Question

$guys \: plz \: help \: me \: in \: q26.dont \: spam$

Answer 1

$hii \: buddy$




Radius (r) of the tank
$= 1 \div 2 \times 17.5$


$= 35 \div 4$


Let d meters be the depth of the tank dug out,


Therefore, Volume of the soil dug out
$= \pi \: r{}^{2} h$


$= \pi \: (35 \div 4) {}^{2} \times d \: \: m{}^{3}$



Internal radius of the embankment
$= 35 \div 4$


Since, the width of the embankment
$= 4 \: m$


Therefore, the external radius of the embankment
$= (35 \div 4 + 4) \: \: m$


$= 51 \div 4 \: \: m$




Height of the embankment
$= 2m$



Therefore, the volume of the soil used in forming the embankment around the circular tank
$= \pi(r1 {}^{2} - r2 {}^{2} )h$




$= \pi(51 \div 4 + 35 \div 4)(51 \div 4 - 35 \div 4) \times 2 \: m {}^{3}$



$= \pi \times 86 \div 4 \times 4 \times 2 \: m {}^{3}$



$= \pi \times 86 \times 2 \: m {}^{3}$



According to the given information ,


$= \pi \times 35 \div 4 \times 35 \div 4 \times d$



$= \pi \times 86 \times 2$



$= > d = 172 \times 16 \div 35 \times 35$


$= > d = 2.247 \: m$



$= > d = 2.25 \: m(approximately)$






$thanks$





$i \: hope \: it \: helps....$




$garu1678$