A long thin metallic rod of length l, cross sectional area S and Young's modulus γ rests on a smooth horizontal surface. A horizontal force P is applied at one end of the rod. Then the elongation produced in the rod is
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a
Zero
b
PlSγ
c
2PlSγ
d
Pl2Sγ
answer is D.
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Detailed Solution
(4)a = Acceleration of the rod, a = P/m. If we observe the rod from its frame, then pseudo force ma acts at the COM C of the rod. So under the action of P and ma, elongation produced in the rod is δ=P× ACSγ = P. l2Sγ=Pl2Sγ.