A cubical wooden block of side $30$ cm has a cylindrical hole of radius $7$ cm drilled completely through it…

A cubical wooden block of side $30$ cm has a cylindrical hole of radius $7$ cm drilled completely through it (perpendicular to a face). The volume of wood remaining is (use $\pi=\dfrac{22}{7}$):
  1. $22380 \text{ cm}^{3}$
  2. $25380 \text{ cm}^{3}$
  3. $27000 \text{ cm}^{3}$
  4. $4620 \text{ cm}^{3}$

Solution

Volume of cube $= 30^{3} = 27000 \text{ cm}^{3}$. Volume of cylindrical hole $= \pi r^{2} h = \dfrac{22}{7} \times 49 \times 30 = 22 \times 7 \times 30 = 4620 \text{ cm}^{3}$. Remaining $= 27000 - 4620 = 22380 \text{ cm}^{3}$.

Asked in: IMO

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