The relative density of a metal may be found by hanging a block of 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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$5.00 \pm 6 %$

$5.00 \pm 11 %$

$5.00 \pm 10 %$

$5.00 \pm 5 %$

answer is A.

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Detailed Solution

$d = \frac{W_{1}}{W_{1} - W_{2}} = \frac{5}{5 - 4} = 5$

$W_{1} - W_{2} = (1.00 \pm 0.10) N$

$\frac{Δ d}{d} \times 100 = (\frac{0.05}{5.00} + \frac{0.10}{1.00}) \times 100 = 11 %$

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