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. A symmetric system consisting of a disc (top), solid sphere (bottom-left), and spherical shell (bottom-right) arranged with vertical axis PQ. The moment of inertia of the given system about PQ is fracmathrmx15mathrmI,$\frac{\mathrm{x}}{15}\mathrm{I},$ where I is the moment of inertia of the disc about its diameter. The value of x is
Numerical Answer Type:
Enter a numerical valueAnswer: 199 to 199+4 marks
Solution & Explanation
### Related Formula
Parallel Axis Theorem:
I_textaxis = I_textcom + M R^2$$I_{\text{axis}} = I_{\text{com}} + M R^2$$
Standard Moments of Inertia about center of mass:
- Disc about diameter: I_textdisc,dia = fracMR^24$I_{\text{disc,dia}} = \frac{MR^2}{4}$
- Solid sphere: I_textsphere = frac25MR^2$I_{\text{sphere}} = \frac{2}{5}MR^2$
- Spherical shell: I_textshell = frac23MR^2$I_{\text{shell}} = \frac{2}{3}MR^2$
### Core Logic
The axis of rotation PQ passes through the center of the top disc (A) along its diameter.
- Top disc (A):
I_A = fracMR^24$$I_A = \frac{MR^2}{4}$$
- Bottom-left solid sphere (B): Center lies at distance R$R$ from the axis PQ.
I_B = I_textcom + M R^2 = frac25MR^2 + MR^2 = frac75MR^2$$I_B = I_{\text{com}} + M R^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$
- Bottom-right spherical shell (C): Center lies at distance R$R$ from the axis PQ.
I_C = I_textcom + M R^2 = frac23MR^2 + MR^2 = frac53MR^2$$I_C = I_{\text{com}} + M R^2 = \frac{2}{3}MR^2 + MR^2 = \frac{5}{3}MR^2$$
### Step 1: Calculate Total System Moment of Inertia
Sum the contributions:
I_textPQ = I_A + I_B + I_C$$I_{\text{PQ}} = I_A + I_B + I_C$$I_textPQ = fracMR^24 + frac75MR^2 + frac53MR^2$$I_{\text{PQ}} = \frac{MR^2}{4} + \frac{7}{5}MR^2 + \frac{5}{3}MR^2$$
To add the fractions, find a common denominator (60$60$):
I_textPQ = left( frac15 + 84 + 10060 right) MR^2 = frac19960 MR^2$$I_{\text{PQ}} = \left( \frac{15 + 84 + 100}{60} \right) MR^2 = \frac{199}{60} MR^2$$
### Step 2: Express in terms of standard Disc Moment
We are given I = fracMR^24 implies MR^2 = 4I$I = \frac{MR^2}{4} \implies MR^2 = 4I$. Substitute this in the expression:
I_textPQ = frac19960 (4I) = frac19915 I$$I_{\text{PQ}} = \frac{199}{60} (4I) = \frac{199}{15} I$$
Comparing with I_textPQ = fracx15 I$I_{\text{PQ}} = \frac{x}{15} I$ yields x = 199$x = 199$.
### Pattern Recognition
Sees: Composite body consisting of three standard symmetric shapes about a tangent/offset axis.
Shortcut: Sum the central inertia terms and the offset terms separately. Offset masses are only B and C, so the offset sum is 2MR^2$2MR^2$. The central sum is (1/4 + 2/5 + 2/3)MR^2$(1/4 + 2/5 + 2/3)MR^2$. Adding these directly yields the combined fractional factor of 199/60$199/60$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
A symmetric system consisting of a disc (top), solid sphere (bottom-left), and spherical shell (bottom-right) arranged with vertical axis PQ.
Keywords:#moment of inertia of disc sphere shell system#JEE Main 2025 Morning Q21#Rotational Dynamics JEE Main 2025#parallel axis theorem composite shapes#moment of inertia#parallel axis theorem#disc#solid sphere#spherical shell
More System of Particles and Rotational Motion Previous-Year Questions
Qjee_main_2026_21_jan_morningRigid Body Dynamics
A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ____. (g acceleration due to gravity)
A uniform rod suspended horizontally by two strings attached at its ends.
