Home
Class 8
PHYSICS
The time period of a pendulum on the sur...

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.

Text Solution

Verified by Experts

`T alpha sqrt((1)/(g))`
`(T_(E))/(T_(M)) = sqrt((g_(M))/(g_(E)))`
`(2)/(5) = sqrt((g_(M))/(g))`
`(4)/(25) = (g_(M))/(g)`
`g_(M) = (4)/(25) xx 9.8 ( :' g = 9.8 ms^(-2))`
`= 1.6 s^(-2)`
Promotional Banner

Topper's Solved these Questions

  • WAVE MOTION AND SOUND

    PEARSON IIT JEE FOUNDATION|Exercise TEST YOUR CONCEPTS (VERY SHORT ANSWER TYPE QUESTIONS )|19 Videos

  • WAVE MOTION AND SOUND

    PEARSON IIT JEE FOUNDATION|Exercise TEST YOUR CONCEPTS (SHORT ANSWER TYPE QUESTIONS )|7 Videos

  • WAVE MOTION AND SOUND

    PEARSON IIT JEE FOUNDATION|Exercise Level 2|20 Videos

  • STARS AND THE SOLAR SYSTEM

    PEARSON IIT JEE FOUNDATION|Exercise COMPETITION CORNER|37 Videos

Similar Questions

Explore conceptually related problems

The acceleration due to gravity near the surface of moon is-

If g_(e) is acceleration due to gravity on earth and g_(m) is acceleration due to gravity on moon, then

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) )

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

Given, acceleration due to gravity on surface of earth is g. if g' is acceleration due to gravity at a height h above the surface of earth, then

A simple pendulum makes 20 oscillations in one second on the surface of the earth. Determine the time period of the simple pendulum on the surface of a planet where the acceleration due to gravity is one fourth of the acceleration due to gravity on the surface of the earth .