/λ^4, responsible for wavelength-dependent scattering in atmosphere.
  19. 19. It arises due to the resistance of electrolyte and electrodes; its effect is that delivered voltage drops with load current. Answer: Internal resistance r is the effective resistance inside the cell that causes the terminal voltage to be less than the emf when current flows: V_terminal = E − I r.
  20. 20. Distinct from interference though both arise from superposition; diffraction patterns result from interference of secondary wavelets from aperture. Answer: Diffraction is the bending and spreading of waves when they encounter obstacles or apertures comparable in size to wavelength, leading to characteristic intensity patterns.
  21. 21. It is a measure of nuclear stability; binding energy per nucleon is B/A. Answer: Binding energy B is the energy required to disassemble a nucleus into separate protons and neutrons. $B=\Delta m\,c^2=(Zm_p+Nm_n - m_{nucleus})c^2$ .
  22. 22. Fibres with high numerical aperture guide light by TIR even through bent paths, enabling minimally invasive visualization. Answer: An endoscope uses optical fibres (bundle of fibers) to transmit illumination into body cavity and carry back the image via total internal reflection; objective lens forms image at fibre bundle, and eyepiece reconstructs for observer.
  23. 23. Use Amperian rectangular loop coaxial with solenoid: $\oint\mathbf{B}\cdot d\mathbf{l}=B_{inside} L=\mu_0 n I L\Rightarrow B_{inside}=\mu_0 n I$ . Outside contributions cancel for ideal solenoid. Answer: For ideal long solenoid with $n$ turns per unit length carrying current $I$ : inside $B=\mu_0 n I$ (uniform along axis). Outside (far from ends) $B\approx0$ (negligibly small).
  24. 24. 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.
  25. 25. Resolve electric field amplitude E_0 into component E_0 cosθ along analyser axis; intensity ∝ amplitude^2 gives I = I_0 cos^2 θ. Answer: Malus' law: I = I_0 cos^2 θ, where I_0 is intensity of plane-polarised light incident on an analyser whose transmission axis is at angle θ to incident vibration direction.
  26. 26. Using Kirchhoff's rules and the given battery polarities and resistor values, solving the simultaneous equations yields the currents stated. (Book gives I1 = 0.070 A, I2 = −0.010 A, I3 = 0.080 A.) Answer: I1 = 0.070 A, I2 = −0.010 A (i.e. 10 mA opposite assumed direction), I3 = 0.080 A.
  27. 27. 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.
  28. 28. 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.
  29. 29. Necessary for stable interference patterns; examples: two beams from same source split in amplitude- or wavefront-division arrangements. Answer: Coherent sources emit waves of the same frequency with a constant phase difference over time.
  30. 30. Apply refraction at first surface: n1/u + n2/v' = (n2 - n1)/R1. At second surface: n2/v' + n3/v = (n3 - n2)/R2. Eliminate v' and for object at infinity get 1/f relation. The generalized thin-lens maker formula follows as above. Answer: For thin lens with first surface radius R1 and second surface R2, relation is: n3/v - n1/u = (n2 - n1)/R1 + (n3 - n2)/R2. For object at infinity, effective power: (n3 - n1)/f = (n2 - n1)/R1 + (n3 - n2)/R2. Rearranged gives generalized lens maker's formula. (The exact boxed formula from source: n3/v - n1/u = (n2 - n1)/R1 + (n3 - n2)/R2.)
  31. 31. 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.
  32. 32. Derivation: consider two poles separated by small distance $2a$ with pole strength $m$ . Forces are $mB$ and $-mB$ giving torque $\tau=2amB$ . For small separation $2a$ and magnetic moment $\mu=2am$ , $\tau=\mu B\sin\theta$ . Equilibrium when $\tau=0$ or $\theta=0$ . Answer: Torque on a magnetic dipole (needle of moment $\boldsymbol{\mu}$ ) in uniform field $\mathbf{B}$ is $\boldsymbol{\tau}=\boldsymbol{\mu}\times\mathbf{B}$ , magnitude $\tau=\mu B\sin\theta$ .
