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Punjab Physics 2015 Paper 1 — Solved Past Paper with Answers
All 18 MCQs from Punjab Physics 2015 Paper 1, solved with the correct answer highlighted and a full explanation for every question. This is a free MDCAT Punjab / UHS past paper — no signup, no ads. Practise it interactively in timed mode, drill more with free MDCAT MCQs, or browse all Punjab / UHS papers.
Attempt this paper interactively →Q1. The electrostatic force of repulsion between two electrons at a distance 1m is:
- A. 2.3x10-24N
- B. 2.3x107-28N
- C. 2.3x10-28N
- D. 2.3x10-30N✓
Explanation: The explanation will be added soon!
Q2. A charged conductor has charge on its?
- A. Inner-surface
- B. Outer-surface✓
- C. Middle point
- D. Surrounding space
Explanation: Excess charge on a conductor resides on the outer surface due to electrostatic repulsion, as mentioned above. This ensures that the electric field inside the conductor is zero and the charges are in a stable equilibrium state.
Why the other options are wrong
- A. For a charged conductor, in the case of excess charge, the charge resides on the outer surface. This is due to the principle of electrostatic equilibrium, which causes charges to repel each other and distribute themselves evenly on the outer surface of the conductor. Therefore, the charge does not reside on the inner surface, making option A incorrect.
- C. The charge does not concentrate at the middle point of a charged conductor. The distribution of charge is such that it remains on the outer surface, as explained above. Therefore, option C is incorrect.
- D. While a charged conductor can create an electric field in the surrounding space due to its charge distribution, the charge itself resides on the outer surface of the conductor. The charge does not exist freely in the surrounding space but rather is bound to the conductor's surface due to electrostatic forces. Therefore, option D is incorrect.
Q3. Ampere second stands for the unit of:
- A. Charge✓
- B. Emf
- C. Energy
- D. Power
Explanation: The unit of charge is the coulomb (C), but it can also be expressed as ampere-second (A·s). This is because 1 coulomb is equivalent to 1 ampere of current flowing for 1 second. So, an ampere-second (A·s) is a measure of the quantity of electricity transferred by a current of 1 ampere in 1 second.
Why the other options are wrong
- B. The unit of electromotive force is the volt (V), not ampere-second (A·s). Electromotive force is a measure of the energy per unit charge that is supplied by a source of electrical energy, such as a battery or generator.
- C. The unit of energy is the joule (J), not ampere-second (A·s). Energy is the capacity to do work, and in the context of electricity, it can be related to the amount of charge moved through a potential difference (voltage).
- D. The unit of power is the watt (W), not ampere-second (A·s). Power is the rate at which work is done or the rate at which energy is transferred, and it is typically measured in joules per second (J/s) or watts.
Q4. The relation between current I and angle of deflection θ in a moving coil galvanometer is:
- A. I α θ✓
- B. I α I/θ
- C. I α sinθ
- D. I α cosθ
Explanation: In a moving coil galvanometer, the deflection of the coil is proportional to the current passing through it. Therefore, the current (I) is proportional to the angle of deflection (theta).
Why the other options are wrong
- B. This relation is incorrect. It suggests that the current is inversely proportional to the angle of deflection, which is not the case in a moving coil galvanometer.
- C. This relation is incorrect. The current is directly proportional to the angle of deflection, not the sine of the angle.
- D. This relation is also incorrect. The current is directly proportional to the angle of deflection, not the cosine of the angle.
Q5. Lorentz force is given by:
- A. Option A
- B. Option B
- C. Option C
- D. Option D✓
Explanation: The explanation will be added soon!
Q6. Which one of the following is not present in an A.C generator?
- A. Armature
- B. Commutator✓
- C. Magnet
- D. Slip ring
Explanation: A commutator is a component found in DC generators, not in AC generators. It consists of a rotating switch or contacts that reverse the direction of the current in the armature coils every half-turn, converting the alternating current induced in the armature coils into direct current.
Why the other options are wrong
- A. The armature is a crucial component of an AC generator. It consists of coils of wire mounted on a rotating shaft. The armature rotates within a magnetic field, inducing an alternating current in the coils through electromagnetic induction.
- C. In an AC generator, a magnet or an electromagnet is used to create a magnetic field. This magnetic field interacts with the rotating armature, inducing an alternating current in the armature coils.
