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(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$. Since $f$ and $\mu$ are constant, we have $l \propto \sqrt{

Vedclass · Accepted Answer

$\frac{1}{\sqrt{8}}$

Vedclass · Answer

(D) The frequency of a sonometer wire is given by $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$. Since $f$ and $\mu$ are constant, we have $l \propto \sqrt{T}$.Thus, $\frac{l_{	ext{air}}}{l_{	ext{water}}} = \sqrt{\frac{T_{	ext{air}}}{T_{	ext{water}}}}$.In air, the tension $T_{	ext{air}} = mg$. When immersed in water, the buoyant force $F_B$ acts upwards, so $T_{	ext{water}} = mg - F_B$.Given specific gravity $\sigma = 8$, the density of iron $ho = 8 ho_w$. The buoyant force $F_B = V ho_w g = \frac{m}{ho} ho_w g = \frac{m}{8 ho_w} ho_w g = \frac{mg}{8}$.Therefore, $T_{	ext{water}} = mg - \frac{mg}{8} = \frac{7}{8} mg = \frac{7}{8} T_{	ext{air}}$.Substituting this into the ratio: $\frac{l_{	ext{air}}}{l_{	ext{water}}} = \sqrt{\frac{T_{	ext{air}}}{\frac{7}{8} T_{	ext{air}}}} = \sqrt{\frac{8}{7}}$.Given $l_{	ext{air}} = \frac{1}{\sqrt{7}} 	imes 1 \ m = \frac{1}{\sqrt{7}} \ m$.Then $l_{	ext{water}} = l_{	ext{air}} 	imes \sqrt{\frac{7}{8}} = \frac{1}{\sqrt{7}} 	imes \sqrt{\frac{7}{8}} = \frac{1}{\sqrt{8}} \ m$.

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