Question

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 1

Correct Answer - 18
Let number of loops in steel are `P` and in aluminium number of loops are `Q`.
`:. (P)/(2l_(s)) sqrt((T)/(mu_(s))) = (Q)/(2l_(Al)) = sqrt((T)/(mu_(Al))) :. (P)/(Q) = (4)/(3) …..(1)`
For minimum frequency `P = 4`
`f=(4)/(2xx(80)/(100))sqrt((40)/(10^(-6)xx7.6xx10^(3)))`
`f = 180 Hz`