Question

the m.i of a disc is about 2 units. its m.i. about axis through a point on its rim and in the plane of the disc is

Answer 1

Moment of inertia I of the disc = MR²/2 = 2 units.
(about an axis through its center and perpendicular to its plane)

Moment of inertia I2 of the disc about axis through a point on the rim and perpendicular to the plane:
     = I + M R² = 3/2 M R² = 6 units.

MOI of the disc about its diameter = I /2 = MR² / 4

MOI about axis through a point on its rim (axis being the tangent to the disc) 
   = MR²/4 + M R²  = 5 M R² / 4

Answer 2

⛦Hᴇʀᴇ Is Yoᴜʀ Aɴsᴡᴇʀ⚑
▬▬▬▬▬▬▬▬▬▬▬▬☟

▶Let ‘r’ be the radius of the disc.

♦ Given,

➧ moment of inertia of the disc about PQ, IPQ
➾ 2 units

➧ Moment of inertia of the disc about AB, IAB
➾ ½ Mr²


➧ We need the MI of the disc about RS or IRS.

♦ Now

➧ From theorem of perpendicular axes,

IAB = 2IPQ

➾ ½ Mr² = 2

➾ Mr² = 4

➧ Now, by theorem of parallel axes
IRS

➾ IPQ + Mr²
➾ 2 + 4
➾ 6 units ...✔

_________
Thanks...✊