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Q.

The rod AB has a cross sectional area 5 mm2. Young's modulus of its material is 2×1010 N/m2. If mass of the rod is 10 Kg, strain produced at its mid point is

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a

0.075%

b

0.25%

c

0.75%

d

0.5%

answer is A.

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Detailed Solution

Acceleration a=100−5010m/s2=5 m/s2Tension produced at its mid section isgiven by, T−50=5×5⇒T=75 N∴ strain =stressyoung's modulus=75/5×10−62×1010=75105×100%⇒ strain=0.075%
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