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The time period of a simple pendulum is ...

The time period of a simple pendulum is 2s. It its length is increased by 4 times, then its period becomes

A

16 s

B

12 s

C

8 s

D

4 s

Text Solution

Verified by Experts

The correct Answer is:
D
As, `Tpropsqrt(l)implies(T_(1))/(T_(2))=sqrt((l_(1))/(l_(2)))`
`implies(2)/(T_(2))=sqrt((l)/(4l))impliesT_(2)=4s`
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Knowledge Check

  • Time period of simple pendulum is

    A
    `T = 2pi sqrt(l/g)`
    B
    `T = 2pi sqrt(g/l)`
    C
    `T = sqrt(2 pi (l)/(g))`
    D
    `T = (1)/(2pi) sqrt(g/l)`

  • The time period of simple pendulum is T. If its length is increased by 2%, the new time period becomes

    A
    `0.98T`
    B
    `1.02T`
    C
    `0.99T`
    D
    `1.01T`

  • The time period of a simple pendulum of infinte length is

    A
    infinite
    B
    `2pisqrt(R//g)`
    C
    `2pi sqrt(g//R)`
    D
    `(1)/(2pi) sqrt(R//g)`

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