Home
Class 11
PHYSICS
Show that the expression of the time per...

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

Text Solution

Verified by Experts

` T = 2pi sqrt((l)/(g)) `
Dimensionally `[T] = sqrt(([L])/([LT^(-2)])) = [T]`
As in the above equation the dimensions of both sides are same . The given formula Is dimensionally correct .
Promotional Banner

Topper's Solved these Questions

  • UNIT AND DIMENSIONS

    DC PANDEY|Exercise Single Correct|20 Videos
  • UNIT AND DIMENSIONS

    DC PANDEY|Exercise More Than One Correct|5 Videos
  • THERMOMETRY,THERMAL EXPANSION & KINETIC THEORY OF GASES

    DC PANDEY|Exercise Level 2 Subjective|9 Videos
  • UNITS, DIMENSIONS & ERROR ANALYSIS

    DC PANDEY|Exercise Medical entrances gallery|32 Videos

Similar Questions

Explore conceptually related problems

The graph of time period (T) of a simple pendulum versus its length (I) is

Show that the time period of a simplw pendulam of infinite length is given by t=2pi sqrt((R)/(g)) where R is the radius of the earth

The periodic time (T) of a simple pendulum of length (L) is given by T=2pisqrt((L)/(g)) . What is the dimensional formula of Tsqrt((g)/(L)) ?

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

Time period T of a simple pendulum of length l is given by T = 2pisqrt((l)/(g)) . If the length is increased by 2% then an approximate change in the time period 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%.

If the time period of a simple pendulum is T = 2pi sqrt(l//g) , then the fractional error in acceleration due to gravity is

The time T of oscillation of a simple pendulum of length l is given by T=2pi. sqrt((l)/(g)) . Find the percentage error in T corresponding to (i) on increase of 2% in the value of l. (ii) decrease of 2% in the value of l.