RBSE Class 12th 2019 Physics-SS-40-1-2019-With Solution Previous Year Papers

Solution:

V = kq / r = (9 × 10⁹ × 4 × 10⁻⁷) / 0.07 = 9 Volt

Answer: Red

Answer: F = B I L sin θ

Answer:

(i) Angle of dip (magnetic inclination): The angle made by the earth's total magnetic field with the horizontal direction in the magnetic meridian is called the angle of dip (δ) at any place.

(ii) At magnetic poles, the angle of dip is 90°.

Answer:

I_rms = I₀ / √2 = 0.707 I₀, where I₀ is the peak value of alternating current.

Answer:

Z = √[R² + (X_L – X_C)²] = √[700² + (700 – 700)²] = √(700²) = 700 Ω

Answer:

Power of lens = ¹⁄_{Focal length (in m)}

Answer:

For a given photosensitive material, there exists a certain minimum cut-off frequency below which no photoelectrons are emitted, howsoever high the intensity of incident radiation. This frequency is called threshold frequency.

Answer:

Here the electrons emitted by the hot filament of an electron gun are accelerated by applying a suitable potential difference V between the cathode and anode.

                    L.T.B.
                       |
    Filament ──► [Electron Gun] ──► Incident electron beam ──► Ni Crystal
                       |                                          |
                    H.T.B.                                     Diffracted
                       |                                      electron beam
                    Anode                                         |
                                                              [Movable Collector]
                                                                     |
                                                              To galvanometer
                                                         (Vacuum chamber)

Experimental arrangement of Davisson and Germer experiment

Answer:

Lyman Series

Answer:

Y = A + B (OR gate output)

Answer:

The original low frequency message signal cannot be transmitted to long distances. Hence, modulation of the low frequency signal is done. Modulation is the phenomenon of superimposing the low frequency message signal (called the modulating signal) on a high frequency wave (called the carrier wave).

Answer:

(A) Angle = 90°

(B) Phase difference = Zero

Answer:

(a) Electric Dipole moment: It measures the strength of an electric dipole. The dipole moment of an electric dipole is a vector whose magnitude is the product of magnitude of both charges and the separation between the two opposite charges, and the direction is along the dipole axis from the negative to the positive charge.

p = q(2a)

-q ──── 2a ──── +q

(b) Equipotential surface: Any surface that has same electric potential at every point on it is called an equipotential surface.

Solution:

For a balanced Wheatstone bridge, the condition is:

\(\frac{P}{Q} = \frac{R}{S}\)

Given: P = 4 Ω, Q = 5 Ω, S = 6 Ω

\(\frac{4}{5} = \frac{R}{6}\)

\(R = \frac{4 \times 6}{5} = \frac{24}{5} = 4.8 \, \Omega\)

Correct Answer: R = 4.8 Ω

Answer: The temperature at which a ferromagnetic substance loses its ferromagnetic properties and becomes paramagnetic is called Curie temperature or Curie point (T_c).

Solution:

Given: Effective length (2l) = 0.4 m, Pole strength (m) = 40 A-m

Magnetic moment (M) = m × (2l)

M = 40 × 0.4 = 16 A-m²

Correct Answer: M = 16 A-m²

Answer:

Lenz's law: The direction of induced current in a circuit is such that it opposes the cause or the change which produces it.

Explanation of conservation of energy:

When a magnet is moved towards or away from a closed coil, the induced current always opposes the motion of the magnet. For example, when the north pole of a magnet is brought closer to a coil, the face of the coil towards the magnet develops north polarity and repels the north pole of the magnet. Work must be done to move the magnet closer against this repulsive force. Similarly, when the north pole is moved away, the coil's face develops south polarity and attracts the north pole, requiring work to move it away. This work done against the magnetic force is converted into electrical energy in the form of induced current.

If Lenz's law were not valid, the induced current would flow in the opposite direction, causing the magnet to accelerate without any work input. This would create a perpetual motion machine, violating the law of conservation of energy. Thus, Lenz's law is a direct consequence of the conservation of energy.

