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A solid iron cuboidal block of dimensions 4.4 m × 2.6 m × 1 m is recast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe. - Mathematics

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प्रश्न

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

योग

उत्तर

Given that a solid iron cuboidal block is recast into a hollow cylindrical pipe.

Length of cuboidal block (I) = 4.4 m

Breadth of cuboidal block (b) = 2.6 m

And height of cuboidal block (h) = 1 m

So, volume of solid iron cuboidal block

= l × b × h

= 4.4 × 2.6 × 1

= 11.44 m3

Also, internal radius of hollow cylindrical pipe (ri) = 30 cm = 0.3 m

And thickness of hollow cylindrical pipe = 5 cm = 0.05 m

So, external radius of hollow cylindrical pipe (re) = ri + Thickness

= 0.3 + 0.05

= 0.35 m

∴ Volume of hollow cylindrical pipe

= Volume of cylindrical pipe with external radius – Volume of cylindrical pipe with internal radius

= `π"r"_"e"^2"h"_1 - π"r"_"i"^2"h"_1`

= `π("r"_"e"^2 - "r"_"i"^2)"h"_1`

= `22/7 [(0.35)^2 - (0.3)^2] xx "h"_1`

= `22/7 [(10.35 + 0.3)(0.35 - 0.3)]"h"_1`

= `22/7 xx 0.65 xx 0.05 xx "h"_1`

= `(0.715 xx "h"_1)/7`

Where, h1 be the length of the hollow cylindrical pipe.

Now, by given condition,

Volume of solid iron cuboidal block = Volume of hollow cylindrical pipe

⇒ 11.44 = `(0.715 xx "h"_1)/7`

∴ h1 = `(11.44 xx 7)/0.715` = 112 m

Hence, required length of pipe is 112 m.

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अध्याय 12: Surface Areas and Volumes - Exercise 12.4 [पृष्ठ १५१]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
अध्याय 12 Surface Areas and Volumes
Exercise 12.4 | Q 9 | पृष्ठ १५१

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