Question

A packet contains silver powder of mass $20.23g \pm 0.01g$. Some of the powder of mass $5.75g \pm 0.01g$ is taken out from it. The mass of the powder left back is:
A) $14.48g \pm 0.00g$
B) $14.48 \pm 0.02g$

Answer 1

Hint: This question is totally based on the understanding of the errors of the quantity. In order to solve the given question, we need to first calculate the mass of the powder left back and then the errors in the mass of powder left. Then finally we can conclude with the final answer.

Complete step by step solution:
The total mass of the powder in the question is given as,$m = (20.23 \pm 0.01)g$
The mass of the powder that is taken out in the question is given as,${m_1} = (5.75 \pm 0.01)g$
Now, the mass of the powder that is left in the packet can be found by,
$\Rightarrow M = m - {m_1} = 20.23 - 5.75 = 14.48g$
Now the change in the value of the mass of the powder left will be equal,
$\Rightarrow \Delta M = \pm (\Delta m + \Delta {m_1}) = \pm (0.01 + 0.01) = \pm 0.02$
Then, we can write that the mass of the powder left, $(14.48 \pm 0.02)g$.

Hence, option (B), i.e. $(14.48 \pm 0.02)g$ is the correct choice for the given question.

Note: We need to know that the errors are always added as we have to find the maximum value. In case of negative errors also we need to add the errors. The calculation of errors in the measurement system does not imply that the dimensions are incorrect. It denotes the errors that are there in the measuring instruments. There are basically three types of errors.
Absolute errors: These errors show the variation between the measured values and the absolute values.
Relative error: It is the ratio of the absolute error to the actual value.
Percentage error: It is the percentage of the absolute errors.