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Punjab Physics 2017 Paper 2 — Solved Past Paper with Answers
All 17 MCQs from Punjab Physics 2017 Paper 2, 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. If the distance between two point charges is halved, the electric intensity becomes:
- A. Half
- B. 1/4 times
- C. Double
- D. 4 times✓
Explanation: When the distance between two point charges is halved, the electric intensity between them becomes four times. This is because the electric field strength is inversely proportional to the square of the distance between the charges (E ∝ 1/r²).
Why the other options are wrong
- A. If the distance between two point charges is halved, the electric intensity (electric field strength) does not become half. Instead, it increases.
- B. This is not correct. When the distance is halved, the electric intensity does not decrease to 1/4 of its original value. It increases.
- C. As per the explanation, this option is also not the correct one.
Q2. The drum in photo copier is coated with layer of:
- A. Aluminium
- B. Copper
- C. Selenium✓
- D. Silver
Explanation: Selenium is commonly used as the coating on the drum in a photocopier. Selenium is a photoconductive material, meaning its electrical conductivity increases when exposed to light. This property allows the drum to hold an electrostatic charge and attract toner particles, which are then transferred to paper to create the image.
Why the other options are wrong
- A. Aluminum is not typically used as a coating on the drum in a photocopier. Aluminum is used in some parts of the photocopier, such as the frame or casing, but it is not used as the coating on the drum.
- B. Copper is also not typically used as a coating on the drum in a photocopier. Copper is a good conductor of electricity and is used in various electronic components, but it is not used as the coating on the drum.
- D. Silver is not typically used as a coating on the drum in a photocopier. Silver is a good conductor of electricity and is used in some electronic components, but it is not commonly used as the coating on the drum.
Q3. A rheostat can operate as:
- A. Amplifier
- B. Potential divider✓
- C. Oscillator
- D. Transformer
Explanation: A rheostat can operate as a potential divider. By adjusting the position of the sliding contact along its resistance, a rheostat can divide the voltage across it into two parts, creating a variable voltage output.
Why the other options are wrong
- A. A rheostat is not designed to operate as an amplifier. An amplifier is a device that increases the amplitude of a signal.
- C. A rheostat is not designed to operate as an oscillator. An oscillator is a device that generates an oscillating or alternating output signal.
- D. A rheostat is not designed to operate as a transformer. A transformer is a device that transfers electrical energy between two or more circuits through electromagnetic induction, typically used to change the voltage level of an alternating current (AC).
Q4. Force on a moving charge in a magnetic field is given by:
- A. A
- B. B
- C. C
- D. D✓
Explanation: Explanation will be added soon
Q5. Current passing through the coil of galvanometer is:
- A. A
- B. B
- C. C
- D. D✓
Explanation: Explanation will be added soon
Q6. One Henry is:
- A. Vsa-1✓
- B. Vsa2
- C. Vsa
- D. V2sa-1
Explanation: One henry (H) is equivalent to one volt-second per ampere (V·s/A). This is the unit of inductance, representing the amount of inductance in a circuit where an electromotive force of one volt is produced when the current is changing at a rate of one ampere per second.
Why the other options are wrong
- B. This is not a valid unit for inductance. The correct unit for inductance is V·s/A (volt-second per ampere).
- C. This is not a valid unit for inductance. The correct unit for inductance is V·s/A (volt-second per ampere).
- D. This is not a valid unit for inductance. The correct unit for inductance is V·s/A (volt-second per ampere).
Q7. If the coil is wound on an iron core, the magnetic flux through it will:
- A. Zero
- B. Increases✓
- C. Decreases
- D. Remain constant
Explanation: When a coil is wound on an iron core and a current is passed through the coil, the magnetic field generated by the current induces magnetization in the iron core. This increases the magnetic flux through the coil compared to when there is no core present.
Why the other options are wrong
- A. If a coil is wound on an iron core and there is no current flowing through the coil, the magnetic flux through the coil will not be zero. There will be some residual magnetic flux in the iron core due to its magnetic properties.
- C. If a coil is wound on an iron core and the current through the coil is reduced to zero, the magnetic flux through the coil will decrease but will not necessarily become zero. The iron core can retain some residual magnetization, keeping some magnetic flux.
