Question

A well with inner diameter 8 m is dug 14 m deep Earth taken out it has been spread evenly all around it to a width of 3 m to form an embankment around the well The height of this embankment is

• A. $6\frac{26}{33}$ m
• B. $7\frac{26}{33}$ m
• C. $4\frac{26}{33}$ m
• D. $5\frac{26}{33}$ m

Answer 1

The correct option is A $6\frac{26}{33}$ m

Given Inner diameter of well $=BC=8m$

And Depth of well $=14m$

then volume of cylindrical well $=\pi r^{2}depth=\pi (\frac{BC}{2}{)}^2\ast (depth)=\frac{\pi \left(B{C}^2\right)\left(depth\right)}{4}=\frac{\pi \left(8^2\right)\left(14\right)}{4}=\frac{\pi \left(64\right)\left(14\right)}{4}=224\pi m^3$

Let height of Embankment $=xm$

Diameter of outer cylindrical shape of Embankment $=AD=AB+BC+CD=3+8+3=14m$

then volume of Embankment $=\pi \ast (R^2-r^2)\ast height=\pi \ast \left(\right(outerdiameter{)}^2-\left(innerdiameter{)}^2\right)\ast x=\pi \ast \left(\right(\frac{AD}{2}{)}^2-\left(\frac{BC}{2}{)}^2\right)\ast x=\frac{\pi x}{4}(A{D}^2-B{C}^2)=\frac{\pi x}{4}({14}^2-8^2)=\frac{\pi x}{4}(196-64)=\frac{132\pi x}{4}=33\pi x$

Now it says that well is dug deep Earth, taken out, and spread evenly all around to make Embankment.

So volume of Embankment would be same that of well.

Volume of Embankment $=$ volume of well

$33\pi x=224\pi$

$x=\frac{224\pi }{33\pi }=\frac{224}{33}=6\frac{26}{33}m$

The height of this embankment is $6\frac{26}{33}m$