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The period of oscillation of a simple pe...

The period of oscillation of a simple pendulum is `T = 2pi sqrt((L)/(g))`. Meaured value of L is `20.0 cm` known of 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wire watch of 1 s resolution. The accuracy in the determination of g is :

Text Solution

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The dimensional of LHS = the dimensional of T = `[M^(0)L^(0) T^(1)]`
The dimensions of RHS = `(("dimensions of length")/("dimensions of accelleration"))^(1//2)` `(because 2pi ` is a dimensionless constant )
`" "= [ (L)/(LT^(-2))]^(1//2) = [T^(2)] ^(1//2) = [T] = [M^(0) L^(0) T^(1)]`
Since the dimensions are same one both the sides, the relation is correct.
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