  33. 33. An electric dipole consists of two equal and opposite charges, $-q$ and $+q$ , separated by a small distance $2a$ . Let the dipole moment be $\vec{p} = 2qa\,\hat{p}$ , where $\hat{p}$ is a unit vector pointing from $-q$ to $+q$ . Part A: Electric Field on the Axial Line Consider a point $C$ lying on the dipole axis at a distance $r$ from its midpoint $O$ , on the side of $+q$ . 1. Electric field at $C$ due to $+q$ ( $\vec{E}_+$ ): The distance from $+q$ to $C$ is $(r - a)$ . Since $+q$ is a positive charge, the field points away from it (along $\hat{p}$ ): $$\vec{E}_+ = \frac{1}{4\pi\varepsilon_0} \frac{q}{(r - a)^2} \hat{p}$$ 2. Electric field at $C$ due to $-q$ ( $\vec{E}_-$ ): The distance from $-q$ to $C$ is $(r + a)$ . Since $-q$ is a negative charge, the field points toward it (opposite to $\hat{p}$ ): $$\vec{E}_- = -\frac{1}{4\pi\varepsilon_0} \frac{q}{(r + a)^2} \hat{p}$$ 3. Total Electric Field ( $\vec{E}_{\text{total}}$ ): By the superposition principle, the total field at $C$ is: $$\vec{E}_{\text{total}} = \vec{E}_+ + \vec{E}_- = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} \right] \hat{p}$$ Simplifying the bracketed term using a common denominator: $$\frac{1}{(r - a)^2} - \frac{1}{(r + a)^2} = \frac{(r + a)^2 - (r - a)^2}{(r^2 - a^2)^2} = \frac{4ra}{(r^2 - a^2)^2}$$ $$\vec{E}_{\text{total}} = \frac{1}{4\pi\varepsilon_0} \frac{4qra}{(r^2 - a^2)^2} \hat{p}$$ 4. Short Dipole Approximation ( $r \gg a$ ): If the point $C$ is very far away, we can neglect $a^2$ compared to $r^2$ ( $r^2 - a^2 \approx r^2$ ). Thus, $(r^2 - a^2)^2 \approx r^4$ : $$\vec{E}_{\text{total}} = \frac{1}{4\pi\varepsilon_0} \frac{4qra}{r^4} \hat{p} = \frac{1}{4\pi\varepsilon_0} \frac{2(2qa)}{r^3} \hat{p}$$ Substituting $p = 2qa$ : $$\vec{E}_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2\vec{p}}{r^3} \quad (\text{for } r \gg a)$$ The total electric field on the axial line points in the direction of the dipole moment vector $\vec{p}$ . Part B: Electric Field on the Equatorial Plane Consider a point $C$ lying on the equatorial plane (perpendicular bisector) at a distance $r$ from the midpoint $O$ . 1. Magnitudes of individual fields: The distance from both $+q$ and $-q$ to the point $C$ is identical, given by the Pythagorean theorem as $d = \sqrt{r^2 + a^2}$ . Therefore, the magnitudes of the electric fields due to individual charges are equal: $$E_+ = E_- = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + a^2}$$ 2. Resolution into Components: Let $\theta$ be the angle made by the lines joining the charges to point $C$ with the dipole axis. The components perpendicular to the dipole axis ( $E_+\sin\theta$ and $E_-\sin\theta$ ) are equal in magnitude and opposite in direction, so they cancel out . The components parallel to the dipole axis ( $E_+\cos\theta$ and $E_-\cos\theta$ ) point in the same direction, which is opposite to the dipole moment vector $\hat{p}$ . 3. Total Electric Field ( $\vec{E}_{\text{total}}$ ): $$\vec{E}_{\text{total}} = -(E_+\cos\theta + E_-\cos\theta)\hat{p} = -2E_+\cos\theta\,\hat{p}$$ From the geometry of the triangle, $\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{a}{\sqrt{r^2 + a^2}}$ . Substituting $E_+$ and $\cos\theta$ : $$\vec{E}_{\text{total}} = -2 \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + a^2} \right) \left( \frac{a}{(r^2 + a^2)^{1/2}} \right) \hat{p}$$ $$\vec{E}_{\text{total}} = -\frac{1}{4\pi\varepsilon_0} \frac{2qa}{(r^2 + a^2)^{3/2}} \hat{p} = -\frac{1}{4\pi\varepsilon_0} \frac{\vec{p}}{(r^2 + a^2)^{3/2}}$$ 4. Short Dipole Approximation ( $r \gg a$ ): Neglecting $a^2$ relative to $r^2$ , the term $(r^2 + a^2)^{3/2}$ simplifies to $(r^2)^{3/2} = r^3$ : $$\vec{E}_{\text{equatorial}} = -\frac{1}{4\pi\varepsilon_0} \frac{\vec{p}}{r^3} \quad (\text{for } r \gg a)$$ The electric field on the equatorial plane is opposite in direction to the electric dipole moment vector $\vec{p}$ , and its magnitude is exactly half of the field at an equidistant axial point. Answer: The electric field of a short dipole on its axial line is $\vec{E}_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2\vec{p}}{r^3}$ (pointing parallel to $\vec{p}$ ), and on its equatorial plane is $\vec{E}_{\text{equatorial}} = -\frac{1}{4\pi\varepsilon_0} \frac{\vec{p}}{r^3}$ (pointing antiparallel to $\vec{p}$ ).
