A body of mass 2mathrm~kg$2\mathrm{~kg}$ moving with velocity of vecv_mathrmin = 3hati +4hatjmathrm~m/s$\vec{v}_{\mathrm{in}} = 3\hat{i} +4\hat{j}\mathrm{~m/s}$ enters into a constant force field of 6mathrm~N$6\mathrm{~N}$ directed along positive z-axis. If the body remains in the field for a period of frac53$\frac{5}{3}$ seconds, then velocity of the body when it emerges from force field is:

### 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

### Related Formula N + Fsintheta = mg implies N = mg - Fsintheta$$N + F\sin\theta = mg \implies N = mg - F\sin\theta$$ Fcostheta = f_k = mu_k N$$F\cos\theta = f_k = \mu_k N$$ W = F cdot S cdot costheta$$W = F \cdot S \cdot \cos\theta$$ ### Core Logic Since the block moves with uniform velocity, horizontal acceleration is zero. Fcos(45^circ) = mu [mg - Fsin(45^circ)]$$F\cos(45^\circ) = \mu [mg - F\sin(45^\circ)]$$ fracFsqrt2 = 0.25 left[25 times 9.8 - fracFsqrt2right]$$\frac{F}{\sqrt{2}} = 0.25 \left[25 \times 9.8 - \frac{F}{\sqrt{2}}\right]$$ fracFsqrt2 = 61.25 - 0.25 fracFsqrt2 implies 1.25 fracFsqrt2 = 61.25$$\frac{F}{\sqrt{2}} = 61.25 - 0.25 \frac{F}{\sqrt{2}} \implies 1.25 \frac{F}{\sqrt{2}} = 61.25$$ ### Step 1: Compute Force and Work Done fracFsqrt2 = frac61.251.25 = 49 implies F = 49sqrt2text N$$\frac{F}{\sqrt{2}} = \frac{61.25}{1.25} = 49 \implies F = 49\sqrt{2}\text{ N}$$ The work done by the external force over displacement S=5text m$S=5\text{ m}$ is: W = F S cos(45^circ) = (49sqrt2) times 5 times frac1sqrt2 = 245text J$$W = F S \cos(45^\circ) = (49\sqrt{2}) \times 5 \times \frac{1}{\sqrt{2}} = 245\text{ J}$$

### Pattern Recognition When uniform velocity is sustained, work done by the external pulling component perfectly matches the work consumed against internal friction dissipation (W = f_k cdot S$W = f_k \cdot S$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula Horizontal equilibrium equation: T_1 costheta_1 = T_2 costheta_2$$T_1 \cos\theta_1 = T_2 \cos\theta_2$$ Vertical balancing equation: T_1 sintheta_1 + T_2 sintheta_2 = mg$$T_1 \sin\theta_1 + T_2 \sin\theta_2 = mg$$ ### Core Logic Substitute T_1 = sqrt3T_2$T_1 = \sqrt{3}T_2$ into horizontal equilibrium: sqrt3T_2 costheta_1 = T_2 costheta_2 implies sqrt3costheta_1 = costheta_2$$\sqrt{3}T_2 \cos\theta_1 = T_2 \cos\theta_2 implies \sqrt{3}\cos\theta_1 = \cos\theta_2$$ ### Step 1: Audit Options Pattern Match Let's check option (2) where theta_1 = 60^circ$\theta_1 = 60^{\circ}$ and theta_2 = 30^circ$\theta_2 = 30^{\circ}$: sqrt3cos(60^circ) = sqrt3 cdot frac12 = fracsqrt32 $$\sqrt{3}\cos(60^{\circ}) = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2} $$ cos(30^circ) = fracsqrt32 $$ cos(30^{\circ}) = \frac{\sqrt{3}}{2} $$ This perfectly matches the horizontal check expression constraints. ### Step 2: Solve for T2 Now map the variables into the vertical component balancing equation: T_2 left[ sqrt3sin(60^circ) + sin(30^circ) ight] = mg$$ T_2 \left[ \sqrt{3}\sin(60^{\circ}) + \sin(30^{\circ}) ight] = mg$$ T_2 left[ sqrt3left(fracsqrt32 ight) + frac12 ight] = mg$$T_2 \left[ \sqrt{3}\left(\frac{\sqrt{3}}{2} ight) + \frac{1}{2} ight] = mg$$ T_2 left[ frac32 + frac12 ight] = mg implies T_2 (2) = mg implies T_2 = fracmg2$$T_2 \left[ \frac{3}{2} + \frac{1}{2} ight] = mg implies T_2 (2) = mg implies T_2 = \frac{mg}{2}$$ ### Pattern Recognition Lami's Theorem or standard rectangular splitting. When tension is scaled by sqrt3$\sqrt{3}$, it directly hints at complementary 30^circ-60^circ$30^{\circ}-60^{\circ}$ geometric alignments. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula For a conical pendulum: T cos theta = Mg$$T \cos \theta = Mg$$ T sin theta = Momega^2 R$$T \sin \theta = M\omega^2 R$$ where R = L sin theta$R = L \sin \theta$. ### Core Logic Substituting R = L sin theta$R = L \sin \theta$ into the centripetal equation:

T sin theta = M omega^2 (L sin theta)$$T \sin \theta = M \omega^2 (L \sin \theta)$$ T = M omega^2 L$$T = M \omega^2 L$$ Given rotational frequency: f = frac3pi\ mathrmrev/s$f = \frac{3}{\pi}\ \mathrm{rev/s}$ Angular velocity: omega = 2pi f = 2pi left(frac3pi ight) = 6\ mathrmrad/s$$\omega = 2\pi f = 2\pi \left(\frac{3}{\pi} ight) = 6\ \mathrm{rad/s}$$ Substituting omega$\omega$ back into the simplified tension equation: T = M (6)^2 L = 36\ ML$$T = M (6)^2 L = 36\ ML$$ ### Pattern Recognition In a conical pendulum, the horizontal projection of tension provides the exact centripetal force. The sintheta$\sin\theta$ terms cancel out, making string tension independent of the semi-vertical angle. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula The maximum velocity limit preventing outer slide breakout on a rough banked plane profile is given by standard centrifugal force equations: v_0 = sqrtrg left(fractantheta + mu1 - mutantheta ight)$$v_{0} = \sqrt{rg \left(\frac{\tan\theta + \mu}{1 - \mu\tan\theta} ight)}$$ ### Core Logic By writing out the components shown in the free body layout

: Nsintheta + fcostheta = fracmv_0^2r$$N\sin\theta + f\cos\theta = \frac{mv_{0}^{2}}{r}$$ Ncostheta - fsintheta = mg$$N\cos\theta - f\sin\theta = mg$$ ### Step 1: Isolate Coefficient of Friction Squaring the velocity boundary relation gives : fracv_0^2rg = fractantheta + mu1 - mutantheta $$\frac{v_{0}^{2}}{rg} = \frac{\tan\theta + \mu}{1 - \mu\tan\theta} $$ Cross multiply to isolate the variable terms : v_0^2 - mu v_0^2tantheta = rgtantheta + mu rg$$v_{0}^{2} - \mu v_{0}^{2}\tan\theta = rg\tan\theta + \mu rg$$ v_0^2 - rgtantheta = mu(rg + v_0^2tantheta)$$v_{0}^{2} - rg\tan\theta = \mu(rg + v_{0}^{2}\tan\theta)$$ mu = fracv_0^2 - rgtanthetarg + v_0^2tantheta$$\mu = \frac{v_{0}^{2} - rg\tan\theta}{rg + v_{0}^{2}\tan\theta}$$ ### Pattern Recognition Maximum limits always balance with plus signs in the numerator fraction. Simply rearrange that template expression directly to map the targeted parameter. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion