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.

4.82 k g f

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9.64 k g f

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2.41 k g f

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None of the above

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Solution

The correct option is A $4.82 k g f$
Weight of body in liquid = Actual weight - upthrust
Upthrust exerted $= V ρ g$
Where, $V =$ volume
ρ= density
$g =$ acc. due to gravity

$U p t h r u s t = 100 \times 10^{- 9} m^{3} \times 1.8 \times 10^{3} k g m^{- 3} \times 10$

$\Rightarrow 0.18 k g f$

So, Weight of body in liquid = Actual weight - upthrust
$\Rightarrow 5 - 0.18 = 4.82 k g f$

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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 upthrust due to 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).