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A uniform elastic rod of cross-section area A, natutal length L and Young's modulus Y is placed on a smooth horizontal surface. Now two horizontal forces (of magnitude F and 3F) directed along the length of rod and in opposite direction act at two of its ends as shown. After the rod has acquired steady state, the extension of the rod will be

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$\frac{4 F}{YA} L$

$\frac{F}{YA} L$

$\frac{3 F}{2 YA} L$

$\frac{2 F}{YA} L$

answer is A.

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Detailed Solution

Tension in rod at a distance x from right edge is

$T = F (3 - 2 \frac{x}{L})$

$∴ Net extension in rod = \int_{0}^{L} \frac{T}{YA} dx = \frac{2 F}{YA} L$

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