A metal wire of length 80 cm, area of cross-section 3 `mm^(2)` and material density 3000 kg `m^(-3)` is joined to another metal wire of length 60 cm, area of cross-sectional 1 `mm^(2)` and material density 9000 `kg-m^(3)`.
The free ends of the two wires are stretched between two rigid supports and a tension of 40 N is produced in the wires. The minimum frequency of the tuning fork which can produce stationary waves with the joint of the wires as a node is

Options

A. `200/3`Hz
B. `400/3Hz`
C. `500/3Hz`
D. `(70C)/3Hz`

Answer

C

Explanation

Given, `l_(1)=80cm=80 times 10^(-2)m`
`" "A_(l)=3mm^(2)=3 times 10^(-6)m^(2)`
`" "rho_(1)=3000 kg.m^(-3)`
`" "l_(2)=60 cm=60 times 10^(-2) m`
`" "A_(2)=1 mm^(2)=10^(-6)m^(2)`
`" "rho_(2)=9000 kg.m^(-3) rArr T=40N`
According to the question,
for stationary wave to be formed, frequency of P th harmonic due ot Ist metal wise = frequency of Q th harmonic due to 2nd metal wire
`" "P times 1/(2l_(1))sqrt(T/(rho_(1)A_(1)))=Q times 1/(2l_(2))sqrt(T/(rho_(2)A_(2)))`
or `" "P/Q=l_(1)/l_(2)sqrt((rho_(1)A_(1))/(rho_(2)A_(2)))=(80 times 10^(-2))/(60 times 10^(-2))sqrt((3000 times 3 times 10^(-6))/(9000 times 10^(-6)))`
or `" "P/Q=4/3`
Hence, p is `4^(th)` harmonic and Q is `3^(rd)` harmonic. So, the minimum frequency of the tuning fork, `E_(0)` produce stationary waves with given wire of 2 metals, frequency of tuning fork = frequency of Pth harmonic/Q th harmonic of 2 respective wire.
`n=P times 1/(2l_(1))sqrt(T/(rho_(1)A_(1)))`
`" "=4 times 1/(2 times 80 times 10^(-2))sqrt(40/(3000 times 3 times 10^(-6)))=500/3 Hz`