1. Volume of the Larger Sphere (V):
Let the radius of the larger sphere be \(r\). The volume of the larger sphere is given by \(V = \frac{4}{3}\pi r^3\).
2. Volume Lost in Process:
10% of the volume is lost in the process. So, the remaining volume is \(90\%\) of the original volume:
Remaining volume =\(0.9 \times \frac{4}{3}\pi r^3\).
3. Volume of Smaller Sphere (V'):
The radius of the smaller spheres is \(\frac{1}{9}\) of the radius of the larger sphere, which is \(\frac{1}{9}r\). The volume of each smaller sphere is given by:
\(V' = \frac{4}{3}\pi \left(\frac{1}{9}r\right)^3 = \frac{4}{3} \cdot \frac{1}{729}\pi r^3\).
4. Number of Smaller Spheres:
The number of smaller spheres that can be formed from the original sphere is:
Number of smaller spheres = \(\frac{0.9 \times \frac{4}{3}\pi r^3}{\frac{4}{3} \cdot \frac{1}{729}\pi r^3} = 0.9 \cdot 729 = 656.1\) (approx).
5. Total Volume of Smaller Spheres:
The total volume of all the smaller spheres is:
Total volume = \(656.1 \times \frac{4}{3} \cdot \frac{1}{729}\pi r^3 = \frac{1752}{729}\pi r^3\)cubic units.
6. Ratio of Volumes:
The ratio of the volume of the larger sphere to the total volume of all the smaller spheres is:
Ratio = \(\frac{\frac{4}{3}\pi r^3}{\frac{1752}{729}\pi r^3} = \frac{2430}{1}\).
7. Paint Needed for Smaller Spheres:
Since 10 litres of paint were needed to paint the larger sphere, the amount of paint needed to paint each smaller sphere is:
Paint needed per smaller sphere = \(\frac{10}{243}\) litres.
8. Total Paint Needed:
The total amount of paint needed to paint all the smaller spheres is:
Total paint needed = \(656.1 \times \frac{10}{243} = 81\) litres.
So, the simplified solution confirms that 81 litres of paint are needed to paint all the smaller spheres.
On the day of her examination, Riya sharpened her pencil from both ends as shown below. 
The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and the length of the entire pencil is 105.6 mm, find the total surface area of the pencil.
Two identical cones are joined as shown in the figure. If radius of base is 4 cm and slant height of the cone is 6 cm, then height of the solid is
From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid.
Use $\pi = \dfrac{22}{7}, \sqrt{5} = 2.2$
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.