A rectangular tank of dimensions $24 m \times 12 m \times 8 m$ is dug inside a rectangular field $600 m$ long and $200 m$ broad. The earth take out is evenly spread over the field. By how much will rise the level of the field rise?

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Solution

Given:
The dimensions of rectangular tank $= 24 m \times 12 m \times 8 m$
$\Rightarrow$ Volume of rectangular tank (Cuboid) $= 24 m \times 2 m \times 8 m$
Dimensions of rectangular field $= A B (l), = 600 m, C D (b) = 200 m$
Let, the rise of the level of the field be $h$ .
Volume of cuboid AEIC= $A E \times E I \times h$
$= 576 \times 200 \times h$
Volume of FHID= $F H \times F I \times h$
$= 188 \times 24 \times h$
Since, volume of earth dug $\Rightarrow 24 m \times 12 m \times 8 m = 576 \times 200 \times h + 188 \times 24 \times h$
$\Rightarrow 24 m \times 12 m \times 8 m = 24 h (4800 + 188)$
$\Rightarrow 24 m \times 12 m \times 8 m = 24 h (4988)$
$\Rightarrow 3 \times 8 = 1247 h$
$\Rightarrow h = \frac{24}{1247}$
$h = 0.0192 m$
Hence, rise of the level of the field is $0.0192 m$ .

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A rectangular tank of dimensions 24 m×12 m×8 m is dug inside a rectangular field 600 m long and 200 m broad. The earth take out is evenly spread over the field. By how much will rise the level of the field rise?

A rectangular tank of dimensions $24 m \times 12 m \times 8 m$ is dug inside a rectangular field $600 m$ long and $200 m$ broad. The earth take out is evenly spread over the field. By how much will rise the level of the field rise?