In the situation shown in figure, the block is in equilibrium and the vessel is placed in a lift. If the lift starts moving down with constant acceleration $a$ , then at equilibrium

the buoyant force will decrease

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the compression in the spring will be more

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the compression in the spring will be less

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the buoyant force will remain same

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Solution

The correct options are
A the buoyant force will decrease
C the compression in the spring will be less
Buoyant force when lift is at rest
$F_{B} = ρ_{l} V (g)$
Buoyant force when lift starts accelerating
$F_{B} = ρ_{l} V (g - a)$ .
Buoyancy force is decreasing.

Spring force when lift is at rest
$F_{s} = m g - F_{B} = ρ_{b} V g - ρ_{l} V g = (ρ_{b} - ρ_{l}) V g$
Spring force when lift starts accelerating
$F_{s} = (ρ_{b} - ρ_{l}) V (g - a)$
Hence the spring force is decreasing.

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