Question

What is the fractional error in g calculated from T = 2 pie square root of l/g? Given that fractional errors in T and l are +-x
nd +-y respectively
I) x+ y
(2)x-y
(3) 2x + y
(4) 2x-y
pls answer fast with explanation...

Answer 1

Answer:

2x-y

Explanation:

$T = 2\pi \sqrt{l/g}$

Apply log on both sides

⇒log T = log √l/g

            = 1/2 log l - 1/2 log g   (∵log aⁿ = n log a)              (2π is constant)

Derivate on both sides

⇒ 1/T ΔT = 1/2 (1/l *Δl + 1/g * Δg)          

(Note: For maximum errors, we should add even though there is a minus sign)

⇒ΔT/T =1/2 ( Δl/l + Δg/g)

Given ΔT/T = x and Δl/l = y

⇒x = 1/2 ( y + Δg/g)

⇒2x = y + Δg/g

∴Δg/g = 2x - y