A cube having a side of 10~cm$10\mathrm{~cm}$ with unknown mass and 200~gm$200\mathrm{~gm}$ mass were hung at two ends of an uniform rigid rod of 27~cm$27\mathrm{~cm}$ long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200~gm$200\mathrm{~gm}$ weight as 25~cm$25\mathrm{~cm}$. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water.
(Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is 1~g/cm³$1\mathrm{~g/cm}^{3}$.)
The unknown mass is ________ kg$\mathrm{kg}$.
Numerical Answer Type:
Enter a numerical valueAnswer: 3 to 3+4 marks
Solution & Explanation
Related Formula
Σ τ = 0 (Rotational Equilibrium)$$\sum \tau = 0 \quad \text{(Rotational Equilibrium)}$$Fₙₑₜ = m g - FB$$F_{\text{net}} = m g - F_B$$FB = ρwater Vsubmerged g$$F_B = \rho_{\text{water}} V_{\text{submerged}} g$$
where,
τ$\tau$ = torque about the wedge point
FB$F_B$ = buoyancy force
Core Logic
Let's list the geometric parameters from the setup:
Length of uniform rigid rod, L = 27~cm$L = 27\mathrm{~cm}$.
Wedge is positioned such that the distance to the 200~g$200\mathrm{~g}$ (0.2~kg$0.2\mathrm{~kg}$) weight is d₁ = 25~cm$d_1 = 25\mathrm{~cm}$.
Therefore, the distance to the unknown mass M$M$ at the other end is d₂ = 27 - 25 = 2~cm$d_2 = 27 - 25 = 2\mathrm{~cm}$.
Calculate the volume of the cube:
Side of the cube, a = 10~cm = 0.1~m$a = 10\mathrm{~cm} = 0.1\mathrm{~m}$.
FB = ρw Vsub g = 1000~kg/m³ × (5 × 10⁻⁴~m³) × g = 0.5 g~N$$F_B = \rho_w V_{\text{sub}} g = 1000\mathrm{~kg/m}^3 \times \left(5 \times 10^{-4}\mathrm{~m}^3\right) \times g = 0.5 g\mathrm{~N}$$
Step 1: Torque Balance Equation
For rotational equilibrium, balance the torques about the wedge point O$O$:
Torque on the left (unknown mass branch): τleft = (M g - FB) × d₂ = (M g - 0.5 g) × 2$\tau_{\text{left}} = \left(M g - F_B\right) \times d_2 = (M g - 0.5 g) \times 2$
Torque on the right (200~g$200\mathrm{~g}$ mass branch): τright = 0.2 g × d₁ = 0.2 g × 25$\tau_{\text{right}} = 0.2 g \times d_1 = 0.2 g \times 25$
τleft = τright$$\tau_{\text{left}} = \tau_{\text{right}}$$(M g - 0.5 g) × 2 = 0.2 g × 25$$(M g - 0.5 g) \times 2 = 0.2 g \times 25$$
2 (M - 0.5) = 5$2 (M - 0.5) = 5$
M - 0.5 = 2.5 M = 3~kg$$M - 0.5 = 2.5 \implies M = 3\mathrm{~kg}$$
Thus, the unknown mass is 3~kg$3\mathrm{~kg}$.
Pattern Recognition
Sees: Rod torque balance + buoyancy force on one end.
Trap: Ensure you measure the distances from the pivot point (the wedge). The unknown mass is at 27 - 25 = 2~cm$27 - 25 = 2\mathrm{~cm}$ from the wedge.
