Standing waves are generated on a sonometer s | vedclass

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(D) The frequency of the sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length of the loop,$T$ is the tension,and

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$1.26$

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(D) The frequency of the sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length of the loop,$T$ is the tension,and $\mu$ is the linear mass density.Since the frequency $f$ remains constant,we have $\frac{1}{L_1} \sqrt{T_1} = \frac{1}{L_2} \sqrt{T_2}$.Given $\frac{L_1}{L_2} = 2.2$,we have $\sqrt{\frac{T_1}{T_2}} = 2.2$,which implies $T_1 = (2.2)^2 T_2 = 4.84 T_2$.When the cylinder is in air,$T_1 = Mg$. When immersed in water,$T_2 = Mg - F_B$,where $F_B$ is the buoyant force.$F_B = V_{cyl} \rho_w g = \frac{M}{\rho_{cyl}} \rho_w g = \frac{Mg}{S.G.}$,where $S.G.$ is the specific gravity.Thus,$Mg = 4.84 (Mg - \frac{Mg}{S.G.})$.Dividing by $Mg$,we get $1 = 4.84 (1 - \frac{1}{S.G.})$.$1 = 4.84 - \frac{4.84}{S.G.}$,which gives $\frac{4.84}{S.G.} = 3.84$.$S.G. = \frac{4.84}{3.84} \approx 1.26$.

Standing waves are generated on a sonometer string loaded with a cylindrical body. If the cylinder is completely immersed in water,the length of the loops changes by a factor of $2.2$. The specific gravity of the material of the cylinder is

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