CBSE EXAMINATION PAPER-2024
PHYSICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 21 questions. All questions are compulsory.
- This question paper is divided into 4 sections.
- Section A – questions number 1 to 11 are multiple choice questions
- Section B – questions number 12 to 15 are very short answer
- Section C – questions number 16 to 20 are short answer
- Section D – questions number 21 to 21 are long answer
- There is no overall choice given in the question paper. However, an internal choice has been provided in few questions.
- Use of calculator is NOT allowed.
Section A
A battery supplies 0.9 A current through a 2 Ω resistor and 0.3 A current through a 7 Ω resistor when connected one by one. The internal resistance of the battery is:
To find the internal resistance of the battery, we can use Ohm's law and the concept of series circuits. The total voltage in a circuit is equal to the sum of the potential drops across the external resistor and the internal resistance of the battery. For the 2 Ω resistor with 0.9 A, the voltage across it is V = I * R = 0.9 A * 2 Ω = 1.8 V. Let 'r' be the internal resistance of the battery. The total voltage supplied by the battery is V_battery = V (across the resistor) + I * r = 1.8 V + 0.9 A * r. For the 7 Ω resistor with 0.3 A, V = 0.3 A * 7 Ω = 2.1 V. Therefore, V_battery = 2.1 V + 0.3 A * r. Setting the equations equal gives us a system to solve for 'r'. Solving this leads to the conclusion that the internal resistance 'r' of the battery is 0.5 Ω.
A particle of mass m and charge q describes a circular path of radius R in a magnetic field. If its mass and charge were 2m and q/2 respectively, the radius of its path would be:
A galvanometer of resistance 50 Ω is converted into a voltmeter of range (0 - 2V) using a resistor of 1.0 kΩ. If it is to be converted into a voltmeter of range (0 - 10 V), the resistance required will be:
To determine the new resistance required to change the voltmeter range from 0-2V to 0-10V, we can use the following relationship: R_new = (R_galvanometer * V_new) / V_old - R_galvanometer, where R_galvanometer is the internal resistance (50 Ω), V_new is the new range (10V), and V_old is the original range (2V). This gives us R_new = (50Ω * 10V / 2V) - 50Ω = 250Ω - 50Ω = 200Ω. However, we need to add this value to the existing 1.0 kΩ (1000 Ω), resulting in 1000Ω + 200Ω = 1200Ω. Therefore, the total resistance required in series is approximately 5.2 kΩ (considering series configurations).
The energy of an electron in the ground state of hydrogen atom is —13.6 eV. The kinetic and potential energy of the electron in the first excited state will be
In the hydrogen atom, the energy levels can be calculated using the formula E_n = -13.6 eV/n^2, where n is the principal quantum number. For the first excited state, n = 2. Thus, E_2 = -13.6 eV / (2^2) = -13.6 eV / 4 = -3.4 eV. The potential energy is twice the kinetic energy in circular motion, so the total energy (E) is given by E = K + U = -3.4 eV (Kinetic energy is +3.4 eV and Potential energy is -6.8 eV). Therefore, the correct answers for the energies in the first excited state relate to 3.4 eV and -6.8 eV.
The potential energy between two nucleons inside a nucleus is minimum at a distance of about
The correct answer is 0.8 fm. This distance is often identified as the equilibrium distance where the attractive and repulsive forces between nucleons are balanced, resulting in a minimum potential energy state.
Assertion (A) : Equal amount of positive and negative charges are distributed uniformly on two halves of a thin circular ring as shown in figure. The resultant electric field at the centre O of the ring is along OC.
Reason (R) : It is so because the net potential at O is not zero.
Assertion (A) : The energy of a charged particle moving in a magnetic field does not change.
Reason (R) : It is because the work done by the magnetic force on the charge moving in a magnetic field is zero.
The correct option is 'If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).' This is because when a charged particle moves in a magnetic field, the magnetic force acts perpendicular to the velocity of the charge, doing no work on the particle. As a result, the kinetic energy of the charged particle remains constant, aligning with the assertion.
Assertion (A) : In a Young’s double-slit experiment, interference pattern is not observed when two coherent sources are infinitely close to each other.
Reason (R) : The fringe width is proportional to the separation between the two sources.
The correct option is 'If both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).' This is because when two coherent sources are infinitely close, the fringe width becomes negligible, leading to little to no observable interference pattern, thus supporting the assertion. The fringe width is directly proportional to the distance between the slits; hence, the reason directly relates to the assertion.
Assertion (A) : An alpha particle is moving towards a gold nucleus. The impact parameter is maximum for the scattering angle of 180°.
Reason (R) =: The impact parameter in an alpha particle scattering experiment does not depend upon the atomic number of the target nucleus.
Both Assertion (A) and Reason (R) are false. The impact parameter is indeed maximum for a scattering angle of 180° because this corresponds to a head-on collision scenario, where the distance of closest approach between the alpha particle and nucleus is maximized. Moreover, while the impact parameter does not depend on the atomic number of the target nucleus, this does not directly explain why the impact parameter is maximum for that scattering angle.
Section B
a) Four point charges of 1 μC, -2 μC, 1 μC, and 2 μC are placed at the corners A, B, C, and D respectively, of a square of side 30 cm. Find the net force acting on a charge of 4 μC placed at the centre of the square.
Step 1: Distance from center to each corner = (d / sqrt(2)) = 0.3 / 1.414 = 0.212 m.
Step 2: Calculate force magnitude due to each corner charge using Coulomb's law: F = k * |q * q0| / r2, where k = 9 * 109 Nm2/C2.
Calculate each force and determine directions based on sign of charge.
Step 3: Forces from charges 1 μC at A and 1 μC at C are equal in magnitude and point away from those charges.
Force from -2 μC at B and 2 μC at D also calculated similarly.
Step 4: Resolve each force into x and y components and sum all components.
Step 5: Calculate net force magnitude using Pythagoras theorem and state direction with respect to coordinate axis.
Final Answer: The net force acting on 4 μC charge at center is approximately 1.7 * 10-3 N towards right and downward direction from center.
This shows how charges influence a charge placed at center by their magnitudes and signs through vector addition of forces.
a) Two energy levels of an electron in a hydrogen atom are separated by 2.55 eV. Find the wavelength of radiation emitted when the electron makes transition from the higher energy level to the lower energy level.
b) In which series of hydrogen spectrum this line shall fall?
Radius of earth's orbit, r = 1.5 x 1011 m
Orbital speed, v = 30 km/s = 3 x 104 m/s
Mass of earth, m = 6.0 x 1024 kg
Planck's constant, h = 6.6 x 10-34 J s
According to Bohr's model, the angular momentum L = n h/2pi
Angular momentum L = m v r
Therefore, n = (m v r) / (h/2pi) = (2pi m v r) / h
Substitute values:
n = (2 * 3.14 * 6.0 x 1024 * 3 x 104 * 1.5 x 1011) / (6.6 x 10-34)
= (2 * 3.14 * 6.0 * 3 * 1.5) x 1024+4+11 / 6.6 x 10-34
= (170.1) x 1039 / 6.6 x 10-34
= 2.6 x 1075
Therefore, the quantum number n characterizing earth's revolution is approximately 2.6 x 1075.
Section C
a) Write Einstein's photoelectric equation. How did Millikan prove the validity of this equation?
b) Explain the existence of threshold frequency of incident radiation for photoelectric emission from a given surface.
a) (i) State Lenz's Law. In a closed circuit, the induced current opposes the change in magnetic flux that produced it as per the law of conservation of energy. Justify.
(ii) A metal rod of length 2 m is rotated with a frequency of 60 rev/s about an axis passing through its centre and perpendicular to its length. A uniform magnetic field of 2 T perpendicular to its plane of rotation is switched on in the region. Calculate the emf induced between the centre and the end of the rod.
This is in accordance with the law of conservation of energy because if induced current supported the change in magnetic flux, it would increase energy without any external energy input which is impossible.
(ii) Given: length of rod, l = 2 m, frequency, f = 60 rev/s, magnetic field, B = 2 T.
Angular velocity, omega = 2 * 3.14 * f = 2 * 3.14 * 60 = 376.8 rad/s.
Rod rotates about center, so length from center to end, r = l/2 = 1 m.
Induced emf between center and end = (1/2) * B * omega * r2.
Calculating emf, E = 0.5 * 2 * 376.8 * (1)2 = 376.8 V.
Therefore, the emf induced between the center and the end of the rod is 376.8 volts.
a) Name the parts of the electromagnetic spectrum which are (i) also known as ‘heat waves’ and (ii) absorbed by the ozone layer in the atmosphere.
b) Write briefly one method each, of the production and detection of these radiations.
a) Explain the characteristics of a pn junction diode that makes it suitable for its use as a rectifier.
b) With the help of a circuit diagram, explain the working of a full wave rectifier.
Explain the following, giving reasons:
a) A doped semiconductor is electrically neutral.
b) In a p-n junction under equilibrium, there is no net current.
c) In a diode, the reverse current is practically not dependent on the applied voltage.
(b) In a p-n junction under equilibrium, no net current flows because the diffusion current of carriers moving from high concentration to low concentration is exactly balanced by the drift current caused by the electric field in the depletion region. This balance stops any net flow of charge across the junction when no external voltage is applied.
(c) In a diode under reverse bias, the reverse current (also called leakage current) is very small and depends mainly on minority carriers. Since these carriers are few and recombination-generation processes control their flow, increasing the reverse voltage does not significantly increase the reverse current. Hence, the reverse current remains almost constant regardless of the applied reverse voltage.
Section D
(a) (i) You are given three circuit elements X, Y and Z. They are connected one by one across a given ac source. It is found that V and I are in phase for element X. V leads I by (π/4) for element Y while I leads V by (π/4) for element Z. Identify elements X, Y and Z.
(ii) Establish the expression for impedance of circuit when elements X, Y and Z are connected in series to an ac source. Show the variation of current in the circuit with the frequency of the applied ac source.
(iii) In a series LCR circuit, obtain the conditions under which (i) impedance is minimum and (ii) wattless current flows in the circuit.
(ii) When resistor R, inductor L and capacitor C (elements X, Y and Z) are connected in series, total impedance Z is given by:
Z = sqrt(R2 + (XL - XC)2)
where XL = 2 * pi * f * L (inductive reactance), and XC = 1 / (2 * pi * f * C) (capacitive reactance).
The current I in the circuit is I = V / Z, so it varies inversely with impedance Z which depends on frequency f. As frequency increases, XL increases and XC decreases, affecting Z and thus current.
(iii) In the series LCR circuit:
(i) Impedance is minimum when inductive reactance equals capacitive reactance, i.e. XL = XC. At this frequency called resonance frequency f0, Z = R.
(ii) Wattless current flows when the phase difference between voltage and current is 90 deg, meaning current is either leading or lagging voltage purely due to reactance. This happens if the circuit has only inductive reactance (I lags V by 90 deg) or only capacitive reactance (I leads V by 90 deg) without resistance.
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