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
  • $F / A Y$
  • B
    $2 F / A Y$
  • C
    $\frac{1}{2}(F / A Y)$
  • D
    $3 F / A Y$

Answer

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

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