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A steel wire of length 60 cm and area of cross section `10^(-6)m^(2)` is joined with a n aluminimum wire of length 45 cm and are of cross section `3xx10^(-6)m^(2)`. The composite string is stretched by a tension of 80 N. Density of steel is `7800kgm^(-3)` and that of aluminimum is `2600kgm^(-3)` the minimum frequency of tuning fork. Which can produce standing wave in it with node at joint isA. 357.3HzB. 375.3 HzC. 337.5 HzD. 325.3 Hz |
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Answer» Correct Answer - 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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