Home/Past Papers/Punjab / UHS/Punjab Physics 2023 Paper 1

Punjab Physics 2023 Paper 1 — Solved Past Paper with Answers

All 17 MCQs from Punjab Physics 2023 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 concept of an electric field was introduced by:

  • A. Henry
  • B. Faraday
  • C. Watt
  • D. Oersted

Explanation: Michael Faraday, a British scientist, introduced the concept of an electric field in the early 19th century. He used the concept to describe the influence that a charged object exerts on its surroundings, even in the absence of other charges.

Why the other options are wrong
  • A. Joseph Henry was an American scientist who made significant contributions to the development of electromagnetism and electromagnetic induction. However, he is not credited with introducing the concept of an electric field.
  • C. James Watt was a Scottish inventor and engineer known for his improvements to the steam engine. While he made important contributions to the field of engineering, he did not introduce the concept of an electric field.Hans Christian.
  • D. Hans Christian Oersted was a Danish physicist who discovered that electric currents create magnetic fields. While his discovery was crucial for understanding the relationship between electricity and magnetism, he did not introduce the concept of an electric field.

Q2. Electric field intensity due to an infinite sheet of charge is:

  • A. Option A
  • B. Option B
  • C. Option C
  • D. Option D

Explanation: For an infinite sheet of charge, the electric field intensity at a distance r from the sheet is given by E = σ/2ε, where σ is the surface charge density and ε is the permittivity of the medium.

Why the other options are wrong
  • A. The correct formula for the electric field intensity due to an infinite sheet of charge is E = σ/2ε, not 2σ/ε.
  • B. This is incorrect. The formula for the electric field intensity due to an infinite sheet of charge does not involve multiplying σ (surface charge density) by ε (permittivity of the medium).
  • D. This is incorrect. The correct formula for the electric field intensity due to an infinite sheet of charge is E = σ/2ε, not σ/ε.

Q3. The value of drift velocity of electrons is of the order of:

  • A. 103ms-1
  • B. 102ms-1
  • C. 10-3ms-1
  • D. 10-2ms-1

Explanation: The drift velocity of electrons in a conductor, under normal conditions, is on the order of 10^-3 m/s. This means that electrons move very slowly through a conductor in the direction opposite to the applied electric field.

Why the other options are wrong
  • A. This value is too high for the drift velocity of electrons in a typical conductor. Drift velocities are generally much lower than this, typically on the order of millimeters per second or less.
  • B. This value is too high for the drift velocity of electrons in a conductor. Drift velocities are typically much lower than this.
  • D. This value is also too high for the drift velocity of electrons in a conductor. Drift velocities are generally much lower than this, typically on the order of millimeters per second or less.

Q4. Formula for shunt resistance Rs is:

  • A. Option A
  • B. Option B
  • C. Option C
  • D. Option D

Explanation: In terms of current, the shunt resistor is used to provide a path for a known fraction of the total current to bypass the main circuit element, such as an ammeter. The total current \(I_t\) in the circuit is the sum of the current flowing through the main circuit element (let's call it \(I\)) and the current flowing through the shunt resistor (\(I_s\)). The relationship between the total current, the current through the main circuit element, and the current through the shunt resistor can be expressed as:It = I + IsThe current through the shunt resistor can be calculated using Ohm's Law:Is = V/Rswhere \(V\) is the voltage across the shunt resistor and \(R_s\) is the resistance of the shunt resistor. The voltage across the shunt resistor can be calculated using the total current and the total resistance in the circuit:V = Rt.ItSubstituting the expressions for \(I_s\) and \(V\) into the equation for total current, we get:It = I + It.Rt/RsSolving for \(I\), which is the current through the main circuit element, we find:I = It - It.Rt/RsThis equation gives the current through the main circuit element in terms of the total current, total resistance, and shunt resistance

Why the other options are wrong
  • B. As per the explanation, this is not the correct option.
  • C. As per the explanation, this is not the correct option.
  • D. As per the explanation, this is not the correct option.

Q5. Voltmeter is connected in the circuit in:

  • A. Perpendicular
  • B. Parallel
  • C. Series
  • D. Anti parallel

Explanation: A voltmeter is connected in parallel to the component or portion of the circuit across which the voltage is to be measured. This is because a voltmeter is designed to measure the potential difference between two points in a circuit, which requires it to be connected in parallel to those points.

Why the other options are wrong
  • A. There's no standard way to connect a voltmeter perpendicular to a circuit. The orientation of the voltmeter is not a standard consideration for its connection.
  • C. Connecting a voltmeter in series would disrupt the current flow in the circuit and prevent the accurate measurement of voltage across the component. Voltmeters are designed to have a very high resistance to minimize this disruption, but they are still connected in parallel to avoid affecting the circuit's current significantly.
  • D. There is no standard connection called "anti-parallel" for a voltmeter. The standard connection for a voltmeter is in parallel with the component or portion of the circuit across which the voltage is to be measured.

Q6. The principle of an A.C. generator is based on:

  • A. Mutual Induction
  • B. Lenz’s law
  • C. Self induction
  • D. Faraday’s law of electromagnetic induction

Explanation: Faraday's law states that a change in magnetic field through a coil of wire induces an emf in the wire. This principle is the fundamental basis for how AC generators produce electricity. As the coil rotates in a magnetic field, the magnetic flux through the coil changes, inducing an emf and generating an alternating current.

Why the other options are wrong
  • A. Mutual induction refers to the process where a change in current in one coil induces a voltage in a nearby coil. While this principle is related to electromagnetic induction, it is not the specific principle on which an AC generator operates.
  • B. Lenz's law is a consequence of Faraday's law and states that the direction of the induced electromotive force (emf) is such that it opposes the change in current that produced it. While Lenz's law is important in understanding the direction of induced currents, it is not the fundamental principle behind the operation of an AC generator.
  • C. Self-induction occurs when a changing current in a coil induces an emf in the same coil, opposing the change in current. While self-induction is related to electromagnetic induction, it is not the primary principle behind the operation of an AC generator.

Q7. When the motor is just started back emf always:

  • A. Becomes zero
  • B. Decreases
  • C. Remains same
  • D. Increases

Explanation: When a motor is not rotating or is just starting to rotate, there is no relative motion between the coils and the magnetic field. As a result, no emf is induced in the coils, and the back emf is zero.

Why the other options are wrong
  • B. The back emf in a motor is proportional to the speed of rotation. When the motor is just starting, the speed is very low or zero, so the back emf is also low or zero. It does not decrease from a higher value; it is simply not present at the start.
  • C. Since there is no back emf when the motor is just starting, it cannot remain the same from a previous state. The back emf is only present when the motor is in motion.
  • D. The back emf does not increase when the motor is just starting. It starts from zero and increases as the motor accelerates and the speed of rotation increases.

Q8. Root mean square value of an alternating voltage is:

  • A. Option A
  • B. Option B
  • C. Option C
  • D. Option D

Explanation: The rms value of an alternating voltage is given by Vo/√2, where Vo is the peak voltage. This relationship holds for a sinusoidal voltage waveform.

Why the other options are wrong
  • A. This is incorrect. The rms value of an alternating voltage is not equal to Vo²/√2. The square of the rms value (Vo²) is equal to the mean of the squares of the voltage values, not the square of the peak voltage.
  • C. This is incorrect. The rms value of an alternating voltage is not equal to Vo²/2. The square of the rms value (Vo²) is equal to the mean of the squares of the voltage values, not half of the square of the peak voltage.
  • D. This is incorrect. The rms value of an alternating voltage is not equal to Vo/2. The rms value is Vo/√2, not Vo/2.

Q9. Power dissipated in a pure inductor is:

  • A. Zero
  • B. Infinite
  • C. Small
  • D. Maximum

Explanation: In a pure inductor, the voltage and current are out of phase by 90 degrees. This means that when the current is at its maximum (or minimum), the voltage across the inductor is zero, and when the voltage is at its maximum (or minimum), the current through the inductor is zero. As a result, the average power dissipated over a complete cycle is zero.

Why the other options are wrong
  • B. The power dissipated in an inductor is not infinite. In an ideal circuit, an inductor stores and releases energy without any net loss.
  • C. The power dissipated in a pure inductor is actually zero, not small. This is because there is no net energy loss in an ideal inductor.
  • D. The power dissipated in a pure inductor is not maximum. In fact, it is zero because there is no resistance in an ideal inductor to dissipate power as heat.

Q10. The value of potential barrier for silicon at room temperature is:

  • A. 0.3V
  • B. 0.5V
  • C. 0.7V
  • D. 0.9V

Explanation: This is the correct value for the potential barrier of silicon at room temperature. Silicon is a semiconductor with a bandgap energy that corresponds to a potential barrier of approximately 0.7V.

Why the other options are wrong
  • A. This value is too low for the potential barrier of silicon at room temperature. The correct value is higher.
  • B. This value is also too low for the potential barrier of silicon at room temperature. The correct value is higher.
  • D. This value is too high for the potential barrier of silicon at room temperature. The correct value is lower.

Q11. The ratio of impurity addition in n intrinsic semiconductor is:

  • A. 1 to 103
  • B. 1 to 104
  • C. 1 to 105
  • D. 1 to 106

Explanation: Explanation will be added soon

Why the other options are wrong
  • A. This ratio is too low. Intrinsic semiconductors are relatively pure and have very few impurities compared to extrinsic (doped) semiconductors.
  • B. This ratio is also too low. The impurity concentration in intrinsic semiconductors is much lower than this.
  • C. While this ratio is closer, it is still too low. The impurity concentration in intrinsic semiconductors is typically even lower than this.

