NEET System of Particles and Rotational Motion Previous Year Questions with Solutions

Apply the parallel axis theorem to find total moment of inertia I_A:

• • Sphere A rotates about its diametrical centroidal axis: I_{A,cm} = (2/5) M R^2.
• • Sphere B's center is at distance d = R + r from axis A: I_{B,about A} = (2/5) m r^2 + m (R + r)^2.

Similarly, apply the parallel axis theorem about vertical axis B:

• • Sphere B rotates about its diametrical centroidal axis: I_{B,cm} = (2/5) m r^2.
• • Sphere A's center is at distance d = R + r from axis B: I_{A,about B} = (2/5) M R^2 + M (R + r)^2.

Subtract equation 2 from equation 1:

• • The centroidal terms I_{A,cm} and I_{B,cm} cancel out identically.
• • I_A - I_B = m(R + r)^2 - M(R + r)^2 = (m - M)(R + r)^2.
• • Matches Option (2) exactly.

For any rigid body of mass M, the total angular momentum about an arbitrary reference point P is given by the sum of spin and orbital angular momentum:

• • \vec{L}_{\text{cm}} = I_{\text{cm}} \vec{\omega} is the intrinsic spin angular momentum about the center of mass.
• • \vec{r}_{\text{cm}/P} \times (M \vec{v}_{\text{cm}}) is the orbital angular momentum of the center of mass relative to point P.

The problem states that the disc is rotating about a fixed vertical axis passing through its centre O:

• • Because the center of mass does not translate, orbital angular momentum vanishes identically for any reference point.

Substitute the result for points A and B:

• • The distance OB = 2 * OA is completely irrelevant.
• • Matches Option (3) exactly.

Since the Sun expands solely due to internal forces with zero external torque acting on it, its total angular momentum is strictly conserved.

• • Angular speed is related to the period of rotation by ω = 2π / T.

For a solid spherical body of uniform density and constant mass M, the moment of inertia is proportional to the square of its radius:

• • Given R₂ / R₁ = 2.

Substitute the given expansion factor and initial period:

• • Matches Option (4) exactly.

Four forces act on the uniform rod in equilibrium: (1) Normal reaction at the smooth wall N_w acting horizontally to the right. (2) Static friction force f from the rough floor acting horizontally to the left. (3) Normal reaction from the floor N_f acting vertically upward. (4) Weight of the rod W = mg = 20 × 10 = 200 N acting downward at the midpoint.

• • The wall is smooth, so there is no vertical friction at the wall.

Let the rod have length L. The angle with the vertical wall is θ = 60°. The top of the rod is at height h = L cos 60°, and the center of mass is at horizontal distance x_cm = (L/2) sin 60° from the base. Taking torque about the base point on the floor:

• • N_w = (mg / 2) tan 60°.

Substitute the given values mg = 200 N and tan 60° = √3:

• • Thus the friction force exerted by the floor on the rod is 100√3 N, matching Option (2).

For uniform density ρ, mass is proportional to the cube of the radius:

• • The smaller sphere has mass m, and the original large sphere has mass 8m.

The Y-axis passes through the center of mass of the large sphere:

• • This is the total moment of inertia before cutting.

The small sphere has its center of mass at x = R, y = 0. By the parallel-axis theorem:

• • Distance d = R between the small sphere’s centroid and the Y-axis.

Subtract the small sphere’s moment of inertia from the total to find the remaining part:

• • Ratio of smaller sphere to rest part:
• • \frac{I_{small}}{I_{rest}} = \frac{\frac{7}{5} m R^2}{\frac{57}{5} m R^2} = \frac{7}{57}.
• • Matches Option (3) with absolute mathematical rigor.

For any rigid body undergoing pure planar rolling without slipping on a stationary surface with linear velocity v and angular velocity ω = v/R, the total velocity of any point on the rim is the vector sum of the translational velocity of the center of mass (v_cm) and the tangential rotational velocity (v_rot = ωR):

At the uppermost point P, the translational velocity is forward (+v î) and the rotational velocity is also in the forward direction (+ωR î = +v î). Adding both vectors yields:

At the bottom contact point Q, the translational velocity is directed forward (+v î), whereas the rotational tangential velocity is directed backward (-ωR î = -v î). Under the condition of pure rolling without slipping, these two components cancel each other out identically:

Comparing the speeds of the two points: v_P = 2v, while v_Q = 0. Therefore, point P moves significantly faster than point Q, matching Option (4).

