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 ± 0.05 N and the weight in water is W2=6.00 ± 0.05 N, then the relative density ρr=W1 /(W1-W2) with the maximum permissible error is

Question

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 ± 0.05 N and the weight in water is W2=6.00 ± 0.05 N, then the relative density ρr=W1 /(W1-W2) with the maximum permissible error is (A) 4.00 ± 0.62 % (B) 4.00 ± 0.82 % (C) 4.00 ± 3.2 % (D) 4.00 ± 5.62 %

Tardigrade Admin · Accepted Answer

Relative density pr=(W1/W1-W2)=(8.00/8.00-6.00)=4.00 (Δ ρr/ρr) × 100=(Δ W1/W1) × 100+(Δ(W1-W2)/W1-W2) × 100 =(0.05/8.00) × 100+(0.05+0.05/2) × 100=5.62 % ∴ ρr=4.00 ± 5.62 %

$4.00 \pm 0.62%$

$4.00 \pm 0.82%$

$4.00 \pm 3.2%$

$4.00 \pm 5.62%$

Q. 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​−W2​) with the maximum permissible error is

4.00±0.62%

4.00±0.82%

4.00±3.2%

4.00±5.62%

Relative density pr​=W1​−W2​W1​​=8.00−6.008.00​=4.00 ρr​Δρr​​×100=W1​ΔW1​​×100+W1​−W2​Δ(W1​−W2​)​×100 =8.000.05​×100+20.05+0.05​×100=5.62% ∴ρr​=4.00±5.62%

$4.00 \pm 0.62%$

$4.00 \pm 0.82%$

$4.00 \pm 3.2%$

$4.00 \pm 5.62%$