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3 years ago

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A platinum sphere with radius 1.21 cm 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.

1 answer:

TEA [102]3 years ago

5 0

Answer:

1.556 N , 0.989 N, 0.567 N

Explanation:

Radius of sphere, r = 1.21 cm = 0.0121 m

density of platinum , d = 2.14 x 10^4 kg/m^3

density of mercury, d' = 13.6 x 10^3 kg/m^3

Volume of sphere, $\frac{4}{3}\pi \times r^{3}= \frac{4}{3}\times 1.34\times \left (0.0121 \right )^{3}$

V = 7.42 x 10^-6 m^3

Weight of sphere = volume of sphere x density of platinum x gravity

W = V x d x g = 7.42 x 10^-6 x 2.14 x 10^4 x 9.8 = 1.556 N

Buoyant force, B = Volume x density of mercury x gravity

B = 7.42 x 10^-6 x 13.6 x 10^3 x 9.8 = 0.989 N

Apparent weight = True weight - Buoyant force

Apparent weight = 1.556 - 0.989 = 0.567 N

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