# NCERT Solutions Class 9 Mathematics Solutions for Surface Areas and Volumes - Exercise 13.8 in Chapter 13 - Surface Areas and Volumes

Question 2 Surface Areas and Volumes - Exercise 13.8

Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm (ii) 0.21 m

(Assume π=22/7)

(i)

Diameter = 28 cm

Radius, r = 28/2 cm = 14 cm

The volume of the solid spherical ball = 4 / 3 \pi r^{3}

\begin{aligned} &\text { Volume of the ball }=4 / 3 \times 22 / 7 \times 14^{3}=34496 / 3\\ &\text { Volume of the ball is } 34496 / 3 \mathrm{cm}^{3} \end{aligned}

(ii)

Diameter = 0.21 m

Radius of the ball = 0.21/2 m = 0.105 m

\begin{array}{l} \text { Volume of the ball }=4 / 3 \pi \mathrm{r}^{3} \\ \text { Volume of the ball }=4 / 3 \times 22 / 7 \times 0.105^{3} \mathrm{m}^{3} \\ \text { Volume of the ball }=0.004851 \mathrm{m}^{3} \end{array}

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