उत्तर:

(a) पूर्ण आंतरिक परावर्तन (Total Internal Reflection):

जब प्रकाश किसी सघन माध्यम से विरल माध्यम में जाता है, तो अंतरापृष्ठ पर प्रकाश का कुछ भाग सघन माध्यम में वापस परावर्तित हो जाता है और कुछ भाग विरल माध्यम में अपवर्तित हो जाता है। इस परावर्तन को आंतरिक परावर्तन कहते हैं। कुछ विशेष परिस्थितियों में, संपूर्ण आपतित प्रकाश सघन माध्यम में वापस परावर्तित हो जाता है। इस घटना को पूर्ण आंतरिक परावर्तन कहते हैं।

(b) प्रकाश का विवर्तन (Diffraction of Light):

प्रकाश का नुकीले कोनों के चारों ओर मुड़ने और अपारदर्शी बाधाओं की ज्यामितीय छाया के भीतर प्रकाश के फैलने की घटना को प्रकाश का विवर्तन कहते हैं। इस प्रकार प्रकाश अपने सरल रेखीय पथ से विचलित हो जाता है। जब छिद्र या बाधा के आयाम प्रकाश की तरंगदैर्ध्य के तुलनीय होते हैं, तो यह विचलन अधिक स्पष्ट हो जाता है।

उत्तर:

(a) मैलस नियम (Malus Law):

इस नियम के अनुसार, "जब पूर्णतः समतल ध्रुवित प्रकाश की एक किरण किसी विश्लेषक पर आपतित होती है, तो विश्लेषक से संचरित प्रकाश की परिणामी तीव्रता I, विश्लेषक और ध्रुवक के संचरण तलों के बीच के कोण (θ) की कोज्या के वर्ग के अनुक्रमानुपाती होती है।"

I = I₀ cos²θ

जहाँ I₀, ध्रुवक P से गुजरने के बाद ध्रुवित प्रकाश की तीव्रता है। [इस नियम को कोज्या वर्ग नियम भी कहते हैं।]

(b) अपवर्तनांक और अपवर्तन कोण की गणना:

ब्रूस्टर के नियम के अनुसार, जब परावर्तित प्रकाश पूर्णतः ध्रुवित होता है, तब:

μ = tan i_p

जहाँ i_p ध्रुवण कोण (आपतन कोण) है।

दिया गया है: i_p = 60°

μ = tan 60° = √3 ≈ 1.73

अतः पदार्थ का अपवर्तनांक (μ) = √3 ≈ 1.73

ब्रूस्टर के नियम से, ध्रुवण कोण और अपवर्तन कोण के बीच संबंध:

i_p + r = 90°

r = 90° - i_p = 90° - 60° = 30°

अतः अपवर्तन कोण (r) = 30°

Solution:

For an electron accelerated through a potential difference V, the de Broglie wavelength λ is given by:

λ = h / √(2meV)

Where h = 6.63 × 10⁻³⁴ Js, m = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, V = 100 V

λ = 6.63 × 10⁻³⁴ / √(2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × 100)

λ = 6.63 × 10⁻³⁴ / √(2.912 × 10⁻⁴⁷)

λ = 6.63 × 10⁻³⁴ / (5.396 × 10⁻²⁴)

λ = 1.228 × 10⁻¹⁰ m = 1.228 Å

Correct Answer: λ = 1.228 Å

Derivation:

Circuit Diagram:

[Circuit diagram: Battery connected across potentiometer wire AB. Cell ε with key K₁ and jockey. Resistance box R and key K₂ in series with cell.]

Procedure:

1. Close key K₁ so that constant current flows through potentiometer wire AB.
2. With key K₂ open, move jockey along AB to find balance point for emf ε of the cell. Let balancing length be l₁. If k is potential gradient, then ε = k l₁.
3. Now introduce resistance R from resistance box and close key K₂. Find balance point for terminal potential difference V of the cell. Let balancing length be l₂. Then V = k l₂.

Dividing the two equations:

ε / V = l₁ / l₂

Let r be internal resistance of the cell. When cell is shunted with resistance R, current I flows. From Ohm's law:

ε = I(R + r) and V = IR

Therefore, ε / V = (R + r) / R

Equating with above: l₁ / l₂ = (R + r) / R

Solving for r:

r = R (l₁ / l₂ - 1) = R (l₁ - l₂) / l₂

Formula: Internal resistance, r = R (l₁ - l₂) / l₂

Meter Bridge (Slide Wire Bridge):

It is the simplest practical application of Wheatstone bridge used to measure an unknown resistance.

Principle: Its working is based on the principle of Wheatstone bridge. When the bridge is balanced:

P / Q = R / S

Circuit Diagram:

[Circuit diagram: A meter bridge wire AB of length 1 m. Unknown resistance X in left gap, known resistance R in right gap. Galvanometer connected between middle point and jockey. Battery connected across ends.]

Method:

1. Connect unknown resistance X in left gap and known resistance R in right gap of the meter bridge.
2. Connect a battery across ends A and B of the wire.
3. Slide the jockey along the wire to find point where galvanometer shows zero deflection (balance point).
4. Let balancing length from end A be l cm. Then length from end B is (100 - l) cm.
5. At balance, X / R = l / (100 - l)
6. Therefore, unknown resistance X = R × l / (100 - l)

Formula: X = R × l / (100 - l)

Bohr's Postulates:

(i) First Postulate (Stationary Orbits): An electron in an atom revolves in certain stable orbits without emitting radiant energy, contrary to classical electromagnetic theory. These are called stationary states, each having a definite total energy.

