### Related Formula For a system in static equilibrium: sum F_x = 0 quad textand quad sum F_y = 0$\sum F_x = 0 \quad \text{and} \quad \sum F_y = 0$ ### Core Logic Let's resolve the tension forces vecT_1$\vec{T}_1$ and vecT_2$\vec{T}_2$ into horizontal and vertical components: - T_1$T_1$ makes 60^circ$60^\circ$ with the horizontal. - T_2$T_2$ makes 30^circ$30^\circ$ with the horizontal. - Downward gravitational force: W = m g = 1 times 10 = 10 \ mathrmN$W = m g = 1 \times 10 = 10 \ \mathrm{N}$. 1. **Horizontal Equilibrium (sum F_x = 0$\sum F_x = 0$):** T_1 cos 60^circ = T_2 cos 30^circ$T_1 \cos 60^\circ = T_2 \cos 30^\circ$ T_1 cdot frac12 = T_2 cdot fracsqrt32 implies T_1 = T_2 sqrt3$T_1 \cdot \frac{1}{2} = T_2 \cdot \frac{\sqrt{3}}{2} \implies T_1 = T_2 \sqrt{3}$ 2. **Vertical Equilibrium (sum F_y = 0$\sum F_y = 0$):** T_1 sin 60^circ + T_2 sin 30^circ = m g = 10$T_1 \sin 60^\circ + T_2 \sin 30^\circ = m g = 10$ T_1 cdot fracsqrt32 + T_2 cdot frac12 = 10$T_1 \cdot \frac{\sqrt{3}}{2} + T_2 \cdot \frac{1}{2} = 10$ ### Step 1: Solve for Tensions Substitute T_1 = T_2 sqrt3$T_1 = T_2 \sqrt{3}$ into the vertical equilibrium equation: (T_2 sqrt3) fracsqrt32 + fracT_22 = 10$(T_2 \sqrt{3}) \frac{\sqrt{3}}{2} + \frac{T_2}{2} = 10$ frac3 T_22 + fracT_22 = 10 implies 2 T_2 = 10 implies T_2 = 5 \ mathrmN$\frac{3 T_2}{2} + \frac{T_2}{2} = 10 \implies 2 T_2 = 10 \implies T_2 = 5 \ \mathrm{N}$ Substitute T_2$T_2$ back to obtain T_1$T_1$: T_1 = 5 sqrt3 \ mathrmN$T_1 = 5 \sqrt{3} \ \mathrm{N}$ Thus, the tension magnitudes are T_1 = 5sqrt3 \ mathrmN$T_1 = 5\sqrt{3} \ \mathrm{N}$ and T_2 = 5 \ mathrmN$T_2 = 5 \ \mathrm{N}$. ### Pattern Recognition Sees: Suspending particle static equilibrium with asymmetric strings. Trap: Associating components with incorrect trigonometry axes or swapping T_1$T_1$ and T_2$T_2$ in options. Shortcut: Since the incline of T_1$T_1$ (60^circ$60^\circ$) is steeper than that of T_2$T_2$ (30^circ$30^\circ$), T_1$T_1$ must carry a larger portion of the load, meaning T_1 > T_2$T_1 > T_2$. From the choices, only (2) satisfies this hierarchy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula For a block sliding down an inclined plane of inclination theta$\theta$: mg sintheta - f_k = ma$mg \sin\theta - f_k = ma$ Where normal reaction is N = mg costheta$N = mg \cos\theta$ and kinetic friction is: f_k = mu_k N = mu_k mg costheta$f_k = \mu_k N = \mu_k mg \cos\theta$ ### Core Logic Substitute f_k$f_k$ into the equation of motion: mg sintheta - mu_k mg costheta = ma$mg \sin\theta - \mu_k mg \cos\theta = ma$ Divide by m$m$: g sintheta - mu_k g costheta = a$g \sin\theta - \mu_k g \cos\theta = a$ ### Step 1: Substitute Values Given a = fracg2$a = \frac{g}{2}$ and theta = 60^circ$\theta = 60^\circ$: g sin 60^circ - mu_k g cos 60^circ = fracg2$g \sin 60^\circ - \mu_k g \cos 60^\circ = \frac{g}{2}$ fracsqrt32 - fracmu_k2 = frac12$\frac{\sqrt{3}}{2} - \frac{\mu_k}{2} = \frac{1}{2}$ sqrt3 - mu_k = 1 implies mu_k = sqrt3 - 1$\sqrt{3} - \mu_k = 1 \implies \mu_k = \sqrt{3} - 1$ ### Pattern Recognition Sees: Block sliding down with acceleration on an incline. Shortcut: The acceleration on an incline is a = g(sintheta - mu_k costheta)$a = g(\sin\theta - \mu_k \cos\theta)$. For theta = 60^circ$\theta = 60^\circ$, this is a = gleft(fracsqrt32 - fracmu_k2right)$a = g\left(\frac{\sqrt{3}}{2} - \frac{\mu_k}{2}\right)$. Equating to g/2$g/2$ gives mu_k = sqrt3 - 1$\mu_k = sqrt{3} - 1$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula vecF = m veca$\vec{F} = m \vec{a}$ vecv = vecu + vecat$\vec{v} = \vec{u} + \vec{a}t$ where, vecF$\vec{F}$ = constant force vector m$m$ = mass of body veca$\vec{a}$ = acceleration vector vecu$\vec{u}$ = initial velocity vector vecv$\vec{v}$ = final velocity vector ### Core Logic Given parameters: - Mass, m = 2mathrm~kg$m = 2\mathrm{~kg}$ - Initial velocity, vecu = 3hati + 4hatjmathrm~m/s$\vec{u} = 3\hat{i} + 4\hat{j}\mathrm{~m/s}$ - Force, vecF = 6hatkmathrm~N$\vec{F} = 6\hat{k}\mathrm{~N}$ (directed along positive z-axis) - Time interval, t = frac53mathrm~s$t = \frac{5}{3}\mathrm{~s}$ Calculate the acceleration vector veca$\vec{a}$: veca = fracvecFm = frac6hatk2 = 3hatkmathrm~m/s^2$\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k}\mathrm{~m/s}^{2}$ ### Step 1: Compute Final Velocity Using the kinematic equation of motion: vecv = vecu + vecat$\vec{v} = \vec{u} + \vec{a}t$ vecv = (3hati + 4hatj) + (3hatk) left(frac53right)$\vec{v} = (3\hat{i} + 4\hat{j}) + (3\hat{k}) \left(\frac{5}{3}\right)$ vecv = 3hati + 4hatj + 5hatkmathrm~m/s$\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$ Thus, the emerging velocity of the body is 3hati + 4hatj + 5hatkmathrm~m/s$3\hat{i} + 4\hat{j} + 5\hat{k}\mathrm{~m/s}$. ### Pattern Recognition Sees: Orthogonal initial velocity and force field direction. Shortcut: Since the force acts entirely along the z-axis, the x and y components of the velocity remain unchanged (3hati + 4hatj$3\hat{i} + 4\hat{j}$). Simply compute the z-component change: v_z = a_z t = left(frac62right) left(frac53right) = 5$v_z = a_z t = \left(\frac{6}{2}\right) \left(\frac{5}{3}\right) = 5$. Result: 3hati + 4hatj + 5hatk$3\hat{i} + 4\hat{j} + 5\hat{k}$. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

### Related Formula F_textthrust = left(fracdmdtright)v$F_{\text{thrust}} = \left(\frac{dm}{dt}\right)v$ P = F cdot v$P = F \cdot v$ ### Core Logic Given the sand dropping rate rule: fracdmdt = C sqrtv quad (textwhere C text is a constant)$\frac{dm}{dt} = C \sqrt{v} \quad (\text{where } C \text{ is a constant})$ To maintain a constant velocity v$v$, the continuous force applied by the conveyor system must balance the rate of gain of momentum of the dropped sand: F = left(fracdmdtright) v = (C sqrtv) cdot v = C v^3/2$F = \left(\frac{dm}{dt}\right) v = (C \sqrt{v}) \cdot v = C v^{3/2}$ Power delivered is the product of force and speed: P = F cdot v = (C