Electronic Science - Online Test

Q1. The circuit shown is a


Answer : Option B
Explaination / Solution:



Q2. Which of the following statements about bar magnets is correct ?
Answer : Option D
Explaination / Solution:

As magnetic monopole does not exist. If we split the bar magnet into two pieces each part will have its own north and south pole.

Q3. In the case of metals the valence and conduction bands have
Answer : Option C
Explaination / Solution:

The materials can be classified by the energy gap between their valence band and the conduction band. The valence band is the band consisting of the valence electron, and the conduction band remains empty. Conduction takes place when an electron jumps from valence band to conduction band and the gap between these two bands is forbidden energy gap. Wider the gap between the valence and conduction bands, higher the energy it requires for shifting an electron from valence band to the conduction band.

  • In the case of conductors, this energy gap is absent or in other words conduction band, and valence band overlaps each other. Thus, electron requires minimum energy to jump from valence band. The typical examples of conductors are Silver, Copper, and Aluminium.
  • In insulators, this gap is vast. Therefore, it requires a significant amount of energy to shift an electron from valence to conduction band. Thus, insulators are poor conductors of electricity. Mica and Ceramic are the well-known examples of insulation material.
  • Semiconductors, on the other hand, have an energy gap which is in between that of conductors and insulators. This gap is typically more or less 1 eV, and thus, one electron requires energy more than conductors but less than insulators for shifting valence band to conduction band.

Q4. Group I lists four types of p − n junction diodes. Match each device in Group I with one of the option in Group II to indicate the bias condition of the device in its normal mode of operation. Group - I Group-II (P) Zener Diode (1) Forward bias (Q) Solar cell (2) Reverse bias (R) LASER diode (S) Avalanche Photodiode
Answer : Option B
Explaination / Solution:

Zener diode and Avalanche diode works in the reverse bias and laser diode works in forward bias. In solar cell diode works in forward bias but photo current is in reverse direction. Thus Zener diode : Reverse Bias Solar Cell : Forward Bias Laser Diode : Forward Bias Avalanche Photo diode : Reverse Bias

Q5. If a charge moves in an electrical field
Answer : Option C
Explaination / Solution:

Ambiguous question. Energy may be gained or lost depending on the sign of the charge and the electric field. It can also remain unchanged if the charge moves perpendicular to the direction of the field.

Q6.
In the circuit given below, the value of R required for the transfer of maximum power to the load having a resistance of 3W is

Answer : Option A
Explaination / Solution:



Q7. A portion of the main program to call a subroutine SUB in an 8085 environment is given below. : : LXI D,DISP LP : CALL SUB : It is desired that control be returned to LP+DISP+3 when the RET instruction is executed in the subroutine. The set of instructions that precede the RET instruction in the subroutine are
Answer : Option C
Explaination / Solution:
No Explaination.


Q8. The transistor used in the circuit shown below has a β of 30 and ICBO is negligible

If the forward voltage drop of diode is 0.7V, then the current through collector will be
Answer : Option D
Explaination / Solution:



Q9.  digit DMM has the error specification as : 0.2% of reading + 10 counts. If a dc voltage of 100V is read on its 200V full scale, the maximum error that can be expected in the reading is 
Answer : Option C
Explaination / Solution:
No Explaination.


Q10. P is a 16-bit signed integer. The 2’s complement representation of P is (F87B)16. The 2’s complement representation of 8*P is 
Answer : Option A
Explaination / Solution:
No Explaination.