All questions of Practice Tests for Electrical Engineering (EE) Exam

During seed germination its store food is mobilized by_______
  • a)
    ABA
  • b)
    Cytokinin
  • c)
    Ethylene
  • d)
    Gibberellin
Correct answer is option 'D'. Can you explain this answer?

Function of Gibberellins -

Internodes elongation

Bolting

Promotes seed germination

Promotes flowing in long day plants and inhibit it in short day plants.

Delay Senescene

Induce parthenocarpy in pome fruits

Break bud and seed dormancy 

- wherein

Commercially used to increase yield of malt from barley.

 

 Gibberellins promotes seed germination and secretes enzyme to break stored food in seed

Plants that act as indicators of SO pollution in the atmosphere are
  • a)
    Mosses
  • b)
    Liverworts
  • c)
    Orchids
  • d)
    Epiphytic lichens
Correct answer is option 'D'. Can you explain this answer?

Shatabdi Malik answered
Lichens are the association of algae nd fungi. They are sensitive to SO2 i. e. will not grow in its presence. These are called pollution indicators they don't grow in polluted areas.

Out of the following six statements how many of them are correct?
1. Electrostatic precipitator can remove more than 99% of particulate matter
2. According to CPCB particulate size 2.5   or less in diameter are most harmful for human health
3. Catalytic converter reduces emission of poisonous gas
4. Amount of biodegradable inorganic matter in sewage water is estimated by measuring BOD
5. Biomagnification is due to the metabolisation of toxic substance by living organisms
6. Eutrophication is natural aging of lake by nutrient enrichment of its water
  • a)
    6  
  • b)
    5  
  • c)
    4  
  • d)
    3
Correct answer is option 'C'. Can you explain this answer?

Gopikas S answered
Option (1),(2),(3),(6) is correct.
incorrect options are (4),(5).
*(4)-Amount of biodegradable organic matter in sewage is estimated by measuring BOD.
*(5)- BIOMAGNIFICATION :
A few toxic substances often present in industrial waste water can undergo biological magnification in the aquatic food chain. It refers to increase in concentration of toxitant at successive tropic levels.
In biomagnification the toxic substances are cannot be metabolised or excreted.

For the amplifier shown below, the drain current changes from 5 mA to 7 mA when the gate voltage D changed from -4.0 V to -3.7 V. The voltage gain of the amplifier is _____
    Correct answer is between '-21,19'. Can you explain this answer?

    The gain of the amplifier is given by:
    Av = - gmrD
    gm: trans conductance of transistor
    rD: effective AC resistance seen by drain terminal
    gm = 6.66 ms
    rD = RD||RL = 6k||6k = 3 kΩ
    Av = -6.66 × 10-3 × 3 × 103
    Av = -19.98 ≈ 20

    The pair of amphoteric hydroxides is
    • a)
      Al(OH)₃, LiOH
    • b)
      B(OH)₃,Be(OH)₂
    • c)
      Be(OH)₂,Mg(OH)₂
    • d)
      Be(OH)₂,Zn(OH)₂
    Correct answer is option 'D'. Can you explain this answer?

    Afifa Aaliya answered
    Beryllium hydroxide and zinc hydroxide are atmospheric in nature because they react with both acid and base.
    So option " D " island correct answer.

    Little leaf disease of Brinjal is caused by an
    • a)
      Algae
    • b)
      Fungus
    • c)
      Mycoplasma
    • d)
      Bacterium
    Correct answer is option 'C'. Can you explain this answer?

    Mahak Agrawal answered
    Little leaf, or witches' broom, is a disease of sweetpotato caused by a mycoplasma-like organism. Affected plants have small, thickened and sometimes chlorotic young leaves which may be mistaken for zinc deficiency.

    The fish Lepidosiren belongs to the country :
    • a)
      South Africa
    • b)
      New Zealand
    • c)
      South America
    • d)
      England
    Correct answer is option 'C'. Can you explain this answer?

    Ramesh Chand answered
    Lepidosiren paradoxa is also called the South American lungfish. It is a freshwater and single species of the lungfish found in swamps and slows moving waters of Amazon, Parguay and lower Parona river basins in South America. They burrow into mud during the dry season. Thus, the correct answer is option'C'.

    Endoenzymes generally act at
    • a)
      Acidic pH
    • b)
      Alkaline pH
    • c)
      Neutral pH
    • d)
      Any pH
    Correct answer is option 'C'. Can you explain this answer?

    Kaneez Fatima answered
    Endoenzymes are the enzymes which functions inside the cell. It is secreted inside the cell and catalyzes the reaction inside the cell. Hexokinase and glycogen synthase are two endoenzyme. The optimum pH for theendoenzyme is neutral.

    Which factor promotes the appearance of the platelet plug?
    • a)
      Collagen
    • b)
      Serotonin
    • c)
      Fibrinogen
    • d)
      Vasodilation
    Correct answer is option 'A'. Can you explain this answer?

