A body of volume $100 c m^{3}$ weighs $5 k g f$ in air. It is completely immersed in a liquid of density $1.8 \times 10^{3} k g m^{- 3}$ . Find the upthrust due to liquid.

0.18 k g f

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0.09 k g f

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0.54 k g f

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0.36 k g f

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Solution

The correct option is A $0.18 k g f$
Upthrust exerted $= V d g$

Where, $V =$ volume

$d =$ density

$g =$ acc. due to gravity

$U p t h r u s t = 100 \times 10^{- 6} m^{3} \times 1.8 \times 10^{3} k g m^{- 3} \times 10$

$\Rightarrow 0.18 k g f$

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Similar questions

Q. A body of volume

100 c m^{3}

weighs

5 k g f

in air. It is completely immersed in a liquid of density

1.8 \times 10^{3} k g m^{- 3}

. Find the weight of the body in liquid.

A body of volume 100 $c m^{3}$ weighs 5 kgf in air. It is completely immersed in a liquid of density 1.8 $\times 10^{3} k g m^{- 3}$ . Find :
(i) the upthrust due to liquid and
(ii) the weight of the body in liquid.
(iii) the upthrust on the body.
State the assumption made in part (i).

Upthrust on a body depends on
A. The acceleration due to gravity, the density of the liquid
B. The volume of the body immersed in the liquid.

A body of volume

V

and density

ρ

is kept completely immersed in a liquid of density

ρ_{L}

. If

g

is the acceleration due to gravity, write expression for the following:

The upthrust on the body

Q. A body weighs

50 g

in air and

40 g

when completely immersed in water. How much it will weigh when immersed completely in a liquid of relative density

1.5

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A body of volume 100cm3 weighs 5kgf in air. It is completely immersed in a liquid of density 1.8×103kgm−3. Find the upthrust due to liquid.

The correct option is A 0.18kgfUpthrust exerted =VdgWhere, V= volumed= density g=acc. due to gravityUpthrust=100×10−6m3×1.8×103kgm−3×10⇒0.18kgf

A body of volume $100 c m^{3}$ weighs $5 k g f$ in air. It is completely immersed in a liquid of density $1.8 \times 10^{3} k g m^{- 3}$ . Find the upthrust due to liquid.

The correct option is A $0.18 k g f$
Upthrust exerted $= V d g$

Where, $V =$ volume

$d =$ density

$g =$ acc. due to gravity

$U p t h r u s t = 100 \times 10^{- 6} m^{3} \times 1.8 \times 10^{3} k g m^{- 3} \times 10$

$\Rightarrow 0.18 k g f$