Question

A platinum sphere with radius 0.0103 m is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting on the sphere, and the sphere\'s apparent weight. The densities of platinum and mercury are 2.14 × 104 kg/m3 and 1.36 × 104 kg/m3, respectively.

Answer 1

r = 0.0103 m

V = Volume of sphere = 4r³/3 = 4 (3.14) (0.0103)³ /3 = 0.000046 m³

r = 0.0103 m V = Volume of sphere = 4 pi r^3/3 = 4 (3.14) (0.0103)^3 /3 = 0.000046 m^3 rho_p = density of platinum = 2.14 times 104 m = mass of platinum = rho_p V = (2.14 times 10^4) (0.000046) = 0.98 kg weight = m g = 0.66 times 9.8 = 9.6 N Buoyant force = F_b = rho_m V g = (1.36 times 10^4) (0.000046) (9.8) = 6.1 N apparent weight = weight - Buoyant force = 9.6-6.1 = 3.5 N

$\rho$ _p = density of platinum = 2.14 x 10⁴

m = mass of platinum = $\rho$ _p V = (2.14 x 10⁴) (0.000046) = 0.98 kg

weight = mg = 0.66 x 9.8 = 9.6 N

Buoyant force = F_b = $\rho$ _m V g = (1.36 x 10⁴) (0.000046) (9.8) = 6.1 N

apparent weight = weight - Buoyant force = 9.6-6.1 = 3.5 N