Apparent weight of the body immersed in a liquid is given as $M g^{'} = M g - V p g .$ What does $V p g$ stands for in the above equation?

Upthrust

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Downthrust

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Velocity

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Gravity

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Solution

The correct option is A Upthrust
Apparent weight of the body = Actual weight in air - upthrust
Thus $V ρ g$ stands for the upthrust acting on the body when it is immersed in the liquid.

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A body of volume V and density ρ is completely immersed in a liquid of density ρ_L . If g is the acceleration due to gravity, which of the following expressions are correct :

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3) apparent weight of the body in liquid = VρLg

4) the loss in weight of the body = V(ρ-ρL)g

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Q. Assertion :The apparent weight of a body of mass m in an elevator moving with a downward acceleration 'a' is

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Apparent weight of the body immersed in a liquid is given as Mg′=Mg−Vpg. What does Vpg stands for in the above equation?

The correct option is A UpthrustApparent weight of the body = Actual weight in air - upthrustThus Vρg stands for the upthrust acting on the body when it is immersed in the liquid.

Apparent weight of the body immersed in a liquid is given as $M g^{'} = M g - V p g .$ What does $V p g$ stands for in the above equation?

The correct option is A Upthrust
Apparent weight of the body = Actual weight in air - upthrust
Thus $V ρ g$ stands for the upthrust acting on the body when it is immersed in the liquid.