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Kpk Physics 2023 — Solved Past Paper with Answers
All 18 MCQs from Kpk Physics 2023, solved with the correct answer highlighted and a full explanation for every question. This is a free MDCAT KPK / ETEA past paper — no signup, no ads. Practise it interactively in timed mode, drill more with free MDCAT MCQs, or browse all KPK / ETEA papers.
Attempt this paper interactively →Q1. Self inductnance (L) in terms of magnetic flux (¢), no of turns(N) and electric current (I) can be written as_.
- A. L=N¢ / I
- B. L=N¢I✓
- C. L = I¢/N
- D. L = N/I¢
Explanation: Self inductance (L) in terms of magnetic flux (¢), no of turns(N) and electric current (I) can be written as: L=N¢/I
Why the other options are wrong
- A. Self inductance (L) in terms of magnetic flux (¢), no of turns(N) and electric current (I) can be written as:L=N¢/I
- C. This is not the exact formula of self inductance.The formula of self inductance is:L=N¢/I
- D. This is not the exact formula of self inductance.The formula of self inductance is:L=N¢/I
Q2. The ratio of power in AC circuit to that of DCcircuit is equal to_
- A. Sin θ
- B. Cos θ
- C. Tan θ✓
- D. Sec θ
Explanation: Explanation will be added soon
Why the other options are wrong
- A. Explanation will be added soon
- B. Explanation will be added soon
- D. Explanation will be added soon
Q3. Pair product cannot take place in vaccum because_is not conserved.
- A. Momentum
- B. Charge
- C. Energy
- D. Both energy and momentum.✓
Explanation: Pair production, also known as pair creation, cannot take place in a vacuum because both energy and momentum are not conserved in such a process. Here’s a detailed explanation:Energy Conservation: For pair production to occur, a photon (or another high-energy particle) must convert into a particle-antiparticle pair (such as an electron and a positron). The energy of the incoming photon must be at least equal to the combined rest mass energy of the particle-antiparticle pair. However, even if this condition is met, energy conservation alone is not sufficient for pair production to occur in a vacuum.Momentum Conservation: The more critical issue in a vacuum is the conservation of momentum. In a vacuum, a single photon cannot simply convert into a particle-antiparticle pair without violating momentum conservation laws. This is because a photon has momentum but no rest mass, whereas the produced particles have both rest mass and momentum. The momentum of the photon cannot be split into two particles without some external object to balance the equation.In other words, if pair production were to occur in a vacuum, the initial photon momentum would need to be equal to the vector sum of the momenta of the resulting particle and antiparticle. Since there is no other object to absorb the excess momentum in a vacuum, this condition cannot be satisfied.Therefore, for pair production to occur, the presence of a nearby object, such as a nucleus or another particle, is required to take up the excess momentum, ensuring both energy and momentum conservation.
Why the other options are wrong
- A. Pair production, also known as pair creation, cannot take place in a vacuum because both energy and momentum are not conserved in such a process. Here’s a detailed explanation:Energy Conservation: For pair production to occur, a photon (or another high-energy particle) must convert into a particle-antiparticle pair (such as an electron and a positron). The energy of the incoming photon must be at least equal to the combined rest mass energy of the particle-antiparticle pair. However, even if this condition is met, energy conservation alone is not sufficient for pair production to occur in a vacuum.Momentum Conservation: The more critical issue in a vacuum is the conservation of momentum. In a vacuum, a single photon cannot simply convert into a particle-antiparticle pair without violating momentum conservation laws. This is because a photon has momentum but no rest mass, whereas the produced particles have both rest mass and momentum. The momentum of the photon cannot be split into two particles without some external object to balance the equation.In other words, if pair production were to occur in a vacuum, the initial photon momentum would need to be equal to the vector sum of the momenta of the resulting particle and antiparticle. Since there is no other object to absorb the excess momentum in a vacuum, this condition cannot be satisfied.Therefore, for pair production to occur, the presence of a nearby object, such as a nucleus or another particle, is required to take up the excess momentum, ensuring both energy and momentum conservation.
- B. Charge has no impact on pair production conservation.
