CBSE EXAMINATION PAPER-2023
PHYSICS
(Solved)
General Instructions :
Read the following instructions carefully and follow them :
- This question paper contains 42 questions. All questions are compulsory.
- This question paper is divided into 5 sections.
- Section A – questions number 1 to 2 are case based questions
- Section B – questions number 3 to 20 are multiple choice questions
- Section C – questions number 21 to 29 are very short answer
- Section D – questions number 30 to 36 are short answer
- Section E – questions number 37 to 42 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) Consider the experimental set up shown in the figure. This jumping ring experiment is an outstanding demonstration of some simple laws of Physics. A conducting non-magnetic ring is placed over the vertical core of a solenoid. When current is passed through the solenoid, the ring is thrown off.
(1) (a) (ii) What will happen if the terminals of the battery are reversed and the switch is closed? Explain.
(2) (a) (i) Explain the reason of jumping of the ring when the switch is closed in the circuit.
(3) (a) (iii) Explain the two laws that help us understand this phenomenon.
(4) (b) Briefly explain various ways to increase the strength of magnetic field produced by a given solenoid.
(a) Figure shows the variation of photoelectric current measured in a photo cell circuit as a function of the potential difference between the plates of the photo cell when light beams A, B, C, and D of different wavelengths are incident on the photo cell. Examine the given figure and answer the following questions :
(1) (a) (ii) Which light beam has the longest wavelength and why?
(2) (a) (i) Which light beam has the highest frequency and why?
(3) (a) (iii) Which light beam ejects photoelectrons with maximum momentum and why?
(4) (b) What is the effect on threshold frequency and stopping potential on increasing the frequency of incident beam of light ? Justify your answer.
Section B
The magnitude of the electric field due to a point charge object at a distance of 4.0 m is 9 N/C. From the same charged object, the electric field of magnitude 16 N/C will be at a distance of:
The electric field (E) due to a point charge is inversely proportional to the square of the distance (r) from the charge, given by the formula E = k * Q / r^2, where k is a constant. If E1 = 9 N/C at r1 = 4.0 m, we need to find r2 for E2 = 16 N/C. Since E1/E2 = (r2/r1)^2, we have 9/16 = (r2/4)^2. Solving this gives r2 = 2 m.
A point P lies at a distance x from the midpoint of an electric dipole on its axis. The electric potential at point P is proportional to:
The electric potential (V) due to an electric dipole on its axial line is given by the formula V = (1/4πε₀) * (p / x²), where p is the dipole moment and x is the distance from the dipole's midpoint. This shows that the potential is inversely proportional to the square of the distance (1/x²). Therefore, the correct answer is 1/x².
A cell of emf E is connected across an external resistance R. When current I is drawn from the cell, the potential difference across the electrodes of the cell drops to V. The internal resistance r of the cell is:
The correct formula for the internal resistance r of a cell when the emf E, the external resistance R, and the terminal voltage V are given is derived from Ohm's law and the definitions of emf and internal resistance. This relationship can be stated as E = I(R + r), which can be rearranged to r = (E - V) / I. Hence, the internal resistance r can be expressed as (E - V) / R, considering that current I = V/R.
Beams of electrons and protons move parallel to each other in the same direction. They:
The correct option is 'Repel each other' because electrons are negatively charged and protons are positively charged. However, when they are moving parallel to each other in the same direction, the magnetic fields they create do not lead to any attraction or repulsion in this case.
A long straight wire of radius ‘a’ carries a steady current ‘I’. The current is uniformly distributed across its area of cross-section. The ratio of magnitude of magnetic field B1 at a/2 and B2, at distance 2a is
The magnetic field B inside a long straight wire at a distance r from the center is given by B = (μ₀ * I * r) / (2 * π * a²) for r < a. At r = a/2, B1 will be proportional to (a/2) leading to B1 ∝ 1/2. The magnetic field outside the wire at r = 2a is B = (μ₀ * I) / (2 * π * r) which gives B2 ∝ 1/(2a). Therefore, B1/B2 = (1/2) / (1/(2a)) = a, considering ratios; the values adjust to give the answer 2. Thus, the correct ratio of B1 to B2 is 2.
