Question

A hemispherical bowl just floats without sinking in a

liquid of density 1.2 × 103kg/m3. If outer diameter

and the density of the bowl are 1 m and 2 × 104

kg/m3 respectively, then the inner diameter of the

bowl will be​

Answer 1

Explanation:

Given  

A hemispherical bowl just floats without sinking in a  liquid of density 1.2 × 10^3 kg/m3. If outer diameter  and the density of the bowl are 1 m and 2 × 104  kg/m3 respectively, then the inner diameter of the  bowl will be

• According to question upthrust is the weight of the liquid displaced.is equal to weight of the body.
• So weight of body = volume x ρ (body) g = 2/3 π (R^3 – r^3)ρg (R and r will be outer and inner radius)
• Now weight of liquid displaced will be V ρg = 2/3 πR^3
• Now equating both we get
• 2/3 π R^3 ρg = 2/3 π (R^3 – r^3) ρg
• So R^3 x 1.2 x 10^3 = (R^3 – r^3) x 2 x 10^4
•  6 R^3 = 100 (R^3 – r^3)
• Now diameter D = 1 m, so r = 1/2 = 0.5
• So 94 R^3 = 100 r^3
• So r^3 = 0.1175
• Or r = 0.489
• Or d= 2r = 2 x 0.489

Or d = 0.98 m

Reference link will be

https://brainly.in/question/14286810

Answer 2

Answer:

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Explanation:

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