(C) Volume of the earth taken out $= (7.5 \times 6 \times 0.8) \, m^3 = 36 \, m^3$.Area of the remaining field $= (18 \times 15) - (7.5 \times 6) \, m

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(C) Volume of the earth taken out $= (7.5 \times 6 \times 0.8) \, m^3 = 36 \, m^3$.Area of the remaining field $= (18 \times 15) - (7.5 \times 6) \, m^2 = 270 - 45 = 225 \, m^2$.Let the rise in the level of the field be $h$ meters.Since the volume of the earth taken out is spread over the remaining area,we have:$225 \times h = 36$.$h = \frac{36}{225} \, m = 0.16 \, m$.Converting to centimeters: $0.16 \times 100 = 16 \, cm$.

Competitive Quantitative Aptitude Mensuration Measurement of Volume and Surface Area

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$A$ field is in the form of a rectangle of length $18\, m$ and width $15\, m$. $A$ pit,$7.5\, m$ long,$6\, m$ broad and $0.8\, m$ deep,is dug in a corner of the field and the earth taken out is evenly spread over the remaining area of the field. The level of the field raised is......$cm$

A
$12$
B
$14$
C
$16$
D
$18$

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Mensuration

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Measurement of Volume and Surface Area

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$A$ field is in the form of a rectangle of length $18\, m$ and width $15\, m$. $A$ pit,$7.5\, m$ long,$6\, m$ broad and $0.8\, m$ deep,is dug in a corner of the field and the earth taken out is evenly spread over the remaining area of the field. The level of the field raised is......$cm$

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$A$ roller is $120 \, cm$ long and has a diameter of $84 \, cm$. If it takes $500$ complete revolutions to level a playground,determine the cost of levelling at the rate of $30 \, \text{paise}$ per $m^2$ (in $Rs.$). (Use $\pi = \frac{22}{7}$)

$A$ circus tent is cylindrical to a height of $3\, m$ and conical above it. If its diameter is $105\, m$ and the slant height of the conical portion is $53\, m$,calculate the length of the canvas $5\, m$ wide required to make the tent. (in $m$)

The outer dimensions of a closed wooden box of $1 \text{ cm}$ thickness are $12 \text{ cm} \times 10 \text{ cm} \times 8 \text{ cm}$. Find the cost of the wood required to make the box if $1 \text{ cm}^3$ of wood costs $₹ 3.00$.

$A$ cone of height $9 \, cm$ with diameter of its base $18 \, cm$ is carved out from a wooden solid sphere of radius $9 \, cm$. The percentage of the wood wasted is

$2\, cm$ of rain has fallen on a $1\, km^2$ of land. Assuming that $50\%$ of the raindrops could have been collected and contained in a pool having a $100\, m \times 10\, m$ base,by what level would the water level in the pool have increased? (in $m$)

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