A solid ball of density $5000 k g m^{- 3}$ weighs $0.5 k g$ in air. It is completely immersed in water of density $1000 k g m^{- 3}$ . The apparent weight of the solid ball in water is $x$ . Find $10 x$ .

4

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46

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5

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4.5

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Solution

The correct option is A $4$
Apparent weight = weight in air - upthrust
upthrust = $V ρ g$
$\Rightarrow \frac{0.5}{5000} \times 1000 \times 10 = 0.1 N$
so, apparent weight $(x) = 0.5 - 0.1 = 0.4 N$
Hence $10 x = 4$

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