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Q. |
Given that the wave number of a circular cavity resonator is 151 (TE011 mode), and the length of the cavity is twice the radius of the cavity, the radius of the circular cavity operating at 5GHz frequency is: |
A. | 2.1 cm |
B. | 1.7 cm |
C. | 2.84 cm |
D. | insufficient data |
Answer» D. insufficient data | |
Explanation: for a circular cavity resonator, wave number is given by √( (p01/a)2 +(π/d)2). p01 for the given mode of resonance is 3.832. |
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