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.

Open in App

Solution

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

Suggest Corrections

Similar questions

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

A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.

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.

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

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.

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