The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of the body in air is W1=8.00±0.05N and the weight in water is W2=6.00±0.05N, then the relative density ρr=W1/W1−W2with the maximum permissible error is

Relative density ρr=W1W1−W2=8.008.00−6.00=4.00Δρrρr×100=ΔW1W1×100+ΔW1−W2W1−W2×100 =0.058.00×100+0.05+0.052×100=5.62%∴ ρr=4.00±5.62%

The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of the body in air is W1=8.00±0.05N and the weight in water is W2=6.00±0.05N, then the relative density ρr=W1/W1−W2with the maximum permissible error is

Relative density ρr=W1W1−W2=8.008.00−6.00=4.00Δρrρr×100=ΔW1W1×100+ΔW1−W2W1−W2×100 =0.058.00×100+0.05+0.052×100=5.62%∴ ρr=4.00±5.62%

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The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of the body in air is $W_{1} = 8.00 \pm 0.05 N$ and the weight in water is $W_{2} = 6.00 \pm 0.05 N$ , then the relative density $ρ_{r} = W_{1} / (W_{1} - W_{2})$ with the maximum permissible error is

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$4.00 \pm 5.62 %$

$4.00 \pm 0.62 %$

$4.00 \pm 3.2 %$

$4.00 \pm 0.82 %$

answer is D.

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Detailed Solution

$\begin{array}{l} Relative density ρ_{r} = \frac{W_{1}}{W_{1} - W_{2}} = \frac{8.00}{8.00 - 6.00} = 4.00 \\ \begin{aligned} \frac{Δ ρ_{r}}{ρ_{r}} \times 100 & = \frac{Δ W_{1}}{W_{1}} \times 100 + \frac{Δ (W_{1} - W_{2})}{W_{1} - W_{2}} \times 100 \\ = \frac{0.05}{8.00} \times 100 + \frac{0.05 + 0.05}{2} \times 100 = 5.62 % \end{aligned} \\ ∴ ρ_{r} = 4.00 \pm 5.62 % \end{array}$