Advertisements
Advertisements
प्रश्न
A set of 56 tuning forks are arranged in series of increasing frequencies. If each fork gives 4 beats with preceding one and the frequency of the last is twice than than that of first, then frequency of the first fork is ____________.
पर्याय
110 Hz
220 Hz
224 Hz
448 Hz
MCQ
रिकाम्या जागा भरा
उत्तर
A set of 56 tuning forks are arranged in series of increasing frequencies. If each fork gives 4 beats with preceding one and the frequency of the last is twice than than that of first, then frequency of the first fork is 220 Hz.
Explanation:
n56 = n1 + (56- 1)4
Also, n56 = 2n1
∴ 2n1 = n1 + 55 x 4
∴ n1 = 220 Hz
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?