Given,

The diameter of the metallic ball (d) = 4.2 cm

Radius of the metallic ball (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{4.2}{2}$ cm = 2.1 cm

The density of the metal = 8.9 g per cm³

We know that,

Density = $\dfrac{\text{Mass}}{\text{Volume}}$

Mass = Density × Volume ...........(1)

By formula,

Volume of the metallic ball (V) = $\dfrac{4}{3}πr^3$

Substituting values we get :

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times (2.1)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 9.261 \\[1em] = \dfrac{88}{21} \times 9.261 \\[1em] = \dfrac{814.968}{21} \\[1em] = \text{38.808 cm}^3$

Substituting value in equation (1), we get :

Mass of the metallic ball = 8.9 × 38.808

= 345.39 g

Hence, the mass of the ball is 345.39 g.

BRIGHT TUTORIALS

CBSE Class IX | Academic Year 2026-2027

9403781999

Excellence in Education

Surface Areas and Volumes — Interactive Study Guide

Formula Master Table

Solid	CSA	TSA	Volume
Cube (a)	4a²	6a²	a³
Cuboid (l,b,h)	2h(l+b)	2(lb+bh+hl)	lbh
Cylinder (r,h)	2πrh	2πr(r+h)	πr²h
Cone (r,h,l)	πrl	πr(l+r)	⅓πr²h
Sphere (r)	4πr²		&frac43;πr³
Hemisphere (r)	2πr²	3πr²	⅔πr³

Common Traps

Cone CSA uses slant height l, NOT height h! l = √(r²+h²)

Hemisphere TSA = CSA + base circle = 2πr² + πr² = 3πr²

Use radius, not diameter! If diameter is given, divide by 2 first.

Quick Self-Check

1. Volume of cube with side 5 cm? (125 cm³)
2. CSA of cylinder with r=7, h=10? (2×22/7×7×10 = 440 cm²)
3. Volume of cone with r=3, h=4? (⅓×π×9×4 = 12π cm³)