- D. Slip rings are present in an AC generator. They are used to collect the alternating current produced in the armature coils. Unlike a commutator, slip rings allow the current to flow continuously in one direction, without changing its direction as in a DC generator.
Q7. The core of transformer is laminated to reduce:
- A. Magnetic loss
- B. Hysteress loss
- C. Eddy current loss✓
- D. Electric loss
Explanation: Laminating the core of a transformer is primarily done to reduce eddy current loss. Eddy currents are induced currents that flow in the core material due to the changing magnetic field. These currents lead to energy loss in the form of heat, which is minimized by laminating the core into thin layers perpendicular to the direction of the magnetic field.
Why the other options are wrong
- A. Laminating the core of a transformer helps reduce eddy current loss, not magnetic loss. Magnetic loss, also known as core loss or iron loss, occurs due to the magnetic properties of the core material and is minimized by using high-quality magnetic materials and proper core design.
- B. Laminating the core of a transformer also helps reduce hysteresis loss. Hysteresis loss occurs due to the repeated magnetization and demagnetization of the core material and is minimized by using materials with low hysteresis, such as silicon steel, and by proper core design.
- D. There is no specific term "electric loss" in the context of transformers. The losses in a transformer are primarily magnetic losses (hysteresis and eddy current losses) and copper losses (due to the resistance of the winding wires). Laminating the core helps reduce magnetic losses, specifically eddy current loss, not electric loss.
Q8. In RLC parallel circuit the resonance frequency is:
- A. Option A
- B. Option B
- C. Option C
- D. Option D✓
Explanation: The explanation will be added soon!
Q9. Electromagnetic waves emitted from radio antenna are:
- A. Stationary
- B. Longitudnal
- C. Transverse✓
- D. Both A & B
Explanation: Electromagnetic waves, including radio waves, are transverse waves. In a transverse wave, the oscillations of the medium (electric and magnetic fields in this case) are perpendicular to the direction of wave propagation. This characteristic allows electromagnetic waves to travel through vacuum, as they do not require a medium for propagation.
Why the other options are wrong
- A. Electromagnetic waves emitted from a radio antenna are not stationary. Stationary waves refer to a specific type of wave pattern that appears to be standing still, characterized by points that appear to be oscillating in place. Electromagnetic waves, including radio waves, are not stationary but propagate through space.
- B. Electromagnetic waves are not longitudinal. Longitudinal waves are waves in which the displacement of the medium is in the same direction as, or the opposite direction to, the direction of propagation of the wave. In contrast, electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to the direction of propagation, making them transverse waves.
- D. Neither option A nor B is correct, so option D is also incorrect. Electromagnetic waves are transverse and not stationary or longitudinal.
Q10. Minority carriers in p-type substances are:
- A. Holes
- B. Electrons✓
- C. Protons
- D. Positrons
Explanation: In p-type semiconductor materials, which are doped with acceptor impurities, the majority carriers are "holes." Holes are essentially vacancies in the valence band created when electrons from the valence band are excited to the conduction band. These holes act as positively charged carriers that can move through the material.
Why the other options are wrong
- A. In p-type semiconductor materials, which are doped with acceptor impurities, the majority carriers are "holes." Holes are essentially vacancies in the valence band created when electrons from the valence band are excited to the conduction band. These holes act as positively charged carriers that can move through the material.
- C. Protons are not carriers in semiconductor materials. Protons are positively charged particles found in atomic nuclei and do not participate in the conduction of electricity in semiconductors.
- D. Positrons are the antimatter counterparts of electrons and are not typically found in semiconductor materials as carriers. Like protons, they do not play a role in the electrical conduction of semiconductors.
Q11. The gain of non-inverting amplifire is:
- A. Option A
- B. Option B
- C. Option C
- D. Option D✓
Explanation: The explanation will be added soon!
Q12. The common emitter current amplification factor β is given by:
- A. Option A
- B. Option B
- C. Option C
- D. Option D✓
Explanation: The explanation will be added soon!
Q13. In compton effect the photon behaves as a:
- A. Wave
- B. Particle✓
- C. Nucleon
- D. Both A & C
Explanation: In the Compton effect, the photon behaves as a particle. The interaction between the X-ray photon and the electron is described by the transfer of energy and momentum as if the photon were a discrete particle.
