Water in a cuboidal tank of base $80$ cm $\times 60$ cm rises by $5$ cm when an iron piece is fully…
Water in a cuboidal tank of base $80$ cm $\times 60$ cm rises by $5$ cm when an iron piece is fully submerged in it. The volume of the iron piece is:
- $24000 \text{ cm}^{3}$
- $2400 \text{ cm}^{3}$
- $240 \text{ cm}^{3}$
- $48000 \text{ cm}^{3}$
Solution
Volume of displaced water $=$ volume of iron piece $= 80 \times 60 \times 5 = 24000 \text{ cm}^{3}$.
Asked in: IMO
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