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
Q26jee_main_2026_22_january_morningMoment of Inertia
A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is \_\_\_\_ kg.m ^2$^{2}$
A.0.36$0.36$
B.0.72$0.72$
C.0.18$0.18$
D.0.63$0.63$
Solution
### Related Formula
I = frac75(m_1 R_1^2 + m_2 R_2^2)$$I = \frac{7}{5}(m_1 R_1^2 + m_2 R_2^2)$$
### Core Logic
Substitute the given mass and radius values into the standard moment of inertia formula for spheres about their common tangent at the contact point:
I = frac75[5(10)^2 + 10 times (20)^2] times 10^-4$$I = \frac{7}{5}[5(10)^{2} + 10 \times (20)^{2}] \times 10^{-4}$$I = 63 times 10^-2 text kg m^2 = 0.63 text kg m^2$$I = 63 \times 10^{-2} \text{ kg m}^2 = 0.63 \text{ kg m}^2$$
### Pattern Recognition
Sees: Two touching solid spheres + moment of inertia about tangent at contact point.
Shortcut: Apply parallel/perpendicular axis theorem adjustments directly via standard formula summation.
Check: Calculations yield 0.63 text kg m^2$0.63 \text{ kg m}^2$, matching option (4). ✓
### Chapter Mix
Class 11 Physics: Rotational Motion
Q49jee_main_2026_22_january_morningMoment of Inertia of Circular Discs
A circular disc has radius R_1$R_{1}$ and thickness T_1$T_{1}$. Another circular disc made of the same material has radius R_2$R_{2}$ and thickness T_2$T_{2}$. If the moment of inertia of both discs are same and fracR_1R_2 = 2$\frac{R_{1}}{R_{2}} = 2$ then fracT_1T_2 = frac1alpha$\frac{T_{1}}{T_{2}} = \frac{1}{\alpha}$. The value of alpha$\alpha$ is \_\_\_\_.
Numerical Answer.Answer: 16 to 16
Solution
### Related Formula
I = fracmR^22, quad m = pi R^2 T rho$$I = \frac{mR^2}{2}, \quad m = \pi R^2 T \rho$$
### Core Logic
Solution disc moment of inertia diagram for Q49 - JEE Main 2026 Morning
Mass of discs:
m_1 = pi R_1^2 T_1 rho, quad m_2 = pi R_2^2 T_2 rho$$m_1 = \pi R_1^2 T_1 \rho, \quad m_2 = \pi R_2^2 T_2 \rho$$
Moments of inertia:
I_1 = fracm_1 R_1^22, quad I_2 = fracm_2 R_2^22$$I_1 = \frac{m_1 R_1^2}{2}, \quad I_2 = \frac{m_2 R_2^2}{2}$$
Equating I_1 = I_2$I_1 = I_2$:
fracpi R_1^2 T_1 rho R_1^22 = fracpi R_2^2 T_2 rho R_2^22 implies fracT_1T_2 = left(fracR_2R_1right)^4 = left(frac12right)^4 = frac116$$\frac{\pi R_1^2 T_1 \rho R_1^2}{2} = \frac{\pi R_2^2 T_2 \rho R_2^2}{2} \implies \frac{T_1}{T_2} = \left(\frac{R_2}{R_1}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$$
Therefore, alpha = 16$\alpha = 16$.
### Pattern Recognition
Sees: Equal moment of inertia for two discs of different radii and thicknesses.
Shortcut: Equate mR^2$mR^2$ expressions and substitute radius ratio fracR_1R_2 = 2$\frac{R_1}{R_2} = 2$.
Check: Numerical answer is 16. ✓
### Chapter Mix
Class 11 Physics: Rotational Motion
A uniform bar of length 12 cm and mass 20 m lies on a smooth horizontal table. Two point masses m and 2 m are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency omega$\omega$. The ratio of v and omega$\omega$ is :
The figure illustrates a uniform bar of length 12 cm with two point masses m and 2m approaching it perpendicularly from opposite directions.
A.33$33$
B.2sqrt88$2\sqrt{88}$
C.66$66$
D.32$32$
Solution
### Related Formula
L_i = L_f$L_i = L_f$I = fracM L_bar^212 + sum m_i r_i^2$$I = \frac{M L_{bar}^2}{12} + \sum m_i r_i^2$$
### Core Logic
Applying angular momentum conservation about the center of mass of the rod:
L_i = m cdot v cdot 4 + 2m cdot v cdot 2$$L_i = m \cdot v \cdot 4 + 2m \cdot v \cdot 2$$
Calculating total moment of inertia I_final$I_{final}$ after collision:
I_final = left( frac20m(12)^212 + m(4)^2 + 2m(2)^2 right)$$I_{final} = \left( \frac{20m(12)^2}{12} + m(4)^2 + 2m(2)^2 \right)$$I_final = (240m + 16m + 8m) = 264m$$I_{final} = (240m + 16m + 8m) = 264m$$
Equating initial and final angular momentum:
4mv + 4mv = 264m cdot omega$$4mv + 4mv = 264m \cdot \omega$$8v = 264 omega implies fracvomega = frac2648 = 33$$8v = 264 \omega \implies \frac{v}{\omega} = \frac{264}{8} = 33$$The figure illustrates a uniform bar of length 12 cm with two point masses m and 2m approaching it perpendicularly from opposite directions.
### Step 1: Final Conclusion
The ratio fracvomega$\frac{v}{\omega}$ is 33.
### Pattern Recognition
Collision on rotating bar: Conserve angular momentum about COM of rod since net external torque about COM is zero during collision.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
More System of Particles and Rotational Motion Questions — jee_main_2025_07_april_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