- C. Pair production, also known as pair creation, cannot take place in a vacuum because both energy and momentum are not conserved in such a process. Here’s a detailed explanation:Energy Conservation: For pair production to occur, a photon (or another high-energy particle) must convert into a particle-antiparticle pair (such as an electron and a positron). The energy of the incoming photon must be at least equal to the combined rest mass energy of the particle-antiparticle pair. However, even if this condition is met, energy conservation alone is not sufficient for pair production to occur in a vacuum.Momentum Conservation: The more critical issue in a vacuum is the conservation of momentum. In a vacuum, a single photon cannot simply convert into a particle-antiparticle pair without violating momentum conservation laws. This is because a photon has momentum but no rest mass, whereas the produced particles have both rest mass and momentum. The momentum of the photon cannot be split into two particles without some external object to balance the equation.In other words, if pair production were to occur in a vacuum, the initial photon momentum would need to be equal to the vector sum of the momenta of the resulting particle and antiparticle. Since there is no other object to absorb the excess momentum in a vacuum, this condition cannot be satisfied.Therefore, for pair production to occur, the presence of a nearby object, such as a nucleus or another particle, is required to take up the excess momentum, ensuring both energy and momentum conservation.
Q4. An electric charge q is Placed at the centre of cube of side ‘a’ the electric flux on its one side will be _
- A. q/4πε0
- B. q/a2ε0
- C. q/6ε0✓
- D. q/ε0
Explanation: Explanation will be added soon
Why the other options are wrong
- A. This option is not correct. Correct option is q/6E^
- B. This option is not correct. Correct option is q/6E^
- D. This option is not correct. Correct option is q/6E^
Q5. A/zX=A-4/z-2Y+Q The given reaction is completed by_
- A. Alpha 𝜶✓
- B. Beta β
- C. Gamma γ
- D. Sigma Σ
Explanation: The given nuclear reaction equation A/zX=A-4/z-2Y+Q is completed by an alpha particle emission. Here’s the reasoning:- The initial nucleus X has a mass number A and an atomic number Z .- The final nucleus Y has a mass number A-4 and an atomic number Z-2 .Therefore, the reaction is completed by the emission of an alpha particle.
Why the other options are wrong
- B. Beta β is not the right answer.
- C. Gamma γ is not the right answer
- D. Sigma Σ is not the right option.
Q6. By increasing applied anode voltage, wavelength (A) of X rays_
- A. Increases
- B. Remains unchanged
- C. Decreases✓
- D. Becomes infinite
Explanation: By increasing the applied anode voltage, the wavelength (λ) of the produced X-rays decreases.As,K.E = eVWhen the anode voltage in an X-ray tube is increased, the kinetic energy of the electrons striking the anode increases. This higher kinetic energy results in the production of higher-energy X-rays.E=hc/YSince higher energy corresponds to shorter wavelength, increasing the applied anode voltage leads to the generation of X-rays with shorter wavelengths.
Why the other options are wrong
- A. By increasing the applied anode voltage, the wavelength (λ) of the produced X-rays decreases.
- B. By increasing the applied anode voltage, the wavelength (λ) of the produced X-rays decreases.
- D. By increasing the applied anode voltage, the wavelength (λ) of the produced X-rays decreases.
Q7. The seismometer works on the principle of_
- A. Electromagnetism✓
- B. Maxwell equations
- C. Electromagnetic induction
- D. Ampere’s Law
Explanation: A seismometer uses the principle of electromagnetism to convert the relative motion (caused by inertia) between a suspended mass and the frame into an electrical signal that can be measured and analyzed.
Why the other options are wrong
- B. This option is wrong as a seismometer uses the principle of electromagnetism to convert the relative motion (caused by inertia) between a suspended mass and the frame into an electrical signal that can be measured and analyzed
- C. This option is wrong as a seismometer uses the principle of electromagnetism to convert the relative motion (caused by inertia) between a suspended mass and the frame into an electrical signal that can be measured and analyzed
- D. This option is wrong as a seismometer uses the principle of electromagnetism to convert the relative motion (caused by inertia) between a suspended mass and the frame into an electrical signal that can be measured and analyzed
Q8. 98238 U contains_ number of nucleons.
- A. 146
- B. 238✓
- C. 92
- D. 330
Explanation: Explanation will be added soon
Why the other options are wrong
- A. This is the no of neutrons only.
- C. This is no of protons.
- D. It is wrong answer.
Q9. For one complete cycle, the simple mean value of AC cycle is_
- A. Zero✓
- B. Vm /2
- C. 2Vm
- D. 0.707Vm
Explanation: The mean average value of alternating current (or emf) during a half, cycle is given by Im = 0.636 I0 or Em = 0.636 E0During the next half cycle, the mean value of ac will be equal in magnitude but opposite in direction.For this reason the average value of ac over a complete cycle is always zero. So the average value is always defined over a half cycle of ac.
