A sonometer wire stretched by weight 'w' is i | vedclass

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(C) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit leng

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$1:2$

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(C) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.Since the wire is in unison with the same tuning fork,the frequency $f$ remains constant.For the first case,tension $T_1 = w$,so $f = \frac{1}{2L_1} \sqrt{\frac{w}{\mu}}$.For the second case,the weight is increased by $3w$,so the new tension $T_2 = w + 3w = 4w$. Thus,$f = \frac{1}{2L_2} \sqrt{\frac{4w}{\mu}}$.Equating the two expressions for $f$:$\frac{1}{2L_1} \sqrt{\frac{w}{\mu}} = \frac{1}{2L_2} \sqrt{\frac{4w}{\mu}}$$\frac{1}{L_1} = \frac{1}{L_2} \sqrt{4}$$\frac{1}{L_1} = \frac{2}{L_2}$Therefore,$\frac{L_1}{L_2} = \frac{1}{2}$.

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