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A block of mass 2kg is suspended by a spring of force constant $K = 10 N/M$ . Another identical spring is fixed below 1m from mass 2kg. Initially both the springs are unstetched and mass released from rest then

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Maximum extension in the upper spring 2.82m

Equilibrium position of the block from initial released positions ‘0.75m’

Maximum extension in the upper spring 1.41m

Equilibrium position of the block from initial released positions ‘1.5m’

answer is A, D.

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Detailed Solution

decreases in PE of block = Gain in PE of springs

$mgh = \frac{1}{2} k x^{2} + \frac{1}{2} k {(x + 1)}^{2}$

$2 \times 10 \times (x + 1) = \frac{1}{2} \times 10 \times x^{2} + \frac{1}{2} 10 {(x + 1)}^{2}$

$x = \frac{2 \pm \sqrt{4 + 24}}{4} = 1.82 m$

$x_{\max} = 1 + 1.82 = 2.82 m$

At equilibrium $mg = Ky + K (y - 1)$

$20 = 10 y + 10 (y - 1)$

$20 = 10 y + 10 y - 10$

$20 = 20 y - 10 \Rightarrow 10 = 20 y \Rightarrow y = \frac{10}{20} = 0.5 m$

$y = 1.5 m$

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