Shortcut: Since g$g$ appears in both gravity and buoyancy terms, it cancels out immediately. Balancing torque simplifies directly to resolving mass differences: 2(M - 0.5) = 0.2 × 25 = 5$2(M - 0.5) = 0.2 \times 25 = 5$. This yields M = 3$M = 3$ instantly! ✓
Chapter Mix
Class 11 Physics: Rotational Motion
Class 11 Physics: Mechanical Properties of Fluids
Rotational balance with buoyancy forceRotational balance with buoyancy forceRotational balance with buoyancy force
Keywords:#Wedge balance torque with buoyancy#JEE Main 2025 Evening Q25#Rotational Motion torque pivot#Archimedes principle torque balance RankBit
More Rotational Motion Previous-Year Questions — Page 4
Q49jee_main_2026_24_january_eveningMoment of Inertia of Continuous Bodies
A uniform solid cylinder of length L and radius R has moment of inertia about its axis equal to I₁$I_{1}$ . A small co-centric cylinder of length L/2 and radius R/3 carved from this cylinder has moment of inertia about its axis equals to I₂$I_{2}$ . The ratio I₁/I₂$I_{1}/I_{2}$ is
Numerical Answer.Answer: 162 to 162
Solution
Related Formula
I = (1)/(2) M R²$$I = \frac{1}{2} M R^2$$M = ρ · V = ρ · π R² L$$M = \rho \cdot V = \rho \cdot \pi R^2 L$$
Core Logic
Moment of Inertia of Continuous Bodies diagram for Q49 - JEE Main 2026 Evening
I₂ = (1)/(2) m ((R)/(3))²$$I_2 = \frac{1}{2} m \left(\frac{R}{3}\right)^2$$I₂ = (1)/(2) ( (M)/(18) ) ( (R²)/(9) )$$I_2 = \frac{1}{2} \left( \frac{M}{18} \right) \left( \frac{R^2}{9} \right)$$I₂ = (1)/(324) M R²$$I_2 = \frac{1}{324} M R^2$$
Step 3: Finding the Ratio
(I₁)/(I₂) = ((1)/(2) M R²)/((1)/(324) M R²) = (324)/(2) = 162$$\frac{I_1}{I_2} = \frac{\frac{1}{2} M R^2}{\frac{1}{324} M R^2} = \frac{324}{2} = 162$$
Pattern Recognition
For similar geometries, mass scales as R² L$R^2 L$. Inertia scales as M R²$M R^2$ which ultimately means I ∝ R⁴ L$I \propto R^4 L$. Here R arrow R/3$R \rightarrow R/3$ (factor of 1/81$1/81$) and L arrow L/2$L \rightarrow L/2$ (factor of 1/2$1/2$), so I₂$I_2$ is 1/162$1/162$ of I₁$I_1$.
Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q29jee_main_2026_28_january_morningMoment of Inertia and Torque
Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure.
Illustration showing two discs connected by a rod rotating about axis AB.
If the mass of each disc is 600 gm and applied torque between two discs is 43 × 10⁵ ~dyne· cm$43 \times 10^{5} \mathrm{~dyne\cdot cm}$, the angular acceleration of the discs about the given axis AB is ____ rad/s²$\mathrm{rad/s}^{2}$ .
A.22$22$
B.11$11$
C.100$100$
D.27$27$
Solution
Related Formula
τ = I α$\tau = I \alpha$
where τ$\tau$ is the torque, I$I$ is the moment of inertia about the axis of rotation, and α$\alpha$ is the angular acceleration.
Core Logic
First, calculate the total moment of inertia of the system (two discs + one rod) about axis AB using the parallel axis theorem. Let m = 600 ~gm$m = 600 \mathrm{~gm}$ and R = 10 ~cm$R = 10 \mathrm{~cm}$.
In compound systems, decompose into basic shapes (rod, disc). Determine the parallel distance to the required axis for each center of mass. Keep everything in CGS units since torque is given in dyne-cm.
Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q48jee_main_2026_28_january_morningRadius of Gyration
A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is √(n)$\sqrt{n}$ cm. The value of n is ____
We are given k = √(n) ⇒ k² = n$k = \sqrt{n} \Rightarrow k^2 = n$.
n = 265$n = 265$
Pattern Recognition
Whenever radius of gyration is asked, divide out the mass immediately. k² = Icm-factor R² + d²$k^2 = I_{\text{cm-factor}} R^2 + d^2$.
Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q37jee_main_2026_28_january_eveningCross Product and Vector Reversal
When the position vector r=x i+y j+z k$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ changes sign as - r$-\vec{r}$ , which one of the following vector will not flip under sign change?
A.Linear momentum$\text{Linear momentum}$
B.Velocity$\text{Velocity}$
C.Acceleration$\text{Acceleration}$
D.Angular momentum$\text{Angular momentum}$
Solution
Related Formula
v = d rdt$$\vec{v} = \frac{d\vec{r}}{dt}$$p = m v$$\vec{p} = m\vec{v}$$a = d vdt$$\vec{a} = \frac{d\vec{v}}{dt}$$L = r × p$$\vec{L} = \vec{r} \times \vec{p}$$
Core Logic
Under a sign change of coordinates (parity transformation or spatial inversion) r arrow - r$\vec{r} \rightarrow -\vec{r}$.
