Question

ulate the relative errors in measurement of
mass 1.02 g + 0.01g and 9.89g + 0.01g.
(a) + 1% and + 0.2%
(b) +1% and + 0.1%
(c) + 2% and 0.3%
(d) + 3% and + 0.4%​

Answer 1

Answer:b) +-1% and +-0.1%

Explanation:The % accuracy in the measurement of 1.02g is +-0.01g

Therefore,% relative error in measurement of 1.02 g is

+-0.01g/1.02 g*100

= +-1%

Similarly, the % error in the measurement of 9.89g = +-0.01g/9.89g*100

= +-0.1%

Thus the relative errors are +-1% and +-0.1%.

Hope it helps!

Answer 2

We have to calculate the relative errors in the measurement of mass , 1.02g ± 0.01 g and 9.89 g ± 0.01g

Percentage relative error is given by the ratio of absolute error to the masured value into 100.  

for mass 1.02g ± 0.01 g

measured value mass = 1.02 g

absolute error = 0.01 g

now, the percentage relative error = $\frac{0.01}{1.02}\times100=\frac{1}{1.02}\approx 1$ %

Therefore the relative error of the given mass is ± 1 %.

for mas 9.89 g ± 0.01g

measured value of mass = 9.89g

absolute error = 0.01g

now the percentage relative error = $\frac{0.01}{9.89}\times100=\frac{1}{9.89}\approx 0.1$ %

Therefore the relative error  of the given mass is ± 0.1 %