A) 200 Hz
B) 400 Hz
C) 500 Hz
D) 1000 Hz
Correct Answer: C
Solution :
\[n\propto \sqrt{T}\] Þ \[\frac{\Delta n}{n}=\frac{\Delta T}{2T}\] If tension increases by 2%, then frequency must increases by 1%. If initial frequency \[{{n}_{1}}=n\]then final frequency n2 ? n1 = 5 Þ \[\frac{101}{100}n-n=5\] Þ \[n=500Hz.\] Short trick : If you can remember then apply following formula to solve such type of problems. Initial frequency of each wire (n) \[=\frac{\text{(Number}\ \text{of}\ \text{beats}\ \text{heard}\ \text{per}\ \text{sec)}\times \text{200}}{\text{(per}\ \text{centage}\ \text{change}\ \text{in}\ \text{tension}\ \text{of}\ \text{the}\ \text{wire)}}\] Here \[n=\frac{5\times 200}{2}=500Hz\]
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