    Prince Yasser answered
    Collagen is exposed when a vessel wall is injured.platelets adhere to the site of injury.von willebrand factor helps platelets to adhere to the exposed collagen and to the platelet receptor complex.due to platelet adhesion to the site of injury,they get activated and a signalling cascade initiates.the stimulated platelets release contents of their granules,which generate thrombin formation on their surface leading to platelet aggregate and a hemostatic plug formation.

    Which of the following is not readily absorbed in the small intestine?
    • a)
      Na+
    • b)
      Cl-
    • c)
      K+
    • d)
      Mg2+
    Correct answer is option 'D'. Can you explain this answer?

    Prince Yasser answered
    Because mg requires specific carrier system to be absorbed by the intestinal cells.about 50% of dietary mg is normally absorbed.

    Two RC coupled amplifiers are connected to form a 2-stage amplifier. If the lower and upper cut off frequencies of each individual amplifiers is 100 Hz and 20 kHz. Then the 3-dB bandwidth of the two-state amplifier is ____ Hz.
      Correct answer is between '12700,12720'. Can you explain this answer?

      Navya Sarkar answered
      Concept:
      For multistage amplifier consisting of n identical amplifying states, the lower cut off fL (multi) is expressed as

       
      Where f1 = lower cut off frequency of individual state.
      Upper cut off frequency is given by:

       
      = 20 × 0.643
      = 12.87 kHz
      Bandwidth = 
      = 12714.6 Hz.

      The electron of H-atom transits from n = 1 to n = 4 by absorbing energy. If the energy of n = 1 state is – 21.8 × 10-19 Joule then its energy in n = 4 state will be :
      • a)
        – 21.8 × 10-19 Joule
      • b)
        – 5.45 × 10-19 Joule
      • c)
        – 2.725 × 10-19 Joule
      • d)
        – 1.362 × 10-19 Joule
      Correct answer is option 'D'. Can you explain this answer?

      Swara Dey answered
      -13.6 eV, then the energy required to excite the electron to n = 4 state can be calculated using the formula:

      En = -13.6 eV/n^2

      For n = 1, En = -13.6 eV/1^2 = -13.6 eV

      For n = 4, En = -13.6 eV/4^2 = -0.85 eV

      The energy required to excite the electron from n = 1 to n = 4 is the difference between the energies of these two states:

      ΔE = En - E1 = (-0.85 eV) - (-13.6 eV) = 12.75 eV

      Therefore, the energy required to excite the electron from n = 1 to n = 4 is 12.75 eV.

      The two stages of the amplifier are shown in the figure. Both transistors have current gain β of 80 and dynamic emitter resistance r′e of 25Ω. The magnitude of the overall voltage gain of the amplifier is approximately
      • a)
        384
      • b)
        515
      • c)
        720
      • d)
        225
      Correct answer is option 'A'. Can you explain this answer?

      Ishita Patel answered
      Objective approach
      The second stage is emitter follower which has voltage gain of unity.
      The overall voltage gain = gain of CE stage

      rC1 = effective impedance seen by collector of transistor Q1

      Zi(base)Q2 at the base of transistor Q2.
      The impedance Zi (base) Q2 at the base of transistor Q is
      Zi(base)2 = βRE2
      = 80 × 3 k
      Zi (base) 2 = 240 kΩ
      Therefore:

      = 10 k||240 k

      rC1 = 9.6kΩ
      Overall voltage gain

      AV− = 3 84
       
      Subjective approach:
      Draw the equivalent h-parameter model
       rπ = β re = 80 x 25 = 2k


      Apply KVL in the base-emitter loop of transistor Q2

      Apply KCL at node VB2

      800 ib1 = -255  ib2

      Which of the following is not a necessary component in a clamper circuit?
      • a)
        Diode
      • b)
        Capacitor
      • c)
        Resistor
      • d)
        Independent DC Supply
      Correct answer is option 'D'. Can you explain this answer?

      Tanya Chauhan answered
      In a clamper circuit, the primary purpose is to shift the DC level of the input signal without altering the shape of the signal. This is achieved by adding a DC bias to the input signal. The necessary components in a clamper circuit include a diode, a capacitor, and a resistor.

      - Diode: The diode is an essential component in a clamper circuit as it allows the flow of current in one direction while blocking it in the opposite direction. It helps in shifting the DC level of the input signal by allowing the capacitor to charge or discharge through it.

      - Capacitor: The capacitor is another necessary component in a clamper circuit. It is connected in parallel to the diode and is charged or discharged depending on the input signal. The capacitor stores the charge and releases it during the negative half-cycle of the input signal, thereby shifting the DC level of the signal.

      - Resistor: The resistor is also an important component in a clamper circuit. It is connected in series with the diode and capacitor. The resistor limits the current flowing through the circuit and helps in stabilizing the voltage across the capacitor.

      - Independent DC Supply: The independent DC supply is not a necessary component in a clamper circuit. It is not required because the purpose of a clamper circuit is to shift the DC level of the input signal using the existing components. The diode, capacitor, and resistor work together to achieve this without the need for an additional DC supply.

