Home
Class 11
PHYSICS
The least count of a stop watch is (1)/(...

The least count of a stop watch is `(1)/(5)` second. The time of 20 oscillations of a pendulum is measured to be 25 seconds. How much will be the percentage error in the measurement of time ?

Text Solution

Verified by Experts

Error in measuring 25 s `= (1)/(5) s = 0.2` sec. `" "therefore ` percentage error = `(0.2)/(25)xx 100= 0.8` %
The final absolute error in this type of questions is taken to be equal to the least count of the measurement intrument.
Promotional Banner

Topper's Solved these Questions

  • PHYSICAL WORLD, UNITS AND DIMENSIONS & ERRORS IN MEASUREMENT

    ALLEN |Exercise BEGINNER S BOX-1|2 Videos
  • PHYSICAL WORLD, UNITS AND DIMENSIONS & ERRORS IN MEASUREMENT

    ALLEN |Exercise BEGINNER S BOX-2|4 Videos
  • MISCELLANEOUS

    ALLEN |Exercise Question|1 Videos
  • SEMICONDUCTORS

    ALLEN |Exercise Part-3(Exercise-4)|50 Videos

Similar Questions

Explore conceptually related problems

The least count of a stop watch is (1//5)s . The time 20 oscillations of a pendulum is measured to be 25 s . The maximum percentage error in this measurement is

The period of oscillation of a simple pendulum is given by T=2pi sqrt((l)/(g)) . The length l of the pendulum is about 0.5s. The time of 100 oscillations is measured with a watch of 1 s resolution. Calcualte percentage error in measurment of g.

The periodic time of simple pendulum is T=2pi sqrt((l)/(g)) . The length (l) of the pendulum is about 100 cm measured with 1mm accuracy. The periodic time is about 2s. When 100 oscillations are measured by a stop watch having the least count 0.1 second. Calcaulte the percentage error in measurement of g.

Periodic time for second pendulum is 1 second.

A sphere has mass of (20+-0.4)kg and radius of (10+-0.1) m. Find the maximum percentage error in the measurement of density.

The period of oscillation of a simple pendulum is given by T=2pisqrt((l)/(g)) where l is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillation is measrued by a stop watch of least count 0.1 s. The percentage error is g is

Time for 20 oscillations of a pendulum is measured as t_(1)= 39.6 s, t_(2) = 39.9 s and t_(3) = 39.5 s . What is the precision in the measurements ? What is the accuracy of the measurement ?

While measuring the acceleration due to gravity by a simple pendulum , a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period . His percentage error in the measurement of g by the relation g = 4 pi^(2) ( l // T^(2)) will be