Answers

2014-11-24T22:25:54+05:30
External radius(R)= 10-2= 5cm
Internal radius(r)= (5-2)= 3cm
l=14cm

Volume of the pipe= π(R²-r²)×l
                           =π(25-9)×14
                           = \frac{22}{7}×16×14
                           =704cm³

Therefore weight of the pipe= volume×density
                                        =704cm³×10g/cm³
                                        =7040g
                                        =7.04kg
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  • Brainly User
2014-11-24T22:37:58+05:30
R=5, r=3, l=14. Volume, V=Π(R²-r²)l=Π(16)(14)=Π(224)=22(32)=704. Mass, m=10(704)=7040gms. Mass of rod is 7.04 kgs
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