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Standard XI

Mathematics

Polynomial Functions

A uniform pla...

Question

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

\frac{F}{A Y}

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\frac{2 F}{A Y}

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\frac{1}{2} (\frac{F}{A Y})

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\frac{3 F}{A Y}

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Solution

The correct option is A $\frac{F}{A Y}$
Compressive strain $= \frac{S t r e s s}{Y o u n g^{'} s m o d u l u s} = \frac{\frac{F}{A}}{Y} = \frac{F}{A Y}$

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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

The correct option is A FAYCompressive strain =StressYoung′s modulus=FAY=FAY

The correct option is A $\frac{F}{A Y}$
Compressive strain $= \frac{S t r e s s}{Y o u n g^{'} s m o d u l u s} = \frac{\frac{F}{A}}{Y} = \frac{F}{A Y}$