Part CCSIR NET June 2024the-jacobian-adds-a-power-of-r

The jacobian adds a power of r

Consider a solid torus of constant density $\rho$ , formed by revolving the disc $(y - b)^{2} + z^{2} \le a^{2}, x = 0$ about the z-axis, where 0 < a < b. Then the moment of inertia of the solid torus about the z-axis is

A. $2\pi^{2}a^{2}b^{2}(4b^{2} + 3a^{2})\rho$
B. $(\pi^{2}/2)a^{2}b(4b^{2} + 3a^{2})\rho$ ✓ Correct
C. $(\pi^{2}/2)a^{2}b(4a^{2} + 3b^{2})\rho$
D. $2\pi^{2}a^{2}b^{2}(4a^{2} + 3b^{2})\rho$

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

The jacobian adds a power of r

More on this topic