191. An antenna in free space receives 2 µW of power when the incident electric field is 20 V/m rms. The effective aperture of the antenna is—

1. 1. 0.005 m²

   
   
   

   2. 0.05 m²

   
   
   

   3. 1.885 m²

   
   
   

   4. 3.77 m²

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Pr =	|Erms|2	Ae
   	2η0

   
   Given η₀ = 120 πΩ, E_rms = 20 V/m,
   P_r = 20 × 10^–6 watt
   Now,
   

   Ae =	2 × 10−6 × 2 × 120∏
   	20 × 20 × 10−6

   
   = 3.77 m²

   Correct Option: D

   Pr =	|Erms|2	Ae
   	2η0

   
   Given η₀ = 120 πΩ, E_rms = 20 V/m,
   P_r = 20 × 10^–6 watt
   Now,
   

   Ae =	2 × 10−6 × 2 × 120∏
   	20 × 20 × 10−6

   
   = 3.77 m²

---

192. The vector H in the far field of an antenna satisfies—

1. 1. ∇. H = 0 and ∇ × H = 0

   
   
   

   2. ∇. H ≠ 0 and ∇ × H ≠ 0

   
   
   

   3. ∇. H = 0 and ∇ × H ≠ 0

   
   
   

   4. ∇. H ≠ 0 and ∇ × H = 0
      

   
   
   

---

1. View Hint View Answer Discuss in Forum

   H = ∇ × A
   ∴ ∇ × H = ∇ × (∇ × A) = 0
   ∇. H = ∇. (∇ × A) = 0

   Correct Option: C

   H = ∇ × A
   ∴ ∇ × H = ∇ × (∇ × A) = 0
   ∇. H = ∇. (∇ × A) = 0

---

---

193. The polarization of wave with electric field vector E = E₀ e^{j zωt + βz)} (ax + ay) is—

1. 1. Linear

   
   
   

   2. Elliptical

   
   
   

   3. Left hand circular

   
   
   

   4. Right hand circular

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Given E = E₀ e^{j (ωt – βz)} [ax + ay]
   [As only space quadrature and no time quadrature]
   Hence the polarization is linear.

   Correct Option: A

   Given E = E₀ e^{j (ωt – βz)} [ax + ay]
   [As only space quadrature and no time quadrature]
   Hence the polarization is linear.

---

194. The wavelength of a wave with propagation constant (0.1 π + j 0.2 π) m– 1 is—

1. 1. 2/√0.05 m

   
   
   

   2. 10 m

   
   
   

   3. 20 m

   
   
   

   4. 30 m

   
   
   

---

1. View Hint View Answer Discuss in Forum

   On comparing, given equation
   γ = 0.1 π + j 0.2 π with
   standard equation γ = α + jβ
   β = 0.2 π
   Now,
   

   λ =	2∏	=	2∏	= 10 m
   	β		.2∏

   Correct Option: B

   On comparing, given equation
   γ = 0.1 π + j 0.2 π with
   standard equation γ = α + jβ
   β = 0.2 π
   Now,
   

   λ =	2∏	=	2∏	= 10 m
   	β		.2∏

---

---

195. All transmission line sections in figure shown below has a characteristic impedance, R₀ + j₀. The input impedance Z_in equals—
     

1. 1. R₀

   
   
   

   2. 2	R0
      3

   
   
   

   3. 3	R0
      2

   
   
   

   4. 2 R₀

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   NA

---