MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is true regarding access lists applied to an interface? You can apply only one access list on any interface One access list may be configured, per direction, for each layer 3 protocol configured on an interface You can apply two access lists to any interface You can place as many access lists as you want on any interface until you run out of memory. You can apply only one access list on any interface One access list may be configured, per direction, for each layer 3 protocol configured on an interface You can apply two access lists to any interface You can place as many access lists as you want on any interface until you run out of memory. ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a short circuit test on circuit breaker, time to reach the peak re-striking voltage is 50 µs with the peak restriking voltage of 100 kV. Determine average RRRV (Rate of rise re-striking voltage)? 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 1 x 10⁶ kV/sec 4 x 10⁶ kV/sec 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 1 x 10⁶ kV/sec 4 x 10⁶ kV/sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP RRRV = Peak value of restriking voltage/time to reach the peak re-striking voltageRRRV = 100*1000/(50 μs)RRRV = 2*10⁶ kV/sec
MGVCL Exam Paper (30-07-2021 Shift 2) Tree Topology Web layout Page Orientation Paper size Font style Web layout Page Orientation Paper size Font style ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Encrypted security payload extension header is new in IPv4 IPv6 IPv5 IP IPv4 IPv6 IPv5 IP ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A short-shunt dc compound generator supplies 50 A at 300 V. If the shunt field resistance, series field resistance and armature resistance are 30 Ω, 0.03Ω, and 0.05 Ω respectively, determine the e.m.f. generated. 308.5 V 306.5 V 302.5 V 304.5 V 308.5 V 306.5 V 302.5 V 304.5 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Vt = 300 VRse = 0.03 Ω, Rsh = 30 Ω, Ra = 0.05 ΩI_L = 50 AVsh = V + I_L*Rse = 300 + 50*0.03 = 301.5 VIsh = Vsh/Rsh = 301.5/30 = 10.05 AIa = Ish + I_L = 60.05 AEg = Vt + I_L*Rse + IaRaEg = 300 + 50*0.03 + 60.05*0.05 = 304.5025 V
MGVCL Exam Paper (30-07-2021 Shift 2) If the velocity of electromagnetic wave in free space is 3 × 10⁸ m/s, then the velocity in a medium with εr of 4.5 and μr of 2 would be 27 × 10⁸ m/s 3 × 10⁸ m/s 1 × 10⁸ m/s 9 × 10⁸ m/s 27 × 10⁸ m/s 3 × 10⁸ m/s 1 × 10⁸ m/s 9 × 10⁸ m/s ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Speed of EM wave in medium = C/√(εr*μr)= (300000000)/√9= 300000000/3= 100000000= 1*10⁸ m/s
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