(ii) Second Postulate (Quantization of Angular Momentum): The electron revolves around the nucleus only in those orbits for which the angular momentum is an integral multiple of h/2π, where h is Planck's constant (6.63 × 10^-34 J·s).

Thus, angular momentum L = mvr = nh/2π, where n = 1, 2, 3... (principal quantum number).

(iii) Third Postulate (Energy Transitions): An electron can jump from a higher energy orbit to a lower energy orbit, emitting a photon of energy equal to the energy difference between the two states. The frequency of the emitted photon is given by hν = E_i – E_f, where E_i and E_f are the initial and final energies.

(a) Acceptor impurity: Gallium (Ga) (it is a trivalent element).

(b) Symbol of Zener diode:

Anode ──►|── Cathode
       (Zener diode)

Solution:

In free space, the relation between electric field (E) and magnetic field (B) for an electromagnetic wave is: c = E / B, where c is the speed of light.

Given: E = 300 V/m, c = 3 × 10⁸ m/s.

Therefore, B = E / c = 300 / (3 × 10⁸) = 1 × 10^-6 T.

Answer: B = 1 × 10^-6 T (or 1 μT).

(a) Rutherford-Soddy Law (Radioactive Displacement Laws):

• When a radioactive nucleus emits an α-particle, its atomic number decreases by 2 and mass number decreases by 4.
• When a radioactive nucleus emits a β-particle, its atomic number increases by 1 but mass number remains the same.
• Emission of a γ-particle does not change the mass number or atomic number.

(b) Derivation of Decay Equation:

The law states that the rate of decay is proportional to the number of undecayed nuclei present at that time.

Let N be the number of undecayed nuclei at time t. Then, dN/dt = –λN, where λ is the decay constant.

Rearranging: dN/N = –λ dt.

Integrating both sides: ∫ dN/N = –λ ∫ dt ⇒ ln N = –λt + C.

At t = 0, N = N₀ (initial number), so C = ln N₀.

Thus, ln (N/N₀) = –λt ⇒ N = N₀ e^–λt.

This is the exponential decay equation.

(c) Exponential Decay Curve:

N
|
|    *
|   * *
|  *   *
| *     *
|*       *
|__________ t

(d) Ratio of Half-life and Mean Life:

Half-life (T_1/2) = ln 2 / λ = 0.693 / λ.

Mean life (τ) = 1 / λ.

Therefore, ratio T_1/2 : τ = 0.693 : 1, or T_1/2 = 0.693 τ.

Answer: The possible value of common base current amplification factor α of a transistor is 0.9.

Explanation: The common base current gain (α) is defined as the ratio of collector current to emitter current (α = I_C/I_E). For a transistor, α is always less than 1 (typically between 0.9 and 0.99). Among the given options, only 0.9 satisfies this condition. Values like 9, 49, and 99 are greater than 1 and are not possible for α.

Series LCR Circuit

Suppose a resistance R, an inductance L, and a capacitance C are connected in series to a source of alternating emf given by:

E = E₀ sin ωt

Phasor Diagram

1. Voltage across R (V_R): V_R = I R is in phase with current I. So phasors V_R and I are in the same direction. Amplitude: V_R = I₀ R.
2. Voltage across L (V_L): V_L = I X_L leads the current I by π/2 rad. So phasor V_L lies π/2 rad anticlockwise with respect to phasor I. Amplitude: V_L = I₀ X_L.
3. Voltage across C (V_C): V_C = I X_C lags behind the current I by π/2 rad. So phasor V_C lies π/2 rad clockwise with respect to phasor I. Amplitude: V_C = I₀ X_C.

Phasor diagram for a series LCR circuit (when X_L > X_C):

As V_L and V_C are in opposite directions, their resultant is (V_L – V_C). By parallelogram law, the resultant of V_R and (V_L – V_C) must be equal to the applied emf E₀, given by the diagonal of the parallelogram.

E₀² = (V_R)² + (V_L – V_C)² = (I₀ R)² + (I₀ X_L – I₀ X_C)² = I₀² [R² + (X_L – X_C)²]

∴ I₀ = E₀ / √[R² + (X_L – X_C)²]

Impedance of the Circuit

Clearly, √[R² + (X_L – X_C)²] is the effective resistance of the series LCR circuit which opposes or impedes the flow of current through it and is called its impedance. It is denoted by Z and its SI unit is ohm (Ω).

