Home
Class 12
PHYSICS
The least count of a stop watch is 0.2 s...

The least count of a stop watch is 0.2 s, The time of 20 oscillations of a pendulum is measured to be 25s. The percentage error in the time period is

A

`16%`

B

`0.8%`

C

`1.8%`

D

`8%`

Text Solution

Verified by Experts

`Deltat = 0.2 s`
`t =25s`
`T = (t)/(N) rArr (DeltaT)/(T) = (Deltat)/(t) = (0.2)/(25) = 0.8%` .
Promotional Banner

Topper's Solved these Questions

  • EXPERIMENTAL PHYSICS

    RESONANCE|Exercise PARTIII|3 Videos

  • EXPERIMENTAL PHYSICS

    RESONANCE|Exercise Exercise -2 PART -1|1 Videos

  • EXPERIMENTAL PHYSICS

    RESONANCE|Exercise Exercise -1 PART|1 Videos

  • ELECTROSTATICS

    RESONANCE|Exercise HLP|39 Videos

  • GEOMATRICAL OPTICS

    RESONANCE|Exercise Advance level Problems|35 Videos

Similar Questions

Explore conceptually related problems

The least count of a stop watch is 0.2 second. The time of 20 oscillations of a pendulum is measured to be 25 seconds.The percentage error in the time period is

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 least count of a stop watch is 0.2 second. The time of 20 oscillations of a pendulum is measured to be 25 seconds. The maximum percentage error in this measurement is found to be 0.x%. What is the value of x?

The least count of a stopwatch is 0.2 s. A student found the time of 25 oscillations of a simple pendulum to be 50 second. What will be the percentage error in the measurement of time?

The time period of a seconds pendulum is ______ s.

A student measures the value of g with the help of a simple pendulum using the formula g = (4pi^(2)L)/(T^(2)) . He measures length L with a meter scale having least count 1 mm and finds it 98.0 cm . The time period is measured with the help of a watch of least count 0.1s . The time of 20 oscillations is found to be 40.4 s. The error Deltag in the measurment of g is (" in" m//s^(2)) .

The time period of an oscillating simple pendulum is given as T=2pisqrt((l)/(g)) where l is its length and is about 1m having 1mm accuracy. Its time period is 2s . The time for 100 oscillations is measured by a stopwatch having least count 0.1s . The percentage error in the measurement of g is

In an experiment of simple pendulum, time period measured was 50s for 25 vibrations when the length of the simple pendulum was taken to be 100cm . If the lest count of stop watch is 0.1s and that of metre scale is 0.1 cm , calculate the maximum possible percentage error in the measurement of value of g .

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