For a uniform thin straight rod of mass M and length L, the moment of inertia about an axis perpendicular to the rod and passing through its center of mass (midpoint) is given by integration as:

Substituting the given values I = 2400 g cm² and M = 400 g directly into the formula (in CGS units):

Simplifying the equation:

Taking the square root of 72:

Angular acceleration α⃗ is defined as the time rate of change of angular velocity ω⃗:

Angular velocity ω⃗ is an axial vector perpendicular to the orbital plane of rotation, pointing along the axis of rotation by the right-hand screw rule. Since α⃗ is the derivative of ω⃗ with respect to scalar time t, it must be parallel (if speeding up) or antiparallel (if slowing down) to ω⃗:

In circular motion, the linear acceleration consists of centripetal acceleration (radially inwards) and tangential acceleration (tangent to the circle), whereas angular acceleration governs rotational rate and is purely axial:

The radius of gyration k of a rigid body of mass M about a specified axis of rotation is the effective radial distance at which the entire mass could be concentrated without altering its moment of inertia:

For a uniform solid sphere of mass M and radius R about its own diametrical axis:

For a thin hollow spherical shell of identical mass M and radius R about its diametrical axis:

Evaluating the ratio k₁ / k₂:

None of the four printed options contained the square root sign. The closest intended option was Option (2) 3:5, which is the ratio of k₁² : k₂² (or I₁ : I₂). Because of this clerical typographical omission, the National Testing Agency (NTA) dropped Question 29 and awarded full marks to all candidates.

For a uniform rod of length $L = 200\text{ cm}$ and mass $M = 500\text{ g} = 0.5\text{ kg}$: - Center of mass of the rod is at its geometric center: $x_{\text{cm}} = 100\text{ cm}$. The gravitational force of the rod $M g = 0.5 g$ acts downward at this point. - The wedge (fulcrum) is located at $x_w = 40\text{ cm}$. - A mass $m_1 = 2\text{ kg}$ is suspended at $x_1 = 20\text{ cm}$. - An unknown mass $m$ is suspended at $x_2 = 160\text{ cm}$.

Measuring perpendicular distances from the fulcrum ($x_w = 40\text{ cm}$): - For $m_1 = 2\text{ kg}$: $d_1 = 40 - 20 = 20\text{ cm}$ (to the left, producing CCW torque) - For the rod weight: $d_{\text{rod}} = 100 - 40 = 60\text{ cm}$ (to the right, producing CW torque) - For unknown mass $m$: $d_2 = 160 - 40 = 120\text{ cm}$ (to the right, producing CW torque)

For the rod to remain in rotational equilibrium, the sum of counter-clockwise moments must equal the sum of clockwise moments about the wedge:

Cancel $g$ and substitute numerical values:

For a uniform circular ring of mass $M$, the linear mass density is $\lambda = \frac{M}{2\pi R}$. A complete ring spans $360^\circ$. The fraction of the ring removed by an arc of $90^\circ$ is:

The axis of rotation passes through the geometric center of the original ring and is perpendicular to its plane. For every infinitesimal mass element $dm$ on the circular arc, its perpendicular distance from this axis is identically equal to the radius $R$:

Substitute $M_{\text{remaining}} = \frac{3}{4}M$:

Place the origin ($$x = 0$$) at the position of the 5 kg particle ($$m_1 = 5\text{ kg}$$). The rigid light rod extends along the x-axis, placing the 10 kg particle ($$m_2 = 10\text{ kg}$$) at coordinate $$x = L = 1\text{ m} = 100\text{ cm}$$.

For a two-particle discrete mass system along a straight line: $$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

$$x_{\text{cm}} = \frac{(5\text{ kg} \times 0) + (10\text{ kg} \times 100\text{ cm})}{5\text{ kg} + 10\text{ kg}}$$ $$x_{\text{cm}} = \frac{1000\text{ kg}\cdot\text{cm}}{15\text{ kg}} = \frac{200}{3}\text{ cm} = 66.666...\text{ cm}$$

Rounding to the nearest whole integer gives $$67\text{ cm}$$, matching Option (3).

The torque (moment of force) $$\vec{\tau}$$ acting on a particle having position vector $$\vec{r}$$ with respect to the origin is defined as the cross product: $$\vec{\tau} = \vec{r} \times \vec{F}$$ Note that the cross product is anti-commutative ($$\vec{r} \times \vec{F} = -(\vec{F} \times \vec{r})$$), so the order of vectors is critical.

Given: $$\vec{r} = 2\hat{k} \text{ m}$$ $$\vec{F} = 3\hat{j} \text{ N}$$

Recall the right-handed Cartesian cyclic cross product relations: - $$\hat{i} \times \hat{j} = \hat{k}$$ - $$\hat{j} \times \hat{k} = \hat{i}$$ - $$\hat{k} \times \hat{i} = \hat{j}$$ In reverse (counter-cyclic) order: - $$\hat{k} \times \hat{j} = -\hat{i}$$ Therefore: $$\vec{\tau} = (2\hat{k}) \times (3\hat{j}) = (2 \times 3)(\hat{k} \times \hat{j}) = 6(-\hat{i}) = -6\hat{i} \text{ N m}$$

$$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 2 \\ 0 & 3 & 0 \end{vmatrix} = \hat{i}(0 - 6) - \hat{j}(0 - 0) + \hat{k}(0 - 0) = -6\hat{i}\text{ N m}$$ This confirms Option (3).