Q12. SI unit of current gain of transistor is:

  • A. Coulomb
  • B. Ampere
  • C. No unit
  • D. Farad

Explanation: Current gain is a dimensionless quantity, representing the ratio of output current to input current in a transistor. It is expressed without any specific unit.

Why the other options are wrong
  • A. The coulomb is the SI unit of electric charge, not current gain. Current gain is a dimensionless quantity representing the ratio of output current to input current in a transistor.
  • B. The ampere is the SI unit of electric current, not current gain. Current gain is a dimensionless ratio and is not measured in amperes.
  • D. The farad is the SI unit of capacitance, not current gain. Current gain is a ratio and does not have units of capacitance.

Q13. When platinum wire is heated it appears cherry red at temperature:

  • A. 500°C
  • B. 900°C
  • C. 1100°C
  • D. 1300°C

Explanation: This is the correct temperature. At around 900°C, platinum wire begins to emit a red glow, known as cherry red, due to its high melting point and emissivity.

Why the other options are wrong
  • A. This temperature is too low for platinum wire to appear cherry red. At 500°C, platinum would not yet exhibit the characteristic cherry red glow.
  • C. While platinum does continue to glow as it gets hotter, at 1100°C, it would typically appear more white than cherry red.
  • D. At 1300°C, platinum would appear white-hot, not cherry red. The cherry red color is typically observed at lower temperatures, around 900°C.

Q14. A photocell is based on:

  • A. Photoelectric effect
  • B. Polarization
  • C. Time dilation
  • D. Compton effect

Explanation: The photoelectric effect is the phenomenon where electrons are emitted from a material when it is exposed to light or electromagnetic radiation of sufficient frequency. Photocells, also known as photoelectric cells, utilize this effect to convert light energy into electrical energy.

Why the other options are wrong
  • B. Polarization refers to the orientation of electric fields in light waves. While polarization can affect how light interacts with materials, it is not the basis for how a photocell operates.
  • C. Time dilation is a concept from the theory of relativity and is not related to the operation of a photocell.
  • D. The Compton effect describes the scattering of photons by charged particles, which results in a change in the wavelength of the photons. While this effect is related to the interaction of photons with matter, it is not the underlying principle behind the operation of a photocell.

Q15. Normally an electron can reside in excited state for about:

  • A. 10-2 s
  • B. 10-4 s
  • C. 10-6 s
  • D. 10-8s

Explanation: This is the correct duration. Electrons in excited states typically have lifetimes on the order of nanoseconds (10^-9 seconds) to picoseconds (10^-12 seconds), depending on the specific electronic transition involved.

Why the other options are wrong
  • A. This duration is too long for the typical lifetime of an electron in an excited state. Electrons in excited states typically relax to lower energy states or return to the ground state much faster.
  • B. While this duration is shorter than 10^-2 seconds, it is still too long for the typical lifetime of an electron in an excited state. The actual lifetime is typically shorter.
  • C. This duration is shorter than 10^-4 seconds, but it is still too long for the typical lifetime of an electron in an excited state. The actual lifetime is typically shorter.

Q16. Dead time of the counter is:

  • A. Option A
  • B. Option B
  • C. Option C
  • D. Option D

Explanation: The dead time of a counter refers to the minimum time interval required between two consecutive detections or events to ensure that each event is counted separately. Its value is 10-4.

Why the other options are wrong
  • A. The dead time of a counter refers to the minimum time interval required between two consecutive detections or events to ensure that each event is counted separately. Its value is 10-4.
  • B. The dead time of a counter refers to the minimum time interval required between two consecutive detections or events to ensure that each event is counted separately. Its value is 10-4.
  • D. The dead time of a counter refers to the minimum time interval required between two consecutive detections or events to ensure that each event is counted separately. Its value is 10-4.

Q17. The building blocks of protons and neutrons are called:

  • A. Quarks
  • B. Electrons
  • C. Protons
  • D. Ions

Explanation: Quarks are elementary particles that combine to form composite particles called hadrons. Protons and neutrons are examples of hadrons, each composed of three quarks.

Why the other options are wrong
  • B. Electrons are elementary particles that are not part of the structure of protons and neutrons. They are a different type of particle.
  • C. Protons are composite particles made up of three quarks. They are not the building blocks of protons themselves.
  • D. Ions are atoms or molecules that have a net electric charge due to the loss or gain of electrons. They are not related to the structure of protons and neutrons.

More Punjab / UHS Solved Papers