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Ex.13.8 Q3 Surface Areas and Volumes Solution - NCERT Maths Class 9

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Question

The diameter of a metallic ball is \(4.2\;\rm{ cm}\). What is the mass of the ball, if the density of the metal is \(8.9\) g per\(\begin{align}\,c{m^3} \end{align}\)?

 Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-8 | Question 3

Text Solution

Reasoning:

Volume of the sphere \(\begin{align} = \frac{4}{3} \end{align}\) \(\pi r^{3}\) and mass is the product of density times volume.

What is the known?

Diameter of the sphere and density of the metal per \(\begin{align}{m^3} \end{align}\).

What is the unknown?

Mass of the ball.

Steps:

Diameter

Radius \(\begin{align}\,r = \frac{{4.2}}{2} = 2.1\,\,cm \end{align}\)

Volume of the sphere

\[\begin{align}&= \frac{4}{3} \times \frac{{22}}{7} \times {(2.1)^3}\\ &= 38.808\,\,c{m^3}\\ &= \frac{4}{3}\pi {r^3} \end{align}\]

Density = \(8.9\) g per \(\begin{align}c{m^3} \end{align}\)

Mass\(=\)volume\(×\)density.

\[\begin{align} &= 38.808\,\, \times \,8.9\\ &= 345.39\,g \end{align}\]

Answer:

Mass of the ball \(\begin{align} = 345.39\,g \end{align}\)