91Ó°ÊÓ

the sphere is a geometrical shape with a set of three-dimensional points with space lying the same distance

Here, tension and force are equal to the weight that the ball carries. for the second case, weight is subtracted from the buoyant force. the buoyant force has a sphere diameter which is D and the liquid with density is p.

Thus, $F_B=pgV=pg\frac{4}{3}\mathrm{Ï€}{R}^3=pg\frac{4}{3}\mathrm{Ï€}{\left(\frac{D}{2}\right)}^3=\frac{1}{6}{\mathrm{Ï€\pm è²\mu \P Ù}}^3$

The frequency and the string are stretched by tension T

The linear density $\mathrm{μ}$and the anti mode number is m which is provided as

$f=m\frac{v}{2L}=\frac{m\sqrt{T}}{2L\sqrt{\mathrm{μ}}}$

after the substitution, the denoted mass or m will be

$\frac{3\sqrt{Mg}}{2L\sqrt{\mathrm{μ}}}=\frac{5\sqrt{Mg-{\displaystyle \frac{1}{6}}{\mathrm{Ï€\pm è²\mu \P Ù}}^3}}{2L\sqrt{\mathrm{μ}}}$

The simplification of the equation is

The square on both sides are

$9M=25(M-\frac{1}{6}{\mathrm{Ï€\pm è\P Ù}}^3)⇒\frac{1}{6}{\mathrm{Ï€\pm è\P Ù}}^3=\frac{16}{25}M$

The diameter can be expressed as,

role="math" localid="1649066728962" $D=\sqrt[3]{\frac{16·6M6}{25\mathrm{Ï€\pm è}}}M⇒\boxed{D=\sqrt[3]{\frac{16·6M6M}{25\mathrm{Ï€\pm è}}}}$

The step we have is

$D=\sqrt[3]{\frac{16·6·1.5}{25\mathrm{π}·1000}}=\boxed{3.93cm}$