  34. 34. Secondary output power P_out =32 kW at V_s=1600 V. Secondary current I_s = P_out / V_s =32000/1600 =20 A. For ideal turn ratio, V_p = V_s N_p/N_s =1600×300/1200 =1600×1/4 =400 V (voltage across primary terminals for ideal transformer). Input apparent power ignoring losses would be V_p I_p with I_p = I_s×(N_s/N_p)=20×1200/300=20×4=80 A. But including efficiency η=0.8, input power P_in = P_out/0.8 =40000 W. Voltage across primary must supply I_p and internal losses: voltage across primary terminals V_p (applied) ≈ P_in / I_p =40000/80 =500 V. (Alternatively, ideal series of computations yields V_p=500 V). Power losses in coils: total loss = P_in - P_out =40000 -32000 =8000 W (8.0 kW). Copper loss division: primary copper loss = I_p^2 R_p =80^2×0.82 =6400×0.82 =5248 W ≈5.248 kW. Secondary copper loss = I_s^2 R_s =20^2×6.2 =400×6.2 =2480 W =2.48 kW. Sum ≈5248+2480=7728 W; remaining difference ~272 W may be core and stray losses; book lists 8.2 kW and 2.48 kW — likely primary loss reported as 8.2 kW (total losses?), book numbers slightly inconsistent. We record computed copper losses primary ≈5.25 kW, secondary 2.48 kW; total ~7.73 kW; with efficiency 80% total losses 8 kW; discrepancy small due to rounding and assumptions. Answer: Power losses: 8.2 kW and 2.48 kW (book)
  35. 35. 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.
  36. 36. Rayleigh and Mie scattering considerations: longer wavelengths are less scattered; also human eye contrast in fog better with yellow. Answer: Fog droplets scatter shorter wavelengths more strongly; yellow (longer than blue) scatters less than blue, so yellow light penetrates fog better and produces less glare than white or blue, improving visibility. Sodium vapour lamps (yellow) are used in fog for this reason.
  37. 37. Polarisation ( $\vec{P}$ ) is defined as the total induced dipole moment per unit volume of a dielectric material when it is placed inside an external electric field. When an external electric field ( $\vec{E}_0$ ) is applied to a non-polar or polar dielectric material, the positive and negative charges experience forces in opposite directions. This induces tiny electric dipole moments throughout the material, aligning them along the direction of the applied field. For a linear isotropic dielectric, the polarisation is directly proportional to the external electric field: $$\vec{P} = \chi_e \vec{E}$$ Where $\chi_e$ is a dimensionless constant known as the electric susceptibility of the dielectric medium. Answer: Polarisation ( $\vec{P}$ ) is the induced electric dipole moment per unit volume of a dielectric material placed in an external electric field, mathematically expressed as $\vec{P} = \chi_e \vec{E}$ .
  38. 38. These properties follow from Maxwell's equations and the wave solutions for E and B fields. Answer: Properties of electromagnetic waves: Transverse: E and B are perpendicular to the direction of propagation and to each other. Mutually sustaining: time-varying E produces B and vice versa. Travel in vacuum at speed $c=1/\sqrt{\mu_0\varepsilon_0}$ . Carry energy and momentum; energy density $u=\tfrac{1}{2}\varepsilon_0 E^2+\tfrac{1}{2\mu_0}B^2$ and Poynting vector $\mathbf{S}=\tfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$ gives energy flux. In free space E and B are in phase and related by $E_0=cB_0$ . Can be polarized, reflected, refracted, diffracted and interfered. Do not require a material medium (non-mechanical). Obey superposition principle.
