Question

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 ________ .

Answer 1

Answer:

Step by step explanation:

Time period (T) is the time taken by a pendulm in completing one oscillation.

In this question the relation between time period of a pendulum with the length (l) and gravity or gravitational acceleration (g) is given:

                                                            (equation 1)

Here LHS = T

        RHS =

The dimention of Time period (T) =  

So, the dimension of LHS =           (equation 2)

The dimension of Length of pendulum (l) =

The dimention of gravity (g) =

2π is constant so it is dimension less =

So the dimention of RHS

                        (equation 3)

By comparing equation 2 and 3.

LHS = RHS

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Answer 2

${\huge{\mathfrak\red{hy \:mate}}}$

${\huge{\purple{ answer :}}}$

Time period (T) is the time taken by a pendulm in completing one oscillation.

In this question the relation between time period of a pendulum with the length (l) and gravity or gravitational acceleration (g) is given:

                                                            (equation 1)

Here LHS = T

        RHS =

The dimention of Time period (T) =  

So, the dimension of LHS =           (equation 2)

The dimension of Length of pendulum (l) =

The dimention of gravity (g) =

2π is constant so it is dimension less =

So the dimention of RHS

                        (equation 3)

By comparing equation 2 and 3.

LHS = RHS

_______________________________

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