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 & Explanation
### 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
Keywords:#rod suspended#two strings#tension cut
More System of Particles and Rotational Motion Previous-Year Questions — Page 2
Q13jee_main_2025_02_april_morningMoment of Inertia and Torque
A square Lamina OABC of length 10mathrm~cm$10\mathrm{~cm}$ is pivoted at 'O'. Forces act on the Lamina as shown in the figure. If the Lamina remains stationary, then the magnitude of F$F$ is:
A square lamina OABC with multiple force vectors acting on its vertices, pivoted at O.
A.20mathrm~N$20\mathrm{~N}$
B. 0 (zero)
C.10mathrm~N$10\mathrm{~N}$
D.10sqrt2mathrm~N$10\sqrt{2}\mathrm{~N}$
Solution
### Related Formula
tau_O = F cdot r_perp$$\tau_O = F \cdot r_{\perp}$$sum tau_O = 0 quad text(for rotational equilibrium)$$\sum \tau_O = 0 \quad \text{(for rotational equilibrium)}$$
### Core Logic
Let the side length of the square lamina be l = 10mathrm~cm$l = 10\mathrm{~cm}$.
The lamina is pivoted at point O(0,0)$O(0,0)$ and remains stationary under rotational equilibrium. Therefore, the net torque about O$O$ must be zero.
Evaluating torque contributions about point O$O$:
- Forces acting directly at pivot O$O$ produce zero torque.
- Forces whose lines of action pass through O$O$ produce zero torque.
- The 10mathrm~N$10\mathrm{~N}$ force perpendicular to side OA$OA$ produces torque:
tau_1 = 10 times l quad text(Counter-Clockwise)$$\tau_1 = 10 \times l \quad \text{(Counter-Clockwise)}$$
- The unknown force F$F$ acting perpendicular to side OC$OC$ produces torque:
tau_2 = F times l quad text(Clockwise)$$\tau_2 = F \times l \quad \text{(Clockwise)}$$
Setting sum tau_O = 0$\sum \tau_O = 0$:
10 cdot l - F cdot l = 0 implies F = 10mathrm~N$$10 \cdot l - F \cdot l = 0 \implies F = 10\mathrm{~N}$$
### Step 1: Final Conclusion
The magnitude of the force $
### Step 1: Final Conclusion
The magnitude of the force $F is $ is $10\mathrm{~N}$.
### Pattern Recognition
In pivoted laminas, focus on the pivot and disregard any force vector whose line of action passes through the pivot. For symmetric placements, equate Clockwise torque = Counter-Clockwise torque directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Q16jee_main_2025_02_april_morningMoment of Inertia and Torque
Moment of inertia of a rod of mass 'M' and length 'L' about an axis passing through its center and normal to its length is 'α'. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is:
A.alpha$\alpha$
B.alpha / 4$\alpha / 4$
C.alpha / 8$\alpha / 8$
D.alpha / 2$\alpha / 2$
Solution
### Related Formula
I = frac112 M L^2$$I = \frac{1}{12} M L^2$$
### Core Logic
Initially, the moment of inertia is:
alpha = fracM L^212$$\alpha = \frac{M L^2}{12}$$
When cut into two equal parts, each smaller rod has:
- Mass, m = fracM2$m = \frac{M}{2}$
- Length, l = fracL2$l = \frac{L}{2}$
When joined symmetrically as a cross, the target axis passes through their joint intersection perpendicular to their plane.
For each rod, this axis passes through its individual center of mass and is perpendicular to its length.
Thus, the total moment of inertia of the cross is the sum of the moments of inertia of the two rods:
I_textcross = I_1 + I_2 = 2 times left(frac112 m l^2right) = frac16 m l^2$$I_{\text{cross}} = I_1 + I_2 = 2 \times \left(\frac{1}{12} m l^2\right) = \frac{1}{6} m l^2$$
Substituting m = fracM2$m = \frac{M}{2}$ and l = fracL2$l = \frac{L}{2}$:
I_textcross = frac16 times left(fracM2right) times left(fracL2right)^2 = frac16 times fracM2 times fracL^24 = fracM L^248$$I_{\text{cross}} = \frac{1}{6} \times \left(\frac{M}{2}\right) \times \left(\frac{L}{2}\right)^2 = \frac{1}{6} \times \frac{M}{2} \times \frac{L^2}{4} = \frac{M L^2}{48}$$
Comparing with alpha$\alpha$:
I_textcross = frac14 left(fracM L^212right) = fracalpha4$$I_{\text{cross}} = \frac{1}{4} \left(\frac{M L^2}{12}\right) = \frac{\alpha}{4}$$
### Step 1: Final Conclusion
The moment of inertia of the cross is $
### Step 1: Final Conclusion
The moment of inertia of the cross is $\alpha / 4.
### Pattern Recognition
Since mass scales linearly ($.
