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(B) $1$. Initial state: The total mass is $M = 2 + 2 = 4 \, kg$. The spring extension at equilibrium is $x_0 = \frac{Mg}{k} = \frac{4 	imes 10}{200}

Vedclass · Accepted Answer

$70 \, cm$

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(B) $1$. Initial state: The total mass is $M = 2 + 2 = 4 \, kg$. The spring extension at equilibrium is $x_0 = \frac{Mg}{k} = \frac{4 	imes 10}{200} = 0.2 \, m = 20 \, cm$.$2$. After burning the thread: The lower mass falls freely. The upper mass oscillates about a new equilibrium position. The new equilibrium position is $x_{new} = \frac{mg}{k} = \frac{2 	imes 10}{200} = 0.1 \, m = 10 \, cm$ below the natural length of the spring.$3$. The amplitude of oscillation for the upper mass is $A = x_0 - x_{new} = 20 \, cm - 10 \, cm = 10 \, cm$.$4$. The upper mass starts from the original equilibrium position ($20 \, cm$ below natural length) and moves to its highest position (upper extreme),which is $10 \, cm$ above the new equilibrium position ($10 \, cm$ below natural length). Thus,it travels $10 \, cm$ to reach the new equilibrium and another $10 \, cm$ to reach the upper extreme.$5$. Time taken to reach the upper extreme is $t = \frac{T}{2} = \frac{1}{2} 	imes 2\pi \sqrt{\frac{m}{k}} = \pi \sqrt{\frac{2}{200}} = \frac{\pi}{10} \, s$.$6$. In this time,the lower body falls a distance $h = \frac{1}{2}gt^2 = \frac{1}{2} 	imes 10 	imes (\frac{\pi}{10})^2 = 5 	imes \frac{10}{100} = 0.5 \, m = 50 \, cm$.$7$. The upper body is at $10 \, cm$ below the natural length. The lower body is at $20 \, cm$ (original equilibrium) $+ 10 \, cm$ (thread length) $+ 50 \, cm$ (free fall) $= 80 \, cm$ below the natural length.$8$. The distance between them is $80 \, cm - 10 \, cm = 70 \, cm$.

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