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

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

Text Solution

Verified by Experts

The dimensional of LHS = the dimensional of T = `[M^(0)L^(0) T^(1)]`
The dimensions of RHS = `(("dimensions of length")/("dimensions of accelleration"))^(1//2)` `(because 2pi ` is a dimensionless constant )
`" "= [ (L)/(LT^(-2))]^(1//2) = [T^(2)] ^(1//2) = [T] = [M^(0) L^(0) T^(1)]`
Since the dimensions are same one both the sides, the relation is correct.
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)|51 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 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) . 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 = 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 = 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 ?

In an experiment to determine the value of acceleration due to gravity g using a simple pendulum , the measured value of lenth of the pendulum is 31.4 cm known to 1 mm accuracy and the time period for 100 oscillations of pendulum is 112.0 s known to 0.01 s accuracy . find the accuracy in determining the 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

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 length of a simple pendulum is about 100 cm known to have an accuracy of 1mm. Its period of oscillation is 2 s determined by measuring the time for 100 oscillations using a clock of 0.1s resolution. What is the accuracy in the determined value of g?