MGVCL Exam Paper (30-07-2021 Shift 3)
A conductor having surface density is embedded in a dielectric medium of permittivity. The electric field in the medium is E. If it is known that the pressure p on the conductor surface is equal to the electric energy density in the medium, then p (in SI unit)is given by

σ²/(4π)
σ²/(2πε)
σ²/(2ε)
σ/(4πε)

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MGVCL Exam Paper (30-07-2021 Shift 3)
Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 20 Ω and the resistance of the resistor connected to the sound core was 40 Ω. Calculate the distance of the fault point from the test end.

100 m
200 m
300 m
400 m

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MGVCL Exam Paper (30-07-2021 Shift 3)
Find the current through the resistor and the capacitor for the current shown in figure The switch is closed at t = O and initial change in the capacitor is zero.

I(t) = 10e⁵⁰⁰⁰ᵗ A.
I(t) = 5e⁵⁰⁰⁰ᵗ A.
I(t) = 5(1 - e⁵⁰⁰⁰ᵗ) A
I(t) = 10(1 - e⁵⁰⁰⁰ᵗ) A

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MGVCL Exam Paper (30-07-2021 Shift 3)
1. મોટેભાગે 'જાણે' શબ્દ હોય ત્યારે,ઉત્પ્રેક્ષા અલંકાર બને છે.
૨. જયારે ઉપમેય અને ઉપમાન એક જ હોય ત્યારે ઉપમા અલંકાર બને છે.
સરખાવવામાં આવેલ બે શબ્દોની વચ્ચે 'જયારે', 'જેવો', 'જેવી' જેવા શબ્દો આવે ત્યારે રૂપક અલંકાર બને.
જયારે ટીકા કે નિંદા કે વ્યંગના રૂપે પ્રશંસા કરાય ત્યારે અતિશયોક્તિ અલંકાર બને.
ઉપરોક્ત વિધાનોમાંથી કયું/ક્યાં વિધાનો સાચા છે.

1, 3
માત્ર 4
માત્ર 1
2, 4

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MGVCL Exam Paper (30-07-2021 Shift 3)
In a 3-phase, 4-wire, 400/230 V system, a lamp (L₁) of 100 W is connected to one phase and neutral and a lamp (L₂) of 150 W is connected to the second pahse and neutral. If the neutral wire is disconnected accidently, what will be the voltage across a 150 W lamp (L₂)?

180 V
120 v
160 V
240 V

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MGVCL Exam Paper (30-07-2021 Shift 3)
The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’.

Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A.
Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A.
Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A.
Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A.

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