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The relative density of a metal may be found by hanging a block of the metal from a spring balance and noting that in air the balance reads ( 5.00 $\pm$ 0.05) N while in water it reads ( 4.00 $\pm$ 0.05)N. The relative density would be quoted as:

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( 5.00 $\pm$ 0.05)

5.00 $\pm$ 11%

( 500 $\pm$ 0.10)

5.00 $\pm$ 6%

answer is B.

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Detailed Solution

$R D = \frac{W_{1}}{W_{1} - W_{2}} = \frac{5}{5 - 4} = 5 l o s s o f w t = W_{1} - W_{2} = 5 - 4 \pm (0.05 + 0.05) = 1 \pm 0.1 \frac{Δ ρ}{ρ} \times 100 = (\frac{0.05}{5} + \frac{0.1}{1.00}) \times 100 = 11 %$

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