Home
Class 11
PHYSICS
The period of oscillation of a simple pe...

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 :

A

`1.13% `

B

`1.03%`

C

`1.33% `

D

`1.30%`

Text Solution

Verified by Experts

The correct Answer is:
A
`T=2pisqrt(l/g)`
`T^2 =4pi^2 [l/g]`
`g = 4pi^2 [l/T^2]`
`(Deltag)/g =(Deltal)/(l)=(2DeltaT)/T`
`=[(1xx10^(-3))/1 +(2(10 xx10^(-3)))/(1.95)]xx100`
`=1.13%`
Promotional Banner

Similar Questions

Explore conceptually related problems

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 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 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 :

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 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 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

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.