MCQ
A uniform plank of Young’s modulus $Y $ is moved over a smooth horizontal surface by a constant horizontal force $F.$ The area of cross section of the plank is $A.$ The compressive strain on the plank in the direction of the force is
  • $F/AY$
  • B
    $2F/AY$
  • C
    $\frac{1}{2}(F/AY)$
  • D
    $3F/AY$

Answer

Correct option: A.
$F/AY$
a
(a) $Y = \frac{{F/A}}{{{\rm{Strain}}}} \Rightarrow {\rm{strain}} = \frac{F}{{Ay}}$

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