A block of density $2000 k g / m^{3}$ and mass 10 kg is suspended by a spring of force constant 100 N/m. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density $1000 k g / m^{3}$ . If the block is in equilibrium position

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The elongation of the spring is 1 cm

The magnitude of buoyant force acting on the block is 50 N

The spring potential energy is 12.5 J

Magnitude of spring force on the block is greater than the weight of the block

answer is B, C.

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Detailed Solution

W = mg = 100 N

$Upthrust = {Vρ}_{1} g = \frac{10}{2000} \times 1000 \times 10 = 50 N$

$∴$ spring force needed for equilibrium

Kx = W-Upthrust = 50 N

$or elongation x = \frac{50}{100} = 0.5 m$

$U = \frac{1}{2} {kx}^{2} = \frac{1}{2} \times 100 \times (0.5)^{2} = 12.5 J$

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