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The time period of oscillation of a simp...

The time period of oscillation of a simple pendulum is given by `T=2pisqrt(l//g)`
The length of the pendulum is measured as `1=10+-0.1` cm and the time period as `T=0.5+-0.02s`. Determine percentage error in te value of g.

Text Solution

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The correct Answer is:
0.09
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Knowledge Check

  • The time period of oscillation of a simple pendulum is given by T=2pisqrt((l)/(g)) . The length of the pendulum is measured as l=10pm0.01cm and the time period as T=0.5pm0.02s . The percentage error in the value of g is

    A
    `5%`
    B
    `8%`
    C
    `7%`
    D
    None of these

  • Time period T of a simple pendulum of length l is given by T = 2pisqrt((l)/(g)) . If the length is increased by 2% then an approximate change in the time period is

    A
    0.02
    B
    `1%`
    C
    `(1)/(2)%`
    D
    none of these

  • The time period of a simple pendulum is 1.2 s. If the length of the pendulum is doubled, the new time period will be

    A
    1.0s
    B
    1.4s
    C
    1.7s
    D
    2.4s

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