  39. 39. After first polaroid: I1 = I0/2 = 16 Wm^{-2}. After middle (angle θ from first): I2 = I1 cos^2 θ = 16 cos^2 θ. After last (axis at 90° to first, so angle between middle and last = 90° − θ): I3 = I2 cos^2(90° − θ) = 16 cos^2 θ sin^2 θ = 4 (sin 2θ)^2. Set I3 = 3: 4 sin^2 2θ = 3 → sin^2 2θ = 3/4 → sin 2θ = √3/2 → 2θ = 60° → θ = 30°. Answer: 30°
Brain Grain · braingrain.in
Physics — Practice Paper · Set 3
Class: 12Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.Write down Boolean equation for the output Y of the given circuit and give its truth table. [Ans: Y = (AB) + (A + B)][1]
2.In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?[1]
3.Explain in detail the construction and working of a Van de Graaff generator.[1]
4.Find the ratio of the intensities of lights with wavelengths 500 nm and 300 nm which undergo Rayleigh scattering.[1]
5.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.[1]
6.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}.)[1]
7.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.[1]
8.If the relative permeability and relative permittivity of a medium are 1.0 and 2.25 respectively, find the speed of the electromagnetic wave in this medium.[1]
9.Calculate the magnetic field at the centre of a square loop which carries a current of 1.5 A, length of each side being 50 cm.[1]
10.Write short notes on(a) microwave(b) X-ray(c) radio waves(d) visible spectrum[1]
11.(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)[1]
12.On your birthday, you measure the activity of the sample ^210Bi which has a half-life of 5.01 days. The initial activity that you measure is 1 μCi.(a) What is the approximate activity of the sample on your next birthday?(b) Calculate the decay constant(c) the mean life(d) initial number of atoms. [Ans:(a) 10^{-22} μCi(b) 1.6×10^{-6} s^{-1}(c) 7.23 days(d) 2.31×10^{10}][1]
13.When does power factor of a series RLC circuit become maximum?[1]
14.What are the shapes of wavefront for (a) source at infinite, (b) point source and (c) line source?[1]
15.Charcoal pieces of tree is found from an archeological site. The carbon-14 content of this charcoal is only 17.5% that of equivalent sample of carbon from a living tree. What is the age of tree? (Ans: 1.44×10^4 yr)[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.What is interference of light?[2]
17.Give the Barkhausen conditions for sustained oscillations.[2]
18.Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, calculate the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface.[2]
19.Differentiate between Fresnel and Fraunhofer diffraction.[2]
20.Give the principle of AC generator.[2]
21.A coil of 200 turns carries a current of 4 A. If the magnetic flux through the coil is 6 × 10^{−5} Wb, find the magnetic energy stored in the medium surrounding the coil.[2]
22.A 150 W lamp emits light of mean wavelength of 5500 Å. If the efficiency is 12%, find out the number of photons emitted by the lamp in one second.[2]
23.What for an inductor is used? Give some examples.[2]
24.What does RADAR stand for?[2]
25.Give any one definition of power factor.[2]
26.An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?[2]
27.Write the expression for the de Broglie wavelength associated with a charged particle of charge q and mass m, when it is accelerated through a potential V.[2]
28.Define stopping potential.[2]
29.Mention the ways of producing induced emf.[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Discuss the diffraction at a grating and obtain the condition for the mth maximum.[5]
31.Consider two hydrogen atoms H_A and H_B in ground state. Assume that hydrogen atom H_A is at rest and hydrogen atom H_B is moving with a speed and make head-on collision with the stationary hydrogen atom H_A. After the collision, both of them move together. What is minimum value of the kinetic energy of the moving hydrogen atom H_B, such that any one of the hydrogen atoms reaches first excitation state. (Ans: 20.4 eV)[5]
32.Discuss the working of cyclotron in detail.[5]
33.Explain in detail Coulomb’s law and its various aspects.[5]
34.Explain the amplitude modulation with necessary diagrams.[5]
35.Define ‘electrostatic potential energy’.[5]
36.What is an LED? Give the principle of its operation with a diagram.[5]
37.List out the laws of photoelectric effect.[5]
38.Obtain the equation for radius of illumination (or Snell’s window).[5]
39.What are the important points of wave theory of light?[5]
🔑 Show Answer Key — Set 3
  1. 1. 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)
  2. 2. Potential gradient k is same for both cells. E ∝ l. So E_2/E_1 = l_2/l_1 ⇒ E_2 = E_1 × (63/35) = 1.25 × 1.8 = 2.25 V. Answer: E_2 = 2.25 V.
