JEE Physics Electromagnetic Waves MCQs Set 14

Answer: (a) 36.6 m

Explanation:
1. \( \lambda = c/f \).
2. \( \lambda = 3\times10^8/8.2\times10^6 = 36.6 \) m.

Teacher's Note:
1. Speed of all EM waves in vacuum is c = 3 × 10⁸ m/s.
2. Convert MHz to Hz before dividing.

Answer: (b) \( 0.25\times10^4 \)

Explanation:
1. 1 mm \( = 10^{-3} \) m and \( \lambda = 4\times10^{-7} \) m.
2. Number of waves \( = 10^{-3}/4\times10^{-7} = 2.5\times10^3 = 0.25\times10^4 \).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (c) 10 m

Explanation:
1. \( \lambda = c/f \).
2. \( \lambda = 3\times10^8/30\times10^6 = 10 \) m.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (b) \( R\propto h^{1/2} \)

Explanation:
1. Range \( d = \sqrt{2Rh} \) where R is the radius of the Earth.
2. Hence \( d\propto\sqrt h \).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (a) \( d^2/2R \)

Explanation:
1. The line-of-sight range is \( d = \sqrt{2Rh} \).
2. Squaring, \( h = d^2/2R \).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (d) 40 lakh

Explanation:
1. Area covered \( = \pi d^2 = 2\pi Rh = 2\pi\times6.37\times10^6\times100 \approx 4\times10^9 \) m² \( = 4000 \) km².
2. Population \( = 4000\times1000 = 4\times10^6 \), i.e. 40 lakh.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (c) ultraviolet region

Explanation:
1. \( \nu = E/h = 11\times1.6\times10^{-19}/6.62\times10^{-34} \).
2. \( \nu \approx 2.7\times10^{15} \) Hz, which lies in the ultraviolet region.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (c) \( E/\sqrt2 \)

Explanation:
1. Intensity \( I\propto P \) and \( I\propto E^2 \), so \( E\propto\sqrt P \).
2. \( E' = E\sqrt{50/100} = E/\sqrt2 \).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (d) \( \dfrac{hc}{eV} \)

Explanation:
1. The maximum photon energy equals the electron energy: \( eV = hc/\lambda_{min} \).
2. \( \lambda_{min} = hc/eV \).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (b) \( 1.83\times10^{-22}\text{ Wh} \)

Explanation:
1. \( E = hc/\lambda = 6.6\times10^{-34}\times3\times10^8/3\times10^{-7} = 6.6\times10^{-19} \) J.
2. 1 Wh = 3600 J, so \( E = 6.6\times10^{-19}/3600 = 1.83\times10^{-22} \) Wh.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (a) \( 2.21\times10^{-10}\text{ J m}^{-3} \)

Explanation:
1. Average electric energy density \( = \tfrac14\varepsilon_0E_0^2 \).
2. \( = 0.25\times8.85\times10^{-12}\times100 = 2.21\times10^{-10} \) J/m³.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (b) \( I/c \)

Explanation:
1. Momentum delivered per unit area per unit time is \( I/c \) for complete absorption.
2. So pressure \( = I/c \).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (c) \( 6\times10^{-8}\text{ T} \)

Explanation:
1. \( B_0 = E_0/c \).
2. \( B_0 = 18/3\times10^8 = 6\times10^{-8} \) T.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (a) \( E_0k = B_0\omega \)

Explanation:
1. \( E_0/B_0 = c = \omega/k \).
2. So \( E_0 k = B_0\omega \).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (b) \( y = \text{constant} \)

Explanation:
1. Wavefronts are surfaces perpendicular to the direction of propagation.
2. The direction of propagation is y, so wavefronts are planes of y = constant.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (b) \( 1.8\times10^{8}\text{ m s}^{-1} \)

Explanation:
1. \( v = c/\sqrt{\mu_r\varepsilon_r} \).
2. \( v = 3\times10^8/\sqrt{1.3\times2.14} = 3\times10^8/1.668 \approx 1.8\times10^8 \) m/s.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (a) 2 A

Explanation:
1. Displacement current in a capacitor equals the conduction current: \( I_d = C\,dV/dt \).
2. \( I_d = 2\times10^{-12}\times10^{12} = 2 \) A.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (b) \( 10^{6}\text{ V s}^{-1} \)

Explanation:
1. \( I_d = C\,dV/dt \).
2. \( dV/dt = 1/10^{-6} = 10^6 \) V/s.

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (c) along the x-direction

Explanation:
1. The phase \( (t - x/c) \) shows the wave advances in the positive x-direction.
2. Propagation is along the x-axis (E is along y, B along z).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.

Answer: (c) \( kr/2 \)

Explanation:
1. \( E\cdot2\pi r = -d\Phi/dt = \pi r^2 k \) (magnitude).
2. \( E = kr/2 \).

Teacher's Note:
1. Recall the relevant standard result before substituting numbers.
2. Check units and orders of magnitude at the end.