Z = √[R² + (X_L – X_C)²]

Impedance triangle (when X_L > X_C):

The relationship between R, (X_L – X_C), and Z can be represented by a right-angled triangle where:

• Base = R
• Perpendicular = (X_L – X_C)
• Hypotenuse = Z

Solution:

Consider a rod of length L rotating in a uniform magnetic field B perpendicular to the plane of rotation. The rod rotates about one end O with angular velocity ω.

Point O will be at a higher potential than point A.

Suppose the rod completes one revolution in time T.

Area swept in one rotation = πL²

Change in flux in one rotation = B × πL²

Induced emf = Rate of change of magnetic flux

= Change in flux / Time

= BπL² / T

= BπL² / (2π/ω)

= (BωL²) / 2

Therefore, |e| = (1/2) BωL²

Solution:

Gauss's Law Statement: Gauss's theorem states that the total electric flux through a closed surface is (1/ε₀) times the net charge enclosed by the closed surface.

Mathematically, it can be expressed as:

∮ E · dS = q_enclosed / ε₀

Electric field due to uniformly charged infinite non-conducting sheet:

Consider a thin, infinite plane sheet of charge with uniform surface charge density σ. We wish to calculate its electric field at a point P at distance r from it.

Using a cylindrical Gaussian surface passing through the sheet, with its axis perpendicular to the sheet:

Flux through the curved surface = 0 (since E is parallel to the surface)

Flux through each flat end = E × A (where A is area of the end)

Total flux = 2EA

Charge enclosed = σA

By Gauss's law: 2EA = σA / ε₀

Therefore, E = σ / (2ε₀)

The electric field is perpendicular to the sheet and directed away from it for positive charge.

Solution:

Net electric flux from the shaded region can be calculated using Gauss's law. The flux depends on the net charge enclosed within the shaded region. Without the specific diagram, the general approach is:

Φ = q_enclosed / ε₀

where q_enclosed is the total charge inside the shaded region.

Numerical Example

Given: σ = 4 + 2 μC/m² = 6 μC/m² = 6 × 10⁻⁶ C/m²

Using the formula: E = σ/(2ε₀)

E = (6 × 10⁻⁶) / (2 × 8.854 × 10⁻¹²)

E = (6 × 10⁻⁶) / (1.7708 × 10⁻¹¹)

E ≈ 0.339 × 10⁶ V/m = 3.39 × 10⁵ V/m

Definition of Capacitor: A capacitor is an arrangement of two conductors separated by an insulating medium (or dielectric medium) that is used to store electric charge and electric energy. The capacitance of an insulated charged conductor is considerably increased when we place an earthed connected conductor near it. Such a system of two conductors is called a capacitor.

Series Combination of Capacitors: When the negative plate of one capacitor is connected to the positive plate of the second, and the negative of the second to the positive of the third, and so on, the capacitors are said to be connected in series.

Circuit Diagram:

   +Q -Q    +Q -Q    +Q -Q
   |  |     |  |     |  |
---| C₁ |---| C₂ |---| C₃ |---
   |  |     |  |     |  |
   +  -     +  -     +  -
   |← V₁ →|← V₂ →|← V₃ →|
   |←──────── V ────────→|

Characteristics of Series Combination:

• Charge (Q) across each capacitor remains the same.
• Potential difference (V) across each capacitor is different.

Derivation for Equivalent Capacitance (C_s):

The potential differences across the individual capacitors are:

V₁ = Q/C₁, V₂ = Q/C₂, V₃ = Q/C₃

For the series circuit, the sum of these potential differences must equal the applied potential difference V:

V = V₁ + V₂ + V₃ = Q/C₁ + Q/C₂ + Q/C₃

Or, V = Q (1/C₁ + 1/C₂ + 1/C₃)

If C_s is the equivalent capacitance of the series combination, then V = Q/C_s.

Therefore, Q/C_s = Q (1/C₁ + 1/C₂ + 1/C₃)

Hence, the relation for equivalent capacitance in series is:

1/C_s = 1/C₁ + 1/C₂ + 1/C₃

Given Figure: The figure shows three capacitors connected between points A and B. Two capacitors of 2 µF each are connected in series, and this series combination is connected in parallel with another capacitor of 2 µF.

        A
        |
        ├─── 2µF ───┐
        │           │
        │           ├─── B
        │           │
        └─── 2µF ───┘
        │
        └─── 2µF ───┘

Step 1: First, find the equivalent capacitance of the two 2 µF capacitors connected in series.