A.mg/2$mg/2$
B.mg/4$mg/4$
C.mg/3$mg/3$
D.mg$mg$
Solution
### Related Formula
tau = Ialpha$\tau = I\alpha$Sigma F_y = m a_CM, y$$\Sigma F_y = m a_{CM, y}$$a_CM, y = alpha fracl2$$a_{CM, y} = \alpha \frac{l}{2}$$
### Core Logic
Immediately after one string is cut, the rod starts rotating about the point where the remaining string is attached.
Taking torque about the end where the string is attached (this point has instantaneous acceleration but initially zero vertical velocity):
tau_textend = I_textend alpha$$\tau_{\text{end}} = I_{\text{end}} \alpha$$
Gravity provides the torque: tau = mg left(fracl2right)$\tau = mg \left(\frac{l}{2}\right)$.
### Step 1: Calculate Angular Acceleration
Moment of inertia about the end is I = fracml^23$I = \frac{ml^2}{3}$.
mg fracl2 = fracml^23 alpha$$mg \frac{l}{2} = \frac{ml^2}{3} \alpha$$alpha = frac3g2l$$\alpha = \frac{3g}{2l}$$A uniform rod suspended horizontally by two strings attached at its ends.
### Step 2: Calculate Force and Tension
The acceleration of the center of mass (CM) is downwards:
a_c = alpha fracl2 = left(frac3g2lright) left(fracl2right) = frac3g4$$a_{c} = \alpha \frac{l}{2} = \left(\frac{3g}{2l}\right) \left(\frac{l}{2}\right) = \frac{3g}{4}$$
Applying Newton's second law for translational motion of the CM in vertical direction:
mg - T = m a_c$$mg - T = m a_{c}$$T = mg - m a_c = mg - m left(frac3g4right) = fracmg4$$T = mg - m a_{c} = mg - m \left(\frac{3g}{4}\right) = \frac{mg}{4}$$
### Pattern Recognition
Classic 'cut string' rigid body problem. Always take torque about the pivot/hinge point to find alpha$\alpha$, then relate the center of mass linear acceleration a = r_cm alpha$a = r_{cm} \alpha$ to find the unknown tension using F_textnet = ma$F_{\text{net}} = ma$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Qjee_main_2026_21_jan_morningMoment of Inertia
Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is fracx2 ML^2text kg m^2$\frac{x}{2} ML^{2}\text{ kg m}^{2}$. The value of x is
An upside down T-shaped arrangement of two identical rods. Point P is at the end of the vertical rod.
Numerical Answer.Answer: 17 to 17
Solution
### Related Formula
I_textend = fracML^23$$I_{\text{end}} = \frac{ML^2}{3}$$I_textparallel axis = I_textcm + Md^2 = fracML^212 + Md^2$$I_{\text{parallel axis}} = I_{\text{cm}} + Md^2 = \frac{ML^2}{12} + Md^2$$
### Core Logic
Let the rods be Rod 1 (vertical, passing through P at its end) and Rod 2 (horizontal, attached at the other end of Rod 1).
An upside down T-shaped arrangement of two identical rods. Point P is at the end of the vertical rod.
For Rod 1 (length L$L$, mass M$M$):
The axis passes through its end perpendicular to its length.
I_1 = fracML^23$I_1 = \frac{ML^2}{3}$
For Rod 2 (length L$L$, mass M$M$):
The axis passes parallel to Rod 2's center of mass axis, at a distance L$L$ from it (since it's attached to the bottom end of Rod 1).
I_2 = I_textcm + M d^2 = fracML^212 + M(L)^2$I_2 = I_{\text{cm}} + M d^2 = \frac{ML^2}{12} + M(L)^2$
### Step 1: Total Moment of Inertia
I = I_1 + I_2 = fracML^23 + left(fracML^212 + ML^2right)$$I = I_1 + I_2 = \frac{ML^2}{3} + \left(\frac{ML^2}{12} + ML^2\right)$$I = frac4ML^2 + ML^2 + 12ML^212$$I = \frac{4ML^2 + ML^2 + 12ML^2}{12}$$I = frac1712 ML^2$$I = \frac{17}{12} ML^2$$
We are given that I = fracx12 ML^2$I = \frac{x}{12} ML^2$ (Correction from source PDF text: the source question text says fracx2 ML^2$\frac{x}{2} ML^2$, but the solution uses fracx12 ML^2$\frac{x}{12} ML^2$. Following the solution steps: x=17$x=17$ is consistent if the denominator is 12. Let's assume the question asked for fracx12$\frac{x}{12}$ or x=17/6$x=17/6$, but the official answer gives 17. Our output will state 17).
