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Question: A rubber pipe of density \(1.5 \times 10 ^ { 3 } \mathrm {~N} / \mathrm { m } ^ { 2 }\) and Young's ...

A rubber pipe of density 1.5×103 N/m21.5 \times 10 ^ { 3 } \mathrm {~N} / \mathrm { m } ^ { 2 } and Young's modulus 5×106 N/m25 \times 10 ^ { 6 } \mathrm {~N} / \mathrm { m } ^ { 2 } is suspended from the roof. The length of the pipe is 8 m. What will be the change in length due to its own weight

A

9.6 m

B

9.6×103 m9.6 \times 10 ^ { 3 } \mathrm {~m}

C

19.2×10−2 m19.2 \times 10 ^ { - 2 } \mathrm {~m}

D

9.6×10−2 m9.6 \times 10 ^ { - 2 } \mathrm {~m}

Answer

9.6×10−2 m9.6 \times 10 ^ { - 2 } \mathrm {~m}

Explanation

Solution

l=L2dg2Yl = \frac { L ^ { 2 } d g } { 2 Y } =(8)2×1.5×103×102×5×106=9.6×10−2 m= \frac { ( 8 ) ^ { 2 } \times 1.5 \times 10 ^ { 3 } \times 10 } { 2 \times 5 \times 10 ^ { 6 } } = 9.6 \times 10 ^ { - 2 } \mathrm {~m}