In the arrangement shown, the springs are light and have stiffness $k_{1} = 100 N / m$ and $k_{2} = 200 N / m$ and the pulleys are ideal. An $8.0 k g$ block suspended from the lower pulley is initially held at rest maintaining the strings straight and springs relaxed. Now the force supporting the block is gradually reduced to zero. Acceleration due to gravity is $10 m / s^{2}$ . If the block descends by a distance $(5 n)$ cm in the process of reduction of the force, find ‘n’.

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answer is 3.

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Detailed Solution

$2 k_{1} x_{1} = k_{2} x_{2} = \frac{m g}{2} \frac{(\frac{x_{1}}{2} + x_{2})}{2} = x, x = 15$

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