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A shot-putt is a metallic sphere of radius 8 cm. If the density of the metal is 3 g per cm3, find the mass of the shot-putt.​

Answer»

\underline{\underline{\sf{\maltese\:\:Question}}}

  • A shot-putt is a metallic sphere of radius 8 cm. If the DENSITY of the metal is 3 g per cm³ , FIND the mass of the shot-putt

\underline{\underline{\sf{\maltese\:\:Given}}}

  • Radius of metallic sphere = 8 cm
  • Density = 3 g per cm³

\underline{\underline{\sf{\maltese\:\:To\:Find}}}

  • The Mass of the shot-putt

\underline{\underline{\sf{\maltese\:\:Answer}}}

  • Mass = 1478.847 g

\underline{\underline{\sf{\maltese\:\:Calculations}}}

Density = Mass/Volume

⇒ Density = Mass/Volume of sphere

⇒ Density = Mass/[4/3 πr³]

⇒ Density = Mass/[4/3 × π × r³]

Since π = 22/7

⇒ Density = Mass/[4/3 × 22/7 × r³]

⇒ Density = Mass/[(4 × 22)/(7× 3) × r³]

⇒ Density = Mass/[88/21 × r³]

Given R = 4.9 cm

⇒ Density = Mass/[88/21 × (4.9 cm)³]

⇒ Density = Mass/[88/21 × (4.9 × 4.9 × 4.9) cm³]

⇒ Density = Mass/[88/21 × 117.649 cm³]

⇒ Density = Mass/[4.190 × 117.649 cm³]

⇒ Density = Mass/[492.949 cm³]

⇒ Density = Mass/492.949 cm³

Given Density = 3 g/cm³

⇒ 3 g/cm³ = Mass/492.949 cm³

Multiplying both Sides by 492.949 cm³

⇒ 3 g/cm³ × 492.949 cm³ = Mass/492.949 cm³ × 492.949 cm³

⇒ 3 g × 492.949  = Mass

⇒ 1478.847 g  = Mass

Switch Sides

⇒ Mass = 1478.847 g

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\underline{\underline{\sf{\maltese\:\:Diagram}}}

\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\qbezier(2.3,0)(2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,2.121)(0,2.3)\qbezier(-2.3,0)(-2.121,-2.121)(0,-2.3)\qbezier(2.3,0)(2.121,-2.121)(-0,-2.3)\put(0,0){\line(1,0){2.3}}\put(0.5,0.3){\bf\large 8\ cm}\end{picture}

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