      Therefore, the correct answer is option 'D' - Independent DC Supply.

      For the given circuit for a 20 Vpeak sinusoidal input vi, what is the value of vi at which the clipping begins?
      • a)
        5 V
      • b)
        0 V
      • c)
        -5 V
      • d)
        Clipping doesn’t occur
      Correct answer is option 'C'. Can you explain this answer?

      It is not possible to determine the value of vi at which the clipping begins without additional information about the circuit. The clipping voltage depends on the specific characteristics and parameters of the circuit, such as the diode threshold voltage or the supply voltage.

      Cytosterility in the pollen of Maize is due to
      • a)
        Extra-chromosomal inheritance
      • b)
        Complementary genes
      • c)
        Nuclear inheritance
      • d)
        Hypostasis
      Correct answer is option 'A'. Can you explain this answer?

      Kaneez Fatima answered
      All 10 chromosomes of maize (Zea mays, 2n = 2x = 20) were recovered as single additions to the haploid complement of oat (Avena sativa, 2n = 6x = 42) among F1 plants generated from crosses involving three different lines of maize to eight different lines of oat. In vitro rescue culture of more than 4,300 immature F1 embryos resulted in a germination frequency of 11% with recovery of 379 F1 plantlets (8.7%) of moderately vigorous growth. Some F1 plants were sectored with distinct chromosome constitutions among tillers of the same plant and also between root and shoot cells. Meiotic restitution facilitated development of un-reduced gametes in the F1. Self-pollination of these partially fertile F1 plants resulted in disomic additions (2n = 6x + 2 = 44) for maize chromosomes 1, 2, 3, 4, 6, 7, and 9. Maize chromosome 8 was recovered as a monosomic addition (2n = 6x + 1 = 43). Monosomic additions for maize chromosomes 5 and 10 to a haploid complement of oat (n = 3x + 1 = 22) were recovered several times among the F1plants. Although partially fertile, these chromosome 5 and 10 addition plants have not yet transmitted the added maize chromosome to F2offspring. We discuss the development and general utility of this set of oat-maize addition lines as a novel tool for maize genomics and genetics.

      Sexual inter-species hybridization is a powerful tool for understanding genome structure and interaction in higher plants. Breaking through inter-species incompatibility also enables chromosome engineering with horizontal (species-to-species) transfer of characters to which plant researchers and breeders have no access when utilizing only intraspecies gene pools. The more remotely related the parental genomes, the more the gene pool may be enriched.

      Most of the favored host plants for chromosome engineering are allopolyploids or amphidiploids. Although in rare cases pure diploid species do tolerate alien chromosome additions, e.g. rye (Secale cereale; Kynast, 1986), the redundancy provided by homoeologs in polyploids is expected to better compensate for the loss of genetic information from alien substitution or translocation. Once stabilized as a disomic addition, the alien chromosome pair is transmitted consistently in the majority of the plants (Riley, 1960).

      The equivalent focal length of a Huygen's eye piece
      • a)
        3f ∕ 2
      • b)
        3f ∕ 4
      • c)
        2f ∕ 3
      • d)
        4f ∕ 3
      Correct answer is option 'A'. Can you explain this answer?

      Sai Nikhil answered
      The equivalent focal length of huygens eye piece is 3f/2 the distance between eye piece and object pieces is 2f and focal lengths are in the ratio 1:3 so, the we get 3f/2

      Mendel’s law of independent assortment always holds good for genes situated on the :
      • a)
        Non-homologous chromosomes
      • b)
        Homologous chromosomes
      • c)
        Extra nuclear genetic element
      • d)
        Same chromosome
      Correct answer is option 'A'. Can you explain this answer?

      Aravind Chavan answered
      Non-homologous Chromosome The law of independent assortment holds true as long as two different genes are on separate chromosomes. When the genes are on separate chromosome, the two alleles of one gene (A and a) will segregate into gametes independently of the two alleles of the other gene (B and b). Equal numbers of four different gametes will form AB, aB, Ab, ab. But if the two genes are on the same chromosome, then they will be linked and will segregate together during meiosis, producing only two kinds of gametes.

      Homologous chromosomes are similar but not identical. Each carries the same gene insame order but the alleles for each trait may not be the same. Extra nuclear genetic elements are also called as plasmids and shows the pattern of maternal inheritance.

      Assuming the diode to be ideal, the current I in the circuit shown is ________ mA.
        Correct answer is between '0.2,0.25'. Can you explain this answer?

        Sonal Tiwari answered
        The above circuit can be redrawn as:
        Assume diode is off:
        Voltage across 20 k resistor connected to left of diode is 
        This will turn on the diode 
        the equivalent circuit is now:
        Now the two 20k resistors are in parallel whose equivalent is 10k
        this 10 k is in series with other 10k resistor 
        9V divides equally among them
        V0 = 4.5V 
        Cuurent I = 4.5/20 mA
        = 0.225 mA

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