JEE Main 2025 · 7 April · morning · Numerical answer · +4 / −1
A, B and C are disc, solid sphere and spherical shell respectively with same radii and masses. These masses are placed as shown in figure.
The moment of inertia of the given system about PQ axis is x/15 I, where I is the moment of inertia of the disc about its diameter. The value of x is ____________.
Answer and solution
Answer 199
Solution
Physics sir · 3 steps
Write everything in terms of MR² = 4I, and add MR² for B and C because their centres sit R from PQ.
- Set the unit. For a disc about its diameter, I = 1/4MR², so MR² = 4I. The axis PQ is a diameter of disc A, so I_A = I.
- Parallel axis. B and C touch PQ at their edges, so each centre is a distance R from the axis. Each gets its own centre-of-mass moment of inertia plus MR².
Body About its own diameter Distance from PQ About PQ A, disc 1/4MR² 0 I B, solid sphere 2/5MR² R 7/5MR² = 28/5I C, spherical shell 2/3MR² R 5/3MR² = 20/3I I_(PQ) = I + 28/5I + 20/3I = (15 + 84 + 100)/15I = 199/15I
- So. Comparing with x/15I gives x = 199.
The common slip: 214. Taking the disc's moment of inertia about PQ as 1/2MR² = 2I gives 214. PQ lies along a diameter of the disc, not through its centre perpendicular to its face, so the disc contributes only I.
The idea to keep. When the answer must be in terms of one body's moment of inertia, first express MR² in that unit. Then use the parallel axis theorem for every body whose centre is off the axis.
Answer from the NTA final answer key. Solution by PrepWiser.