Question

Ammeter 'A' has 5 divisions between the marks 0 and 1A while ammeter 'B'
has 10 divisions between the marks 0 and 1 A. Which will give more precise
reading ?

Answer 1

Given:

Ammeter 'A' has 5 divisions between the marks 0 and 1A while ammeter 'B' has 10 divisions between the marks 0 and 1 A.

To find:

Which instrument is more precise?

Calculation:

In this type of questions, it is best to calculate the least count of the instrument. The instrument with lower least-count will be more precise.

For Instrument A:

$\sf \: Least \: Count = \dfrac{range}{divisions}$

$\sf \implies\: Least \: Count = \dfrac{1 - 0}{5}$

$\sf \implies\: Least \: Count = \dfrac{1}{5}$

$\sf \implies\: Least \: Count = 0.2$

For Instrument B :

$\sf \: Least \: Count = \dfrac{range}{divisions}$

$\sf \implies\: Least \: Count = \dfrac{1 - 0}{10}$

$\sf \implies\: Least \: Count = \dfrac{1}{10}$

$\sf \implies\: Least \: Count = 0.1$

Since Instrument B has lower least count, it will be more precise.

Answer 2

$\underline{ \underline{\bf{\large{\sf{Information \: given \: to \: us:-}}}}}$

• Ammeter 'A' has 5 divisions
• Marks in which Ammeter was in between 0 and 1A
• Ammeter 'B' has 10 divisions
• Marks in which Ammeter'B' was in between 0 and 1A

$\underline{ \underline{\bf{\large{\sf{Our \: need \: to \: calculate:-}}}}}$

• Which Ammeter would be giving us more precise reading than other one.

$\underline{ \underline{\bf{\large{\sf{Solving \: way:-}}}}}$

• First of all simply we would use the Formula for calculating the least count of an instrument for Ammeter'A' in order to calculate which instrument would give us more precise reading. The formula for calculating the least count is range / divisions and apply all the values and solve it.

• After that in the same way we would again use the formula for calculating the least count which is range / divisions. This is for Ammeter'B'. And substitute the values and solve it.

• At last we would compare the least count of Ammeter'B' with Ammeter'A'. In this the instrument which would have lower least count would be precise than the other one.

$\underline{ \underline{\bf{\large{\sf{Required \: Solution:-}}}}}$

$\red\bigstar \: \underline {\sf{Formula \: for \: least \: count:-}}$

• $\red{ \boxed{ \sf{Least \: count \: = \: \frac{Range}{Division}}}}$

$\red\bigstar \: \underline {\sf{Again:-}}$

• Marks 0 and 1A is the range
• Divisions are 5

$\red\bigstar \: \underline {\sf{Substituting \: the \: the \: values \: in \: least \: count\: formula \: for \: Ammeter'A':-}}$

$\large {\bf{{ \longmapsto}}} \: \mathrm{least \: count \: = \dfrac{1 \: - \: 0}{5} }$

$\red\bigstar \: \underline {\sf{We \: gets,}}$

$: \large {\bf{{ \longmapsto}}} \: \mathrm{ least \: count = \dfrac{1}{5} }$

$\red\bigstar \: \underline {\sf{Now,}}$

$: \large {\bf{{ \longmapsto}}} \: \mathrm{ least \: count \: = \: \dfrac{ \cancel{1}}{ \cancel5} }$

$\red\bigstar \: \underline {\sf{Thus,}}$

$: \large {\bf{{ \longmapsto}}} \: \boxed {\mathrm{ \: least \: count \: = \: 0.2}}$

$\red\bigstar \: \underline {\sf{Substituting \: the \: the \: values \: in \: least \: count\: formula \: for \: Ammeter'B':-}}$

$\large {\bf{{ \longmapsto}}} \: \mathrm{ least \: count \: =\dfrac{ 1 \: - \: 0}{10} }$

$\red\bigstar \: \underline {\sf{We \: gets,}}$

$: \large {\bf{{ \longmapsto}}} \: \mathrm{ least \: count \: = \:\dfrac{1}{10} }$

$\red\bigstar \: \underline {\sf{Now,}}$

$: \large {\bf{{ \longmapsto}}} \: \mathrm{ least \: count \: = \: \dfrac{ \cancel{1}}{ \cancel10} }$

$\red\bigstar \: \underline {\sf{Thus,}}$

$: \large {\bf{{ \longmapsto}}} \: \boxed {\mathrm{ \: least \: count \: = \: 0.1}}$

$\red \bigstar \: \underline{ \sf{Therefore, \: Instrument \: or \: Ammeter'B' \: would \: be \: more \: precise}}$