8.Mechanical Properties of Solids
easy

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

A

$F/AY$

B

$2F/AY$

C

$\frac{1}{2}(F/AY)$

D

$3F/AY$

Solution

(a) $Y = \frac{{F/A}}{{{\rm{Strain}}}} \Rightarrow {\rm{strain}} = \frac{F}{{Ay}}$

Standard 11
Physics

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