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Question: The mass of a beaker is found to be \( 10.1 \pm 0.1 \) gm when empty and \( 17.3 \pm 0.1 \) gm when ...

The mass of a beaker is found to be 10.1±0.110.1 \pm 0.1 gm when empty and 17.3±0.117.3 \pm 0.1 gm when partially filled with a liquid. What is the best value of mass of liquid along with limits of accuracy.
(A) 7.2±0.2gm7.2 \pm 0.2gm
(B) 7.2±0.1gm7.2 \pm 0.1gm
(C) 7.2±0.3gm7.2 \pm 0.3gm
(D) 7.1±0.2gm7.1 \pm 0.2gm

Explanation

Solution

Hint
In this type of question, we first need to find the actual value of the quantity. This can be done by considering the absolute values of the measurements i.e. without any error. After that we need to find the error in the final result. This can be done by using the relative error formula.

Complete step by step answer
Here we know that the mass of the liquid is the difference between the empty beaker and the filled beaker. This comes out to be 17.310.1=7.217.3 - 10.1 = 7.2.
Now we will use the relative error formula to find the error in this value.
Δmm=±[Δm1m1+Δm2m2]\dfrac{{\Delta m}}{m} = \pm [\dfrac{{\Delta {m_1}}}{{{m_1}}} + \dfrac{{\Delta {m_2}}}{{{m_2}}}]
Δm7.2=±[0.110.1+0.117.3] Δm=±0.1  \dfrac{{\Delta m}}{{7.2}}\, = \, \pm [\dfrac{{0.1}}{{10.1}}\, + \,\dfrac{{0.1}}{{17.3}}] \\\ \Delta m\, = \, \pm 0.1 \\\
Therefore the final value will be : 7.2±0.17.2\, \pm \,0.1 gm. So, the answer with the correct option is option (B).

Note
The relative error formula is very useful in calculating the percentage increase in a dependent quantity when some other independent quantity changes. In case there exists a squared or cubic relation, the power comes down to the multiple of the relative error. For eg if A=B2A = {B^2}, then ΔAA=±2ΔBB\dfrac{{\Delta A}}{A}\, = \, \pm 2\dfrac{{\Delta B}}{B}.