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How is the length of seconds pendulum r...

How is the length of seconds pendulum related with acceleration due to gravity of any planet?

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To find the relationship between the length of a second pendulum and the acceleration due to gravity on any planet, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Pendulum**: - A second pendulum is defined as a pendulum that has a time period of 2 seconds. This means it takes 2 seconds to complete one full oscillation. 2. **Formula for Time Period**: ...
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If the length of a seconds pendulum on a planet is 2 m then the acceleration due to gravity on the surface of that planet is "______" (Take the acceleration due to gravity on the surface of the Earth =9.8 m s^(-2) )

How time period of a simple pendulum depends on the acceleration due to gravity ?

Knowledge Check

  • Given that T stands for time and l stands for the length of simple pendulum . If g is the acceleration due to gravity , then which of the following statements about the relation T^(2) = ( l// g) is correct?

    A
    It is correct both dimensionally as well as numerically.
    B
    It is neither dimensionally correct nor numerically.
    C
    It is dimensionally correct but not numerically.
    D
    It is numerically correct but not dimensionally.

  • If 'L' is length of sample pendulum and 'g' is acceleration due to gravity then the dimensional formula for ((l)/(g))^(1/2) is same that for

    A
    frequency
    B
    velocity
    C
    time period
    D
    wavelength

    • Similar Questions

    Explore conceptually related problems

    The time period of a simple pendulum is given by T = 2pi sqrt((l)/(g)) where 'l' is the length of the pendulum and 'g' is the acceleration due to gravity at that place. (a) Express 'g' in terms of T (b) What is the length of the seconds pendulum ? (Take g = 10 m s^(-2) )

    Calculate the length of the second's pendulum on the surface of moon where acceleration due to gravity is one sixth as that on earth.

    The time period of a pendulum on the surface of the moon is 5 s. If it is a seconds pendulum on earth and the acceleration due to gravity on the Earth is 9.8 m s^(-2) , find the acceleration due to gravity on the surface of the moon.

    The time period of a pendulum on the surfacce of the moon is 5s. If it is a seconds pendulum on earth and the acceleratio due to gravity on the earth is 9.8 m s^(-2) , find the acceleration due to gravity on the surface of the moon.

    A simple pendulum designed on the moon as a seconds pendulum is taken to a planet where the acceleration due to gravity on the surface is twice that on the Earth . If g_("earth") : g_("moon") =6 :1 find the period of oscillation of the pendulum on the planet mentionedabove.

    If 'L' is length of simple pendulum and 'g' is acceleration due to gravity then the dimensional formula for (l/g)^(1/2) is same as that for

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