System of Particles and Rotational Motion Previous Year Questions — JEE Main Physics

58 past-year System of Particles and Rotational Motion questions from JEE Main (Physics) — page 12.

Q48 (2026)

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is $\sqrt{n}$ cm. The value of n is ____
### Related Formula $$I = I_{\text{cm}} + md^2$$ $I = mk^2$ $$I_{\text{cm (solid sphere)}} = \frac{2}{5}mR^2$$ ### Core Logic Use the parallel axis theorem to find the moment of inertia about the new axis, then equate it to $mk^2$ to find the radius of gyration. ### Step 1: Applying Parallel Axis Theorem As per the given solution steps: $$mk^2 = \frac{2}{3}mR^2 + md^2$$ $$k^2 = \frac{2}{3}R^2 + d^2$$ ### Step 2: Substituting Values Given $R = 10 \mathrm{~cm}$ and $d = 15 \mathrm{~cm}$: $$k^2 = \frac{2}{3}(10)^2 + (15)^2$$ $$k^2 = \frac{200}{3} + 225$$ $$k^2 = \frac{2}{5}R^2 + d^2$$ $$k^2 = \frac{2}{5}(10)^2 + (15)^2$$ $$k^2 = 40 + 225 = 265$$ ### Step 3: Final Answer We are given $k = \sqrt{n} \Rightarrow k^2 = n$. $n = 265$ ### Pattern Recognition Whenever radius of gyration is asked, divide out the mass immediately. $k^2 = I_{\text{cm-factor}} R^2 + d^2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

Q37 (2026)

When the position vector $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ changes sign as $-\vec{r}$ , which one of the following vector will not flip under sign change?
  1. $\text{Linear momentum}$
  2. $\text{Velocity}$
  3. $\text{Acceleration}$
  4. $\text{Angular momentum}$
### Related Formula $$\vec{v} = \frac{d\vec{r}}{dt}$$ $$\vec{p} = m\vec{v}$$ $$\vec{a} = \frac{d\vec{v}}{dt}$$ $$\vec{L} = \vec{r} \times \vec{p}$$ ### Core Logic Under a sign change of coordinates (parity transformation or spatial inversion) $\vec{r} \rightarrow -\vec{r}$. Then, velocity $\vec{v} = \frac{d\vec{r}}{dt} \rightarrow -\vec{v}$. Linear momentum $\vec{p} = m\vec{v} \rightarrow -\vec{p}$. Acceleration $\vec{a} = \frac{d\vec{v}}{dt} \rightarrow -\vec{a}$. Angular momentum $\vec{L} = \vec{r} \times \vec{p}$. Under the transformation: $\vec{L}' = (-\vec{r}) \times (-\vec{p}) = \vec{r} \times \vec{p} = \vec{L}$. ### Step 1: Final Conclusion Since both position and momentum vectors change sign, their cross product (angular momentum) retains its original sign. It does not flip. ### Pattern Recognition Angular momentum is a pseudovector (or axial vector). True vectors (polar vectors) flip signs under spatial inversion, but pseudovectors (which are cross products of two polar vectors) do not. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Systems of Particles and Rotational Motion

Q50 (2026)

A fly wheel having mass $3 \text{ kg}$ and radius $5 \text{ m}$ is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a $3 \text{ kg}$ mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by $3 \text{ m}$ is ____ J. ($g = 10 \text{ m/s}^{2}$ )
### Related Formula $$\Delta K.E. + \Delta P.E. = 0$$ $$K_{\text{wheel}} = \frac{1}{2} I \omega^2$$ $$I_{\text{disc}} = \frac{m R^2}{2}$$ ### Core Logic {{SOL_IMG1}} By conservation of mechanical energy, the loss in gravitational potential energy of the descending mass equals the gain in kinetic energy of the block and the rotational kinetic energy of the flywheel. $$mg h = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$$ ### Step 1: Relate Velocity and Angular Velocity Since the string does not slip, the linear velocity $v$ of the mass is related to the angular velocity $\omega$ of the wheel by: $$v = \omega R \Rightarrow \omega = \frac{v}{R}$$ ### Step 2: Energy Conservation The flywheel is treated as a solid disc/cylinder (from the $I = mR^2/2$ usage in the solution): $$mg h = \frac{1}{2} \left(\frac{M R^2}{2}\right) \omega^2 + \frac{1}{2} m v^2$$ Given $M = 3 \text{ kg}$ (wheel), $m = 3 \text{ kg}$ (block), $h = 3 \text{ m}$. Notice that the solution text specifies the flywheel mass as $m$ and block mass also as $m$, both being $3\text{ kg}$. Let's follow the PDF exactly: $$mg \times 3 = \frac{1}{2} \left(\frac{m R^2}{2}\right) \omega^2 + \frac{1}{2} m v^2$$ ### Step 3: Solve for Velocity Squared Substitute $\omega R = v$: $$mg \times 3 = \frac{1}{4} m v^2 + \frac{1}{2} m v^2$$ $$3mg = \frac{3}{4} m v^2$$ $$v^2 = 4g = 4 \times 10 = 40 \text{ (m/s)}^2$$ ### Step 4: Calculate Kinetic Energy of Flywheel $$K.E._{\text{wheel}} = \frac{1}{2} I \omega^2 = \frac{1}{4} m v^2$$ $$K.E._{\text{wheel}} = \frac{1}{4} \times 3 \times 40 = 30 \text{ J}$$ ### Pattern Recognition For a mass pulling a wheel of identical mass (disc), the total $K.E.$ is split between translational ($1/2 mv^2$) and rotational ($1/4 mv^2$). Rotational gets exactly $1/3$ of the total potential energy lost, $30\text{ J}$ out of $90\text{ J}$ total. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Systems of Particles and Rotational Motion Class 11 Physics: Work, Energy and Power
Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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