Then, velocity v = d rdt arrow - v$\vec{v} = \frac{d\vec{r}}{dt} \rightarrow -\vec{v}$.
Linear momentum p = m v arrow - p$\vec{p} = m\vec{v} \rightarrow -\vec{p}$.
Acceleration a = d vdt arrow - a$\vec{a} = \frac{d\vec{v}}{dt} \rightarrow -\vec{a}$.
Angular momentum L = r × p$\vec{L} = \vec{r} \times \vec{p}$.
Under the transformation: L' = (- r) × (- p) = r × p = L$\vec{L}' = (-\vec{r}) \times (-\vec{p}) = \vec{r} \times \vec{p} = \vec{L}$.
Step 1: Final Conclusion
Since both position and momentum vectors change sign, their cross product (angular momentum) retains its original sign. It does not flip.
Pattern Recognition
Angular momentum is a pseudovector (or axial vector). True vectors (polar vectors) flip signs under spatial inversion, but pseudovectors (which are cross products of two polar vectors) do not.
Chapter Mix
Class 11 Physics: Systems of Particles and Rotational Motion
Q50jee_main_2026_28_january_eveningRotation about Fixed Axis
A fly wheel having mass 3 kg$3 \text{ kg}$ and radius 5 m$5 \text{ m}$ is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a 3 kg$3 \text{ kg}$ mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m$3 \text{ m}$ is ____ J. (g = 10 m/s²$g = 10 \text{ m/s}^{2}$ )
Solution for Rotation about Fixed Axis
By conservation of mechanical energy, the loss in gravitational potential energy of the descending mass equals the gain in kinetic energy of the block and the rotational kinetic energy of the flywheel.
mg h = (1)/(2) I ω² + (1)/(2) m v²$$mg h = \frac{1}{2} I \omega^2 + \frac{1}{2} m v^2$$
Step 1: Relate Velocity and Angular Velocity
Since the string does not slip, the linear velocity v$v$ of the mass is related to the angular velocity ω$\omega$ of the wheel by:
v = ω R ⇒ ω = (v)/(R)$$v = \omega R \Rightarrow \omega = \frac{v}{R}$$
Step 2: Energy Conservation
The flywheel is treated as a solid disc/cylinder (from the I = mR²/2$I = mR^2/2$ usage in the solution):
mg h = (1)/(2) ((M R²)/(2)) ω² + (1)/(2) m v²$$mg h = \frac{1}{2} \left(\frac{M R^2}{2}\right) \omega^2 + \frac{1}{2} m v^2$$
Given M = 3 kg$M = 3 \text{ kg}$ (wheel), m = 3 kg$m = 3 \text{ kg}$ (block), h = 3 m$h = 3 \text{ m}$.
Notice that the solution text specifies the flywheel mass as m$m$ and block mass also as m$m$, both being 3 kg$3\text{ kg}$. Let's follow the PDF exactly:
mg × 3 = (1)/(4) m v² + (1)/(2) m v²$$mg \times 3 = \frac{1}{4} m v^2 + \frac{1}{2} m v^2$$3mg = (3)/(4) m v²$$3mg = \frac{3}{4} m v^2$$v² = 4g = 4 × 10 = 40 (m/s)²$$v^2 = 4g = 4 \times 10 = 40 \text{ (m/s)}^2$$
Step 4: Calculate Kinetic Energy of Flywheel
K.E.wheel = (1)/(2) I ω² = (1)/(4) m v²$$K.E._{\text{wheel}} = \frac{1}{2} I \omega^2 = \frac{1}{4} m v^2$$K.E.wheel = (1)/(4) × 3 × 40 = 30 J$$K.E._{\text{wheel}} = \frac{1}{4} \times 3 \times 40 = 30 \text{ J}$$
Pattern Recognition
For a mass pulling a wheel of identical mass (disc), the total K.E.$K.E.$ is split between translational (1/2 mv²$1/2 mv^2$) and rotational (1/4 mv²$1/4 mv^2$). Rotational gets exactly 1/3$1/3$ of the total potential energy lost, 30 J$30\text{ J}$ out of 90 J$90\text{ J}$ total.
Chapter Mix
Class 11 Physics: Systems of Particles and Rotational Motion
Class 11 Physics: Work, Energy and Power
More Rotational Motion Questions — jee_main_2025_08_april_evening
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.
