A 7.8-kg solid sphere, made of metal whose density is 2500 kg/m³, is suspended by a cord. When the sphere is immersed in water (of density 1000 kg/m³), what is the tension in the cord? A) 46 N B) 61 N C) 76 N D) 92 N E) 110 N

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Quizplus · Accepted Answer

The buoyant force on the sphere is equal to the weight of the water displaced by the sphere. The volume of the sphere is:$$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{7.8	ext{ kg}}{2500	ext{ kg/m}^3}ight)^3 = 4.19	imes10^{-3}	ext{ m}^3$$The weight of the water displaced by the sphere is:$$F_b = ho_w g V = (1000	ext{ kg/m}^3)(9.81	ext{ m/s}^2)(4.19	imes10^{-3}	ext{ m}^3) = 41.1	ext{ N}$$The tension in the cord is equal to the weight of the sphere minus the buoyant force:$$T = W - F_b = (7.8	ext{ kg})(9.81	ext{ m/s}^2) - 41.1	ext{ N} = 46.1	ext{ N}$$

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