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प्रश्न
Consider the situation shown in figure. The force F is equal to the m2 g/2. If the area of cross section of the string is A and its Young modulus Y, find the strain developed in it. The string is light and there is no friction anywhere.
उत्तर
Given:
Force (F) = m2g/2
Area of cross-section of the string = A
Young's modulus = Y
Let a be the acceleration produced in block m2 in the downward direction and T be the tension in the string.
From the free body diagram:
\[\text{m}_2 \text{g - T = m}_2 \text{a} . . . \left( \text{i} \right)\]
\[\text{ T - F = m}_1 \text{a . . . (ii)}\]
From equations (i) and (ii), we get:
\[a = \frac{\text{m}_2 \text{g - F}}{\text{m}_1 + \text{m}_2}\]
\[\text{ Applying F }= \frac{\text{m}_2 \text{g}}{2}\]
\[ \Rightarrow a = \frac{\text{m}_2 \text{g}}{2\left( \text{m}_1 + \text{m}_2 \right)}\]
Again, T = F + m1a
On applying the values of F and a, we get:
\[ \Rightarrow \text{ Strain }= \frac{∆ \text{L}}{\text{L}} = \frac{\text{F}}{\text{AY}}\]
\[ \Rightarrow \text{ Strain } = \frac{\left( \text{m}_2^2 + 2 \text{m}_1 \text{m}_2 \right)\text{g}}{2\left( \text{m}_1 + \text{m}_2 \right) \text{AY}}\]
\[ = \frac{\text{m}_2 g\left( 2 \text{m}_1 + \text{m}_2 \right)}{2\text{AY} \left( \text{m}_1 + \text{m}_2 \right)}\]
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