In a sonometer wire,the tension is maintained | vedclass

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(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(A) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density. Since $l$ and $\mu$ are constant,$n \propto \sqrt{T}$.Initially,the tension $T_1 = M g = 50.7 	imes 10 = 507 \, N$.When the mass is submerged in water,the buoyant force $F_B = V ho_{water} g = 0.0075 	imes 1000 	imes 10 = 75 \, N$.The new tension $T_2 = M g - F_B = 507 - 75 = 432 \, N$.Using the ratio $\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$,we get $\frac{260}{n_2} = \sqrt{\frac{507}{432}}$.$\frac{260}{n_2} = \sqrt{1.1736} \approx 1.0833$.$n_2 = \frac{260}{1.0833} \approx 240 \, Hz$.

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