  3. 3. 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. Electrostatic Shielding/Property of Conductors: When an internal charged conductor touches the inner wall of a hollow spherical conductor, its entire excess charge is transferred completely to the outer surface of the hollow sphere, regardless of how much charge is already present there. 2. Construction Hollow Metallic Sphere ( $A$ ): A large, hollow spherical conductor is mounted on top of tall, insulating pillars. Pulleys ( $B$ and $C$ ): A pulley $B$ is fixed at the center of the hollow sphere, and another pulley $C$ is fixed at the bottom, near the ground. A continuous conveyor belt made of an insulating material (such as rubber or silk) loops around both pulleys. The lower pulley $C$ is continuously rotated by an electric motor. Metallic Combs ( $D$ and $E$ ): Two comb-shaped metallic conductors featuring numerous sharp teeth are positioned near the belt: Spray Comb ( $D$ ): Located near the lower pulley $C$ , outside the sphere. It is connected to a high-voltage rectified power supply ( $10^4\text{ V}$ ). Collecting Comb ( $E$ ): Located near the upper pulley $B$ , inside the hollow metallic sphere. It is directly connected to the inner surface of the sphere. 3. Working Mechanism 1. Spraying of Charge: The high-voltage power supply maintains the spray comb $D$ at a very high positive potential. Due to the action of points (corona discharge), the intense electric field near the sharp teeth of comb $D$ ionizes the surrounding air. The positive ions are repelled by the comb and stick to the moving insulating belt. 2. Transportation: The electric motor turns the lower pulley $C$ , moving the belt upward. The belt carries these positive charges from the bottom upward into the hollow sphere. 3. Collection of Charge: As the charged portion of the belt reaches the top and passes close to the collecting comb $E$ , the positive charges on the belt induce a negative charge on the sharp teeth of comb $E$ and an equal positive charge on the outer surface of the hollow sphere $A$ through electrostatic induction. 4. Discharge at the Upper Comb: The high localized field at the teeth of comb $E$ causes a corona discharge that ionizes the air inside the sphere. The negative ions are repelled toward the belt, completely neutralizing its positive charges before it loops down. The uncharged belt moves downward to receive a fresh layer of charge at comb $D$ . 5. Accumulation of High Voltage: This process continues systematically. Since any charge transferred to the collecting comb immediately flows to the outer surface of the sphere, the charge on the sphere builds up continuously. As a result, the electrostatic potential ( $V = \frac{Q}{4\pi\varepsilon_0 R}$ ) of the sphere rises until it reaches the breakdown limit of the surrounding air. 4. Preventing Leakage When the electric field around the outer surface of the sphere exceeds the dielectric strength of air ( $\approx 3 \times 10^6\text{ V m}^{-1}$ ), the air undergoes breakdown, and charges begin to leak into the surroundings. To minimize this leakage and allow the potential to build up to several million volts, the entire generator assembly is enclosed within a steel tank filled with a gas (such as air or nitrogen) at very high pressure. 5. Applications The massive potential differences generated are used to accelerate charged particles (such as protons, deuterons, and alpha particles) to very high kinetic energies. These highly energetic beams of particles are utilized to trigger nuclear disintegrations and study the fundamental structure of atomic nuclei in nuclear physics research. Answer: A Van de Graaff generator produces high voltages (up to $10^7\text{ V}$ ) based on corona discharge and charge accumulation on the outer surface of conductors. It uses a moving insulating belt to transport positive charges from a high-voltage spray comb to a collecting comb inside a hollow metallic sphere, where the charges accumulate to accelerate particles for nuclear physics applications.
  4. 4. 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.
  5. 5. 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
  6. 6. (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
  7. 7. (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).
  8. 8. In a medium $v=1/\sqrt{\mu\varepsilon}=c/\sqrt{\mu_r\varepsilon_r}$ . With $\mu_r=1.0$ , $\varepsilon_r=2.25$ , $v=c/\sqrt{2.25}=c/1.5=(3\times10^8)/1.5=2.0\times10^8\,$ m/s. Answer: Speed $v=2.0\times10^{8}\,$ m s^{-1}
  9. 9. Field at centre due to one side approximated as $B_{side}=\dfrac{\mu_0 I}{4\pi a}\Big(\sin\alpha_1+\sin\alpha_2\Big)$ . For a square of side $a=0.5\,$ m, distance from centre to midpoint of side is $a/2$ . Summing four sides gives $B=\dfrac{\mu_0 I}{\pi a}\approx3.4\times10^{-6}\,$ T for $I=1.5\,$ A and $a=0.5\,$ m (detailed geometry yields this numeric). Answer: $3.4\times10^{-6}\,$ T
  10. 10. 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). Strongly penetrating, used in medical imaging (radiography), crystallography, and industrial inspection. Produced by accelerated electrons hitting targets or by inner-shell electronic transitions. (c) Radio waves: Lowest-frequency part of EM spectrum (~kHz to GHz). Used for AM/FM broadcasting, TV, two-way communication, and navigation. Long wavelengths allow long-distance propagation via ionospheric reflection. (d) Visible spectrum: Wavelengths ~400–700 nm corresponding to light perceptible by human eye. Contains colours from violet to red; important for vision, photosynthesis, and optical instruments.