### Pattern Recognition
For composite shapes, calculate I$I$ for each simple shape separately about the desired axis using Parallel Axis Theorem, then sum them up.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Qjee_main_2025_02_april_eveningTorque and Moment of Inertia
A wheel of radius 0.2 mathrm~m$0.2 \mathrm{~m}$ rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 mathrm~N$10 \mathrm{~N}$ as shown in figure. The established torque produces an angular acceleration of 2 mathrmrad / mathrms^2$2 \mathrm{rad} / \mathrm{s}^{2}$ . Moment of inertia of the wheel is ________ kg m^2$m^2$ .
(Acceleration due to gravity = 10mathrmm / mathrms^2$= 10\mathrm{m} / \mathrm{s}^2$)
The diagram displays a circular wheel rotating about its center under a tangential pulling force of 10 N.
Numerical Answer.Answer: 1 to 1
Solution
### Related Formula
1. Torque (tau$\tau$) produced by a tangential pulling force:
tau = F cdot R$$\tau = F \cdot R$$
2. Newton's second law for rotation:
tau = I cdot alpha$$\tau = I \cdot \alpha$$
### Core Logic
We are given:
- Radius of the wheel R = 0.2 \ mathrmm$R = 0.2 \ \mathrm{m}$
- Applied force F = 10 \ mathrmN$F = 10 \ \mathrm{N}$
- Angular acceleration alpha = 2 \ mathrmrad/s^2$\alpha = 2 \ \mathrm{rad/s^2}$
### Step 1: Calculate torque and moment of inertia
First, find the torque tau$\tau$:
tau = F cdot R = 10 \ mathrmN times 0.2 \ mathrmm = 2 \ mathrmN cdot m$$\tau = F \cdot R = 10 \ \mathrm{N} \times 0.2 \ \mathrm{m} = 2 \ \mathrm{N \cdot m}$$
Next, calculate the moment of inertia I$I$ using tau = Ialpha$\tau = I\alpha$:
I = fractaualpha = frac2 \ mathrmN cdot m2 \ mathrmrad/s^2 = 1 \ mathrmkg cdot m^2$$I = \frac{\tau}{\alpha} = \frac{2 \ \mathrm{N \cdot m}}{2 \ \mathrm{rad/s^2}} = 1 \ \mathrm{kg \cdot m^2}$$
Thus, the moment of inertia is 1 mathrm~kgcdot m^2$1 \mathrm{~kg\cdot m^2}$.
### Pattern Recognition
Sees: Pulley/wheel torque with basic rotational dynamics.
Trap: Attempting to integrate gravity (g = 10text m/s^2$g = 10\text{ m/s}^2$) into mass equations. Gravity is a redundant distractor here because the pulling tension force is explicitly defined!
Shortcut: Directly compute torque as tau = F R$\tau = F R$ and divide by the angular acceleration alpha$\alpha$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q3jee_main_2025_02_april_eveningMoment of Inertia of Ring
The moment of inertia of a circular ring of mass M$M$ and diameter r$r$ about a tangential axis lying in the plane of the ring is:
A.frac12 M r^2$\frac{1}{2} M r^2$
B.frac38 M r^2$\frac{3}{8} M r^2$
C.frac32 M r^2$\frac{3}{2} M r^2$
D.2 M r^2$2 M r^2$
Solution
### Related Formula
1. Moment of inertia of a circular ring of mass M$M$ and radius R$R$ about a diametrical axis in its plane:
I_textdia = frac12 M R^2$$I_{\text{dia}} = \frac{1}{2} M R^2$$
2. Parallel Axis Theorem:
I = I_textcm + M d^2$$I = I_{\text{cm}} + M d^2$$
where d$d$ is the perpendicular distance between the center of mass axis and the parallel axis.
### Core Logic
The axis of rotation is tangential and lies in the plane of the ring. Thus, the distance from the center of mass axis (which is also diametrical and in-plane) is equal to the radius R$R$.
Using the parallel axis theorem:
I_texttangent = I_textdia + M R^2 = frac12 M R^2 + M R^2 = frac32 M R^2$$I_{\text{tangent}} = I_{\text{dia}} + M R^2 = \frac{1}{2} M R^2 + M R^2 = \frac{3}{2} M R^2$$
### Step 1: Express in terms of diameter r$r$
The question specifies the diameter of the ring is r$r$.
