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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 `l=10+-0.1` cm and the time period as `T=0.5+-0.02s`. Determine percentage error in te value of g.

A

`5%`

B

`8%`

C

`7%`

D

None of these

Text Solution

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The correct Answer is:
B
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Knowledge Check

  • 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 T of oscillation of as simple pendulum of length l is given by T=2pi sqrt(l/g) . The percentage error in T corresponding to an error of 2% in the value of l is

    A
    0.02
    B
    0.01
    C
    0.03
    D
    0.012

  • In a simple pendulum the period of oscillation (T) is related to the length of the pendulum (L) as

    A
    `L//T =`Constant
    B
    `L^(2)//T^(2) =` Constant
    C
    `L//T^(2) =`Constant
    D
    `LT =`Constant

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