Elastic Limit :: Discussion


10.The diameter of brass rod is 4mm. Young’s modulus of brass is 9×109 N/m2. The force required to stretch 0.1% of its length is?
A.360Ï€N
B.36N
C.36π×105 N
D.144π×103N
Answer:  Option  A
Explanation:

∆l/l=0.1/100
F=YA∆l/l=(Y×πr2×∆l)/l
F=(9×109×π×(2×10(-3))2×0.1)/100 N=360πN.

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Published by:Michael Daani

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