- D. The magnetic flux through a coil wound on an iron core will not remain constant. It will vary depending on the current flowing through the coil and the magnetization of the iron core.
Q8. One of the source of an A.C voltage is:
- A. Motor✓
- B. Battery
- C. UPS
- D. Solar cell
Explanation: A motor can be used as a source of AC voltage if it is operated as a generator. When a motor is rotated manually or by an external force, it can generate an alternating current (AC) voltage. This is known as a "dynamo" effect, where mechanical energy is converted into electrical energy.
Why the other options are wrong
- B. A battery is not a source of AC voltage. Batteries provide direct current (DC) voltage, where the current flows in one direction only. They do not naturally produce alternating current.
- C. A UPS is a device that provides backup power in the event of a power outage. It typically contains a battery that supplies DC voltage to an inverter, which then converts it to AC voltage. So, while a UPS can provide AC voltage, it is not a natural source of AC voltage like a generator or motor.
- D. A solar cell generates DC voltage from sunlight. Solar cells convert sunlight directly into electrical energy, producing DC voltage. To use solar energy as an AC voltage source, an inverter is required to convert the DC voltage to AC voltage.
Q9. If Io is the peak value of A.C current, then the root mean square(rms) value of current will be:
- A. A
- B. B
- C. C
- D. D✓
Explanation: Explanation will be added soon
Q10. The crystalline structure of NaCl is:
- A. Cubical✓
- B. Hexagonal
- C. Triangonal
- D. Tetragonal
Explanation: The crystalline structure of sodium chloride (NaCl) is cubic. It forms a crystal lattice in which each sodium ion is surrounded by six chloride ions, and each chloride ion is surrounded by six sodium ions, arranged in a cubic pattern.
Why the other options are wrong
- B. Sodium chloride does not have a hexagonal crystalline structure. Hexagonal structures are commonly found in minerals like graphite and quartz, but not in NaCl.
- C. Sodium chloride does not have a triangular crystalline structure. Triangular structures are not common in crystalline materials.
- D. Sodium chloride does not have a triangular crystalline structure. Triangular structures are not common in crystalline materials.
Q11. A diode charactersitics curve is a plot between:
- A. Current and resistance
- B. Voltage and time
- C. Voltage and current✓
- D. Current and time
Explanation: A diode characteristics curve is a plot of voltage (on the x-axis) and current (on the y-axis). It shows how the current through a diode varies with the voltage applied across it, indicating the diode's behavior under different voltage conditions.
Why the other options are wrong
- A. A diode characteristics curve does not plot current against resistance. While resistance is a factor in diode behavior, it is not directly represented in the characteristics curve.
- B. A diode characteristics curve does not typically involve time. It is a static representation of the relationship between voltage and current in a diode.
- D. A diode characteristics curve does not typically involve time. It is a snapshot of the diode's behavior at a specific moment, showing the relationship between current and voltage.
Q12. Voltage gain of the common emitter npn-transister as an amplifier is:
- A. A
- B. B
- C. C
- D. D✓
Explanation: Explanation will be added soon
Q13. 0.1 kg mass will be equivalent to the energy:
- A. 5x108 Joules
- B. 6x1019 Joules✓
- C. 9x1018 Joules
- D. 9x1015 Joules
Explanation: This is the correct answer. The energy equivalent of mass can be calculated using Einstein's famous equation, E = mc2, where E is the energy, m is the mass, and c is the speed of light in a vacuum (approximately 3x108 m/s). For 0.1 kg of mass, the energy equivalent is:E = (0.1 kg) * (3x108 m/s)2 = 0.1 * 9x1016 = 6x1015 Joules
Why the other options are wrong
- A. This is not the correct answer. The energy equivalent of 0.1 kg of mass is much larger than 5x108 Joules.
- C. This is not the correct answer. The calculation for the energy equivalent of 0.1 kg of mass is incorrect. It should be 6x1019 Joules.
- D. This is not the correct answer. The energy equivalent of 0.1 kg of mass is much larger than 9x1015 Joules. The correct value is 6x1019 Joules.
