In an experiment, weight of a body is air is found to be (5.00±0.05)N and its weight in water is found to be (4.00±0.05)N then relative density of body is

w1=(5±0.05),w2=(4±0.05)N and w1−w2=(1±0.1) R.D =w1w1−w2=51=5% R.D =Δw1w1×100+Δw1−w2w1−w2×100=0.055×100+0.101×100=1+10=11%∴ R.D =(5±11%)

In an experiment, weight of a body is air is found to be (5.00±0.05)N and its weight in water is found to be (4.00±0.05)N then relative density of body is

w1=(5±0.05),w2=(4±0.05)N and w1−w2=(1±0.1) R.D =w1w1−w2=51=5% R.D =Δw1w1×100+Δw1−w2w1−w2×100=0.055×100+0.101×100=1+10=11%∴ R.D =(5±11%)

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In an experiment, weight of a body is air is found to be $(5.00 \pm 0.05) N$ and its weight in water is found to be $(4.00 \pm 0.05) N$ then relative density of body is

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$5 \pm 20 %$

$5 \pm 1 %$

$5 \pm 10 %$

$5 \pm 11 %$

answer is A.

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Detailed Solution

$\begin{array}{l} w_{1} = (5 \pm 0.05), w_{2} = (4 \pm 0.05) N \\ and (w_{1} - w_{2}) = (1 \pm 0.1) \\ R.D = \frac{w_{1}}{(w_{1} - w_{2})} = \frac{5}{1} = 5 \\ % R.D = (\frac{{Δw}_{1}}{w_{1}} \times 100) + \frac{Δ (w_{1} - w_{2})}{(w_{1} - w_{2})} \times 100 \\ = (\frac{0.05}{5} \times 100) + (\frac{0.10}{1} \times 100) \\ = 1 + 10 = 11 % \\ ∴ R.D = (5 \pm 11 %) \end{array}$