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A constant force F₀ 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 area of cross-section is S. the strain produced in the string in the direction of force is

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$\frac{F_{0}}{SY}$

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

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

$\frac{F_{0}}{2 SY}$

answer is C.

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Detailed Solution

Given,

$a = \frac{F_{0}}{m}$

$T = m_{x} a = (\frac{m}{l}) x (\frac{F_{0}}{m}) = (\frac{F_{0}}{l}) x$

$Δ l = \int_{0}^{L} \frac{Tdx}{SY}$

$Δ l = (\frac{F_{0}}{l SY}) \frac{l^{2}}{2}$

$Strain = \frac{∆ l}{l} = \frac{F_{0}}{2 SY}$

Hence the correct answer is $\frac{F_{0}}{2 SY} .$

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