Home
Class 12
PHYSICS
The period of oscillation of a simple pe...

The period of oscillation of a simple pendulum is `T = 2pisqrt((L)/(g))`. Meaured value of `L` is `20.0 cm` know to `1mm` accuracy and time for `100` oscillation of the pendulum is found to be `90 s` using a wrist watch of `1 s` resolution. The accracy in the determinetion of `g` is :

A

`1%`

B

`5%`

C

`2%`

D

`3%`

Text Solution

Verified by Experts

The correct Answer is:
D
`T=2pisqrt((L)/(g)tog=(4pi^(2)L)/(T^(2))=(4pi^(2)Ln^(2))/(t^(2))(T=(t)/(n))`
Maximum percentage error in `g`
`(Deltag)/(g)xx100=(DeltaL)/(L)xx100+2(Deltat)/(t)xx100`
`(0.1)/(20.0)xx100+2xx(1)/(90)xx100=2.72%=3%`
Promotional Banner

Topper's Solved these Questions

  • EXPERIMENTAL PHYSICS

    NARAYNA|Exercise Level-v(Single answer)|23 Videos

  • EXPERIMENTAL PHYSICS

    NARAYNA|Exercise Multiple answer|6 Videos

  • ELECTROSTATICS AND GAUSS LAW

    NARAYNA|Exercise Intergers type question|11 Videos

  • MAGNETISM

    NARAYNA|Exercise LEVEL-II (H.W)|24 Videos

Similar Questions

Explore conceptually related problems

The period of oscillation of a simple pendulum is T = 2pisqrt(L//g) . Measured value of L is 20.0 cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?

The time period of a simple pendulum is given by T = 2pisqrt(l/g) . The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is,

The period of oscillation of a simple pendulum is T = 2 pi sqrt((L)/(g)) .L is about 10 cm and is known to 1mm accuracy . The period of oscillation is about 0.5 s . The time of 100 oscillation is measured with a wrist watch of 1 s resolution . What is the accuracy in the determination of g ?

The time period of a simple pendulum is given by T = 2 pi sqrt(L//g) , where L is length and g acceleration due to gravity. Measured value of length is 10 cm known to 1 mm accuracy and time for 50 oscillations of the pendulum is 80 s using a wrist watch of 1 s resloution. What is the accuracy in the determination of g ?

The period of oscillation of a simple pendulum is T = 2pi sqrt(L/g) . Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :

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

The time period of a pendulum is given by T = 2 pi sqrt((L)/(g)) . The length of pendulum is 20 cm and is measured up to 1 mm accuracy. The time period is about 0.6 s . The time of 100 oscillations is measured with a watch of 1//10 s resolution. What is the accuracy in the determination of g ?

The time period of oscillation of simple pendulum is given by t = 2pisqrt(I//g) What is the accurancy in the determination of 'g' if 10cm length is knownj to 1mm accuracy and 0.5 s time period is measured form time of 100 oscillations with a wastch of 1 sec. resolution.