JEE Main 2017 · 9 Apr · Shift 1 · Physics Q28

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PhysicsSingle correct+4 / -1Question 28 of 90

A signal is to be transmitted through a wave of wavelength λ, using a linear antenna. The length l of the antenna and effective power radiated $P_{eff}$ will be given respectively as :

(K is a constant of proportionality)

\lambda

P_{eff}=K\left(\dfrac1\lambda\right)^2

\dfrac\lambda8

P_{eff}=K\left(\dfrac1\lambda\right)

\dfrac\lambda{16}

P_{eff}=K\left(\dfrac1\lambda\right)^3

\dfrac\lambda5

P_{eff}=K\left(\dfrac1\lambda\right)^{1/2}

Answer and worked solution

Official answer: A

The power radiated by a linear antenna goes as $\left(\frac l\lambda\right)^2$ , so for a given antenna $P_{eff}\propto\frac1{\lambda^2}$ . Only (1) has that dependence; it gives the antenna length as of the order of λ.

Topics:Propagation of EM waves

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JEE Main 2017 · 9 Apr · Shift 1, Physics question 28