A body of volume $100 c m^{3}$ weights $1 k g f$ in air. Its weight in water is $X g f$ . Find $\frac{X}{100}$

10

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9

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Solution

The correct option is B $9$
As volume is ${100 c m}^{3}$ and weighs 1000 gm as $(1 k g f = 9.8 N = 1 k g \times g (9.8))$ , so its density is ${10 g m / c m}^{3}$ .

So the body sinks in water. It displaces ${100 c m}^{3}$ of water, whose density is ${1 g m / c m}^{3}$ .

So the weight of water displaced = $100 \times 1 = 100 g m .$

As the displaced water acts as buoyant force, the weight of body in water is (1000-100) = 900 gms = X.

Hence, the answer is X/100 = 9.

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Similar questions

100 c m^{3}

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A body of volume 100 $c m^{3}$ weighs 1 kgf in air.

Find : (i) its weight in water and (ii) its relative density.

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