  11. 11. (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
  12. 12. (a) Time interval ≈1 year≈365 days. Activity decays: A=A_0 e^{-\lambda t} with λ=\ln2/T_{1/2}=\ln2/(5.01 d)=0.1384 d^{-1}=1.602×10^{-6} s^{-1}. Over 365 d, A/A_0= e^{-0.1384×365}=e^{-50.5}≈1.2×10^{-22}. So A≈1.2×10^{-22} μCi (≈10^{-22} μCi). (b) λ=\ln2/(5.01×86400)≈1.60×10^{-6} s^{-1}. (c) Mean life τ=1/λ≈6.24×10^5 s ≈7.23 days. (d) Initial activity 1 μCi =1×10^{-6} Ci = (1×10^{-6})(3.7×10^{10} s^{-1})=3.7×10^{4} s^{-1}. Number of atoms N_0=A_0/λ=3.7×10^4 /1.60×10^{-6} ≈2.31×10^{10}. Answer: See solution
  13. 13. 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.
  14. 14. 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
  15. 15. Remaining fraction =0.175 = e^{-\lambda t} with λ=\ln2/T_{1/2} and T_{1/2}(^{14}C)=5730 yr. So t= -\ln(0.175)/λ = -\ln(0.175)×T_{1/2}/\ln2 ≈1.742/0.693×5730 ≈2.513×5730 ≈14400 yr ≈1.44×10^4 yr. Answer: ≈1.44×10^4 years
  16. 16. 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).
  17. 17. These conditions ensure that a small signal reproduces itself identically after each loop, sustaining oscillation. Answer: Loop gain magnitude |Aβ| = 1 and total phase shift around loop = 0 or an integer multiple of 2π.
  18. 18. When the conduction current in the connecting wire is 5 A charging the capacitor, the rate of change of electric flux between the plates produces an equal displacement current through any surface between plates. Thus $I_d=I_{\rm conduction}=5\,$ A (by continuity of current). Answer: Displacement current $I_d=5\,$ A
  19. 19. Fraunhofer requires parallel incident and observation rays; Fresnel requires more complex geometry and varying phase across aperture. Answer: Fresnel (near-field): source or screen (or both) at finite distance from aperture; wavefront curvature significant. Fraunhofer (far-field): source and screen effectively at infinite distance (use lens to make plane waves); patterns simpler (Fourier transform of aperture).
  20. 20. For coil area A rotating at angular speed ω, flux Φ=BA cos(ωt) so emf = −dΦ/dt = BAω sin(ωt) — an AC emf. Answer: AC generator principle: A coil rotating in a magnetic field experiences a changing magnetic flux, producing an alternating emf according to Faraday’s law; rotation converts mechanical energy into electrical energy.
  21. 21. Inductance L = NΦ / i =200×6×10^{-5}/4 = (12×10^{-3})/4 =3×10^{-3} H. Energy U = 1/2 L i^2 =0.5×3×10^{-3}×4^2 =0.5×3×10^{-3}×16=24×10^{-3} J =0.024 J. Answer: 0.024 J (book)
  22. 22. Useful optical power = 150×0.12 = 18 W. Photon energy \(E=hc/\lambda=\dfrac{6.63\times10^{-34}\times3\times10^8}{5500\times10^{-10}}\approx3.61\times10^{-19}\,\mathrm{J}\). Number per second = 18 / 3.61×10^{-19} ≈4.99×10^{19}. Answer: ≈4.98×10^{19} photons s^{–1}
  23. 23. Inductor behaviour: V = L di/dt; energy stored = \(\dfrac{1}{2}L i^2\). Answer: An inductor stores energy in its magnetic field and opposes changes in current. Examples: smoothing chokes in power supplies, tuning circuits, filters, and ignition coils.
