Two wires are fixed on a sonometer with their tensions are in the ratio 8:1, the length are in the ratio 36,35, the diameters in the ratio 4:1 and densities in the ratio 1:2. If the note of higher pitch has a frequency of 360Hz, then the frequency of other string will be

Options

A. 5
B. 10
C. 15
D. 20

Answer

B

Explanation

The frequency of fundamental note is `upsilon=(1)/(lD)sqrt((T)/(pirho))`
where l is the length, D is the diameter, T is the tension and `rho` is the density of the wire.
`therefore (upsilon_(1))/(upsilon_(2))=(l_(2))/(l_(1))(D_(2))/(D_(1))sqrt((T_(1))/(T_(2))(rho_(2))/(rho_(1)))`
Substituting the given values, we get
`(upsilon_(1))/(upsilon_(2))=(35)/(36)xx(1)/(4)sqrt((8)/(1)xx(2)/(1))=(35)/(36)or36upsilon_(1)=35upsilon_(2)" "...(i)`
`therefore upsilon_(2)gtupsilon_(1)`
`therefore upsilon_(2)` is the note of the higher pitch.
From (i), `upsilon_(1)=(35)/(36)upsilon_(2)=(35)/(36)xx360=350Hz`
`therefore` Beat frequency = `upsilon_(2)-upsilon_(1)=360-350=10Hz`