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A steel wire of length 60 cm and area of...

A steel wire of length 60 cm and area of cross-section `10^(-6)m^(2)` is joined with an aluminium wire of length 45 cm and area of cross-section `3 xx 10^(-6) m^(2)`. The composite string is stretched by tension of 80 N. Density of steel is 7800 kg `m^(-3)` and that of aliminimum is `2600 kg m^(-3)`. The minimum frequency of tuning fork when can produce standing wave in it with node at the joint is :-

A

357.3Hz

B

375.3 Hz

C

337.5 Hz

D

325.3 Hz

Text Solution

Verified by Experts

The correct Answer is:
C
Mass per unit length `mu=(m)/(l)=(rhoAl)/(l)=rhoA`
`mu_(s)=mu_(Al)=78xx10(-4)kg//m`
`therefore` speed of wave is same both wire
`V=sqrt((T)/(mu))=sqrt((80xx10^(4))/(78))=(2xx10^(2))/(sqrt(3.9))`
`v_(min)=(V)/(lamda_(max))=(200)/(sqrt(3.9)xx0.3)[(lamda_(max))/(2)=15cm" for C as a node"] =337.5Hz`
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