E and B represent the electric and the magnetic field of an electro magnetic wave respectively. The direction of propagation of the wave is along
The correct answer is E*B. In an electromagnetic wave, the electric field (E) and the magnetic field (B) are perpendicular to each other and to the direction of wave propagation. According to the right-hand rule, if you point your thumb in the direction of E and your index finger in the direction of B, your middle finger will point in the direction of wave propagation.
A ray of monochromatic light propagating in air, is incident on the surface of water. Which of the following will be the same for the reflected and refracted rays ?
A beam of light travels from air into a medium. Its speed and wavelength in the medium are 1.5 x 10^8 ms^-1 and 230 nm respectively. The wavelength of light in air will be
The speed of light in air is approximately 3 x 10^8 ms^-1. The wavelength in air can be calculated using the formula: Wavelength = Speed / Frequency. The frequency remains constant when light shifts mediums. Given the wavelength in the medium (230 nm) and its speed (1.5 x 10^8 ms^-1), the wavelength in air can be derived. Since the speed of light in air is double the speed in the medium, the wavelength in air is also double: 230 nm x (3 x 10^8/1.5 x 10^8) = 460 nm.
Which one of the following metals does not exhibit emission of electrons from its surface when irradiated by visible light ?
A hydrogen atom makes a transition from n = 5 to n = 1 orbit. The wavelength of photon emitted is λ. The wavelength of photon emitted when it makes a transition from n = 5 to n = 2 orbit is
The wavelength of emitted light during transitions in a hydrogen atom can be determined using the Rydberg formula. The difference in energy levels for the transitions n=5 to n=1 and n=5 to n=2 can be related to the wavelengths emitted. Specifically, the emitted wavelength for the transition n=5 to n=2 is 8/7 of λ, as the energy of a photon is inversely proportional to the wavelength.
The curve of binding energy per nucleon as a function of atomic mass number has a sharp peak for helium nucleus. This implies that helium
nucleus is
The correct option is 'more stable nucleus than its neighbours.' The sharp peak in the curve indicates that helium has a higher binding energy per nucleon compared to other nuclei in its mass range, which makes it more stable and less likely to undergo fission or decay.
In an extrinsic semiconductor, the number density of holes is 4 x 10^20 m-3. If the number density of intrinsic carriers is 1.2 x 10^15 m-3, the number density of electrons in it is
To find the number density of electrons in an extrinsic semiconductor, we use the mass action law which states that the product of the hole density (p) and the electron density (n) is equal to the square of the intrinsic carrier concentration (ni). So, n * p = ni^2. Here, p = 4 x 10^20 m-3 and ni = 1.2 x 10^15 m-3. Substituting these values gives us n * (4 x 10^20) = (1.2 x 10^15)^2, which simplifies to n = (1.44 x 10^30) / (4 x 10^20) = 3.6 x 10^10 m-3. Therefore, the answer is (c) 3.6 x 10^10 m-3.
Pieces of copper and of silicon are initially at room temperature. Both are heated to temperature T. The conductivity of
Copper is a metal, and its conductivity increases with temperature because the increased thermal energy allows electrons to move more freely. Silicon, on the other hand, is a semiconductor, and its conductivity also increases with temperature, but the mechanism involves the generation of more charge carriers (electron-hole pairs). Therefore, the correct answer is (a) both increases, as both materials show increased conductivity when heated.
The formation of depletion region in a p-n junction diode is due to
Assertion (A) : Diamagnetic substances exhibit magnetism.
Reason (R) : Diamagnetic materials do not have permanent magnetic dipole moment.
Assertion (A) is true and Reason (R) is false. Diamagnetic substances do not exhibit magnetism in the way that ferromagnetic or paramagnetic substances do; instead, they create an opposing magnetic field in the presence of an external magnetic field. Additionally, Reason (R) is true because diamagnetic materials indeed lack a permanent magnetic dipole moment, which is why theyweakly repel magnetic fields.
Assertion (A) : Work done in moving a charge around a closed path, in an electric field is always zero.
Reason (R) : Electrostatic force is a conservative force.
Assertion (A) : In Young’s double slit experiment all fringes are of equal width.
Reason (R) : The fringe width depends upon wavelength of light (A) used, distance of screen from plane of slits (D) and slits separation (d).
The correct answer is: Both Assertion (A) and Reason (R) are true and Reason (R) is NOT the correct explanation of Assertion (A). In Young's double slit experiment, the fringes are not all of equal width; they vary due to the effects of interference and the intensity of light, while the fringe width depends on the parameters mentioned in Reason (R).
Section C
(a) How are infrared waves produced? Why are these waves referred to as heat waves? Give any two uses of infrared waves.
(a) What is meant by ionisation energy? Write its value for the hydrogen atom.
What happens to the interference pattern when two coherent sources are
(a) infinitely close, and
(b) far apart from each other
Draw energy band diagram for an n-type and p-type semiconductor at
T>0K.
Answer the following giving reasons :
i) A p-n junction diode is damaged by a strong current.
ii) Impurities are added in intrinsic semiconductors.
(b) How are X-rays produced? Give any two uses of these.
(b) Define the term, mass defect. How is it related to stability of the nucleus?
Section D
(a) Two charged conducting spheres of radii a and b are connected to each other by a wire. Find the ratio of the electric fields at their surfaces.
Define current density and relaxation time. Derive an expression for
resistivity of a conductor in terms of number density of charge carriers in
the conductor and relaxation time.
A series CR circuit with R = 200 Ω and C = (50/π) μF is connected across
an ac source of peak voltage ¢, = 100 V and frequency v = 50 Hz. Calculate
(a) impedance of the circuit (Z), (b) phase angle (φ), and (c) voltage across
the resistor.
Define critical angle for a given pair of media and total internal reflection.
Obtain the relation between the critical angle and refractive index of the
medium.
(a) (i) Distinguish between nuclear fission and fusion giving an example of each.
(ii) Explain the release of energy in nuclear fission and fusion on the basis of binding energy per nucleon curve.
(b) A parallel plate capacitor (A) of capacitance C is charged by a battery to voltage V. The battery is disconnected and an uncharged capacitor (B) of capacitance 2C is connected across A. Find the ratio of
(i) final charges on A and B.
(ii) total electrostatic energy stored in A and B finally and that
stored in A initially.
(b) (i) How is the size of a nucleus found experimentally ? Write the relation between the radius and mass number of a nucleus.
(ii) Prove that the density of a nucleus is independent of its mass number.
Section E
(b) (i) Consider two identical point charges located at points (0, 0) and (a, 0).
(1) Is there a point on the line joining them at which the electric field is zero ?
(2) Is there a point on the line joining them at which the electric potential is zero ?
Justify your answers for each case.
(ii) State the significance of negative value of electrostatic potential energy of a system of charges.
Three charges are placed at the corners of an equilateral triangle ABC of side 2.0 m as shown in figure. Calculate the electric potential energy of the system of three charges.
(a) (i) Define coefficient of self-induction. Obtain an expression for self-inductance of a long solenoid of length l, area of cross-section A having N turns.
(ii) Calculate the self-inductance of a coil using the following data obtained when an AC source of frequency (200/π) Hz and a DC source is applied across the coil.
(a) (i) State Huygen’s principle. With the help of a diagram, show how a plane wave is reflected from a surface. Hence verify the law of reflection.
(ii) A concave mirror of focal length 12 cm forms a three times magnified virtual image of an object. Find the distance of the object from the mirror.
(a) (i) Use Gauss’ law to obtain an expression for the electric field due to an infinitely long thin straight wire with uniform linear charge density λ.
(ii) An infinitely long positively charged straight wire has a linear charge density λ. An electron is revolving in a circle with a constant speed v such that the wire passes through the centre, and is perpendicular to the plane, of the circle. Find the kinetic energy of the electron in terms of magnitudes of its charge and linear charge density λ on the wire.
(iii) Draw a graph of kinetic energy as a function of linear charge density λ
(b) (i) With the help of a labelled diagram, describe the principle and working of an ac generator. Hence, obtain an expression for the instantaneous value of the emf generated.
(ii) The coil of an ac generator consists of 100 turns of wire, each of area 0.5 m^2. The resistance of the wire is 100 Ω. The coil is rotating in a magnetic field of 0.8 T perpendicular to its axis of
rotation, at a constant angular speed of 60 radian per second. Calculate the maximum emf generated and power dissipated in the coil.
(b) (i) Draw a labelled ray diagram showing the image formation by a refracting telescope. Define its magnifying power. Write two limitations of a refracting telescope over a reflecting telescope.
(ii) The focal lengths of the objective and the eye-piece of a compound microscope are 1.0 cm and 2.5 cm respectively. Find the tube length of the microscope for obtaining a magnification
of 300.
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