EE MCQ

SSC JE Electrical 2019 with solution SET-1
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For a three-layer stranded wire, the number of strands in the first, second, and third layer respectively are:

The total number of the conductor in stranded cable is given by

N = 3x2 − 3x + 1

Where

x = no. of layer

For layer 1

N1 = 3(1)2 − 3(1) + 1

N1 = 1

For layer 2

N2 = 3(2)2 − 3(2) + 1

N2 = 7

In layer 2 no. of conductor = N2 − N1 = 7 − 1 = 6

For layer 3

N3 = 3(3)2 − 3(3) + 1

N3 = 19

In layer 3 no. of conductor = N3 − N2 = 19 − 7 = 12

Related Question

SSC JE Electrical 2019 with solution SET-1
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Which of the following fuel has the least calorific value?

In the above option, the wood has the lowest calorific value. Its value is around 17000 – 20000 kilojoule/kg.

The calorific value of LPG is 55000 kilojoule/kg.

The calorific value of Kerosene oil is 45000 kilojoule/kg.

The calorific value of hydrogen oil is 150000 kilojoule/kg.

SSC JE Electrical 2019 with solution SET-1
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P = ρQHg is the water power equation. Head ‘H’ is measured in meter ‘g’ gravity constant is measured in meter/second square, then the measuring unit of ‘Q’, the flow rate of water is:

P = ρ QHg is the water power equation.

Where

H = heead = meter

g = gravity =m/s2

ρ =density = (kg/m3)

P = power = watt = N-m/sec

Q = water flow rate = m3/sec

SSC JE Electrical 2019 with solution SET-1
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With the current direction marked in the circuit shown, the net voltage applied is

Net voltage = −V2 + V1

Net voltage = −(V2 − V1)