Question

Question on least count.

Answer 1

If time period of oscillation of a simple pendulum is measured as 2.5 second using a stopwatch with least count 1/2 second then the permissible error in measurement is:

Answer 2

The error is dependent on the scale of time!

In this case, \[{0.5 \over 2.5} * 100 = \rm percent ~~error\]

Answer 3

Can you explain the concept more detailed? I would be really grateful.

Answer 4

@Mashy

Answer 5

@Vincent-Lyon.Fr

Answer 6

if the least count of your clock is 0.5 second, then ur clock will go as follows

0 , 0.5, 1, 1.5 and so on

now imagine when you stop your clock it turns out to be 2.5s (like in your problem)
you know for sure its not EXACTLY 2.5s
so what is the range of times ?

well.. u see even if when you hit the stop button, if the time was 2.4s ... it would round off and show you 2.5..
so the minimum time could be 2.25 if its less than that, your clock would show 2.0 else and more than that your clock would show 2.5

similarly even if the time was 2.75, your clock would show 2.5 (if it is more than 2.75, your clock would show 3.0)

thus now you know ur time is between 2.25 to 2.75
hence your measurement is 2.5 + or - 0.25 s

Is that right @Vincent-Lyon.Fr :D :D ??? or am i horribly wrong?

Answer 7

That's a great explanation Mashy

Answer 8

I have never met this concept of 'least count' before.
Sorry, not used here.

Answer 9

Thank you :)