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(D) The longitudinal strain is given by $	ext{Strain} = \frac{\Delta L}{L} = \frac{F}{AY}$.Since the area of cross-section $A$ and Young's modulus $Y

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$3$

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(D) The longitudinal strain is given by $	ext{Strain} = \frac{\Delta L}{L} = \frac{F}{AY}$.Since the area of cross-section $A$ and Young's modulus $Y$ are the same for both wires,the strain is directly proportional to the tension $F$ in the wire.For the lower wire,the tension $F_2$ is due to the $1 \ kg$ load: $F_2 = m_2 g = 1 \ kg 	imes 10 \ ms^{-2} = 10 \ N$.For the upper wire,the tension $F_1$ is due to both the $2 \ kg$ load and the $1 \ kg$ load: $F_1 = (m_1 + m_2) g = (2 \ kg + 1 \ kg) 	imes 10 \ ms^{-2} = 30 \ N$.The ratio of the longitudinal strain of the upper wire to that of the lower wire is:$\frac{	ext{Strain}_1}{	ext{Strain}_2} = \frac{F_1 / AY}{F_2 / AY} = \frac{F_1}{F_2} = \frac{30 \ N}{10 \ N} = 3$.

Similar Questions

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