A solid iron cuboidal block of dimensions 4.4 | vedclass

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Question

(A) Given that,a solid iron cuboidal block is recast into a hollow cylindrical pipe.Volume of the cuboidal block $= l 	imes b 	imes h = 4.4 \, m 	i

Vedclass · Accepted Answer

$112$

Vedclass · Answer

(A) Given that,a solid iron cuboidal block is recast into a hollow cylindrical pipe.Volume of the cuboidal block $= l 	imes b 	imes h = 4.4 \, m 	imes 2.6 \, m 	imes 1 \, m = 11.44 \, m^3$.Internal radius of the hollow cylindrical pipe $(r_i) = 30 \, cm = 0.3 \, m$.Thickness of the pipe $= 5 \, cm = 0.05 \, m$.External radius $(r_e) = r_i + 	ext{thickness} = 0.3 \, m + 0.05 \, m = 0.35 \, m$.Volume of the hollow cylindrical pipe $= \pi (r_e^2 - r_i^2) h_1$,where $h_1$ is the length of the pipe.Volume $= \frac{22}{7} 	imes (0.35^2 - 0.3^2) 	imes h_1 = \frac{22}{7} 	imes (0.1225 - 0.09) 	imes h_1 = \frac{22}{7} 	imes 0.0325 	imes h_1$.Since the volume remains constant: $11.44 = \frac{22}{7} 	imes 0.0325 	imes h_1$.$h_1 = \frac{11.44 	imes 7}{22 	imes 0.0325} = \frac{80.08}{0.715} = 112 \, m$.Thus,the length of the pipe is $112 \, m$.

$A$ solid iron cuboidal block of dimensions $4.4 \, m \times 2.6 \, m \times 1 \, m$ is recast into a hollow cylindrical pipe of internal radius $30 \, cm$ and thickness $5 \, cm$. Find the length of the pipe in meters.

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