The problem asks for the number of smaller spheres (balls) that can be formed from a larger solid sphere. This can be determined by comparing the volumes of the two spheres.
The volume ($V$) of a sphere is calculated using the formula:
$V = \frac{4}{3}\pi r^3$
where $r$ is the radius of the sphere.
Let $R$ denote the radius of the large solid sphere and $r$ denote the radius of the smaller balls.
The volume of the large sphere is:
$V_{large} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (4)^3 \text{ cm}^3$
The volume of one small ball is:
$V_{small} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2)^3 \text{ cm}^3$
The total number of small balls is the ratio of the volume of the large sphere to the volume of a small ball:
$ \text{Number of balls} = \frac{V_{large}}{V_{small}} $
$ \text{Number of balls} = \frac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} $
The $\frac{4}{3}\pi$ terms cancel out, simplifying the calculation to:
$ \text{Number of balls} = \frac{R^3}{r^3} = \left(\frac{R}{r}\right)^3 $
Substitute the given values for $R$ and $r$:
$ \text{Number of balls} = \left(\frac{4}{2}\right)^3 = (2)^3 = 8 $
Thus, 8 balls of radius 2 cm can be made from the solid sphere of radius 4 cm.
A cylindrical tube, open at both ends, is made of a metal sheet which is 0.5 cm thick. Its outer radius is 4 cm and length is 2 m. How much metal (in cm 3) has been used in making the tube?
The volume of a right circular cone is 308 cm 3 and the radius of its base is 7 cm. What is the curved surface area (in cm 2) of the cone? (Take π = \(\frac{22}{7} \) )
The slant height and radius of a right circular cone are in the ratio 29 ∶ 20. If its volume is 4838.4 π cm 3, then its radius is:
Six cubes, each of edge 2 cm, are joined end to end. What is the total surface area of the resulting cuboid in cm 2?
A solid cube of side 8 cm is dropped into a rectangular container of length 16 cm, breadth 8 cm and height 15 cm which is partly filled with water. If the cube is completely submerged, then the rise of water level (in cm) is: