Question

Two tuning forks P and Q when set vibrating gives 4 beats/ second. If the prong of fork P is filed, the beats are reduced to 2/*. What is the frequency of P, if that of Q is 250 Hz.?

(A) 246 Hz

(B) 250 Hz

(C) 254 Hz

(D) 252 Hz

Answer 1

The correct option (A) 246Hz

Explanation:

There are 4 beats between P & ɸ

∴ Possible frequencies of p are 250 ± 4 Hz.

When prong of p is filed, frequency becomes greater than original. If we assume original frequency of p is 254, then on filing, its frequency is greater than 254. The beats between P & ɸ will be more than 4. But given is that the beats are reduced to 2.

Hence 254 is not answer. Hence frequency is (250 – 4) = 246 Hz