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GATE’22 EE: Daily Practices Quiz
GATE’22 EE: Daily Practices Quiz 09-Oct-2021
Each question carries 2 marks
Negative marking: 0.66 mark
Total Questions: 10
Total marks: 20
Time: 20 min.
Q1. A co-axial cable of inner radius a and outer radius b. The space between conductor is filled with dielectric of permittivity ∈ and length of cable 4m. what is the capacitance of the cable?
(c) (π∈)/(2 ln(b/a) )
Q2. In a two-unit insulator string, voltage across the unit near to line conductor is 18.5 kV and string efficiency is 75%. The voltage of line conductor will be:
(a) 37 kV
(b) 48.07 kV
(c) 27.75 kV
(d) 30 kV
Q3. A step-down chopper connected to a d.c. supply, supplies a separately excited d.c. motor at a duty ratio of α. The ripple factor in the terminal voltage of the motor assuming continuous current conduction is-
Q5. In a three-phase induction motor, the ratio power across air gap, rotor copper loss and gross mechanical power output is
(a) PG:sPG:(1-s) PG
(b) PG:sPG:(s-1) PG
(c) PG:(1-s) PG:sPG
(d) PG:sPG:(1+s) PG
Q7. Consider a network function H(s)= (2(s+3))/((s+2)(s+4)). What is the steady state response due to step input?
Q8. Which of the following is a non-maskable interrupt of 8085 microprocessors?
(c) RST 6.5
(d) RST 7.5
Q9. For a JFET I_DSS = 8 mA and peak voltage V_p = –8V, what will be the drain current for gate to source voltage of –2V?
(a) 4.5 mA
(b) 8 mA
(c) 16 mA
(d) 12 mA
Q10. Determine the deflection factor (in V/m) of a CRO, when the deflection sensitivity of the CRO is 5 m/V.
Sol. In a three-phase induction motor, the ratio power across air gap, rotor copper loss and gross mechanical power output is-
P_G= Air gap power
sP_G = rotor copper loss
P_G (1-s) = gross mechanical power
So, ratio is-
P_G:sP_G ∶(1-s) P_G
Sol. Peak current=I_0+(△i_c)/2=5+1.6/2=5.8 A
Sol. steady state response=lim┬(s→0)〖sC(s)=〗 lim┬(s→0)= s×(2(s+3))/((s+2)(s+4))×1/s=3/4
Sol. The hardware vectored interrupts are classified in to maskable and non-maskable interrupts.
TRAP is non-maskable interrupt.
RST 7.5, RST 6.5, RST 5.5 and INTR are maskable interrupt.
Sol. I_D=I_DSS (1-V_GS/V_P )^2=8×10^(-3) (1-(-2)/(-8))^2=4.5 mA
Sol. Deflection factor=1/(deflection sensitivity)=1/5=0.2 V/m.