Therefore, the radius R$R$ is:
R = fracr2$$R = \frac{r}{2}$$
Substitute R = fracr2$R = \frac{r}{2}$ into the moment of inertia formula:
I_texttangent = frac32 M left(fracr2right)^2 = frac38 M r^2$$I_{\text{tangent}} = \frac{3}{2} M \left(\frac{r}{2}\right)^2 = \frac{3}{8} M r^2$$
Thus, the moment of inertia is frac38 M r^2$\frac{3}{8} M r^2$.
### Pattern Recognition
Sees: Circular ring tangential axis in-plane.
Trap: Directly using radius r$r$ instead of diameter r$r$. Read variables carefully: often r$r$ is radius, but here it is explicitly given as diameter!
Shortcut: Standard in-plane tangent is frac32MR^2$\frac{3}{2}MR^2$. Substitute R = fracr2$R = \frac{r}{2}$ to get frac38Mr^2$\frac{3}{8}Mr^2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q3jee_main_2025_02_april_morningMoment of Inertia and Torque
A cord of negligible mass is wound around the rim of a wheel supported by spokes with negligible mass. The mass of wheel is 10mathrm~kg$10\mathrm{~kg}$ and radius is 10mathrm~cm$10\mathrm{~cm}$ and it can freely rotate without any friction. Initially the wheel is at rest. If a steady pull of 20mathrm~N$20\mathrm{~N}$ is applied on the cord, the angular velocity of the wheel, after the cord is unwound by 1mathrm~m$1\mathrm{~m}$, would be:
A wheel rotating with a pull force of 20 N acting on the cord.
A.20mathrm~rad/s$20\mathrm{~rad/s}$
B.30mathrm~rad/s$30\mathrm{~rad/s}$
C.10mathrm~rad/s$10\mathrm{~rad/s}$
D.0mathrm~rad/s$0\mathrm{~rad/s}$
Solution
### Related Formula
W_F = F cdot s$W_F = F \cdot s$K_R = frac12 I omega^2$$K_R = \frac{1}{2} I \omega^2$$I = M R^2 quad text(Moment of inertia of a ring/rim)$$I = M R^2 \quad \text{(Moment of inertia of a ring/rim)}$$
### Core Logic
The work done by the constant pulling force F = 20mathrm~N$F = 20\mathrm{~N}$ through distance s = 1mathrm~m$s = 1\mathrm{~m}$ is completely converted into the rotational kinetic energy of the wheel.
Work done by force:
W_F = F cdot s = 20 times 1 = 20mathrm~J$$W_F = F \cdot s = 20 \times 1 = 20\mathrm{~J}$$
Since the spokes are of negligible mass, all mass M = 10mathrm~kg$M = 10\mathrm{~kg}$ is distributed on the outer rim of radius R = 10mathrm~cm = 0.1mathrm~m$R = 10\mathrm{~cm} = 0.1\mathrm{~m}$. The wheel acts as a thin ring:
I = M R^2 = 10 times (0.1)^2 = 0.1mathrm~kgcdot m^2$$I = M R^2 = 10 \times (0.1)^2 = 0.1\mathrm{~kg\cdot m^2}$$
Using the work-energy theorem:
W_F = Delta K_R = frac12 I omega^2$$W_F = \Delta K_R = \frac{1}{2} I \omega^2$$20 = frac12 times 0.1 times omega^2$$20 = \frac{1}{2} \times 0.1 \times \omega^2$$40 = 0.1 omega^2 implies omega^2 = 400 implies omega = 20mathrm~rad/s$$40 = 0.1 \omega^2 \implies \omega^2 = 400 \implies \omega = 20\mathrm{~rad/s}$$
### Step 1: Final Conclusion
The angular velocity of the wheel after the cord is unwound by 1mathrm~m$1\mathrm{~m}$ is 20mathrm~rad/s$20\mathrm{~rad/s}$.
### Pattern Recognition
Work done in unwinding a string is F cdot s$F \cdot s$. Under pure rotation with zero friction, this is exactly frac12 I omega^2$\frac{1}{2} I \omega^2$. Always identify the mass distribution (here, rim with massless spokes acts as a thin cylinder/ring I = MR^2$I = MR^2$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Class 11 Physics: Work, Energy and Power
More System of Particles and Rotational Motion Questions — jee_main_2025_07_april_morning
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