The surface area of a pipe,open at both ends, | vedclass

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Question

(C) Let $r$ be the radius and $l$ be the length of the pipe.Given that the surface area of the pipe (curved surface area) is $2 \pi r l = 628 \, m^2$.

Vedclass · Accepted Answer

$1570$

Vedclass · Answer

(C) Let $r$ be the radius and $l$ be the length of the pipe.Given that the surface area of the pipe (curved surface area) is $2 \pi r l = 628 \, m^2$.Taking $\pi \approx 3.14$,we have $2 	imes 3.14 	imes r 	imes l = 628$,which simplifies to $r 	imes l = 100$.Given that the difference between the length and radius is $15 \, m$,we have $l - r = 15$,so $l = r + 15$.Substituting $l$ in the equation $r 	imes l = 100$,we get $r(r + 15) = 100$,which is $r^2 + 15r - 100 = 0$.Factoring the quadratic equation: $(r + 20)(r - 5) = 0$.Since the radius cannot be negative,$r = 5 \, m$.Then,$l = 5 + 15 = 20 \, m$.If the pipe is closed at one end,it acts as a cylinder with volume $V = \pi r^2 l$.$V = 3.14 	imes (5)^2 	imes 20 = 3.14 	imes 25 	imes 20 = 3.14 	imes 500 = 1570 \, m^3$.

The surface area of a pipe,open at both ends,is equal to $628 \, m^2$. The difference between its length and its radius is $15 \, m$. If the pipe were closed at one end,what amount of water could it hold (in $, m^3$)?

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