Question

In a certain obervation we got,l = 23.2cm,r =1.32cm and time taken for 10 oscillations was 10.0 s. Find, maximum percentage error in determinaton of 'g'.

Answer 1

Given:

l = 23.2cm

r = 1.32cm

Number of oscillations = n = 10

Time taken for 10 oscillations = 10.0 sec

To Find:

The maximum percentage error in determination of 'g' , acceleration due to gravity.

Solution:

We can infer that the error in measuring length , radius and time are as follows:

• Δ l = 0.1cm
• Δr = 0.01cm
• Δt = 0.1 sec

Now considering radius of the bob,

Equation of 'g' will be,

• g -  $4\pi^{2} \frac{L }{T^{2} }$ = $4\pi ^{2} \frac{L + r}{(t/n)^{2} }$

Therefore, maximum percentage error in g will be,

• Δg/g x 100 =  (Δl + Δr )/ ( l+ r ) x100 + 2(Δt/t)x100
• Δg/g x 100 = {(0.1 + 0.01)/(23.2 + 1.32 ) + 2 ( 0.1/10)} x 100

• Δg/g x 100 = 2.4%

The maximum percentage error in determination of 'g' is 2.4%