A tank 4 , m long, 2.5 , m wide, and 6 , m de | vedclass

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(B) Volume of the earth dug out $= 4 	imes 2.5 	imes 6 \, m^3 = 60 \, m^3$.Area of the field where the earth is spread $= (	ext{Total area of field

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(B) Volume of the earth dug out $= 4 	imes 2.5 	imes 6 \, m^3 = 60 \, m^3$.Area of the field where the earth is spread $= (	ext{Total area of field}) - (	ext{Area of the tank opening})$.Area of the field $= 10 	imes 9 = 90 \, m^2$.Area of the tank opening $= 4 	imes 2.5 = 10 \, m^2$.Area of the remaining portion $= 90 - 10 = 80 \, m^2$.Let the rise in the level of the field be $h$.Volume of earth spread $= 	ext{Area of remaining portion} 	imes h$.$60 = 80 	imes h$.$h = \frac{60}{80} = 0.75 \, m$.Since $1 \, m = 100 \, cm$, the rise in level $= 0.75 	imes 100 = 75 \, cm$.

$A$ tank $4 \, m$ long, $2.5 \, m$ wide, and $6 \, m$ deep is dug in a field $10 \, m$ long and $9 \, m$ wide. If the earth dugout is evenly spread over the field, the rise in level of the field will be (in $cm$):

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