A pen stand is made of wood in the sphere of cuboid with three conical depressions to hold the
pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm radius of each of the
depression is 0.5 cm and the depth is 1.4 cm, then the volume of wood in the entire stand is

$$\begin{gathered} \mathrm{The} \mathrm{dimensions} \mathrm{of} \mathrm{a} \mathrm{penstand} \mathrm{which} \mathrm{is} \mathrm{in} \mathrm{the} \mathrm{shape} \mathrm{of} \mathrm{cuboid} \mathrm{are} 15\mathrm{cm}, 10 \mathrm{cm}, 3.5 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{cuboid} = 15 \times 10 \times 3.5 {\mathrm{cm}}^3 = 525 {\mathrm{cm}}^3 \\ \mathrm{In} \mathrm{a} \mathrm{pen} \mathrm{stand} 3 \mathrm{conical} \mathrm{depressions} \mathrm{are} \mathrm{there} \mathrm{to} \mathrm{hold} \mathrm{the} \mathrm{pens} \\ \mathrm{Radius} \mathrm{of} \mathrm{each} \mathrm{depression} = 0.5 \mathrm{cm} \\ \mathrm{Depth} = 1.4 \mathrm{cm} \\ \mathrm{Volume} \mathrm{of} \mathrm{depression}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}=\frac{1}{3}\times \frac{22}{7}\times \frac{1}{2}=\frac{11}{30} {\mathrm{cm}}^3 \\ \mathrm{Volume} \mathrm{of} 3 \mathrm{depressions} = 3\left(\frac{11}{30}\right) {\mathrm{cm}}^3 = \frac{11}{10} {\mathrm{cm}}^3 \\ \mathrm{Volume} \mathrm{of} \mathrm{wood} \mathrm{in} \mathrm{the} \mathrm{entire} \mathrm{penstand} = \mathrm{volume} \mathrm{of} \mathrm{cuboid} – \mathrm{volume} \mathrm{of} 3 \mathrm{depressions} = \\ 525 - \frac{11}{10}=523.9 {\mathrm{cm}}^3 \end{gathered}$$