v^3/2) cdot v = C v^5/2$P = F \cdot v = (C v^{3/2}) \cdot v = C v^{5/2}$ Squaring both sides of the expression: P^2 propto v^5$P^2 \propto v^5$ ### Pattern Recognition In variable mass problems involving dropping dust/sand at rest onto a moving frame, the thrust force always simplifies to v cdot fracdmdt$v \cdot \frac{dm}{dt}$, which makes power scale as v^2 cdot fracdmdt$v^2 \cdot \frac{dm}{dt}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula For combined system performing simple harmonic motion without relative slipping: T = 2pi sqrtfracm_texttotalk$T = 2\pi \sqrt{\frac{m_{\text{total}}}{k}}$ a = -omega^2 x = -frackM+m x$a = -\omega^2 x = -\frac{k}{M+m} x$ ### Core Logic Let's analyze each statement: - **Statement (A)**: Since both blocks perform SHM together, the combined mass is (M + m)$(M + m)$. The spring constant is k$k$. Thus, the time period of small oscillation is: T = 2pi sqrtfracM+mk$T = 2\pi \sqrt{\frac{M+m}{k}}$ This is correct. (A is True) - **Statement (B)**: When the system is displaced by x$x$, the restoring spring force on the combined system is F = -kx$F = -kx$. The common acceleration of the combined mass is: a = -frackxM+m implies |a| = frack|x|M+m$a = -\frac{kx}{M+m} \implies |a| = \frac{k|x|}{M+m}$ This matches the expression (taking magnitude). (B is True) - **Statement (C)**: The upper block of mass m$m$ moves solely due to the static frictional force f$f$ acting on it. Thus: f = m a = m left( frackxM+m right) = fracmkxM+m$f = m a = m \left( \frac{kx}{M+m} \right) = \frac{mkx}{M+m}$ Statement (C) claims the frictional force is fracmmu|x|M+m$\frac{m\mu|x|}{M+m}$, which is incorrect because friction is determined by acceleration, not by coefficient of friction mu$\mu$ during static grip. (C is False) - **Statement (D)**: For no slipping to occur, the maximum frictional force required at peak amplitude A$A$ must be less than or equal to the limiting static friction f_L = mu mg$f_L = \mu mg$: f_textmax = fracmkAM+m le mu mg$f_{\text{max}} = \frac{mkA}{M+m} \le \mu mg$ frackAM+m le mu g implies A le fracmu(M+m)gk$\frac{kA}{M+m} \le \mu g \implies A \le \frac{\mu(M+m)g}{k}$ Thus, the maximum amplitude is fracmu(M+m)gk$\frac{\mu(M+m)g}{k}$. (D is True) - **Statement (E)**: The maximum static frictional force between the blocks is f_L = mu mg$f_L = \mu mg$, not mu (M+m)g$\mu (M+m)g$. (E is False) ### Step 1: Conclusion Only statements A, B, and D are correct. Hence, the correct option is (1).

### Pattern Recognition In stacked blocks with springs, always identify the force driving the non-spring-loaded block. Here, mass m$m$ is driven purely by friction, so f = m cdot a$f = m \cdot a$. Slipping begins when this required force exceeds f_textlimit = mu m g$f_{\text{limit}} = \mu m g$. This simple boundary matches the derivation of maximum amplitude perfectly! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion: Friction Class 11 Physics: Oscillations: Simple Harmonic Motion