Question

15. A student measures the distance traversed in free
fall of a body, initially at rest in a given time. He
uses this data to estimate g, the acceleration due
to gravity. If the maximum percentage errors in
measurement of the distance and the time are
e1 and e2 respectively, the maximum percentage
error in the estimation of g is
(1) e2-e1
(2) e1+2e2
(3) e1+e2
(4) e1-2e2​

Answer 1

Answer:

e1+2e2

Explanation:

s=1\2gt^2

s/t^2=g

Delta S/s×100+2 Delta t/t=Delta g/g ×100

e1+2e2

Answer 2

Hello Dear,

◆ Answer -

(2) e1 + 2e2

◆ Explaination -

According to Newton's 2nd eqn of kinematics -

s = ut + 1/2 at²

Here, u = 0 & a = g

s = 1/2 gt²

g = 2s/t²

So percentage error in measurement of g is -

∆g/g % = ∆s/s % + 2 ∆t/t %

∆g/g % = e1 + 2e2

Hope this helps you...