Why the other options are wrong
- B. This is not the right option as average value of complete AC cycle is always zero.
- C. This is not the right option as average value of complete AC cycle is always zero.
- D. This is not the right option as average value of complete AC cycle is always zero.
Q10. “mho” is a unit of_
- A. Resistivity✓
- B. Conductivity
- C. Resistance
- D. Conductance
Explanation: Explanation will be added soon
Why the other options are wrong
- B. “mho” is a unit of conductivity.G = I/V
- C. SI unit of resistivity is ohm (Ω)
- D. The SI unit of conductance is the siemens (S)
Q11. The function of unipolar junction transistors depends on_
- A. Majority carriers✓
- B. Minority carriers
- C. Both A and B
- D. None of these
Explanation: Unipolar junction transistors operate based on the majority charge carriers.
Why the other options are wrong
- B. Unipolar junction transistors operate based on the majority charge carriers.
- C. Unipolar junction transistors operate based on the majority charge carriers and not in minority cahrge carriers.
- D. Unipolar junction transistors operate based on the majority charge carriers.
Q12. The super conductor has a_resistance.
- A. Moderate
- B. High
- C. Zero✓
- D. Infinite
Explanation: In a superconductor, the electric resistance is equal to zero. This is why an electric current can circulate forever in a superconducting ring even when the battery has been unplugged! This is how magnetic fields are created in MRIs.
Why the other options are wrong
- A. In a superconductor, the electric resistance is equal to zero.
- B. In a superconductor, the electric resistance is equal to zero.
- D. In a superconductor, the electric resistance is equal to zero.
Q13. How much voltage is required to make 2A flow through a resistance of 80 Ω?
- A. 16AΩ-1
- B. 16AΩ-1
- C. 160AΩ✓
- D. 16A-1Ω-1
Explanation: How much voltage is required to make 2A flow through a resistance of 80 Ω?as V = IR,PUTTING VALUESV = 2 * 80V = 160V or 160AΩ
Why the other options are wrong
- A. Caluculation and units are wrong.
- B. Calculation and units are wrong
- D. Calculation and units are wrong.
Q14. Which of the given is a boson?
- A. Photon✓
- B. Neutino
- C. Proton
- D. Pion
Explanation: Photon is an example of boson.
Why the other options are wrong
- B. Nutrino is fermion.
- C. Proton is Subatomic particle.
- D. Pion is an example of meson.
Q15. The reverse process of x-rays Production car be related to_
- A. Compton effect
- B. Photoelectric effect✓
- C. Pair production
- D. None of these
Explanation: In photoelectric effect, the photon having the energy greater than the work function of the material when strikes the material, the electron gets knocked out from the inner shell of the atom. Thus, we can say that it is the reverse process of production of continuous X-ray
Why the other options are wrong
- A. Compton effect is not recverse process of xray production.
- C. Pair production is the anihilation of matter.
- D. photoelectric is the reverse process of production of continuous X-ray
Q16. Volt per meter is a non Sl unit of _
- A. Electric force
- B. Electric Power
- C. Voltage
- D. Electric intensity✓
Explanation: V/m is equilent to N/C which is the unit of electric field or electric intensity
Why the other options are wrong
- A. SI unit of force is N.
- B. The SI unit of electric power is watt (W).
- C. The SI unit of Voltage is volt(V).
Q17. Charge to mass ratio of electron is_that of proton.
- A. Equal to
- B. Greater than✓
- C. Less than
- D. Zero compare to
Explanation: Charge to mass ratio of electron is greater than that of proton because proton is 1837 times heavier than electron and e/m have inverse relation with mass.
Why the other options are wrong
- A. Charge to mass ratio of electron is greater than that of proton.
- C. Charge to mass ratio of electron is greater than that of proton.
- D. Charge to mass ratio of electron is greater than that of proton.
Q18. According to wein"s displacement law with rise in temperature of black body,the product λmaxT is_
- A. Increased
- B. Decreased
- C. Constant
- D. None of these✓
Explanation: The product of λmaxT is always constant.λmaxT = 2.9 × 10−3 m K
Why the other options are wrong
- A. It remains constant not increased
- B. It remains constant not decreased
- C. The product of λmaxT is always constant.
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