A body of mass 20.00 g has volume 5.0 cm 3. The maximum possible error in the measurement of mass and volume respectively are 0.01 and 0.1 cm 3 . The percentage error in the density will be nearest to

Question

A body of mass 20.00 g has volume 5.0 cm 3. The maximum possible error in the measurement of mass and volume respectively are 0.01 and 0.1 cm 3 . The percentage error in the density will be nearest to (A) 1% (B) 2% (C) 11% (D) 25%

Tardigrade Admin · Accepted Answer

ρ=M V . Hence (Δ ρ/ρ)=(Δ M/M)+(Δ V/V) =(0.01/20.00)+(0.1/5.0) =0.0205 ≅ 2 %

Q. A body of mass $20.00 g$ has volume $5.0 c m^{3}$ . The maximum possible error in the measurement of mass and volume respectively are $0.01$ and $0.1 c m^{3} .$ The percentage error in the density will be nearest to

1%

2%

11%

25%

$ρ = M V .$ Hence $\frac{Δ ρ}{ρ} = \frac{Δ M}{M} + \frac{Δ V}{V}$
$= \frac{0.01}{20.00} + \frac{0.1}{5.0}$
$= 0.0205 ≅ 2%$

Q. A body of mass 20.00g has volume 5.0cm3. The maximum possible error in the measurement of mass and volume respectively are 0.01 and 0.1cm3. The percentage error in the density will be nearest to

1%

2%

11%

25%

ρ=MV. Hence ρΔρ​=MΔM​+VΔV​ =20.000.01​+5.00.1​ =0.0205≅2%

Q. A body of mass $20.00 g$ has volume $5.0 c m^{3}$ . The maximum possible error in the measurement of mass and volume respectively are $0.01$ and $0.1 c m^{3} .$ The percentage error in the density will be nearest to

$ρ = M V .$ Hence $\frac{Δ ρ}{ρ} = \frac{Δ M}{M} + \frac{Δ V}{V}$
$= \frac{0.01}{20.00} + \frac{0.1}{5.0}$
$= 0.0205 ≅ 2%$