Grade 12th passMechanics

The least count of a stopwatch is 1/5 second. The time of 20 oscillation of a pendulum is measured to be 25 second s. The maximum percentage error in the measurement of time will be

Anupam Bhowmik

9 Years agoGrade 12th pass

5 Answers

Vikas TU

9 Years ago

The least count is 1/5 sec.

Time taken = 25 sec

Percentage error -= least count/time taken * 100

= 100/125

=4/5

= 0.8

Harshad Bharat Bhere

8 Years ago

Least count=1/5 sNo. of oscillations= 20Time taken= 25s%age error × 100= Least count÷ time taken × 100 =1/5÷ 125 × 100 =0.8%

Harshad Bharat Bhere

8 Years ago

Least count=1/5 sNo. of oscillations= 20Time taken= 25s%age error × 100= Least count÷ time taken × 100 =1/5÷ 125 × 100 =0.8%

Navya vats

7 Years ago

We have to calculate the max % error in the measurement so, firstly , least count of stop wash = 1/5 or 0.2 as you want to take then , . Given time take =25sec now we know %error =dalta A / A *100 so by doing this we get the value 0.8

Rishi Sharma

6 Years ago

Hello students,
Number of oscillations =20
Total time taken for 20 oscillation =25sec
Time period for one oscillation =25/20=1.25sec
Least count of watch =0.2sec
The measurement of time period 1.25 the measuring watch can measure upto 0.2s thus the watch can measure 1.2s but the uncertainity remainsin the 0.05s.
1.25s=1.2+0.05
Thus, error =0.05sec
Percentage error =(0.05/1.25)×100=4%

I hope the solution will solve all your doubts.
Thank You,
All the Best for the Exams.