216. The input impedance of a loss-less transmission line is 100 ohms when terminated in a short-circuit, and 64 ohms when terminated in an open circuit. The characteristic impedance of the line is—

1. 1. 80 ohms

   
   
   

   2. 164 ohms

   
   
   

   3. 36 ohms

   
   
   

   4. 64 ohms
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   Z₀ = √Z_in·Z_L = √100 × 64 = 80 ohm

   Correct Option: A

   Z₀ = √Z_in·Z_L = √100 × 64 = 80 ohm

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217. A rectangular loop in the x-z plane bounded by the lines
     x = 0, x = a, z = 0 and z = b, in a time-varying magnetic field given by
     B = B0 cos ωt ay
     where B0 is a constant, ω is the angular frequency and – ay is the unit vector in the y-direction. The emf induced in the loop is given by—

1. 1. ab B₀ cos ωt

   
   
   

   2. ab B₀ ω sin ωt

   
   
   

   3. B₀ ω sin ωt

   
   
   

   4. Zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   B = B₀ cos ωt + iy
   Again | e (t) | = d/ dt (ab B₀ cos ωt)
   = ab B₀ ω sin ωt
   

   Correct Option: B

   B = B₀ cos ωt + iy
   Again | e (t) | = d/ dt (ab B₀ cos ωt)
   = ab B₀ ω sin ωt
   

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218. Kirchhoff's current law for direct currents is implicit in the expression—

1. 1. ∇ · D = ρ

   
   
   

   2. ∫ J · n ds = 0

   
   
   

   3. ∇ · B = 0

   
   
   

   4. ∇ × H = J + ∂ D/∂t
      

   
   
   

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1. View Hint View Answer Discuss in Forum

   ∫_s J.ds implies that the net current coming out of a closed surface is zero; which is the KCL for direct current.
   

   Correct Option: B

   ∫_s J.ds implies that the net current coming out of a closed surface is zero; which is the KCL for direct current.
   

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219. A circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field 0.1 Wb/m2. When the radius of the loop is 2 m, the emf induced in the loop will be—

1. 1. 0.8 π V

   
   
   

   2. 0.04 π V

   
   
   

   3. 0.2 π V

   
   
   

   4. Zero

   
   
   

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1. View Hint View Answer Discuss in Forum

   The area of the loop is A = π r²
   Flux linking the area, φ = B × πr²
   

   Induced emf, e(t) =	dφ
   	dt

   
   e (t) = d/ dt {B πr²} = B π 2r (dr/ dt)
   or    e (t) = 0.1 × π × 2 × 2 × 2 = 0.8 πV
   

   Correct Option: A

   The area of the loop is A = π r²
   Flux linking the area, φ = B × πr²
   

   Induced emf, e(t) =	dφ
   	dt

   
   e (t) = d/ dt {B πr²} = B π 2r (dr/ dt)
   or    e (t) = 0.1 × π × 2 × 2 × 2 = 0.8 πV
   

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220. Given that L = 1 µH/m, C = 11.11 pf/m the group delay time is—

1. 1. Nil

   
   
   

   2. 0.33 n sec

   
   
   

   3. 0.3 µ sec

   
   
   

   4. 33 sec

   
   
   

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1. View Hint View Answer Discuss in Forum

   Group delay time td = β/ ω
   β = ω√LC

   Correct Option: B

   Group delay time td = β/ ω
   β = ω√LC

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