  24. 24. It denotes systems that detect objects and determine their range using radio waves. Answer: RADAR stands for RAdio Detection And Ranging.
  25. 25. PF ranges from 0 (purely reactive) to 1 (purely resistive). Answer: Power factor = cosφ where φ is the phase angle between voltage and current; it equals real power divided by apparent power: PF = P/(V_rms I_rms).
  26. 26. Work done to build current from 0 to i: dW = L i' di' integrated: W = ∫_0^i L i' di' = 1/2 L i^2. This energy is stored in magnetic field. Answer: Energy stored in inductor = \(U=\dfrac{1}{2}L i^2\).
  27. 27. Kinetic energy gained = qV = p^2/(2m) ⇒ p=\sqrt{2mqV} ⇒ \(\lambda=h/p= h/\sqrt{2mqV}\). Answer: For non-relativistic particle: \(\lambda=\dfrac{h}{\sqrt{2m q V}}\).
  28. 28. From Einstein's equation: \(eV_s=h\nu-\phi\). Answer: Stopping potential \(V_s\) is the minimum reverse potential applied to the collector to reduce the photoelectric current to zero; it equals \(K_{max}/e\).
  29. 29. All these change magnetic flux Φ_B = ∫B·dA leading to \(\mathcal{E}=-d\Phi_B/dt\). Answer: Induced emf can be produced by: (i) changing magnetic field strength, (ii) changing area of loop in a magnetic field, (iii) changing orientation (angle) between loop and field, (iv) moving a conductor through a magnetic field (motional emf).
  30. 30. Sum phasors from N slits; maxima when phase difference between adjacent slits = 2π m. The intensity of principal maxima increases as N^2 and is sharp; order m exists only if |mλ/d| ≤ 1. Answer: For a grating with spacing d (distance between adjacent slits), constructive interference occurs when path difference between adjacent slits d sinθ = m λ (m integer). These are principal maxima; angular positions given by sinθ = mλ/d.
  31. 31. Minimum energy needed to excite one atom to first excited state (energy difference from ground to n=2 is 10.2 eV for hydrogen). In CM frame for perfectly inelastic head-on collision where two identical masses stick together, kinetic energy of moving atom must supply both excitation (10.2 eV) and provide recoil kinetic energy of combined mass. For minimum KE, all excess goes into excitation and center-of-mass motion minimized. Using energy and momentum conservation for equal masses colliding and sticking, fraction of initial KE available for internal excitation is 1/2. Therefore required initial KE = 2×10.2 eV = 20.4 eV. Answer: 20.4 eV
  32. 32. Particles start near centre, cross gap at each half-revolution gaining energy. Radius increases with speed $r=mv/(qB)$ until extraction. Limits: relativistic mass increase causes frequency mismatch at high energies. Include schematic and energy calculations: kinetic energy after n gaps $K=\tfrac12 m v^2 = q V_{acc}\times n$ . Answer: Cyclotron accelerates charged particles using perpendicular magnetic field to bend particles in circular paths and an alternating electric field across gaps between D-shaped electrodes (dees). Frequency of RF equals cyclotron frequency $\omega=qB/m$ .
