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?
Text Solution
Verified by Experts
`g=4pi^(2)L//T^(2)` Here, `T=t/n` and `Delta T=(Delta t)/n`. Therefore, `(Delta T)/T=(Delta t)/t`. The errors in both L and t are the least count errors. Therefore, `(Delta g//g)=(Delta L//L)+2(Delta T//T)` `=0.1/20.0+2(1/90)=0.027` thus, the percentage error in g is `100(Delta g//g)=100 (Delta L//L)+2xx100 (Delta T//T)` `=3 %`
Topper's Solved these Questions
UNITS AND MEASUREMENT
NCERT|Exercise EXERCISE|33 Videos
UNITS AND MEASUREMENT
NCERT|Exercise QUESTION|1 Videos
THERMODYNAMICS
NCERT|Exercise EXERCISE|10 Videos
WAVES
NCERT|Exercise EXERCISE|27 Videos
Similar Questions
Explore conceptually related problems
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 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 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 ?
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 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 an accuray of 1 mm. its period of oscillation oscillations using a clock of 0.1 resolution. What is the accuray in the determined 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 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?
