Home
Class 12
PHYSICS
In an experiment to determine the period...

In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical bobs of radii `r_(1)` and `r_(2)`. The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be `5xx10^(-4)` s, the difference in radii, `|r_(1)-r_(2)|` is best given by :

A

1 cm

B

0.1 cm

C

0.5 cm

D

0.01 cm

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the difference in radii of the spherical bobs based on the given relative difference in their periods, we can follow these steps: ### Step 1: Understand the formula for the period of a simple pendulum The period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. ### Step 2: Differentiate the period with respect to the length Taking the logarithm of both sides, we have: \[ \log T = \log(2\pi) + \frac{1}{2} \log L - \frac{1}{2} \log g \] Differentiating both sides gives: \[ \frac{dT}{T} = \frac{1}{2} \frac{dL}{L} \] ### Step 3: Relate the change in period to the change in length From the differentiation, we can express the relative change in the period \( \frac{\Delta T}{T} \) in terms of the relative change in length \( \frac{\Delta L}{L} \): \[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} \] ### Step 4: Substitute the known values We know that the relative difference in the periods is given as: \[ \frac{\Delta T}{T} = 5 \times 10^{-4} \] Since the length \( L \) of the pendulum is 1 m, we can substitute \( L = 1 \): \[ 5 \times 10^{-4} = \frac{1}{2} \frac{\Delta L}{1} \] ### Step 5: Solve for \( \Delta L \) Rearranging the equation gives: \[ \Delta L = 2 \times 5 \times 10^{-4} = 10 \times 10^{-4} = 10^{-3} \text{ m} \] ### Step 6: Convert to centimeters Since \( 1 \text{ m} = 100 \text{ cm} \): \[ \Delta L = 10^{-3} \text{ m} = 0.1 \text{ cm} \] ### Conclusion Thus, the difference in radii \( |r_1 - r_2| \) is \( 0.1 \text{ cm} \). ### Final Answer The difference in radii \( |r_1 - r_2| \) is \( 0.1 \text{ cm} \). ---

To solve the problem of finding the difference in radii of the spherical bobs based on the given relative difference in their periods, we can follow these steps: ### Step 1: Understand the formula for the period of a simple pendulum The period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. ...
Promotional Banner

Topper's Solved these Questions

  • OSCILLATIONS

    DISHA PUBLICATION|Exercise Exercise-1 : Concept Builder (TOPIC 1: Displacement, Phase, Velocity, Acceleration and Energy in S.H.M.)|27 Videos
  • OSCILLATIONS

    DISHA PUBLICATION|Exercise Exercise-1 : Concept Builder (TOPIC 2: Time Period, Frequency, Simple Pendulum and Spring Pendulum)|31 Videos
  • NUCLEI

    DISHA PUBLICATION|Exercise EXERCISE - 2 (CONCEPT APPLICATOR)|30 Videos
  • PHYSICAL WORLD, UNITS AND MEASUREMENTS

    DISHA PUBLICATION|Exercise Exercise - 2 : Concept Applicator|30 Videos
DISHA PUBLICATION-OSCILLATIONS -Exercise-2 : Concept Applicator
  1. In an experiment to determine the period of a simple pendulum of lengt...

    Text Solution

    |

  2. A point particle of mass 0.1 kg is executing SHM of amplitude 0.1m. Wh...

    Text Solution

    |

  3. A particle performs SHM on x- axis with amplitude A and time period ...

    Text Solution

    |

  4. A mass is suspended separately by two different springs in successive ...

    Text Solution

    |

  5. Two particles execute SHM of same amplitude and frequency on parallel ...

    Text Solution

    |

  6. If a simple pendulum has significant amplitude (up to a factor of1//e ...

    Text Solution

    |

  7. A mass (M) is suspended from a spring of negligible mass. The spring i...

    Text Solution

    |

  8. A bent tube of uniform cross-section area A has a non-viscous liquid o...

    Text Solution

    |

  9. A particle performs SHM about x=0 such that at t=0 it is at x=0 and mo...

    Text Solution

    |

  10. A particle of mass m = 2 kg executes SHM in xy plane between points A ...

    Text Solution

    |

  11. A simple harmonic motion along the x-axis has the following properties...

    Text Solution

    |

  12. Two simple harmonic are represented by the equation y(1)=0.1 sin (100p...

    Text Solution

    |

  13. A uniform cylinder of length L and mass M having cross-sectional area ...

    Text Solution

    |

  14. A uniform cylinder of length L and mass M having cross-sectional area ...

    Text Solution

    |

  15. A body executes simple harmonic motion under the action of a force F1 ...

    Text Solution

    |

  16. A simple pendulum attached to the ceiling of a stationary lift has a t...

    Text Solution

    |

  17. Two bodies of masses 1 kg and 4 kg are connected to a vertical spring,...

    Text Solution

    |

  18. A particle of mass m oscillates with a potential energy U=U(0)+alpha x...

    Text Solution

    |

  19. Two simple pendulums of length 0.5 m and 20 m respectively are given s...

    Text Solution

    |

  20. A uniform pole of length l = 2 L is laid on smooth horizontal table as...

    Text Solution

    |

  21. A particle of mass m executes simple harmonic motion with amplitude a ...

    Text Solution

    |