Two blocks with masses 100g and 200g are attached to the ends of springs A and B as shown in figure. The energy stored in A is E. The energy stored in B, when spring constants k_A$k_{A}$, k_B$k_{B}$ of A and B, respectively satisfy the relation 4k_A = 3k_B$4k_{A} = 3k_{B}$, is :

### Related Formula F = -fracdVdx$$F = -\frac{dV}{dx}$$ ### Core Logic The magnitude of the force is given by the absolute value of the slope of the potential energy versus position curve: |F| = |textslope of V-x graph|$|F| = |\text{slope of V-x graph}|$. Comparing the slopes of different regions from the graph: - Region BC has the steepest slope (largest magnitude). - Region AB has a moderate positive slope. - Region DE has a moderate negative slope, but less steep than AB. - Region CD is perfectly flat (slope = 0). ### Step 1: Ranking Forces Therefore, magnitudes of the force follow the order: F_BC > F_AB > F_DE > F_CD$$F_{BC} > F_{AB} > F_{DE} > F_{CD}$$ ### Pattern Recognition In a V-x$V-x$ graph, steeper slopes always mean a stronger restoring or repulsive force. A flat line means zero force (equilibrium). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula W = int F cdot dx = F cdot Delta x quad (textif force is constant)$$W = \int F \cdot dx = F \cdot \Delta x \quad (\text{if force is constant})$$ v = fracdxdt, quad a = fracdvdt$$v = \frac{dx}{dt}, \quad a = \frac{dv}{dt}$$ F = m a$F = m a$ ### Core Logic Given displacement equation: x(t) = 1t^2 + 1t + 1$$x(t) = 1t^2 + 1t + 1$$ Velocity is the first derivative: v(t) = 2t + 1$v(t) = 2t + 1$ Acceleration is the second derivative: a(t) = 2 text m/s^2$$a(t) = 2 \text{ m/s}^2$$ Since acceleration is constant, the force is also constant: F = m a = 2 times 2 = 4 text N$$F = m a = 2 \times 2 = 4 \text{ N}$$ ### Step 1: Finding Displacement Interval We need the displacement during the interval t=2$t=2$ to t=3$t=3$: x(3) = 3^2 + 3 + 1 = 9 + 3 + 1 = 13 text m$$x(3) = 3^2 + 3 + 1 = 9 + 3 + 1 = 13 \text{ m}$$ x(2) = 2^2 + 2 + 1 = 4 + 2 + 1 = 7 text m$$x(2) = 2^2 + 2 + 1 = 4 + 2 + 1 = 7 \text{ m}$$ Total displacement S = x(3) - x(2) = 13 - 7 = 6 text m$S = x(3) - x(2) = 13 - 7 = 6 \text{ m}$ ### Step 2: Final Conclusion Work done: W = F cdot S = 4 text N times 6 text m = 24 text J$$W = F \cdot S = 4 \text{ N} \times 6 \text{ m} = 24 \text{ J}$$ ### Pattern Recognition If x(t)$x(t)$ is a quadratic in t$t$, acceleration is constant. Work done simply becomes ma times Delta x$ma \times \Delta x$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics Class 11 Physics: Work, Energy and Power

### Related Formula W = int_r_1^r_2 vecF cdot dvecr$$W = \int_{r_1}^{r_2} \vec{F} \cdot d\vec{r}$$ oint vecF_textcons cdot dvecr = 0$$\oint \vec{F}_{\text{cons}} \cdot d\vec{r} = 0$$ ### Core Logic Evaluating Statement I: Work done by a force is defined as W = +int_r_1^r_2 vecF cdot dvecr$W = +\int_{r_1}^{r_2} \vec{F} \cdot d\vec{r}$. The negative sign represents potential energy change (Delta U = -W$\Delta U = -W$), not work done itself. Thus Statement I is incorrect. Evaluating Statement II: For a conservative force, work done depends ONLY on initial and final positions, independent of the path taken. Thus Statement II is incorrect. ### Step 1: Final Conclusion Both Statement I and Statement II are false. ### Pattern Recognition Conservative force definition: Work done is path-independent (W$W$ depends only on end points). Formula for work is +int vecF cdot dvecr$+\int \vec{F} \cdot d\vec{r}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power

### Related Formula V = sqrt2gl(1 - cos theta)$$V = \sqrt{2gl(1 - \cos \theta)}$$ V_textmin = sqrt5gR$$V_{\text{min}} = \sqrt{5gR}$$ ### Core Logic Bob A gains kinetic energy converting from potential energy when it falls from angle 60^circ$60^{\circ}$. Since the collision is elastic and masses are identical, Bob A will transfer its entire momentum to Bob B. Bob B will then use this velocity to complete the vertical circular loop. ### Step 1: Velocity of Bob A before collision Velocity of Bob A at the lowest point: V_A = sqrt2gl(1 - cos theta)$$V_{A} = \sqrt{2gl(1 - \cos \theta)}$$ V_A = sqrt2 times 10 times 1 left(1 - frac12right) = sqrt10 text m/s$$V_{A} = \sqrt{2 \times 10 \times 1 \left(1 - \frac{1}{2}\right)} = \sqrt{10} \text{ m/s}$$ ### Step 2: Velocity of Bob B after collision Due to elastic collision between identical masses, velocities are exchanged. V_B = V_A = sqrt10 text m/s$$V_{B} = V_{A} = \sqrt{10} \text{ m/s}$$ ### Step 3: Condition for completing loop To just complete the circular track of radius R$R$, the minimum velocity required at the lowest point is: V_B = sqrt5gR$$V_{B} = \sqrt{5gR}$$ sqrt10 = sqrt5 times 10 times R$$\sqrt{10} = \sqrt{5 \times 10 \times R}$$ 10 = 50R$10 = 50R$ R = frac15 text m$$R = \frac{1}{5} \text{ m}$$ ### Pattern Recognition Sees: "elastic collision" + "identical masses" → velocities swap instantly. "just completes circular path" → velocity is exactly sqrt5gR$\sqrt{5gR}$ at the bottom. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Work, Energy and Power Class 11 Physics: Systems of Particles and Rotational Motion