A well with inner radius $4 m$ is dug $14 m$ deep. Earth taken out of it has been spread evenly all around a width of $3 m$ it to form an embankment. Find the height of the embankment.

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Solution

The height of the embankment $= \frac{V o l u m e o f t h e e a r t h d u g o u t}{A r e a o f t h e e m b a n k m e n t} = \frac{\frac{22}{7} \times 4^{2} \times 15}{\frac{22}{7} \times (7^{2} - 4^{2})} = \frac{16 \times 14}{33} = 6.78 m$

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Q. Question 4
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

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A well with inner radius 4 m is dug 14 m deep. Earth taken out of it has been spread evenly all around a width of 3 m it to form an embankment. Find the height of the embankment.

The height of the embankment =VolumeoftheearthdugoutAreaoftheembankment=227×42×15227×(72−42)=16×1433=6.78m

A well with inner radius $4 m$ is dug $14 m$ deep. Earth taken out of it has been spread evenly all around a width of $3 m$ it to form an embankment. Find the height of the embankment.

The height of the embankment $= \frac{V o l u m e o f t h e e a r t h d u g o u t}{A r e a o f t h e e m b a n k m e n t} = \frac{\frac{22}{7} \times 4^{2} \times 15}{\frac{22}{7} \times (7^{2} - 4^{2})} = \frac{16 \times 14}{33} = 6.78 m$