Question

Twenty four tuning forks are arranged in such a way that each fork produces 6 beats/s with the preceding fork. If the frequency of the last tuning fork is double than the first fork, then the frequency of the second tuning fork is……… 

(A) 132 

(B) 138 

(C) 276 

(D) 144

Answer 1

 The correct option (B) 138

Explanation:

Total no of forks = 24

each fork produces 6 beats per second with preceding fork.

freq of last fork = 2 × freq of first fork

Let frequency of 1^st fork be f₁

∴ freq of 2^nd fork = f₁ + 6

= f₁ + 6(2 – 1)

freq of 3^rd fork   = f₁ + 6 + 6

 = f₁ + 6(3 – 1)

similarly

freq of 24^th fork  = f₁ + 6(24 – 1)

= f₁ + 138

also freq of 24^th fork = 2 × freq of 1^st fork

∴ f₁ + 138 = 2 f₁

∴ f₁ = 138 Hz