JEE MAIN - Physics (2022 - 26th June Evening Shift - No. 22)
A set of 20 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first, then the frequency of last fork is _________ Hz.
پاسخ دهید
152
توضیح
Given $${v_{20}} = 2{v_1}$$
Also $${v_{20}} = 4 \times 19 + {v_1}$$
So $${v_{20}} = 152\,Hz$$