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The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is `3.5s ?` Take `g=9.8ms^(-2)` on the surface of the earth.

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Step by step text solution for The acceleration due to gravity on the surface of the moon is 1.7ms^(-2). What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is 3.5s ? Take g=9.8ms^(-2) on the surface of the earth. by PHYSICS experts to help you in doubts & scoring excellent marks in Class 11 exams.

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The acceleration due to gravity on the surface of the moon is 1.7ms^(-2) . What is the time period of as simple pendulum on the moon, if its time period on the earth is 3.5s ? [g = 9.8ms^(-2)]

A simple pendulum is set up at an unknown planet.The acceleration due to gravity on the surface is 6.2ms^(-2) .Find the time period of the pendulum on the surface of the planet,if its time period on the surface of the earth is 3.5s ,Take g =9.8ms^(-2)

Knowledge Check

  • The time period of a simple pendulum on the surface of the earth is 4s. Its time period on the surface of the moon is

    A
    4s
    B
    8s
    C
    10s
    D
    12s

  • The value of acceleration due to gravity on the surface of moon is__________ m s^(-2)

    A
    ` 274.1`
    B
    ` 0.610`
    C
    ` 9.81`
    D
    ` 1.625`

  • If acceleration due to gravity on the moon is one-sixth that on the earth, then the length and time period of seconds pendulum on the surface of moon are

    A
    6 m, 1.5 s
    B
    1/6 m, 2 s
    C
    1/6 m, 0.5 s
    D
    6 m, 2 s

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