Physics Question Paper - Part 1
Part - A
One mark questions.
Choose the correct options : 20x1=20
(i) In a purely capacitive AC circuit, the current :
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(ii) The electromagnetic waves with highest frequency are :
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(iii) A convex lens of glass (n = 1.5) is immersed in water (n = 1.33). Its focal length :
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(iv) The phenomenon of interference is based on :
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(v) In YDSE, if slit separation is halved, fringe width :
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(vi) A diminished virtual image can be produced only with a :
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(vii) Resolving power of a telescope increases when :
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(viii) Momentum of a photon of wavelength λ is :
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(ix) The electric potential at a point on the axial line of an electric dipole is :
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(x) A parallel plate capacitor is charged. If the plates are pulled apart, the capacitance :
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(xi) Kirchhoff's junction rule (first law) is based on conservation of :
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(xii) The resistance of a conductor increases with temperature because :
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(xiii) A proton and an alpha particle enter a uniform magnetic field with the same kinetic energy. The ratio of their radii (rp : rα) is :
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(xiv) Magnetic susceptibility is negative for :
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(xv) SI unit of mutual inductance is :
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True / False :
(xvi) The depletion region in a p-n diode contains no mobile charge carriers.
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(xvii) The work function of a metal depends on intensity of incident radiation.
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(xviii) Balmer series of hydrogen lies in ultraviolet region.
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(xix) Nuclear density is independent of mass number A.
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(xx) In an n-type semiconductor, holes are majority charge carriers.
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Part - B
Two marks questions : 7x2=14
2. Write any four properties of electromagnetic waves ?
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- They do not require any material medium for propagation.
- They travel in a vacuum with the speed of light (c = 3×10⁸ m/s).
- They are transverse in nature.
- They carry energy and momentum, and exert radiation pressure.
3. State Lenz's Law. On which fundamental principle of conservation is this law based ?
The instantaneous current from an AC source is given by the equation I = 10 sin(314t). What is the RMS (Root Mean Square) value and the frequency (f) of the current ?
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Answer (OR part):
Comparing I = 10 sin(314t) with standard equation I = I₀ sin(ωt):
Peak current (I₀) = 10 A.
RMS Current (I_rms) = I₀ / √2 = 10 / 1.414 ≈ 7.07 A.
Angular frequency (ω) = 314 rad/s. Since ω = 2πf, f = 314 / 2π ≈ 50 Hz.
4. Define mass defect and binding energy per nucleon.
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Mass Defect (Δm): The difference between the sum of the masses of individual nucleons making up a nucleus and the actual rest mass of the nucleus.
Binding Energy per Nucleon: The total binding energy of a nucleus divided by its mass number (A). It represents the average energy required to remove a single nucleon from the nucleus.
5. Explain Gaussian surface with its importance.
A point charge of +10 µC is placed at the center of a cubic surface of edge 0.1 m. Calculate the net electric flux passing through the entire cube. What is the flux passing through just one face of the cube ? (Given : ε₀ = 8.854 × 10⁻¹² C²N⁻¹m⁻²)
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Answer (OR part):
Total flux through the entire cube (Φ_total) = q / ε₀ = (10 × 10⁻⁶ C) / (8.854 × 10⁻¹² C²N⁻¹m⁻²) ≈ 1.13 × 10⁶ N·m²/C.
Flux through one face = Φ_total / 6 ≈ 1.88 × 10⁵ N·m²/C.
6. Define the 'drift velocity' of electrons in a conductor. How is it related to the electric current (I) flowing through it ?
A copper wire of cross-sectional area 1.0 × 10⁻⁷ m² carries a current of 1.5 A. Calculate the drift velocity of the electrons. (Given : Number density of free electrons in copper, n = 8.5 × 10²⁸ m⁻³; charge on an electron, e = 1.6 × 10⁻¹⁹ C)
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Answer (OR part):
V_d = I / (n e A)
V_d = 1.5 / (8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷)
V_d ≈ 1.1 × 10⁻³ m/s or 1.1 mm/s.
7. State Ampere's circuital law. Use it to find the magnetic field at a point inside a long current-carrying solenoid.
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Ampere's Law: The line integral of the magnetic field (B) around any closed path is equal to µ₀ times the net steady current (I) enclosed by the path. (∮ B·dl = µ₀I)
Solenoid: Taking a rectangular Amperian loop inside the solenoid, ∮ B·dl = B × L = µ₀ (n L I), where n is the number of turns per unit length. Therefore, B = µ₀nI.
8. Why is the core of a transformer laminated ? What is the function of a step-up transformer ?
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The core is laminated to minimize energy losses due to eddy currents.
