### 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

### 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

### 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

### Related Formula omega = 2pi fracN60$\omega = 2\pi \frac{N}{60}$ omega_f = omega_i + alpha t$\omega_f = \omega_i + \alpha t$ tau = I alpha$\tau = I \alpha$ I_textdia = frac14 m R^2$I_{\text{dia}} = \frac{1}{4} m R^2$ where, N$N$ = rotational speed in rpm alpha$\alpha$ = angular acceleration tau$\tau$ = torque applied I_textdia$I_{\text{dia}}$ = moment of inertia of a solid disk about its diameter axis ### Core Logic Given parameters: - Mass, m = 1mathrm~kg$m = 1\mathrm{~kg}$ - Initial speed, N_i = 1800mathrm~rpm implies omega_i = frac1800 times 2pi60 = 60pimathrm~rad/s$N_i = 1800\mathrm{~rpm} \implies \omega_i = \frac{1800 \times 2\pi}{60} = 60\pi\mathrm{~rad/s}$ - Final speed, N_f = 2100mathrm~rpm implies omega_f = frac2100 times 2pi60 = 70pimathrm~rad/s$N_f = 2100\mathrm{~rpm} \implies \omega_f = \frac{2100 \times 2\pi}{60} = 70\pi\mathrm{~rad/s}$ - Time, t = 40mathrm~s$t = 40\mathrm{~s}$ - Torque, tau = 25pimathrm~Ncdot m$\tau = 25\pi\mathrm{~N\cdot m}$ First, calculate the angular acceleration alpha$\alpha$: omega_f = omega_i + alpha t implies 70pi = 60pi + alpha (40)$\omega_f = \omega_i + \alpha t \implies 70\pi = 60\pi + \alpha (40)$ alpha = frac10pi40 = fracpi4mathrm~rad/s^2$\alpha = \frac{10\pi}{40} = \frac{\pi}{4}\mathrm{~rad/s}^{2}$ Now relate torque to moment of inertia: tau = I alpha implies 25pi = left( frac14 m R^2 right) left( fracpi4 right)$\tau = I \alpha \implies 25\pi = \left( \frac{1}{4} m R^2 \right) \left( \frac{\pi}{4} \right)$ 25pi = frac14 (1) R^2 fracpi4 implies 25pi = R^2 fracpi16$25\pi = \frac{1}{4} (1) R^2 \frac{\pi}{4} \implies 25\pi = R^2 \frac{\pi}{16}$ R^2 = 400 implies R = 20mathrm~m$R^2 = 400 \implies R = 20\mathrm{~m}$ ### Step 1: Compute Diameter The question asks for the diameter of the disk (D$D$): D = 2R = 2 times 20 = 40mathrm~m$D = 2R = 2 \times 20 = 40\mathrm{~m}$ ### Pattern Recognition Sees: Torque acting on a rotating disk increasing its speed. Trap: The axis of rotation is the *diameter* axis, not the normal geometric center axis. This means the moment of inertia is I = frac14 m R^2$I = \frac{1}{4} m R^2$, not frac12 m R^2$\frac{1}{2} m R^2$! Check this detail carefully. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion

### Related Formula sum tau = 0 quad text(Rotational Equilibrium)$\sum \tau = 0 \quad \text{(Rotational Equilibrium)}$ F_textnet = m g - F_B$F_{\text{net}} = m g - F_B$ F_B = rho_textwater V_textsubmerged g$F_B = \rho_{\text{water}} V_{\text{submerged}} g$ where, tau$\tau$ = torque about the wedge point F_B$F_B$ = buoyancy force ### Core Logic Let's list the geometric parameters from the setup: - Length of uniform rigid rod, L = 27mathrm~cm$L = 27\mathrm{~cm}$. - Wedge is positioned such that the distance to the 200mathrm~g$200\mathrm{~g}$ (0.2mathrm~kg$0.2\mathrm{~kg}$) weight is d_1 = 25mathrm~cm$d_1 = 25\mathrm{~cm}$. - Therefore, the distance to the unknown mass M$M$ at the other end is d_2 = 27 - 25 = 2mathrm~cm$d_2 = 27 - 25 = 2\mathrm{~cm}$. Calculate the volume of the cube: - Side of the cube, a = 10mathrm~cm = 0.1mathrm~m$a = 10\mathrm{~cm} = 0.1\mathrm{~m}$. - Volume, V = a^3 = 1000mathrm~cm^3 = 10^-3mathrm~m^3$V = a^3 = 1000\mathrm{~cm}^3 = 10^{-3}\mathrm{~m}^3$. When the system is balanced, half the volume of the cube is submerged in water: - Submerged volume, V_textsub = fracV2 = 500mathrm~cm^3 = 5 times 10^-4mathrm~m^3$V_{\text{sub}} = \frac{V}{2} = 500\mathrm{~cm}^3 = 5 \times 10^{-4}\mathrm{~m}^3$. - Buoyancy force: F_B = rho_w V_textsub g = 1000mathrm~kg/m^3 times left(5 times 10^-4mathrm~m^3right) times g = 0.5 gmathrm~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): tau_textleft = left(M g - F_Bright) times d_2 = (M g - 0.5 g) times 2$\tau_{\text{left}} = \left(M g - F_B\right) \times d_2 = (M g - 0.5 g) \times 2$ - Torque on the right (200mathrm~g$200\mathrm{~g}$ mass branch): tau_textright = 0.2 g times d_1 = 0.2 g times 25$\tau_{\text{right}} = 0.2 g \times d_1 = 0.2 g \times 25$ tau_textleft = tau_textright$\tau_{\text{left}} = \tau_{\text{right}}$ (M g - 0.5 g) times 2 = 0.2 g times 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 implies M = 3mathrm~kg$M - 0.5 = 2.5 \implies M = 3\mathrm{~kg}$ Thus, the unknown mass is 3mathrm~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 = 2mathrm~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 times 25 = 5$2(M - 0.5) = 0.2 \times 25 = 5$. This yields M = 3$M = 3$ instantly! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Rotational Motion Class 11 Physics: Mechanical Properties of Fluids