A tank 3, m long,2, m wide and 1.5, m deep is | vedclass

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(C) Volume of the earth dug out from the tank $= (3 	imes 2 	imes 1.5) \, m^3 = 9 \, m^3$.The area of the field is $(22 	imes 14) \, m^2 = 308 \, m

Vedclass · Accepted Answer

$2.98$

Vedclass · Answer

(C) Volume of the earth dug out from the tank $= (3 	imes 2 	imes 1.5) \, m^3 = 9 \, m^3$.The area of the field is $(22 	imes 14) \, m^2 = 308 \, m^2$.The area of the tank is $(3 	imes 2) \, m^2 = 6 \, m^2$.The area over which the earth is spread $= (308 - 6) \, m^2 = 302 \, m^2$.Let the rise in the level of the field be $h$ meters.Volume of earth spread = Area $	imes$ height$9 = 302 	imes h$$h = \frac{9}{302} \, m$.To convert into $cm$,multiply by $100$:$h = \frac{9}{302} 	imes 100 \, cm = \frac{900}{302} \, cm \approx 2.98 \, cm$.

$A$ tank $3\, m$ long,$2\, m$ wide and $1.5\, m$ deep is dug in a field $22\, m$ long and $14\, m$ wide. If the earth dug out is evenly spread out over the field,the level of the field will rise by nearly (in $cm$)

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