A hollow metal sphere of outside diameter of 10 cm weighing 5gm/cm3 in air floats half submerged in fresh water. Determine the radius of the cavity
Explanations & Calculations
w=mg=43π(ro3−ri3)dgu=wu=vρg=12.43πro3ρg43π([10cm]3−ri3)×5gcm−3×g=23π×[10cm]3×1gcm−3×gri=9.7 cm\qquad\qquad \begin{aligned} \small w&=\small mg\\ \small &=\small \frac{4}{3}\pi(r_o^3-r_i^3)dg\\ \\ \small u&=\small w\\ \\ \small u&=\small v\rho g\\ &=\small \frac{1}{2}.\frac{4}{3}\pi r_o^3\rho g\\ \\ \small \frac{4}{3}\pi([10cm]^3-r_i^3)\times5gcm^{-3}\times g&=\small \frac{2}{3}\pi\times[10cm]^3\times 1gcm^{-3}\times g\\ \small r_i&=\small 9.7\,cm \end{aligned}wuu34π([10cm]3−ri3)×5gcm−3×gri=mg=34π(ro3−ri3)dg=w=vρg=21.34πro3ρg=32π×[10cm]3×1gcm−3×g=9.7cm
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