Question #4ef82

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Answer 1 · 2018-02-05T08:29:19

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It is revealed from the above figure that surface area of the cyndrical solid with a conical cavity will be

$A=2pirh+pirl+pir^2$

$=>A=pir(2h+l+r)$

Given
Total surface area of the solid $A=904.32dm^2$
Height of the solid $h=16dm$

Radius $r=6dm$

Density $rho=7.5g"/"cm^3=7.5kg"/"dm^3$

The slant height $(l)$ of the cone is not known.

So
$904.32=3.14xx6(2xx16+l+6)$

$=>l+38=(904.32)/(3.14xx6)=48$

$=>l=48-38=10dm$

So height of the conical cavity will be

$h_"cone"=sqrt(l^2-r^2)$

$=>h_"cone"=sqrt(10^2-6^2)=8dm$

Now volume of the solid

$V="volume of cylinder"-"volume of cavity"$

$=pir^2h-1/3pir^2h_"cone"$

$=pir^2(h-h_"cone"/3)$

So weight (mass) of the solid will be

$W=Vxxrho$

$=>W=Vxxrho=pir^2(h-h_"cone"/3)rho$

$=>W=3.14xx6^2(16-8/3)xx7.5kg$

$=>W=3.14xx36xx40/3xx7.5kg=11304kg$