### Pattern Recognition
Since mass scales linearly ($M \propto L), cutting a rod into $), cutting a rod into $n equal segments scales the length by $ equal segments scales the length by $1/n and mass by $ and mass by $1/n. The moment of inertia of each segment scales as $. The moment of inertia of each segment scales as $1/n^3. Reassembling $. Reassembling $n segments linearly sums their contributions, so the final moment of inertia scales as $ segments linearly sums their contributions, so the final moment of inertia scales as $n \times \frac{1}{n^3} = \frac{1}{n^2}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Q6jee_main_2025_07_april_morningCentre of Mass
A rod of length 5 mathrm~L$5 \mathrm{~L}$ is bent right angle keeping one side length as 2 mathrm~L$2 \mathrm{~L}$ . Diagram of an L-shaped rod aligned with the x and y axes, with lengths 2L and 3L respectively. The position of the centre of mass of the system : (Consider mathrmL = 10mathrmcm$\mathrm{L} = 10\mathrm{cm}$ )
### Related Formula
For a continuous system modeled as discrete point masses located at their respective centers of mass:
x_textcom = fracm_1x_1 + m_2x_2m_1 + m_2$$x_{\text{com}} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$$y_textcom = fracm_1y_1 + m_2y_2m_1 + m_2$$y_{\text{com}} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}$$
### Core Logic
Let the uniform linear mass density of the rod be lambda$\lambda$.
- Total length is 5L$5L$.
- One segment of length 2L$2L$ lies on the x-axis. Its mass is 2m = lambda(2L)$2m = \lambda(2L)$ and its center of mass is at (L, 0)$(L, 0)$.
- The remaining segment of length 3L$3L$ lies on the y-axis. Its mass is 3m = lambda(3L)$3m = \lambda(3L)$ and its center of mass is at (0, 1.5L)$(0, 1.5L)$.
### Step 1: Calculate COM Coordinates
Find the coordinates of the system's center of mass:
x_textcom = frac2m(L) + 3m(0)2m + 3m = frac2L5 = 0.4L$$x_{\text{com}} = \frac{2m(L) + 3m(0)}{2m + 3m} = \frac{2L}{5} = 0.4L$$y_textcom = frac2m(0) + 3m(1.5L)2m + 3m = frac4.5L5 = 0.9L$$y_{\text{com}} = \frac{2m(0) + 3m(1.5L)}{2m + 3m} = \frac{4.5L}{5} = 0.9L$$
Given L = 10 mathrm~cm$L = 10 \mathrm{~cm}$:
x_textcom = 0.4 times 10 = 4 mathrm~cm$$x_{\text{com}} = 0.4 \times 10 = 4 \mathrm{~cm}$$y_textcom = 0.9 times 10 = 9 mathrm~cm$$y_{\text{com}} = 0.9 \times 10 = 9 \mathrm{~cm}$$
### Step 2: Vector Form
Expressing in vector notation:
vecr_textcom = 4hatmathbfi + 9hatmathbfj$$\vec{r}_{\text{com}} = 4\hat{\mathbf{i}} + 9\hat{\mathbf{j}}$$
### Pattern Recognition
Sees: L-shaped rod formed by bending a total length L_texttotal$L_{\text{total}}$.
Shortcut: Treat each arm as a point mass at its geometric midpoint. For segments of ratio 2:3$2:3$, the COM divides the distance between their midpoints in the inverse ratio 3:2$3:2$ closer to the heavier segment on the y-axis.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q21jee_main_2025_07_april_morningMoment of Inertia
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.Answer: 199 to 199
Solution
### 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
Q2jee_main_2025_08_april_eveningMoment of Inertia
A rod of linear mass density lambda^prime$\lambda^{\prime}$ and length L$L$ is bent to form a ring of radius R$R$. Moment of inertia of ring about any of its diameter is:
### Related Formula
I_textdia = frac12 M R^2$$I_{\text{dia}} = \frac{1}{2} M R^2$$
where,
I_textdia$I_{\text{dia}}$ = moment of inertia of a ring about its diameter
M$M$ = total mass of the ring
R$R$ = radius of the ring
### Core Logic
Since the linear mass density is lambda^prime$\lambda^{\prime}$ (or lambda$\lambda$ as per the options), the total mass M$M$ of the rod of length L$L$ is:
M = lambda L$M = \lambda L$
When this rod is bent into a ring of radius R$R$, its circumference equals the length of the rod:
2pi R = L implies R = fracL2pi$$2\pi R = L \implies R = \frac{L}{2\pi}$$
Substituting M$M$ and R$R$ into the formula for the moment of inertia about the diameter:
I_textdia = frac12 M R^2 = frac12 (lambda L) left(fracL2piright)^2 = fraclambda L^38pi^2$$I_{\text{dia}} = \frac{1}{2} M R^2 = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$$
### Pattern Recognition
Sees: "Rod of length L$L$ bent to form a ring" → R = fracL2pi$R = \frac{L}{2\pi}$.
Trap: Moment of inertia about the central axis perpendicular to the plane is MR^2$MR^2$, but about its diameter, it is half, i.e., frac12MR^2$\frac{1}{2}MR^2$.
Shortcut: I = frac12 (lambda L) left(fracL2piright)^2 = fraclambda L^38pi^2$I = \frac{1}{2} (\lambda L) \left(\frac{L}{2\pi}\right)^2 = \frac{\lambda L^3}{8\pi^2}$. ✓
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
More System of Particles and Rotational Motion Questions — jee_main_2026_21_jan_morning
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