A) 2 : 1
B) 3 : 2
C) 3 : 4
D) 1 : 3
Correct Answer: A
Solution :
Fundamental frequency in case of string is \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\Rightarrow n\propto \frac{\sqrt{T}}{l}\Rightarrow \frac{n'}{n}=\sqrt{\frac{T'}{T}}\times \frac{l}{l'}\] putting \[T'=T+0.44T=\frac{144}{100}T\] and \[l'=l-0.4l=\frac{3}{5}l\] We get \[\frac{n'}{n}=\frac{2}{1}\].
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