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M and R be the mass and radius of a disc. A small disc of radius R/3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is frac4mathrmxmathrmMR^2 The value of x is
Moment of Inertia diagram for Q25 - JEE Main 2025 Evening
The graphic depicts a flat circular uniform disk of radius R with a smaller circular cut-out cavity of radius R/3 touching the perimeter.
[cite: 197, 198, 199, 200, 208]

Numerical Answer Type:
Enter a numerical value Answer: 9 to 9 +4 marks

Solution & Explanation

### Related Formula I_textdisc = frac12 M R^2 [cite: 833] I = I_textcm + M d^2 quad text(Parallel Axis Theorem) [cite: 848] ### Core Logic Let the original mass density per unit area be sigma. Without any cavity, the moment of inertia is: [cite: 198, 833] I_1 = fracMR^22 [cite: 833] The mass of the removed small section scales directly with its cut-out area profile: [cite: 198, 834] m = fracMpi R^2 times pi left(fracR3right)^2 = fracM9 [cite: 198, 834, 844] The center of mass of the removed disk sits at a distance d = R - fracR3 = frac2R3 away from the primary center O[cite: 198, 205, 848]. Calculating its partial moment of inertia about O via the parallel axis theorem: [cite: 199, 848] I_2 = fracm r^22 + m d^2 = fracfracM9left(fracR3right)^22 + fracM9left(frac2R3 ight)^2 [cite: 848] I_2 = fracMR^2162 + frac4MR^281 = fracMR^2 + 8MR^2162 = frac9MR^2162 = fracMR^218 [cite: 848, 850] Subtracting the removed component from the original configuration: [cite: 851] I = I_1 - I_2 = fracMR^22 - fracMR^218 = frac9MR^2 - MR^218 = frac8MR^218 = frac49MR^2 [cite: 851] Matching this with frac4xMR^2, we find x = 9[cite: 200, 208, 851, 853]. ### Pattern Recognition For uniform planar surfaces, mass always scales squarely with linear dimension changes (r rightarrow fracr3 implies m rightarrow fracm9)[cite: 198, 834, 844]. Always apply the parallel axis theorem to bring the component values to a unified reference point before executing addition or subtraction[cite: 848, 851]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

Reference Study Guides

System Of Particles And Rotational Motion

More System of Particles and Rotational Motion Previous-Year Questions — Page 4

Q11 2025 Rolling Motion
A uniform solid cylinder of mass 'm' and radius 'r' rolls along an inclined rough plane of inclination 45^circ If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be :-
  • A. frac1sqrt2 g
  • B. frac13sqrt2 g
  • C. fracsqrt2 g3
  • D. sqrt2 g

Solution

### Related Formula The linear acceleration a for pure rolling motion down an incline profile is: a = fracgsintheta1 + fracImr^2 ### Core Logic For a uniform solid cylinder, the moment of inertia around its central axis is: I = frac12mr^2 implies fracImr^2 = frac12 ### Step 1: Calculating Acceleration Substitute theta = 45^circ and the cylinder inertial factor into the formula : a = fracgsin 45^circ1 + frac12 = fracfracgsqrt2frac32 a = frac2g3sqrt2 = fracsqrt2g3 ### Pattern Recognition Solid cylinder rolls with acceleration matching frac23 gsintheta. Since sin 45^circ = frac1sqrt2, this simplifies directly to fracsqrt2g3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion
Q17 2025 Torque and Equilibrium
A uniform rod of mass 250mathrmg having length 100mathrmcm is balanced on a sharp edge at 40mathrmcm mark[cite: 150, 151]. A mass of 400mathrmg is suspended at 10mathrmcm mark. To maintain the balance of the rod, the mass to be suspended at 90mathrmcm mark, is [cite: 154, 156]
  • A. 300mathrmg
  • B. 190mathrmg
  • C. 200mathrmg
  • D. 290mathrmg

Solution

### Related Formula For rotational equilibrium, the \sum of all counter-clockwise torques about the pivot point must exactly balance the \sum of all clockwise torques: sum tau_textpivot = 0 implies sum (m_i cdot g cdot x_i) = 0 ### Core Logic The rod is uniform, meaning its mass (250text g) acts exactly at its geometric center of mass, the 50text cm mark[cite: 150, 151]. Let the pivot point be the sharp edge at the 40text cm mark . Calculate the relative lever arms from the pivot [cite: 775, 776, 777]: * 400text g mass at 10text cm mark: lever arm = 40 - 10 = 30text cm (counter-clockwise) * 250text g rod mass at 50text cm mark: lever arm = 50 - 40 = 10text cm (clockwise) * Unknown mass M at 90text cm mark: lever arm = 90 - 40 = 50text cm (clockwise) Setting up the torque balance equation: 400 times 30 = (250 times 10) + (M times 50) 12000 = 2500 + 50M 50M = 9500 implies M = frac950050 = 190text g ### Step 1: Visual Context The structural layout of forces acting on the balanced rod system is shown below:
Torque and Equilibrium balancing diagram for Q17
Torque and Equilibrium balancing diagram for Q17
### Pattern Recognition Never forget to include the weight of a uniform rod itself in equilibrium equations. It is a common oversight to omit the rod's mass, which always acts at its geometric center. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

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