DFCCIL’21 EE: Daily Practices Quiz 31-Aug-2021

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               DFCCIL’21 EE: Daily Practices Quiz 31-Aug-2021

Each question carries 1 mark.
Negative marking: 1/4 mark
Total Questions: 06

Time: 8 Min.

Q1. The closed-loop gain of the system shown in the given figure is

(a) -9/5
(b) -6/5
(c) 6/5
(d) 9/5

Q2.A ramp input applied to a unity feedback system result in 5% steady state for. The type number and zero frequency gain of the system are respectively
(a) 1 and 20
(b) 0 and 20
(c) 0 and 1/20
(d) 1 and 1/20

Q3. P-type and N-type extrinsic semiconductors are formed by adding impurities of valency
(a) 5 and 3 respectively.
(b) 5 and 4 respectively.
(c) 3 and 5 respectively.
(d) 3 and 4 respectively.

Q4. To obtain the minimum value of stress in cables, the ratio (R/r) should be
(a) 2.13
(b) 2.718
(c) 1.96
(d) 1.5

Q5. Reciprocal of permeability is
(a) reluctivity
(b) susceptibility
(c) permittivity
(d) conductance

Q6. A 4 pole, 1200 rpm DC lap wound generator has 1520 conductors. If the flux per
pole is 0.01 weber, the emf of generator is
(a) 608 volts
(b) 304 volts
(c) 152 volts
(d) 76 volts

Solutions

S3. Ans.(c)
Sol. Pentavalent impurities Impurity atoms with 5 valence electrons produce n-type semiconductors by contributing extra electrons. Trivalent impurities Impurity atoms with 3 valence electrons produce p-type semiconductors by producing a “hole” or electron deficiency.

S4. Ans.(b)
Sol. For value of R/r = e; i.e., 2.71828 the minimum value of stress is obtained in a cable.

S5. Ans.(a)
Sol. Reluctivity is reciprocal of permeability.

S6. Ans.(b)
Sol. The emf generated,
E=PϕNZ/(60 A)
P= 4, ϕ =0.01 Weber
A= P =4 (for lap)
N= 1200 rpm, Z = 1520
∴ Emf, E=(4×0.01×1200×1520)/(60×4)
E=304 Volts

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