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)
Rigid Body Dynamics diagram for Q39 - JEE Main 2026 Morning
A uniform rod suspended horizontally by two strings attached at its ends.

Solution & Explanation

### Related Formula tau = Ialpha Sigma F_y = m a_CM, y a_CM, y = alpha fracl2 ### 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 Gravity provides the torque: tau = mg left(fracl2right). ### Step 1: Calculate Angular Acceleration Moment of inertia about the end is I = fracml^23. mg fracl2 = fracml^23 alpha alpha = frac3g2l
Rigid Body Dynamics solution diagram for Q39 - JEE Main 2026 Morning
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 Applying Newton's second law for translational motion of the CM in vertical direction: mg - T = m a_c T = mg - m a_c = mg - m left(frac3g4right) = fracmg4 ### Pattern Recognition Classic 'cut string' rigid body problem. Always take torque about the pivot/hinge point to find alpha, then relate the center of mass linear acceleration a = r_cm alpha to find the unknown tension using F_textnet = ma. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

More System of Particles and Rotational Motion Previous-Year Questions — Page 5

Q11 jee_main_2025_04_april_morning Uniform Circular Motion and Dynamics
If vecL and vecP represent the angular momentum and linear momentum respectively of a particle of mass 'm' having position vector vecr = a(haticosomega t + hatjsinomega t). The direction of force is
  • A. Opposite to the direction of vecr
  • B. Opposite to the direction of vecL
  • C. Opposite to the direction of vecP
  • D. Opposite to the direction of vecLtimesvecP

Solution

### Related Formula Acceleration vector equation via secondary derivation: veca = fracd^2vecrdt^2 Force equation: vecF = mveca ### Core Logic Given position tracking trace: vecr = a(haticosomega t + hatjsinomega t) Velocity vector vecv: vecv = fracdvecrdt = aomega(-hatisinomega t + hatjcosomega t) ### Step 1: Differentiate to find Acceleration veca = fracdvecvdt = aomega^2(-haticosomega t - hatjsinomega t) veca = -omega^2 left[ a(haticosomega t + hatjsinomega t) ight] = -omega^2vecr ### Step 2: Establish Force Direction vecF = mveca = -momega^2vecr The minus sign indicates the net centripetal pulling force aligns explicitly **opposite to the direction of vecr**. ### Pattern Recognition The expression describes a standard uniform circular motion profile. In circular configurations, acceleration and centripetal forces point radially inward, directly opposing the outbound position tracker vector. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics Class 11 Physics: System of Particles and Rotational Motion
Q21 jee_main_2025_04_april_morning Rolling Motion on an Inclined Plane
A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is sqrtfracx5 where x =
Numerical Answer. Answer: 3.5 to 4

Solution

### Related Formula Velocity of a rolling body from mechanical energy conservation: v = sqrtfrac2gh1 + frack^2R^2 ### Core Logic Evaluate radius of gyration factor coefficients: 1. For a circular ring: frack^2R^2 = 1 2. For a solid sphere: frack^2R^2 = frac25 ### Step 1: Compute Velocity Expressions v_textring = sqrtfrac2gh1 + 1 = sqrtgh v_textsphere = sqrtfrac2gh1 + frac25 = sqrtfrac10gh7 ### Step 2: Calculate the Velocity Ratio fracv_textringv_textsphere = fracsqrtghsqrtfrac10gh7 = sqrtfrac710 = sqrtfrac3.55 Matching with the prompt expression format sqrtfracx5 reveals: x = 3.5 Rounding to the nearest integer yields 4. ### Pattern Recognition Objects with lower mass concentration near the center (lower frack^2R^2 like the sphere) convert gravitational potential energy into translational kinetic energy more efficiently, rolling faster than hollow equivalents. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion
Q25 jee_main_2025_07_april_evening Moment of Inertia of a Disc with Cavity
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. Answer: 9 to 9

Solution

### 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
Q4 jee_main_2025_24_jan_evening Rolling Motion
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:
  • A. frac25
  • B. frac52
  • C. frac34
  • D. frac43

Solution

### Related Formula K_textlinear = frac12 m v_textcm^2 K_textrotational = frac12 I omega^2 ### Core Logic For a solid sphere, the moment of inertia about the center of mass is I = frac25mR^2. Since it rolls without slipping, the condition v_textcm = omega R holds. Substituting I and omega into the ratio: fracK_textlinearK_textrotational = fracfrac12 m v_textcm^2frac12 left(frac25mR^2 ight) left(fracv_textcmR ight)^2 fracK_textlinearK_textrotational = frac1frac25 = frac52 ### Pattern Recognition The ratio of translational to rotational kinetic energy for any rolling body is given by fracmR^2I_textcm. For a solid sphere, this becomes frac12/5 = frac52. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion
Q9 jee_main_2025_24_jan_evening Rolling on an Inclined Plane
A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be t_1 and t_2, respectively, then
  • A. t_1 < t_2
  • B. t_1 = t_2
  • C. t_1 = 2t_2
  • D. t_1 > t_2

Solution

### Related Formula t = sqrtfrac2ella_textcm a_textcm = fracg sin theta1 + fracI_textcmMR^2 ### Core Logic For a solid sphere: I_textsolid = frac25MR^2 implies a_1 = fracgsintheta1 + 2/5 = frac57gsintheta. For a hollow sphere: I_texthollow = frac23MR^2 implies a_2 = fracgsintheta1 + 2/3 = frac35gsintheta. Comparing accelerations: a_1 > a_2 Since acceleration of the solid sphere is greater, it takes less time to descend the incline: t_1 < t_2
Sphere rolling down an incline schematic Q9
Sphere rolling down an incline schematic Q9
### Pattern Recognition Smaller moment of inertia mass distribution distribution (more concentrated at the center) yields larger acceleration down an incline, meaning a quicker descent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: System of Particles and Rotational Motion

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