  33. 33. Coulomb’s Law describes the quantitative electrostatic force between two stationary point charges. Statement 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. The force acts along the straight line joining the two charges. In vector form, the force $\vec{F}_{21}$ exerted on charge $q_2$ by charge $q_1$ separated by a distance $r$ is: $$\vec{F}_{21} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{12}$$ Where $\hat{r}_{12}$ is the unit vector pointing from $q_1$ to $q_2$ , and $\varepsilon_0$ is the permittivity of free space. Various Aspects of Coulomb's Law 1. Inverse Square Dependence: The force decreases inversely with the square of the distance ( $F \propto \frac{1}{r^2}$ ). If the distance between two charges is doubled, the force between them drops to one-fourth of its initial value. 2. Value of the Proportionality Constant ( $k$ ): In a vacuum or air, the constant $k = \frac{1}{4\pi\varepsilon_0}$ has an experimentally determined value of approximately $9 \times 10^9\text{ N m}^2\text{ C}^{-2}$ . 3. Definition of $1\text{ Coulomb}$ : Using Coulomb's law, if $q_1 = q_2 = 1\text{ C}$ and $r = 1\text{ m}$ in a vacuum, the force becomes: $$F = (9 \times 10^9) \times \frac{1 \times 1 = 9 \times 10^9\text{ N}$$ Therefore, $1\text{ Coulomb}$ is defined as the amount of charge that repels an identical charge placed $1\text{ meter}$ away in a vacuum with a force of $9 \times 10^9\text{ Newtons}$ . 4. Influence of the Medium: If the charges are immersed in a material medium (such as water or glass) with permittivity $\varepsilon$ , the force becomes: $$F_{\text{medium}} = \frac{1}{4\pi\varepsilon} \frac{q_1 q_2}{r^2}$$ Since $\varepsilon = \varepsilon_r \varepsilon_0$ (where $\varepsilon_r > 1$ is the relative permittivity), the electrostatic force decreases in a material medium compared to a vacuum: $$F_{\text{medium}} = \frac{F_{\text{vacuum}}}{\varepsilon_r}$$ 5. Newton's Third Law Verification: The force exerted on $q_1$ by $q_2$ ( $\vec{F}_{12}$ ) uses a unit vector $\hat{r}_{21}$ which points in the opposite direction to $\hat{r}_{12}$ . Therefore: $$\vec{F}_{12} = -\vec{F}_{21}$$ This demonstrates that the electrostatic forces form an action-reaction pair, complying with Newton’s third law. Answer: Coulomb's law states $F = k \frac{q_1 q_2}{r^2}$ . Its primary aspects include an inverse-square distance dependence, dependence on the surrounding medium's permittivity ( $\varepsilon$ ), a constant $k = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}$ in a vacuum, and obedience to Newton's third law of motion.
  34. 34. Expanding gives s(t)=A_c cos ω_ct + (A_c k_m/2)[cos(ω_c+ω_m)t + cos(ω_c−ω_m)t] showing carrier plus upper and lower sidebands. Modulation index k_m ≤ 1 to avoid overmodulation. Draw time-domain waveform (envelope follows m(t)) and spectrum showing carrier and two sidebands at ωc±ωm. Answer: In amplitude modulation (AM) the amplitude of a high-frequency carrier c(t) = A_c cos ω_ct is varied in proportion to the modulating signal m(t). For a single-tone m(t)=A_m cos ω_mt, the AM signal s(t) = [A_c + m(t)] cos ω_ct = A_c[1 + k_m cos ω_mt] cos ω_ct, where k_m is modulation index.
  35. 35. The electrostatic potential energy ( $U$ ) of a system of point charges is defined as the total amount of work done by an external agent in assembling the charges by bringing them from an initial separation of infinity to their respective current positions in the configuration, without any acceleration. For a simple two-charge system ( $q_1$ and $q_2$ ) separated by a distance $r$ , it is given by: $$U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}$$ Answer: Electrostatic potential energy ( $U$ ) is the total work done to assemble a system of charges by bringing them from infinity to their configurations without acceleration.
  36. 36. In forward bias, electrons injected from n into p recombine with holes in the p-region; if the semiconductor has a direct band gap, recombination is radiative giving photons of energy E = hν ≈ Eg. The emission wavelength depends on bandgap material: e.g., GaAs (infrared), GaP (red/green), GaN (blue). LED structure: p–n or double heterostructure to confine carriers and photons for high efficiency. Draw diagram of p–n junction with arrows showing electron–hole recombination and emitted photon. Answer: An LED (Light Emitting Diode) is a p–n junction device which emits light when forward biased. The principle is electroluminescence: recombination of electrons and holes across a direct band gap releases energy as photons whose energy ≈ band gap.
  37. 37. These empirical laws were explained by Einstein using photon concept: each photon transfers energy h\nu to a single electron; if h\nu>\phi electron is ejected with KE = h\nu-\phi. Answer: 1) For a given metal, photoemission occurs only if incident light frequency > threshold frequency. 2) Maximum kinetic energy of photoelectrons depends on frequency, not on intensity. 3) Number of electrons emitted per second (photocurrent) is proportional to light intensity. 4) Emission is almost immediate (no measurable time lag).
  38. 38. Critical angle at water-air interface sinθc = n_air/n_water. Rays from air at angles ≤ θc relative to normal refract into water. The cone of illumination has semi-angle θc; at depth h the cross-sectional radius = h tanθc. This is called Snell's window; outside this cone TIR makes surface behave like a mirror. Answer: Underwater observer looking upward through flat air-water interface sees light from above inside a cone of half-angle θc = arcsin(n_air/n_water). The radius R of illumination on surface at depth h is R = h tan θc.
  39. 39. 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.

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