In figure shown, the spring constant is k. The mass of the upper block is m and that of the lower back is 3m. The upper block is depressed down from its equilibrium position by a distance δ and released at t = 0.

The minimum value of δ for which the lower block loses contact with the ground is

3 mg/k

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4 mg/k

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2 mg/k

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mg/k

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Solution

The correct option is B
4 mg/k

At equilibrium Position $x = \frac{m g}{k}$ ; Then we compress $δ$

At highest point: $K (δ - \frac{m g}{k}) = 3 m g$ or $δ = \frac{4 m g}{k}$

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In figure shown, the spring constant is k. The mass of the upper block is m and that of the lower back is 3m. The upper block is depressed down from its equilibrium position by a distance δ and released at t = 0. The minimum value of δ for which the lower block loses contact with the ground is

The correct option is B 4 mg/k At equilibrium Position x=mgk ; Then we compress δ At highest point: K(δ−mgk)=3mg or δ=4mgk

In figure shown, the spring constant is k. The mass of the upper block is m and that of the lower back is 3m. The upper block is depressed down from its equilibrium position by a distance δ and released at t = 0.

The minimum value of δ for which the lower block loses contact with the ground is

The correct option is B
4 mg/k

At equilibrium Position $x = \frac{m g}{k}$ ; Then we compress $δ$

At highest point: $K (δ - \frac{m g}{k}) = 3 m g$ or $δ = \frac{4 m g}{k}$