Given,

Diameter of the hemispherical ball (d) = 10.5 cm

Radius of the hemispherical ball (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}$ cm = 5.25 cm

By formula,

Volume of the hemispherical ball (V) = $\dfrac{2}{3}πr^3$

Substituting values we get :

$V = \dfrac{2}{3} \times \dfrac{22}{7} \times (5.25)^3 \\[1em] = \dfrac{44}{21} \times 144.703 \\[1em] = \dfrac{6366.9375}{21} \\[1em] = 303.1875 \\[1em] = \dfrac{303.1875}{1000} (∵ 1000 \text{ cm}^3 = 1 \text{ L}) \\[1em] = 0.3031875 \\[1em] \approx 0.303 \text{ l}$

Hence, the hemispherical bowl can hold 0.303 litres of milk.

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CBSE Class IX | Academic Year 2026-2027

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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)