The diameter of a metallic ball is $4.2 c m .$ What is the mass of the ball, if the density of the metal is $8.9 g / c m^{3}$ ?
[Assume $π = \frac{22}{7}$ ]

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Solution

Radius (r) of the metallic ball
$= (\frac{4.2}{2}) c m = 2.1 c m$

Volume of the metallic ball (Spherical in shape)
$= \frac{4}{3} π r^{3}$
$= [\frac{4}{3} \times \frac{22}{7} \times (2.1)^{3}] c m^{3}$
$= 38.808 c m^{3}$

$Density = \frac{Mass}{Volume}$
$Mass= Density \times Volume$
$= (8.9 \times 38.808) g$
$= 345.3912 g$

Hence, the mass of the ball is $345.3912 g$ (approximately).

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Dalton's Atomic Theory

Standard VII Chemistry

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The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g/cm3?[Assume π=227]

Radius (r) of the metallic ball =(4.22) cm=2.1 cm Volume of the metallic ball (Spherical in shape) =43πr3 =[43×227×(2.1)3] cm3 =38.808 cm3 Density=MassVolume Mass= Density×Volume =(8.9×38.808)g =345.3912 g Hence, the mass of the ball is 345.3912 g (approximately).

The diameter of a metallic ball is $4.2 c m .$ What is the mass of the ball, if the density of the metal is $8.9 g / c m^{3}$ ?
[Assume $π = \frac{22}{7}$ ]