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
As, `g=4pi^(2)(l)/(T^(2))`
So, `(Deltag)/(g)xx100=(Deltal)/(l)xx100+2(DeltaT)/(T)xx100`
`=(0.1)/(20)xx100+2xx(1)/(90)xx100=2.72~=3%`
Promotional Banner

Topper's Solved these Questions

  • PHYSICAL WORLD, UNITS AND MEASUREMENTS

    DISHA PUBLICATION|Exercise Exercise - 1 : Concept Builder (Topicwise)(Topic 1 : Units of Physical Quantities)|20 Videos

  • PHYSICAL WORLD, UNITS AND MEASUREMENTS

    DISHA PUBLICATION|Exercise Exercise - 1 : Concept Builder (Topicwise)(Topic 2 : Errors in Measurements and Significant Figures)|28 Videos

  • OSCILLATIONS

    DISHA PUBLICATION|Exercise Exercise-2 : Concept Applicator|28 Videos

  • RAY OPTICS AND OPTICAL INSTRUMENTS

    DISHA PUBLICATION|Exercise Exercise -2 : Concept Applicator|26 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.

DISHA PUBLICATION-PHYSICAL WORLD, UNITS AND MEASUREMENTS-Exercise - 2 : Concept Applicator
  1. The period of oscillation of a simple pendulum is T = 2pisqrt((L)/(g))...

    Text Solution

    |

  2. A metal sample carrying a current along x axis with density is subject...

    Text Solution

    |

  3. In an experiment of calibartion of voltmeter, a standard cell of emf 1...

    Text Solution

    |

  4. The velocity of water wave v may depend on their wavelength lambda, th...

    Text Solution

    |

  5. If E, M, L and G denote energy, mass, angular momentum and gravitation...

    Text Solution

    |

  6. The potential energy of a point particle is given by the expression V(...

    Text Solution

    |

  7. A physical parameter a can be determined by measuring the parameters b...

    Text Solution

    |

  8. The mass of the liquid flowing per second per unit area of cross-secti...

    Text Solution

    |

  9. a quantity X is given by epsilon(0)L(DeltaV)/(Deltat) where in(0) is ...

    Text Solution

    |

  10. In a certain system of units, 1 unit of time is 5 sec, 1 unit of mass ...

    Text Solution

    |

  11. The frequency (f ) of a wire oscillating with a the length l, in p loo...

    Text Solution

    |

  12. The electric field is given by vecE=(A)/(x^(3))hati+By hatj+Cz^(2)hatk...

    Text Solution

    |

  13. A formula is given as P=(b)/(a)sqrt(1+(k.theta.t^(3))/(m.a)) where P...

    Text Solution

    |

  14. A physical quantity rho is calculated by using the formula rho =(1)/(1...

    Text Solution

    |

  15. The momentum of an electron in an orbit is h//lambda, where h is a con...

    Text Solution

    |

  16. A gas bubble from an explosion under water oscillates with a period T ...

    Text Solution

    |

  17. A student performs an experiment to determine the Young's modulus of a...

    Text Solution

    |

  18. An experiment is performed to obtain the value of acceleration due to ...

    Text Solution

    |

  19. from the following combinations of physical constants (expressed throu...

    Text Solution

    |

  20. A, B, C and D are four different physical quantities having different ...

    Text Solution

    |

  21. A famous relation in Physics relates moving mass m to the rest mass m0...

    Text Solution

    |