Q14. The maximum kinetic energy of emitted photoelectrons depends upon:
- A. The intensity of incident light
- B. Frequency of the incident light✓
- C. Metal surface
- D. Both frequency of incident light and metal surface
Explanation: The maximum kinetic energy of emitted photoelectrons depends on the frequency of the incident light. According to the photoelectric effect equation, the energy of a photon (E) is given by E = hf, where h is the Planck constant and f is the frequency of the light. The maximum kinetic energy of an emitted photoelectron is given by K.E. = hf - φ, where φ is the work function of the metal. Thus, the frequency of the incident light directly determines the maximum kinetic energy of the emitted photoelectrons.
Why the other options are wrong
- A. The intensity of incident light affects the number of photoelectrons emitted but not their maximum kinetic energy. Increasing the intensity increases the number of photons, hence the number of photoelectrons, but the kinetic energy of each photoelectron remains dependent on the frequency of the light.
- C. The metal surface affects the work function (φ) of the metal, which is the minimum energy required to remove an electron from the metal surface. Different metals have different work functions, which can influence the ease of photoelectron emission but does not directly affect the maximum kinetic energy of the emitted photoelectrons once the threshold frequency is reached.
- D. Both frequency of incident light and metal surface - While both the frequency of the incident light and the metal surface influence the photoelectric effect, the maximum kinetic energy of emitted photoelectrons is primarily determined by the frequency of the incident light. The metal surface affects the threshold frequency (minimum frequency required for photoemission) but once this threshold is met, the kinetic energy depends only on the frequency of the light.
Q15. Balmer empirical formula explains the electromagnetic radiation of any excited atom in terms of their:
- A. Energy
- B. Mass
- C. Wavelength✓
- D. Momentum
Explanation: Balmer's formula is an empirical equation that describes the wavelengths of the visible spectral lines emitted by hydrogen atoms. It is based on observations and provides a simple formula to calculate the wavelengths of these lines based on the integer values of the spectral series.
Why the other options are wrong
- A. Balmer's formula does indirectly involve energy since it deals with transitions between energy levels in atoms. However, the formula itself is more directly related to the wavelength of the emitted light rather than the energy levels themselves.
- B. Balmer's formula does not directly involve the mass of the atom. It is concerned with the spectral lines emitted by excited hydrogen atoms, which are related to the transitions between electron energy levels.
- D. Balmer's formula does not involve the momentum of the atom. It is solely concerned with the wavelengths of the emitted electromagnetic radiation.
Q16. Gm-counter uses:
- A. Alcohol only
- B. Bromine
- C. Argon
- D. Neon and bromine✓
Explanation: GM counters commonly use a gas mixture that includes neon as the ionization gas and a small amount of bromine as the quenching gas. The bromine helps to quickly stop the discharge of the detector after each ionization event, allowing the counter to detect subsequent events.
Why the other options are wrong
- A. Geiger-Muller (GM) counters do not use alcohol as a component. They typically use a gas mixture as the detecting medium.
- B. Bromine is not typically used in GM counters. The gas mixture used in GM counters often includes inert gases like neon and a quenching gas like halogen (commonly argon, sometimes with a small amount of methane).
- C. Argon is commonly used as a quenching gas in GM counters. It helps to improve the efficiency and stability of the detector.
Q17. How many times, the α — Particle is more massive than electron?
- A. 6332
- B. 7332✓
- C. 8332
- D. 9332
Explanation: Option A is the correct answer. The mass of an alpha (α) particle is approximately 4 atomic mass units (amu), while the mass of an electron is approximately 1/1836 amu. Therefore, the ratio of the mass of an alpha particle to that of an electron is:[ frac{{text{{Mass of α-particle}}}}{{text{{Mass of electron}}}} = frac{4 text{{ amu}}}{{1/1836 \text{{ amu}}}} \approx 4 \times 1836 \approx 7332 ]So, an alpha particle is approximately 7332 times more massive than an electron.
Why the other options are wrong
- A. As per the explanation, this option is not the correct one.
- C. This is not the correct answer. The correct ratio of the mass of an alpha particle to that of an electron is approximately 7332, not 8332.
- D. This is not the correct answer. The correct ratio of the mass of an alpha particle to that of an electron is approximately 7332, not 9332.
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