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(B) The volume of the cubical tank is $V = 	ext{edge}^3 = 30 	imes 30 	imes 30 = 27000 \, cm^3$.Since $1000 \, cm^3 = 1 \, litre$, the total capaci

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$27$

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(B) The volume of the cubical tank is $V = 	ext{edge}^3 = 30 	imes 30 	imes 30 = 27000 \, cm^3$.Since $1000 \, cm^3 = 1 \, litre$, the total capacity of the tank is $27000 / 1000 = 27 \, litres$.When $2.7 \, litres$ of water is removed, the remaining volume of water is $27 - 2.7 = 24.3 \, litres$.Converting this back to $cm^3$, we get $24.3 	imes 1000 = 24300 \, cm^3$.The base area of the tank is $30 	imes 30 = 900 \, cm^2$.Let the depth of the remaining water be $h$. Then, $	ext{Base Area} 	imes h = 	ext{Remaining Volume}$.$900 	imes h = 24300$.$h = 24300 / 900 = 27 \, cm$.

$A$ cubical metallic tank whose each edge measures $30\, cm$ is completely filled with water. If $2.7\, litres$ of water is taken out of it, what will be the depth of the remaining water in the tank? (in $cm$)

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