a cuboid lead having dimensions 53 × 40 × 15 cm is melted and recast into a cylindrical pipe the outer and inner radius of pipe are 4 and 7/2 cm respectively. find the height of the pipe.
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2
Step-by-step explanation:
Vol. of the cuboid = lbh = 53*40*15
= 3180 sq. cm
Inner Radius = 7/2 cm
Outer Radius = 4 cm
Height = Vol./ πr^2
Vol. of cylinder = Vol. of cuboid = 3180 sq. cm
Therefore, h=3180/π*r*r
Width of the pipe = Inner Radius - Outer Radius
W = 0.5 cm
h = 3180/(22/7)*(7/2)*(7/2)
h = 82.59 cm
#Hopeithelps
Answered by
1
Answer:
Volume of cuboid = 53 × 40 × 15
= 31800 cm³
Volume of hollow-cylinder = πh (R² - r²)
= 22/7 × h (4² - 3.5²)
= 22/7 × h (16 - 12.25)
= 22/7 × h * 3.75
31800 = 22/7 × h × 3.75
222600 = 22 × h × 3.75
222600 = 82.5 × h
h = 222600/82.5
= 2698.18 cm
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