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The time period of a simple pendulum is ...

The time period of a simple pendulum is 4 s and the acceleration due to gravity at the given place is ` 9.8 ms ^(-2)` . Write the following steps in sequential order to find the length of the pendulum.
(A) From the above formula, write the value of l and `l=(T^(2).g)/(4pi^(2))`
(B) Note the given values of the time period (T) and acceleration due to gravity (g).
(C)Write the formula, ` T = 2pi sqrt(l/g)` , where l is the length of the pendulum.
(D). Substitute the given data and get the value of `l`.

A

CDBA

B

ABCD

C

BCAD

D

ADBC

Text Solution

Verified by Experts

The correct Answer is:
C
(B) Note the given values of time period (T) and acceleration due to gravity (g).
( C) Write the formula, `T=2pisqrt(l/g)` where .l. is the length of the pendulum.
(A) From the above formula, write the value of l as `l=(T^(2)g)/(4pi^(2))`.
(D) Substitute the given data and get the value of .l..
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Knowledge Check

  • If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

    A
    `pi^(2)ms^(-2)`
    B
    `9.8ms^(-2)`
    C
    `2pi^(2) ms^(-2)`
    D
    `16 m//s^(-2)`

  • The period of oscillation of a simple pendulum at a given place with acceleration due to gravity 'g' depends on

    A
    the length of the pendulum 'l' only
    B
    both 'l' and 'm' of the pendululm
    C
    mass of the bob of the pendulum only
    D
    l' and 'g'

  • If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

    A
    `(4 pi^(2)Deltal)/(DeltaT^(2))`
    B
    `(Deltal)/(l) - 2(Deltal)/(T)`
    C
    `(Delta l)/(l)+2(Delta T)/(T)`
    D
    None of these

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