Figure shows a metal ball suspended by thread of negligible mass from an upright cylinder that floats partially submerged in water. The cylinder has height 6 cm, face area $11 c m^{2}$ on the top and bottom and density $0.5 g / c m^{3}$ . 4 cm of cylinder’s height is inside the water surface. If density of the metal ball is 8g/cm³ then its radius is equal to: $(ρ_{w} = 1 gm / {cm}^{2})$

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${(\frac{3}{4})}^{1 / 3} cm$

${(\frac{5}{11})}^{1 / 3} cm$

${(\frac{3}{8})}^{1 / 3} cm$

${(\frac{6}{8})}^{1 / 3} cm$

answer is A.

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Detailed Solution

Taking cylinder and the ball as system

$\frac{4}{3} π r^{3} \cdot ρ_{2} \cdot g + A h \cdot ρ_{1} \cdot g = \frac{4}{3} π r^{3} \cdot ρ_{w} \cdot g + A h_{1} \cdot ρ_{w} \cdot g$

$\Rightarrow r = {[\frac{3 A (h_{1} ρ_{w} - {hρ}_{1})}{4 π (ρ_{2} - ρ_{w})}]}^{1 / 3}$

$A = 11 {cm}^{2}, h_{1} = 4 cm; ρ_{w} = 1 gm / {cm}^{3}$

$ρ_{1} = 0.5 gm / {cm}^{3}; ρ_{2} = 8 gm / {cm}^{3}$

$r = [\frac{3 \times 11 (4 \times 1 - 6 \times 0.5)}{4 \times (\frac{22}{7}) \times (8 - 1)}] = {(\frac{3}{8})}^{1 / 3} cm$

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