For series combination: 1/C_s = 1/2 + 1/2 = 1

Therefore, C_s = 1 µF

Step 2: This 1 µF equivalent capacitance is connected in parallel with the third 2 µF capacitor.

For parallel combination: C_eq = C_s + 2 µF = 1 µF + 2 µF = 3 µF

Therefore, the equivalent capacitance between points A and B is 3 µF.

Step-by-step solution:

Capacitors of 4 µF and 4 µF are in parallel, so their equivalent capacitance = 4 + 4 = 8 µF.

Capacitors of 1 µF and 1 µF are in parallel, so their equivalent capacitance = 1 + 1 = 2 µF.

Now, the two 2 µF capacitors are in series, so their equivalent capacitance = (2 × 2) / (2 + 2) = 4/4 = 1 µF.

Similarly, the two 8 µF capacitors are in series, so their equivalent capacitance = (8 × 8) / (8 + 8) = 64/16 = 4 µF.

Now, the capacitors of 1 µF and 4 µF are in parallel, so the total equivalent capacitance between A and B = 1 + 4 = 5 µF.

Answer: The equivalent capacitance between A and B is 5 µF.

Ampere's Circuital Law: It states that the line integral of the magnetic field B around any closed loop is equal to μ₀ (permeability of free space) times the total current I passing through the loop.

Mathematically: ∮ B · dl = μ₀ I

Diagram: Consider a circular loop of radius r around an infinitely long straight wire carrying current I. The magnetic field lines are concentric circles around the wire.

Derivation:

• By symmetry, the magnitude of the magnetic field B is the same at every point on the circular loop.
• The direction of B at any point is tangential to the circle.
• The line integral of B around the loop: ∮ B · dl = ∮ B dl cos 0° = B ∮ dl = B × (2πr)
• Applying Ampere's law: B × 2πr = μ₀ I
• Therefore, the magnetic field at a distance r from the wire is: B = (μ₀ I) / (2πr)

Biot-Savart Law: It states that the magnetic field dB produced by a small current element Idl at a point is:

• Directly proportional to the current I and the length of the element dl.
• Directly proportional to the sine of the angle between the element and the line joining the element to the point.
• Inversely proportional to the square of the distance r from the element.

Mathematically: dB = (μ₀ / 4π) × (I dl sinθ) / r²

Working of a Cyclotron:

• A cyclotron consists of two hollow D-shaped metal chambers (dees) placed in a uniform magnetic field perpendicular to their plane.
• A high-frequency alternating voltage is applied across the dees.
• Charged particles (e.g., protons) are injected at the center. The magnetic field makes them move in a circular path.
• Each time the particle crosses the gap between the dees, the alternating voltage accelerates it, increasing its speed and radius of curvature.
• This process continues until the particle reaches the edge of the dees, where it is extracted.

Diagram: A schematic sketch shows a spiral path of the particle inside the two dees, with the particle gaining energy at each crossing of the gap.

Derivation of Cyclotron Frequency:

• In a magnetic field B, the centripetal force on a particle of charge q, mass m, and velocity v is provided by the magnetic force: qvB = mv²/r
• The radius of the path: r = mv/(qB)
• The time period for one revolution: T = 2πr/v = 2πm/(qB)
• The cyclotron frequency (angular frequency): ω = 2π/T = qB/m
• Thus, the cyclotron frequency is: f = qB/(2πm)

dB ∝ (I dl sin θ) / r²

or dB = (μ₀ / 4π) × (I dl sin θ) / r²

Here, μ₀ = 4π × 10⁻⁷ T m A⁻¹ (or Wb m⁻¹ A⁻¹) is the permeability of free space.

Key Points:

• Cyclotron frequency: ν = qB / (2πm) – independent of speed and radius.
• Resonance condition: Frequency of applied voltage = cyclotron frequency.
• Maximum K.E. = q² B² r₀² / (2m)

Solution:

Given: Radius of curvature, R = 0 cm

Using the relation: R = 2f

Therefore, f = R/2 = 0/2 = 0 cm

Answer: The focal length of the mirror is 0 cm. (A spherical mirror with zero radius of curvature is a plane mirror.)

Solution:

Given: Fringe width, β = 2 mm

For dark fringes, the position of the n^th dark fringe from the central fringe is given by:

\[ x_n = (2n - 1)\frac{\lambda D}{2d} = (2n - 1)\frac{\beta}{2} \]

For the second dark fringe, n = 2:

\[ x_2 = (2 \times 2 - 1) \times \frac{2 \text{ mm}}{2} = (4 - 1) \times 1 \text{ mm} = 3 \text{ mm} \]

Answer: The distance of the second dark fringe from the central fringe is 3 mm.