As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .
Q1. Current gain of a transistor in common base mode is 0.95. Its value in common emitter mode is
Solution
(c) β=α/(1-α)=0.95/(1-0.95)=0.95/0.05=19
(c) β=α/(1-α)=0.95/(1-0.95)=0.95/0.05=19
Q2.In the following circuit of PN junction diodes D1,D2 and D3 are ideal then i is
Solution
(a) Diodes D1 and D3 are forward biased and D2 is reverse biased so the circuit can be redrawn as follows ⇒i=E/R.
(a) Diodes D1 and D3 are forward biased and D2 is reverse biased so the circuit can be redrawn as follows ⇒i=E/R.
Q3. Which one of the following statement is not correct in the case of light emitting diodes
Solution
(c)It emits light only when it is reverse biased
(c)It emits light only when it is reverse biased
Q4. The ionic bond is absent in
Solution
(d) Ionic bond is a type of chemical bond based on electrostatic forces between two oppositely charged ions. Bond between NaCl, CsCl and LiF are ionic, while H2O forms a covalent bond.
(d) Ionic bond is a type of chemical bond based on electrostatic forces between two oppositely charged ions. Bond between NaCl, CsCl and LiF are ionic, while H2O forms a covalent bond.
Q5.Different voltages are applied across a p-n junction and the currents are measured from each value. Which of the following graphs is obtained between voltage and current?
Solution
(a) When the junction diode is forward biased, ie, p-side is kept at higher potential, the current in the diode changes with the voltage applied across the diode. The current increases very slowly till the voltage across the diode crosses a certain value. After this voltage, the diode current increases rapidly even for very small increase in the diode voltage. The current-voltage relation for diode is shown
(a) When the junction diode is forward biased, ie, p-side is kept at higher potential, the current in the diode changes with the voltage applied across the diode. The current increases very slowly till the voltage across the diode crosses a certain value. After this voltage, the diode current increases rapidly even for very small increase in the diode voltage. The current-voltage relation for diode is shown
.
Q6. In a diode valve, the state of saturation can be obtained easily by
Solution
(b)Low filament current and high plate voltage
(b)Low filament current and high plate voltage
Q7.The current gain of a transistor in a common emitter configuration is 40. If the emitter current is 8.2 mA, then base current is
Solution
(b)0.2 mA.
(b)0.2 mA.
Q8.In a P-type semiconductor
Solution
(a) In P-type semiconductors, holes are majority charge carriers
(a) In P-type semiconductors, holes are majority charge carriers
Q9.Bonding in a germanium crystal (semi-conductor) is
Solution
(d) Covalent bonding exists in semi-conductor
(d) Covalent bonding exists in semi-conductor
Q10. The reverse bias in a junction diode is changed from 8 V to 13 V then the value of the current charges from 40 μA to 60 μA. The resistance of junction diode will be
Solution
(b)Re=∆V/∆I=((13-8))/((60-40)×10-6 )=2.5×105 Ω.
(b)Re=∆V/∆I=((13-8))/((60-40)×10-6 )=2.5×105 Ω.