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An aluminium rod has a breaking strain 0.2%. The minimum cross sectional area of the rod in m 2 in order to supporta load of 10 4 N is if young's modulus is 7×10 9 Nm -2

An aluminium rod has a breaking strain 0.2%. The minimum cross sectional area of the rod in m2 in order to supporta load of 104N is if young's modulus is 7×109Nm-2

Grade:11

1 Answers

Vikas TU
14149 Points
7 years ago
From Young’s Modulus formulae:
Y =  Fl’/Al
where l’ = 2l/1000
putting all the values we get,
7 x 10^9 = 10^4 * 2l/1000*A*l
solving we get,
A = 1/1400 m^2.

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