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A string is tied at one end to a vibrating blade, and the other end is passes over a fixed pulley as shown in figure. A sphere of mass $m$ hangs on the other end of the string. The string is vibrating in its $n_{1}$ harmonic. A container of water is raised under the sphere so that the sphere is completely submerged. In this configuration, the string vibrates in its $n_{2}$ harmonic. (Density of water is $ρ_{w a t e r}$ )

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The radius of the sphere is ${\frac{3 m}{4 π ρ_{w a t e r}} [1 - {(\frac{n_{1}}{n_{2}})}^{2}]}^{1 / 3}$

The magnitude of buoyant force is $m g [1 - {(\frac{n_{2}}{n_{1}})}^{2}]$

The radius of the sphere is ${\frac{3 m}{4 π ρ_{w a t e r}} [1 - {(\frac{n_{2}}{n_{1}})}^{2}]}^{1 / 3}$

The magnitude of buoyant force is $m g [1 - {(\frac{n_{1}}{n_{2}})}^{2}]$

answer is A, C.

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Detailed Solution

$\begin{array}{l} T_{1} = m g & - (1) \\ T_{2} = m g - B & - (1) \end{array}$

$f = \frac{n_{1}}{2 / \sqrt{\frac{T_{1}}{μ}}} = \frac{n_{1}}{2 /} \sqrt{\frac{T_{2}}{μ}}$

$T_{2} = {(\frac{n_{1}}{n_{2}})}^{2} T_{1} = {(\frac{n_{1}}{n_{2}})}^{2} m g$

$B = m g - {(\frac{n_{1}}{n_{2}})}^{2} m g = m g [1 - {(\frac{n_{1}}{n_{2}})}^{2}], r = {(\frac{3 B}{4 π ρ_{ϱ} g})}^{1 / 3} .$

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