Question
An aluminium rod, Young’s modulus 7× 109 N m−2, has a breaking strain of 0.2%. The minimum cross sectional area of the rod in m2 in order to support a load of
104 N is
104 N is
Answer: Option D
:
D
Y=FABreakingstrain
or a=FY×Breakingstrain=104×1007×10×0.2
=0.71×10−3=7.1×10−4
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:
D
Y=FABreakingstrain
or a=FY×Breakingstrain=104×1007×10×0.2
=0.71×10−3=7.1×10−4
Was this answer helpful ?
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