The function of a step-up transformer is to increase the alternating voltage (and correspondingly decrease the current) from the primary coil to the secondary coil.
Physics Question Paper - Part 2
Part - C
Three marks questions : 7x3=21
9. State the major observations and conclusions drawn from Rutherford's alpha-particle scattering experiment.
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Observations:
1. Most alpha particles passed straight through the gold foil undeflected.
2. A small fraction deflected by small angles.
3. A very small number (1 in 8000) bounded back (deflected by 180°).
Conclusions:
1. Most of the space in an atom is empty.
2. The positive charge and most of the mass of the atom are densely concentrated in a tiny central volume called the nucleus.
10. What is a p - n junction diode ? Explain its working as a full-wave rectifier with a neat circuit diagram.
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A p-n junction diode is a semiconductor device formed by joining p-type and n-type semiconductor materials, allowing current to flow in only one direction.
Full-wave rectifier working: It utilizes two diodes connected to a center-tapped secondary of a transformer. During the positive half-cycle, Diode D1 is forward-biased and conducts, while D2 is reverse-biased. During the negative half-cycle, D2 is forward-biased and conducts, while D1 is reverse-biased. As a result, current flows through the load resistor in the same direction during both half-cycles, converting AC to pulsating DC. (Students must draw circuit showing transformer, two diodes, and load resistor).
11. State Gauss's Law. Apply it to find the electric field due to a uniformly charged infinite plane sheet.
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Gauss's Law: Total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity (Φ = q_enclosed / ε₀).
Application: Imagine a cylindrical Gaussian surface passing symmetrically through the charged sheet (surface charge density σ). Flux passes only through the two flat end caps (area A), not the curved surface.
Total Flux = EA + EA = 2EA.
Charge enclosed = σA.
By Gauss's Law: 2EA = σA / ε₀ ⇒ E = σ / (2ε₀).
12. Draw a circuit diagram of a Wheatstone bridge. Using Kirchhoff's laws, derive the condition for its balance.
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A Wheatstone bridge consists of four resistors P, Q, R, S connected in a quadrilateral ABCD, with a galvanometer between B and D and a battery between A and C.
Balanced condition: No current flows through the galvanometer (I_g = 0).
Using Kirchhoff's loop rule in loop ABDA: I₁P - I₂R = 0 ⇒ I₁P = I₂R
In loop BCDB: I₁Q - I₂S = 0 ⇒ I₁Q = I₂S
Dividing both equations: P/Q = R/S. (Students must sketch the proper circuit).
13. Using the Biot-Savart law, derive an expression for the magnetic field at the centre of a current-carrying circular loop.
A radio station broadcasts at a frequency of 100 MHz (MegaHertz). What is the wavelength (λ) of the electromagnetic waves broadcast by the station ? (Given : C = 3 × 10⁸ m/s)
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Integrating over the entire circle (∫ dl = 2πR):
B = (µ₀/4π) * (I * 2πR / R²) = µ₀I / (2R).
Answer (OR part):
Using c = fλ ⇒ λ = c / f
λ = (3 × 10⁸ m/s) / (100 × 10⁶ Hz)
λ = 3 meters.
14. Derive the expression for the average power dissipated in a series LCR circuit. Define the term power factor and state the condition for wattless current.
A 200 µF capacitor in series with a 100Ω resistor is connected to a 220 V, 50 Hz AC source. Calculate :
(i) The capacitive reactance (X_C)
(ii) The impedance of the circuit (Z)
(iii) The RMS current (I_rms)
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Average Power P_avg = V_rms * I_rms * cos(φ).
Power factor: It is the cosine of the phase angle (φ) between voltage and current, given by cos(φ) = R/Z.
Wattless current: Current in an AC circuit when average power dissipated is zero (when φ = 90°, purely inductive or capacitive).
Answer (OR part):
(i) X_C = 1 / (2πfC) = 1 / (2 * 3.14 * 50 * 200×10⁻⁶) = 15.9 Ω
(ii) Z = √(R² + X_C²) = √(100² + 15.9²) = √(10000 + 252.81) ≈ 101.2 Ω
(iii) I_rms = V_rms / Z = 220 / 101.2 ≈ 2.17 A
15. Draw a ray diagram for the formation of an image by a compound microscope when the final image is at the least distance of distinct vision (D). Write the expression for its magnifying power.
In a Young's double-slit experiment, the two slits are separated by 0.2 mm, and the screen is placed 1.0 m away. If the wavelength (λ) of the light used is 600 nm, what is the fringe width (β) ?
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(Students need to draw ray diagram showing objective and eyepiece lenses, object placed just outside f_o, forming real image, which acts as object for eyepiece forming large virtual image at D).
Magnifying power formula: M = m_o × m_e = (L/f_o) * (1 + D/f_e).
Answer (OR part):
Given: d = 0.2 mm = 0.2 × 10⁻³ m, D = 1.0 m, λ = 600 nm = 600 × 10⁻⁹ m.
Fringe width (β) = (λD) / d
β = (600 × 10⁻⁹ × 1.0) / (0.2 × 10⁻³) = 3000 × 10⁻⁶ m = 3 × 10⁻³ m = 3 mm.
Part - D
Five marks questions : 3x5=15
16. Read the following paragraph and answer the questions given below. 5x1=5
When a current-carrying conductor is placed in an external magnetic field, it experiences a mechanical force. This force is the vector sum of the magnetic Lorentz forces acting on all the free charge carriers moving within the conductor. The magnitude and direction of this force are given by the vector formula F = I(L × B), where I is the current, L is the length vector of the conductor (in the direction of the current), and B is the magnetic field. The direction of this force can be found using Fleming's Left-Hand Rule. This principle is fundamental to the operation of electric motors.
Questions :
(i) What is the origin of the force on the conductor ?
(ii) Based on the formula F = I(L × B), when will the magnitude of the force on the conductor be zero (minimum) ?
(iii) What is the fundamental principle behind the operation of an electric motor ?
(iv) Which rule is used to determine the direction of the mechanical force on the conductor ?
(v) Write the vector formula given in the passage for the force F on the conductor ?
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(i) The origin of the force is the magnetic Lorentz force acting on the free charge carriers moving within the conductor.
(ii) When the conductor is placed parallel or anti-parallel to the magnetic field (angle is 0° or 180°, so sinθ = 0).
(iii) The fundamental principle is that a current-carrying conductor placed in a magnetic field experiences a mechanical force.
(iv) Fleming's Left-Hand Rule.
(v) F = I(L × B).
17. Draw the electric field lines for : (i) A single positive point charge (+q) (ii) An electric dipole. State three key properties of electric field lines.
A parallel plate capacitor is charged by a battery. After sometime, the battery is disconnected. What will be the effect on the following quantities if the distance between the plates is now doubled ?
(i) Charge on the plates
(ii) Potential difference across the plates
(iii) Energy stored in the capacitor.
Justify your answer in each case.
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(Drawings: Radially outward straight lines for +q; curved lines from + to - for a dipole).
Properties: 1. They originate from positive charges and terminate at negative charges. 2. Two field lines never intersect. 3. Tangent at any point gives the direction of the electric field.
Answer (OR part):
Since the battery is disconnected, the system is isolated.
(i) Charge (Q): Remains the same (constant) due to conservation of charge.
(ii) Potential difference (V): Doubles. Capacitance C = ε₀A/d. If d is doubled, C becomes half. Since V = Q/C, halving C makes V double.
(iii) Energy stored (U): Doubles. U = Q² / 2C. Since Q is constant and C is halved, U becomes twice its original value. (Work was done to pull plates apart).
18. (a) State the assumptions used to derive the Lens Maker's Formula. 2
(b) Derive the Lens Maker's Formula. 3
Using Huygens principle, derive and prove the law of reflection for a plane reflecting surface. 5
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(a) Assumptions: 1. The lens is thin so distance measured from poles can be taken as from the optical center. 2. The aperture of the lens is small. 3. The object is a point object placed on the principal axis.
(b) Derivation steps: Apply refraction formula at single spherical surface for both surfaces: (n₂/v₁ - n₁/u = (n₂-n₁)/R₁) and (n₁/v - n₂/v₁ = (n₁-n₂)/R₂). Add them to get 1/v - 1/u = (n₂₁ - 1)(1/R₁ - 1/R₂). Substitute u=∞ and v=f to get 1/f = (n₂₁ - 1)(1/R₁ - 1/R₂).
Answer (OR part):
Derivation using Huygens Principle:
Assume a plane wavefront AB incident on a plane mirror MN at angle i. The wavefront touches mirror at A, while point B is still at a distance. The time 't' taken by secondary wavelets from B to reach mirror at C is t = BC/v.
During this time, the wavelet from A spreads to radius AE = v*t.
Drawing a tangent from C to this sphere gives the reflected wavefront CE.
In triangles ABC and AEC:
∠B = ∠E = 90°
AC is common.
BC = AE = v*t
By RHS congruence, ΔABC ≅ ΔAEC.
Therefore, ∠BAC (angle of incidence i) = ∠ECA (angle of reflection r). Hence, i = r.
