Q. If an electromagnetic wave is transmitted to the height equal to 150 km with maximum frequency 300 kHz and critical frequency 100 kHz, the skip distance is

A) 426 km
B) 636 km
C) 824 km
D) 849 km
Explanation

(d) Here, $h = 150\ km,\ v = 300\ kHz$,
$v_{c} = 100\ kHz$
$D_{skip} = 2h\sqrt{\left( \frac{v}{v_{c}} \right)^{2} - 1}$
$= 2 \times 150 \times 10^{3}\sqrt{\left( \frac{300 \times 10^{2}}{100 \times 10^{3}} \right) - 1}$
$= 849\ Hz$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.