bitsat 5

BITSAT Help › Optics questions › bitsat 5

A no. of tuning forks are arranged in the order of increasing frequency & any tw successive tuning foks produce 4 beatsper second,whn sounded together. If the last tuning fork has a frequency of octave higher than that of the first tuning fok & the frequecy of the first tuning fork is 256 Hz, then the no. of tuning forks is?

A)63
b)64
c)65
d)66

#Physics #Optics

UP 0 DOWN 0 1 6
Post on FacebookPost anonymously?

6 Answers

1

ANKIT MAHATO ·2009-05-02 23:47:50

65 is the ans

UP 0 DOWN 0
Post on FacebookPost anonymously?

13

deepanshu001 agarwal ·2009-05-03 00:37:22

ankit ...wat is meant by 1 octave higher...?

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

ANKIT MAHATO ·2009-05-03 00:39:50

1 octave higher means .. double .. :)

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

ANKIT MAHATO ·2009-05-03 00:40:03

now this sum is a cake walk :)

UP 0 DOWN 0
Post on FacebookPost anonymously?

13

deepanshu001 agarwal ·2009-05-03 00:42:01

yeah!

UP 0 DOWN 0
Post on FacebookPost anonymously?

1

sushrut17 ·2009-05-03 02:27:18

65

UP 0 DOWN 0
Post on FacebookPost anonymously?

Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω