Two identical stringed instruments have frequency 100 Hz. If tension in one of them is increased by 4% and they are sounded together then the number of beats in one second is

(1) 1

(2) 8

(3) 4

(4) 2

(4) Frequency of vibration in tight string

$n=\frac{p}{2l}\sqrt{\frac{T}{m}}⇒n\propto \sqrt{T}$$\frac{\Delta n}{n}=\frac{\Delta T}{2T}=\frac{1}{2}×\left(4%\right)=2%$

⇒ Number of beats = $\Delta n=\frac{2}{100}×n=\frac{2}{100}×100=2$

Difficulty Level:

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