Why the other options are wrong
- A. In the Compton effect, which is a phenomenon in which X-ray photons are scattered by electrons, the photon behaves more like a particle than a wave. This behavior is evident in the conservation of momentum and energy during the scattering process, which is characteristic of particle-like behavior.
- C. Nucleons are particles found in atomic nuclei, including protons and neutrons. Nucleons are not directly related to the Compton effect, which involves the scattering of X-ray photons by electrons.
- D. Option D is incorrect because the Compton effect primarily demonstrates the particle-like behavior of photons, not their wave-like behavior (option A), nor does it involve nucleons (option C).
Q14. The binding energy per nucleon is maximum for:
- A. Helium
- B. Iron✓
- C. Polonium
- D. Radium
Explanation: Iron (specifically, iron-56) has the highest binding energy per nucleon of all the elements. This means that, on average, each nucleon (proton or neutron) in an iron nucleus is bound more tightly than in any other element. This is why iron is often referred to as the "peak of stability."
Why the other options are wrong
- A. Helium has a low atomic number and is relatively light compared to heavier elements. While helium is stable, its binding energy per nucleon is not as high as that of iron, which is closer to the peak of the binding energy curve.
- C. Polonium is a relatively heavy and unstable element. While it has a high atomic number, its binding energy per nucleon is not as high as that of iron.
- D. Radium is also a relatively heavy and unstable element. While it is known for its radioactivity, its binding energy per nucleon is not as high as that of iron.
Q15. Gamma rays from cobalt-60 are used for the treatment of:
- A. Circulation of blood
- B. Cancer✓
- C. Heart attack
- D. Thyroid glands
Explanation: Gamma rays from cobalt-60 are used in radiation therapy for the treatment of cancer. The high-energy gamma rays can penetrate tissue and damage the DNA of cancer cells, leading to their destruction.
Why the other options are wrong
- A. Gamma rays from cobalt-60 are not used for the treatment of circulation of blood. Treatments for circulation issues typically involve medications, lifestyle changes, or surgical procedures, not gamma radiation.
- C. Gamma rays from cobalt-60 are not used for the treatment of heart attacks. Heart attack treatment involves medications, lifestyle changes, and sometimes surgical procedures to restore blood flow to the heart muscle.
- D. Gamma rays from cobalt-60 are not typically used for the treatment of thyroid gland issues. Treatments for thyroid conditions often involve medications or, in some cases, radioactive iodine therapy, which uses a different radioactive isotope.
Q16. The radius of 10th orbit in hydrogen atom is:
- A. 0.053nm
- B. 0.53nm
- C. 5.3nm
- D. 53nm✓
Explanation: The explanation will be added soon!
Q17. In an electronic transition atom cannot emit:
- A. Infrared radiations
- B. Ultraviolet radiations
- C. Visible radiations
- D. γ-radiations✓
Explanation: Atoms do not emit gamma (γ) radiation during electronic transitions. Gamma radiation is a high-energy electromagnetic radiation emitted from the atomic nucleus during nuclear transitions. Electronic transitions involve changes in the electron cloud surrounding the nucleus and do not result in the emission of gamma radiation.
Why the other options are wrong
- A. Atoms can emit infrared radiation during electronic transitions. When an electron moves from a higher energy level to a lower energy level, it releases energy in the form of electromagnetic radiation. Depending on the energy difference between the levels, this radiation can fall in the infrared region of the electromagnetic spectrum.
- B. Atoms can emit ultraviolet (UV) radiation during electronic transitions. Similar to infrared radiation, when an electron moves to a lower energy level, it releases energy in the form of electromagnetic radiation. If this energy corresponds to the UV region of the spectrum, UV radiation is emitted.
- C. Atoms can emit visible light during electronic transitions. When the energy released during an electronic transition corresponds to the visible region of the electromagnetic spectrum, visible light is emitted. This is the basis for the colors emitted by excited atoms in flames or in gas discharge tubes.
Q18. 1/C_eq = 1/C_1 + 1/C_2 + 1/C_3 + ... + 1/C_n
- A. Series✓
- B. Parallel
- C. Both of them
- D. None of them
Explanation: Correct option is A.For capacitors in series, the total capacitance can be found by adding the reciprocals of the individual capacitance's, and taking the reciprocal of the sum.
Why the other options are wrong
- B. Incorrect as per formula
- C. Incorrect as per formula
- D. Incorrect as per formula
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