Question

226.A wire under a load has a frequency n1, when the load is completely immersed in water, its frequency is n2.The relative density of the material of the load is

Answer 1

W = weight of the load.

$v=\nu*\lambda,\ \nu=frequency,\ \lambda=wavelength\\\\v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{W}{\mu}}\\\\\frac{W_1}{W_2}=\frac{\nu_1^2}{\nu_2^2}\\\\\frac{W_1-W_2}{W_2}=\frac{\nu_1^2-\nu_1^2}{\nu_2^2}\\\\\frac{W_1}{W_1-W_2}=\frac{wt.\ in\ air}{Loss\ of\ wt.\ in\ water}=\frac{\nu_1^2}{\nu_1^2-\nu_2^2}$

Answer 2

Answer:

W = weight of the load...

(Refer to the attachment)