A string of length '/' and radius 'a' is stretched between two fixed points A and B without tension. The middle point of the string is pulled to a small distance 'd' in a direction,
perpendicular to the original length of the string. If the Young's modulus of the material of the string is Y, the energy stored in the string is:

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${πa}^{2} {Yd}^{2} / l$

${πa}^{2} Yd$

$2 {πYa}^{2} d^{4} / l^{3}$

$Y a^{2} l^{2} / d$

answer is C.

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Detailed Solution

$∆ l = 2 \sqrt{d^{2} + \frac{l^{2}}{4}} - l = 2 {(\frac{l^{2}}{4} + d^{2})}^{\frac{1}{2}} - l = 2 \frac{1}{2} {(1 + \frac{4 d^{2}}{l^{2}})}^{\frac{1}{2}} - l = \frac{2 d^{2}}{l} = l [1 + \frac{1}{2} (\frac{4 d^{2}}{l^{2}})] - l = \frac{2 d^{2}}{l} Strain = \frac{∆ l}{l} = \frac{2 d^{2}}{l^{2}} Energy stored = \frac{1}{2} Y {(strain)}^{2} ∆ V = \frac{1}{2} Y {(\frac{2 d^{2}}{l^{2}})}^{2} {πa}^{2} l = 2 {πa}^{2} \frac{{Yd}^{4}}{l^{3}}$