Question

if length of a rod l is 4.0+-0.m and it's mass m is 5.0 plus or minus 0.1 kg then the maximum percentage error in measurement of moment of inertia is ​

Answer 1

Answer:

3.898%

Explanation:

Moment of Inertia of a rod of mass m and lenght l is $\frac{ml^{2} }{12}$

given , m = 5±0.1 kg

l = 4±0 m

max possible length = 4 + 0 = l m

min possible length = 4-0= 4 m

max possible mass = 5+0.1 = 5.1 kg

min. possible mass = 5 -0.1 = 4.9 kg

max moment of Inertia =  1/12 x 5.1 x 4² = 6.8 kgm²

min moment of inertia = 1/12 x 4.9 x 4² = 6.54kgm²

max .error possible = max MI - min MI = 0.26 kgm²

Avg Moment of Inertia, not considering deviation = 1/12 x 5 x 4² = 6.67 kgm²

error % = (error/avg MI) × 100

= (0.26/6.67) x 100

= 3.898 %