JEE Mains · Physics · STD 11 - 14. waves and sound
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 \(\dots \; Hz\).
- A \(151\)
- B \(152\)
- C \(153\)
- D \(154\)
Answer & Solution
Correct Answer
(B) \(152\)
Step-by-step Solution
Detailed explanation
\(f_{1}= f\) \(f _{2}= f +4\) \(f _{3}= f +2 \times 4\) \(f _{4}= f +3 \times 4\) \(f _{20}= f +19 \times 4\) \(f +(19 \times 4)=2 \times f\) \(f =76 \; Hz\) Frequency of last tuning forks \(=2 \; f\) \(=152 \; Hz\)
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