System of Particles and Rotational Motion: NEET previous year questions

With uniform angular acceleration the number of revolutions equals the average angular speed times the time: $\frac{600+1200}{2} = 900$ revolutions per minute (rpm) over $\frac{10}{60}$ min = $\frac{1}{6}$ min, giving $900 \times \frac{1}{6} = 150$ revolutions, option (3). Equivalently, using $\omega_2^2 = \omega_1^2 + 2\alpha\theta$ gives $\theta = 300\pi$ rad, so $\theta/2\pi = 150$.

The ring's mass is $M = mL$ and its radius $r = L/2\pi$ (since $L = 2\pi r$). The axis $yy'$ is tangential to the ring in its plane, so by the parallel axis theorem $I_{yy'} = I_{CM} + Mr^2 = \tfrac{1}{2}Mr^2 + Mr^2 = \tfrac{3}{2}Mr^2 = \tfrac{3}{2}(mL)\left(\tfrac{L}{2\pi}\right)^2 = \dfrac{3mL^3}{8\pi^2}$. The correct expression has $L^3$ and $\pi^2$, eliminating the other options.

Why the other options are wrong:

- (1) Uses $\pi$ instead of $\pi$$^{2}$ in the denominator.
- (3) Wrong power of L and $\pi$.
- (4) Wrong power of L.

By symmetry the two particles sit at mirror-image angles about the vertical, so their tangential velocities have equal and opposite horizontal components while their vertical components add. Writing the particles at $\theta$ and $\pi - \theta$, the resultant works out to $P = 2m\omega r|\cos\theta|$, which is zero at the start ($\theta = \pi/2$), peaks when the particles are diametrically opposite the meeting point ($\theta = 0$), and returns to zero as they collide at $B$. That single smooth hump is option (3). NCERT Class 11, System of Particles and Rotational Motion (motion of the centre of mass).

Why the other options are wrong:

- (1) assumes momentum is conserved as a constant because the SPEEDS are constant, forgetting that momentum is a vector whose direction keeps changing
- (2) inverts the phase - takes $P \propto |\sin\theta|$, i.e. maximum where the particles start rather than where they are farthest apart
- (4) assumes $P$ falls off linearly with angle instead of following the cosine of the half-separation

The spheres touch, so their centres are separated by $d = R + r$. About the axis through $A$: $I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m d^2$; about the axis through $B$: $I_B = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + M d^2$. The two spin terms are common to both and cancel, leaving $I_A - I_B = (m - M)(R + r)^2$, a negative quantity as expected since the heavier sphere sits off-axis in the second case. NCERT Class 11, System of Particles and Rotational Motion (parallel axis theorem).

Why the other options are wrong:

- (1) gets the separation right but flips the sign - attaches the parallel-axis term to the wrong body in each case
- (3) uses $R - r$ as the centre separation, as if the small sphere were embedded in the large one rather than attached externally
- (4) assumes the moment of inertia is independent of the choice of axis, ignoring the parallel axis theorem altogether

The angular momentum of a rigid body about any point splits as $\vec{L} = \vec{L}_{cm} + \vec{r} \times M\vec{v}_{cm}$. Here the centre is FIXED, so $\vec{v}_{cm} = 0$ and the orbital term vanishes for every reference point. Only the spin term $I_{cm}\omega$ survives, which is the same whether you compute about $A$ or $B$, so the ratio is 1 regardless of the distances. NCERT Class 11, System of Particles and Rotational Motion (angular momentum of a rotating rigid body).

Why the other options are wrong:

- (1) assumes $L \propto 1/r^2$, importing an inverse-square intuition that does not apply
- (2) assumes $L$ scales inversely with the distance to the reference point
- (4) assumes $L$ grows in proportion to the distance, as it would for a single particle moving in a straight line

Torque $\tau$ = r × F = (2k̂) × (3ĵ) = 6(k̂ × ĵ) = -6î N m. Chapter: System of Particles and Rotational Motion.

The center of mass $x_{CM}$ is given by $x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Substituting, $x_{CM} = \frac{5 \times 0 + 10 \times 100}{5 + 10} = \frac{1000}{15} \approx 67\ \mathrm{cm}$. Chapter: System of Particles and Rotational Motion.