Answer:

Step-by-step explanation:
Given that,
Mass of the object, m = 29 kg
Torsion constant of the wire, K = 1.14 N-m
Number of cycles, n = 98
Time, t = 74 s
To find,
The rotational inertia of the object.
Solution,
Relationship between the moment of inertia, time period and the torsion constant of the spring is given by :

Where I is the moment of inertia
K is spring constant
Let T Is the time period of oscillation, such that,




So, the rotational inertia of the object is
.