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Kpk Physics 2021 — Solved Past Paper with Answers

All 18 MCQs from Kpk Physics 2021, 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.

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Q1. A one microfarad of a TV is subjected to 4000 V potential difference, The energy Stored in capacitor is_.

  • A. 8 J
  • B. 16J
  • C. 4x10-3J
  • D. 2x10-3 J

Explanation: To calculate the energy stored in a capacitor, you can use the formula:E = 1/2 CV2E = 1/2 (1x10-6)(4000)2E= 8J

Why the other options are wrong
  • B. Calculation is wrong.
  • C. Calculation is wrong.
  • D. Calculation is wrong.

Q2. Whonie one of the given are ohmic materials?

  • A. Semi conductor diode
  • B. Tungsten filament bulb
  • C. Thermistor
  • D. Metals

Explanation: All metals are ohmic conductors and all semiconductors are non-ohmic materials.Ohmic material are those material that conduct electricity and obey ohms law

Why the other options are wrong
  • A. All semi-conductors are non-ohmic materials.
  • B. It is non-ohmic material
  • C. It is non-ohmic material

Q3. A magnetic field B exerts a force of 10 N on a current carrying conductor of length 0.5 m. If 50 A current flows in a conductor L then the strength of field is......

  • A. 10T
  • B. 5.0 T
  • C. 0.4 T
  • D. 0.8 T

Explanation: F = BILB= F/IL= 10/(0.5)(50)= 0.4 T

Why the other options are wrong
  • A. Calculation is wrong.
  • B. Calculation is wrong
  • D. Calculation is wrong

Q4. When a wire is stretched and its radius becomes r / 21 then its resistance will be.

  • A. 16R
  • B. 4R
  • C. 2R
  • D. 0
Why the other options are wrong
  • A. This is the correct option.
  • C. It is not the best option
  • D. 0 is the wrong answer

Q5. Which one of the given device is always connected in series to a P.D. source?

  • A. Potentiometer
  • B. Ammeter
  • C. Voltmeter
  • D. Both A & B

Explanation: Ammeter is used to measure the current flowing through a component/circuit. Since, current remains same in series connection and also the resistance of an ammeter is very small due to which it doesn't affect the current to be measured. So, an ammeter is connected in series to measure current.

Why the other options are wrong
  • A. The potentiometer has two resistances, R2 and R1, connected in series with each other
  • C. Voltmeter is always connected in parallel with PD
  • D. A is wrong.

Q6. The peak value of sinusoidal voltage in AC circut is 50. The rms Value of voltage is roughly equal to _.

  • A. 70 V
  • B. 40 V
  • C. 35 V
  • D. 45 V

Explanation: Vrms=Vpeak/√2=50/√2= 35V

Why the other options are wrong
  • A. Calculation is wrong
  • B. Calculation is wrong
  • D. Calculation is wrong

Q7. The angular momentum of electronic in the 3rd Orbit of H- atom is_.

  • A. h / 2π
  • B. 2h / π.
  • C. 3h / 2π.
  • D. 2h / 3π

Explanation: L=mvr=nh/2πAs n=3so,L=3h/2π

Why the other options are wrong
  • A. L=3h/2π
  • B. L=3h/2π
  • D. L=3h/2π

Q8. A 96 gm sample of radioactove nuclide is Placed container Afar 12 min, only 6 gm of sample is left undecayed. What is the half-life radioactive nuclide?

  • A. 48 hr
  • B. 3 hr
  • C. 48 min
  • D. 3 min

Explanation: Calculate no of half lives:After one half life: 96=> 48After 2 half life: 48=> 24After 3 half life: 24=> 12After 4 half life: 12=> 6total half lives are 4.Half life = Total time/ no of half lives.= 12/4=3 min

Why the other options are wrong
  • A. This is wrong option.
  • B. We need answer in mins.
  • C. This option is wrong.

Q9. The energy of electronic in the excited state n = 3 is _.

  • A. -13.6 eV
  • B. -3.4 eV. This is enery when n=2
  • C. -1.8 eV
  • D. -1.5 eV

Explanation: As,E = -13.6/n2Here, n=3,= -13.6/32= -1.5eV

Why the other options are wrong
  • A. This is enery when n=1
  • B. Wrong calculation.
  • C. Calculation is wrong

Q10. The device which converts A.C into D.C is Called _.

  • A. Diode
  • B. Transformer
  • C. Amplifier
  • D. N-type, substance

Explanation: Diod is used to conver AC to DC.

Why the other options are wrong
  • B. Transformer step up or step down AC voltage.
  • C. An amplifier is an electronic device that increases the voltage, current, or power of a signal
  • D. It is not the best option

Q11. The hall-life of a radioactive material is_.

  • A. T1 / 2 = λ / log2
  • B. T1 / 2 = 1 / λ
  • C. T1 / 2 = log2 / λ
  • D. T1 / 2 = 3λ / log2

Explanation: T1 / 2 = log2 / λ is the right formula.

Why the other options are wrong
  • A. T1 / 2 = log2 / λ is the right formula.
  • B. T1 / 2 = log2 / λ is the right formula.
  • D. T1 / 2 = log2 / λ is the right formula.

Q12. If the K.E of fee electron doubles,its de Broglie wavelength changes by the factor.

  • A. √2
  • B. 1/√2
  • C. 2
  • D. 1/2

Explanation: When the kinetic energy (KE) of a free electron doubles, its de Broglie wavelength changes by a factor of 1/√2. This is because the de Broglie wavelength is proportional to 1/KE, so when KE doubles, the wavelength becomes proportional to 1/2KE

Why the other options are wrong
  • A. This option is not right
  • C. This is not the right option
  • D. Tgis is not the right option as it changes by factor 1/√2.

Q13. The positron has charge which is in magnitude equal to the charge on-------

  • A. Electron
  • B. Proton
  • C. β-particle
  • D. All these

Explanation: Positron has equal cahrge as proton, electron and B-particle have charge.

Why the other options are wrong
  • A. Positron has equal cahrge as proton, electron and B-particle have charge.
  • B. Positron has equal cahrge as proton, electron and B-particle have charge.
  • C. Positron has equal cahrge as proton, electron and B-particle have charge.

Q14. A current in a primary coil changes from 5.0 A to 6.0 in 6.0 sec an induced emf of 10.0 V is produced in the nearby placed secondary coll. The co-efficient of mutual induction is........

  • A. 15H
  • B. 30H
  • C. 50H
  • D. 60H

Explanation: As,E = M dI/dtM = E/dI/dtM = E dt/ dIM = 10(6)/ 1M = 60H

Why the other options are wrong
  • A. Wrong calculations.
  • B. Wrong Calculatinos.
  • C. Wrong Calculations

Q15. In the reaction »zXA > »°ZX-1 +Y + neutron+Q; Y stands for:

  • A. Electron
  • B. positron
  • C. Neutron
  • D. Anti-neutrino

Explanation: This reaction is B-decay and Positron is released in B-decay

Why the other options are wrong
  • A. Electron is not released in B-decay.
  • C. Neutron is mentioned already.
  • D. Positron is more appropriate answer.

Q16. Most of the electrons in the base of NPN transistor flow:

  • A. Out of the base lead
  • B. Into the collector
  • C. Into the emitter
  • D. Into the base supply

Explanation: Majority of Electrons: Most of the electrons injected into the base from the emitter do not recombine in the base. Instead, they are swept into the collector by the electric field of the reverse-biased collector-base junction.Current Flow: This results in a majority of the electrons flowing from the base into the collector, contributing to the collector current.Therefore, in an NPN transistor, most of the electrons in the base flow into the collector.

Why the other options are wrong
  • A. They dont flow back to base lead.
  • C. They will move towards collector
  • D. This is not the right option

Q17. Which one of the following materials is weakly attracted by magnet?

  • A. Ferro-magnetic
  • B. Diamagnetic
  • C. Paramagnetic
  • D. All these

Explanation: Ferromagnetic materials: These materials, such as iron, cobalt, and nickel, have strong magnetic properties. They can become permanently magnetized and are strongly attracted to magnets.Diamagnetic materials: These materials, such as copper, silver, and gold, are weakly repelled by a magnetic field. They do not retain magnetic properties when the external magnetic field is removed.Paramagnetic materials: These materials, such as aluminum, platinum, and certain metal oxides, are weakly attracted by a magnetic field. They have unpaired electrons that align with the magnetic field but do not retain any magnetization when the external field is removed.So, paramagnetic materials are weakly attracted by a magnet.

Why the other options are wrong
  • A. These materials, such as iron, cobalt, and nickel, have strong magnetic properties. They can become permanently magnetized and are strongly attracted to magnets.
  • B. These materials, such as copper, silver, and gold, are weakly repelled by a magnetic field. They do not retain magnetic properties when the external magnetic field is removed.
  • D. Diamagnets are repelled by magntic field

Q18. Let a charge of 1 nano-coulomb lies at the centre of a surface,the number of electric field lines leaving the surface are:

  • A. 1000
  • B. 250
  • C. 155
  • D. 113

Explanation: As Φ = q/E0= 1x10-9 / 8.854x10-12= 1.3x102= 113

Why the other options are wrong
  • A. As Φ = q/E0= 1x10-9 / 8.854x10-12= 1.3x102= 113
  • B. As Φ = q/E0= 1x10-9 / 8.854x10-12= 1.3x102= 113
  • C. As Φ = q/E0= 1x10-9 / 8.854x10-12= 1.3x102= 113

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