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A rectangular tank measuring 5 m × 4.5 m × 2.1 m is dug in the centre of the field measuring 13.5 m by 2.5 m. The earth dug out is evenly spread over the remaining portion of the field. How much is the level of the field raised ?
Answer & Solution
Correct Answer:
Option
C
Volume of earth dug out :
$$\eqalign{ & = \left( {5 \times 4.5 \times 2.1} \right){{\text{m}}^{\text{3}}} \cr & = 47.25\,{{\text{m}}^{\text{3}}} \cr} $$
Area over which earth is spread :
$$\eqalign{ & = \left( {13.5 \times 2.5 - 5 \times 4.5} \right){{\text{m}}^2} \cr & = \left( {33.75 - 22.5} \right){{\text{m}}^2} \cr & = 11.25\,{{\text{m}}^2} \cr} $$
$$\eqalign{ & \therefore {\text{Rise in level}} = \frac{{{\text{Volume}}}}{{{\text{Area}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{47.25}}{{11.25}}} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.2\,m \cr} $$
$$\eqalign{ & = \left( {5 \times 4.5 \times 2.1} \right){{\text{m}}^{\text{3}}} \cr & = 47.25\,{{\text{m}}^{\text{3}}} \cr} $$
Area over which earth is spread :
$$\eqalign{ & = \left( {13.5 \times 2.5 - 5 \times 4.5} \right){{\text{m}}^2} \cr & = \left( {33.75 - 22.5} \right){{\text{m}}^2} \cr & = 11.25\,{{\text{m}}^2} \cr} $$
$$\eqalign{ & \therefore {\text{Rise in level}} = \frac{{{\text{Volume}}}}{{{\text{Area}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{47.25}}{{11.25}}} \right)m \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4.2\,m \cr} $$
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Loginlength of field is 13.5 m so tank length 5 m is okay.