Question

The least count of a stopwatch is $\dfrac{1}{5}s$. The time of 20 oscillations of a pendulum is measured to be $25s$ . What is the maximum percentage error in this measurement?
$A)8$%
$B)0.8$%
$C)4$%
$D)16$%

Answer 1

Here we need to know about the definition of least count of an instrument and the percentage error in a reading taken. There is always some error in taking a reading by any apparatus because any apparatus can give the reading with limited accuracy, there is bound to be some error.
Formula used:
Maximum percentage error $e$ =$(\dfrac{\vartriangle T}{T})\times 100$

Complete answer:
When we take readings by any instrument, there is bound to be some error due to the limitation in accuracy of the apparatus. The least value of a quantity that can be measured accurately by any apparatus is called the least count of the apparatus. For example, in case of a meter scale the least length that can be measured accurately is $0.1cm$ . Therefore, its least count of meter scale is $0.1cm$ . Any length smaller than this value like $0.02cm,0.07cm$ etc. cannot be measured correctly by the meter scale. Now in this problem the least count of the stop watch is given as$\vartriangle T=(1/5)s=0.2s$. So, by this stopwatch, any time interval less than $0.2s$ cannot be measured correctly. Now for a reading $T$ , the maximum percentage error $e$ is defined as
$e=\dfrac{\vartriangle T}{T}\times 100$ . Now putting the values of all the quantities we get
$e=\dfrac{0.2}{25}\times 100=0.8$%
So, the maximum percentage error in this measurement is $0.8$%.

Therefore, the correct option is B.

Note:
To solve this kind of problem we must convert the least count and the reading taken in the same unit, otherwise it will give the wrong answer. In this problem both are taken in seconds. If to determine a quantity we need to measure multiple quantities then the total percentage error will be the sum of the individual percentage errors.