Home
Class 11
PHYSICS
The least count of a stopwatch is 0.2 s....

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?

A

`0.2%`

B

`0.4%`

C

`0.01%`

D

`1.5%`

Text Solution

Verified by Experts

The correct Answer is:
b

The error can be equal to the least count of the stopwatch i.e. 0.2 s = `Deltat` and t = 50 s
`therefore` Percentage error `=((Deltat)/(t))xx100`
`=(0.2)/(50)xx100=0.4%`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • MEASUREMENTS

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos
  • MAGNETISM

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 Videos
  • RAY OPTICS (MIRRORS, LENSES AND OPTICAL INSTRUMENTS)

    MARVEL PUBLICATION|Exercise TEST YOUR GRASP|10 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 maximum percentage error in this measurement is found to be 0.x%. What is the value of x?

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

Knowledge Check

  • 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
    `1.2%`
    B
    `0.8%`
    C
    `1.8%`
    D
    None of these
  • 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

    A
    `8%`
    B
    `1%`
    C
    `0.8%`
    D
    `16%`
  • A student measures time for 20 oscillations of a simple pendulum as 30s, 32s, 35s and 31s. If the minimum division in the measuring clock is 1s, then correct meant time in second is

    A
    `32 pm 3`
    B
    `32 pm 1`
    C
    `32 pm 2`
    D
    `32 pm 5`
  • Similar Questions

    Explore conceptually related problems

    If the length of simple pendulum is increased by 50% , what is the percentage increase in its time period?

    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 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

    A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?

    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)) .