Figure shows an aluminium wire of length `60 cm` joined to a steel wire of length `80 cm` and stretched between two fixed supports. The tension produced is `40 N`. The cross-sectional area of the steel wire is `1.0 mm^(2)` and that of the aluminimum wire is `3.0 mm^(2)` The minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node is `10P` (in `Hz`) the find `P`. Given density of aluminimum is `2.6 g//cm^(3)` and that of steel is `7.8 g//cm^(3)`.

Answer

A

Explanation

`P_s=7.8(gm)/cm^3`
`P_A=2.6(gm)/cm^3`
`m_s=P_sA_s`
`=7.8xx10^-2(gm)/cm`
(m=mass per unit length)
`7.8xx10^-3kg/m`
`m_A=P_A`
`=21.6xx10^-2xx3(gm)/cm`
`=7.8xx10^-2(gm)/cm`
`=7.8xx10^-3(kg)/m`
A node is always placed in the joint. Since aluminium and steel rod has same mass per unit lenght, velocity of wave in both of them is same
`rarr v=sqrt((T/m))`
`=sqrt({40/((7.8xx10^-3))})`
`=sqrt((((4xx10^4)))/7.8)`
`=71.6m/s`
For minimum frequency there would be maximum wavelength. for maximu wavelength minimum no. of loops are to be produced.
`:.` Maximum distance of a loop =20cm
`rarr` wavelength =`lamda=2xx20`
`=40cm=0.4m`=180Hz`