Home
Class 7
PHYSICS
T=2pisqrt(l/g) is the time period of a s...

`T=2pisqrt(l/g)` is the time period of a simple pendulum, then the unit of `4pi^(2)l/T^(2)` in SI system is ______ .

A

`ms^(-1)`

B

`s^(-2)`

C

`ms^(-2)`

D

`s^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`g=4pi^(2)l/T^(2)`
Its unit is `m/s^(2)`. That is m `s^(-2)`.
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS

    PEARSON IIT JEE FOUNDATION|Exercise ASSESSMENT TEST (TEST 1 (SELECT THE CORRECT ALTERNATIVE)|14 Videos

  • HEAT

    PEARSON IIT JEE FOUNDATION|Exercise Test 2|14 Videos

  • LIGHT

    PEARSON IIT JEE FOUNDATION|Exercise ASSESSMENT TEST (TEST 2 (SELECT THE CORRECT ALTERNATIVE))|15 Videos

Similar Questions

Explore conceptually related problems

If T = 2pi sqrt(l/g) is the time period of a simple pendulu, then the unit of 4pi^(2) l/T^(2) in the SI system is ________ .

How does time period (T) of a seconds pendulum vary with length(l)?

If T= 2pisqrt(1/8) , then relative errors in T and l are in the ratio

The graph of l against T for a simple pendulum is

Show that the expression of the time period T of a simple pendulum of length l given by T = 2pi sqrt((l)/(g)) is dimensionally currect

The time period of a pendulum of length 'l' and mass of the bob 'm' is T. Time period of a pendulum with mass of the bob 2m and length l is ______.

The time period of a simple pendulum on the surface of the Earth is T_1 and that on the planet Mars is T_2 . If T_2 < T_1 , then acceleration due to gravity on Mars is

The time period 'T' of a simple pendulum of length 'l' is given by T = 2pisqrt(l/g) .Find the percentage error in the value of 'T' if the percentage error in the value of 'l' is 2%.