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JEE Main 2022 (26 Jun, Shift 2) · Beats

Beats question from JEE Main 2022 (26 Jun, Shift 2)

Q44JEE Main 2022 · 26 JUN · SHIFT 2Easy

Beats

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.

Numerical answer type. Work it out, then open the answer.

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Answer: 152

Last (20th) fork $= f + 19 \times 4 = f + 76$
$f + 76 = 2f \Rightarrow f = 76\text{ Hz}$
Last fork $= 2 \times 76 = 152\text{ Hz}$

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