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Q.

A constant force $2 F_{0}$ is applied on a uniform elastic string placed over a smooth horizontal surface as shown in figure. Young’s modulus of string is Y and cross-sectional area of string is S. The strain produced in the string in the direction of force is

$2 F_{0}$

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a

$\frac{F_{0} Y}{S}$

b

$\frac{F_{0}}{S Y}$

c

$\frac{F_{0}}{2 S Y}$

d

$\frac{F_{0} Y}{2 S}$

answer is B.

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Detailed Solution

Let F be the force at distance x from the front end

$\begin{array}{l} \frac{F}{X} = \frac{2 F_{0}}{L} ∴ F = \frac{2 F_{0} X}{L} \\ S t r a i n = \frac{d δ}{d X} = \frac{2 F_{0} X}{Y A L} = \frac{s t r e e s}{Y}; \int d δ = \int_{0}^{L} \frac{2 F_{0} X}{Y S L} d X; \\ δ = \frac{2 F_{0}}{Y S} {[\frac{x^{2}}{2}]}^{2} = \frac{F_{0} L}{S Y} \end{array}$

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