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Question: An iron rod of length 2m and cross section area of \(50 \mathrm {~mm} ^ { 2 }\), stretched by 0.5 mm...

An iron rod of length 2m and cross section area of 50 mm250 \mathrm {~mm} ^ { 2 }, stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young's modulus of the iron rod is

A

19.6×1010 N/m219.6 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }

B

19.6×1015 N/m219.6 \times 10 ^ { 15 } \mathrm {~N} / \mathrm { m } ^ { 2 }

C

19.6×1018 N/m219.6 \times 10 ^ { 18 } \mathrm {~N} / \mathrm { m } ^ { 2 }

D

19.6×1020 N/m219.6 \times 10 ^ { 20 } \mathrm {~N} / \mathrm { m } ^ { 2 }

Answer

19.6×1010 N/m219.6 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }

Explanation

Solution

Y=MgLAl=250×9.8×250×10−6×0.5×10−3Y = \frac { M g L } { A l } = \frac { 250 \times 9.8 \times 2 } { 50 \times 10 ^ { - 6 } \times 0.5 \times 10 ^ { - 3 } }

=19.6×1010 N/m2= 19.6 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }