Consider the situation shown in the figure. T | vedclass

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(C) Let the acceleration of the system be $a$. The equation of motion for the system is:$T - F = m_1a$ (for mass $m_1$)$m_2g - T = m_2a$ (for mass $m_

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$\frac{m_2g(2m_1 + m_2)}{2AY(m_1 + m_2)}$

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(C) Let the acceleration of the system be $a$. The equation of motion for the system is:$T - F = m_1a$ (for mass $m_1$)$m_2g - T = m_2a$ (for mass $m_2$)Adding these equations: $m_2g - F = (m_1 + m_2)a$Given $F = m_2g/2$,we have $m_2g - m_2g/2 = (m_1 + m_2)a$,so $a = \frac{m_2g}{2(m_1 + m_2)}$.Now,find the tension $T$ in the string using the equation for $m_2$:$T = m_2(g - a) = m_2(g - \frac{m_2g}{2(m_1 + m_2)}) = m_2g(1 - \frac{m_2}{2(m_1 + m_2)}) = m_2g(\frac{2m_1 + 2m_2 - m_2}{2(m_1 + m_2)}) = \frac{m_2g(2m_1 + m_2)}{2(m_1 + m_2)}$.The strain is given by $	ext{Strain} = \frac{	ext{Stress}}{Y} = \frac{T}{AY}$.Substituting $T$: $	ext{Strain} = \frac{m_2g(2m_1 + m_2)}{2AY(m_1 + m_2)}$.Thus,the correct option is $C$.

Consider the situation shown in the figure. The force $F$ is equal to $m_2g/2$. If the area of cross-section of the string is $A$ and its Young's modulus is $Y$,find the strain developed in it. The string is light and there is no friction anywhere.

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