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A tuning fork of frequency 400Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2 , keeping the tension constant and the linear density constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is

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6 Hz

8 Hz

4 Hz

12 Hz

answer is C.

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Detailed Solution

$n = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$

$2 % decrease in length so length becomes 49 cm$

$\frac{400}{n_{2}} = \frac{49}{50} n_{2 =} \frac{400 \times 50}{49} n_{2} - n_{1} = 400 (\frac{50}{49} - 1) \approx 8$

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