Jump to
- Surface Areas and Volumes - Exercise 13.1
- Surface Areas and Volumes - Exercise 13.2
- Surface Areas and Volumes - Exercise 13.3
- Surface Areas and Volumes - Exercise 13.4
- Surface Areas and Volumes - Exercise 13.5
- Surface Areas and Volumes - Exercise 13.6
- Surface Areas and Volumes - Exercise 13.7
- Surface Areas and Volumes - Exercise 13.8
- Surface Areas and Volumes - Exercise 13.9
NCERT Solutions Class 9 Mathematics Solutions for Surface Areas and Volumes - Exercise 13.8 in Chapter 13 - Surface Areas and Volumes
Prev
Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm (ii) 0.21 m
(Assume π=22/7)
Answer:
(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}
Related Questions
Facebook
Whatsapp
Copy Link
Exercises
Chapters
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
Connect with us on social media!
2022 © Quality Tutorials Pvt Ltd All rights reserved