The portion of a ditch 48 m long, 16.5 m wi | vedclass

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(C) Volume of the ditch $= 	ext{length} 	imes 	ext{width} 	imes 	ext{depth} = 48 \ m 	imes 16.5 \ m 	imes 4 \ m = 3168 \ m^3$.The tunnel is cyl

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$2/9$ Part

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(C) Volume of the ditch $= 	ext{length} 	imes 	ext{width} 	imes 	ext{depth} = 48 \ m 	imes 16.5 \ m 	imes 4 \ m = 3168 \ m^3$.The tunnel is cylindrical with diameter $d = 4 \ m$, so radius $r = 2 \ m$, and length $h = 56 \ m$.Volume of earth and stones from the tunnel $= \pi r^2 h = \frac{22}{7} 	imes (2)^2 	imes 56 = 22 	imes 4 	imes 8 = 704 \ m^3$.The portion of the ditch filled $= \frac{	ext{Volume of earth}}{	ext{Volume of ditch}} = \frac{704}{3168}$.Dividing both by $352$, we get $\frac{704 \div 352}{3168 \div 352} = \frac{2}{9}$.

The portion of a ditch $48 \ m$ long, $16.5 \ m$ wide, and $4 \ m$ deep that can be filled with stones and earth available during the excavation of a tunnel, cylindrical in shape, of diameter $4 \ m$ and length $56 \ m$ is (Take $\pi = \frac{22}{7}$)

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