A sand dropper drops sand of mass m(t)$m(t)$ on a conveyer belt at a rate proportional to the square root of speed (v$v$) of the belt, i.e. fracdmdt propto sqrtv$\frac{dm}{dt} \propto \sqrt{v}$ . If P$P$ is the power delivered to run the belt at constant speed then which of the following relationship is true?

### Related Formula By Newton's second law of motion, the net upward force acting on an accelerating balloon system is given by: F_textbuoyant - m_texttotal g = m_texttotal a$$F_{\text{buoyant}} - m_{\text{total}} g = m_{\text{total}} a$$ ### Core Logic Let F$F$ be the constant buoyant force acting upward on the balloon. **Case 1 (Initial upward acceleration)** : F - M g = M a implies F = M(g + a) quad text$$F - M g = M a \implies F = M(g + a) \quad \text{}$$ **Case 2 (After releasing mass x$x$)** : The new total mass becomes (M - x)$(M - x)$, and its acceleration increases to 3a$3a$: F - (M - x)g = (M - x)3a quad text$$F - (M - x)g = (M - x)3a \quad \text{}$$ Substitute the value of F$F$ from Case 1 into Case 2 [cite: 836, 839]: M(g + a) - (M - x)g = (M - x)3a$$M(g + a) - (M - x)g = (M - x)3a$$ M g + M a - M g + x g = 3 M a - 3 x a quad text$$M g + M a - M g + x g = 3 M a - 3 x a \quad \text{}$$ M a + x g = 3 M a - 3 x a$$M a + x g = 3 M a - 3 x a$$ x(g + 3a) = 2 M a$$x(g + 3a) = 2 M a$$ x = frac2 M a3a + g quad text$$x = \frac{2 M a}{3a + g} \quad \text{}$$ ### Step 1: Visual Context The free-body force layout for both accelerating phases is shown below:

### Pattern Recognition Since the upward buoyant force is completely determined by the balloon's volume, it remains constant. Expressing this constant force in terms of the initial conditions allows you to quickly solve for mass changes when acceleration states vary. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula Centripetal force configuration for simplified horizontal rotation layout: T = momega^2 l$T = m\omega^2 l$ ### Core Logic Given values: m = 0.5mathrm~kg$m = 0.5\mathrm{~kg}$, l = 50mathrm~cm = 0.5mathrm~m$l = 50\mathrm{~cm} = 0.5\mathrm{~m}$, T_textmax = 400mathrm~N$T_{\text{max}} = 400\mathrm{~N}$. Equating max tension to centripetal requirement: 400 = 0.5 times omega^2 times 0.5$$400 = 0.5 \times \omega^2 \times 0.5$$ ### Step 1: Compute Angular Velocity 400 = 0.25 omega^2$$400 = 0.25 \omega^2$$ omega^2 = frac4000.25 = 1600$$\omega^2 = \frac{400}{0.25} = 1600$$ omega = sqrt1600 = 40mathrm~rad/s$$\omega = \sqrt{1600} = 40\mathrm{~rad/s}$$ ### Pattern Recognition Ensure units are metric (50mathrm~cm to 0.5mathrm~m$50\mathrm{~cm} \to 0.5\mathrm{~m}$). Direct mapping to horizontal string projection metrics. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula Kinetic friction force: f_k = mu_k N = mu_k m_1 g$$f_k = \mu_k N = \mu_k m_1 g$$ System acceleration: a = fracF_textpull - f_km_texttotal$$a = \frac{F_{\text{pull}} - f_k}{m_{\text{total}}}$$ ### Core Logic Given values: Trolley mass m_1 = 20mathrm~kg$m_1 = 20\mathrm{~kg}$, total system mass component in frame m_texttotal = 26mathrm~kg$m_{\text{total}} = 26\mathrm{~kg}$ (from solution fraction frac60-826$\frac{60-8}{26}$). Applied pulling force F = 60mathrm~N$F = 60\mathrm{~N}$. mu_k = 0.04$\mu_k = 0.04$. Calculate the kinetic friction resisting the trolley: f_k = 0.04 times 20 times 10 = 8mathrm~N$$f_k = 0.04 \times 20 \times 10 = 8\mathrm{~N}$$ ### Step 1: Calculate Acceleration Using the dynamic equation for connected translation systems: a = frac60 - 826 = frac5226 = 2mathrm~ms^-2$$a = \frac{60 - 8}{26} = \frac{52}{26} = 2\mathrm{~ms}^{-2}$$ ### Pattern Recognition Treat connected inline systems as a single collective mass block, balancing external driving forces against collective internal friction resistance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula At the lowest point of a vertical circle, the equation of motion for a mass m$m$ is: T - mg = m r omega^2$$T - mg = m r \omega^2$$ Rearranging to solve for tension T$T$: T = mg + m r omega^2$$T = mg + m r \omega^2$$ ### Core Logic Given data: * Mass, m = 900text g = 0.9text kg$m = 900\text{ g} = 0.9\text{ kg}$ * Radius, r = 1text m$r = 1\text{ m}$ * Frequency, N = 10text rpm = frac1060text rps = frac16text rps$N = 10\text{ rpm} = \frac{10}{60}\text{ rps} = \frac{1}{6}\text{ rps}$ * Angular velocity, omega = 2pi N = 2pi left(frac16right) = fracpi3text rad/s$\omega = 2\pi N = 2\pi \left(\frac{1}{6}\right) = \frac{\pi}{3}\text{ rad/s}$ ### Step 1: Calculate Force Values Now we substitute our parameters into the tension equation: T = (0.9)(9.8) + (0.9)(1)left(fracpi3right)^2$$T = (0.9)(9.8) + (0.9)(1)\left(\frac{\pi}{3}\right)^2$$ T = 8.82 + 0.9 times fracpi^29$$T = 8.82 + 0.9 \times \frac{\pi^2}{9}$$ Since pi^2 = 9.8$\pi^2 = 9.8$: T = 8.82 + 0.1 times 9.8$$T = 8.82 + 0.1 \times 9.8$$ T = 8.82 + 0.98 = 9.80text N$$T = 8.82 + 0.98 = 9.80\text{ N}$$

### Pattern Recognition Always convert Mass to textkg$\text{kg}$ and rotational speed to textrad/s$\text{rad/s}$ first. Using the prompt constraint pi^2 = 9.8$\pi^2 = 9.8$ yields a perfect decimal addition match. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

### Related Formula For a particle in circular motion: * Normal (centripetal) acceleration: a_c = fracv^2r$a_c = \frac{v^2}{r}$ * Tangential acceleration: a_t = fracdvdt$a_t = \frac{dv}{dt}$ ### Core Logic Given a_c = a_t$a_c = a_t$: fracv^2r = fracdvdt$$\frac{v^2}{r} = \frac{dv}{dt}$$ int_v_0^v fracdvv^2 = int_0^t fracdtr$$\int_{v_0}^{v} \frac{dv}{v^2} = \int_{0}^{t} \frac{dt}{r}$$ left[ -frac1v right]_v_0^v = fractr$$\left[ -\frac{1}{v} \right]_{v_0}^{v} = \frac{t}{r}$$ -frac1v + frac1v_0 = fractr implies frac1v = frac1v_0 - fractr$$-\frac{1}{v} + \frac{1}{v_0} = \frac{t}{r} \implies \frac{1}{v} = \frac{1}{v_0} - \frac{t}{r}$$ v = fracv_01 - fracv_0 tr$$v = \frac{v_0}{1 - \frac{v_0 t}{r}}$$ ### Step 1: Relate Velocity to Position and Integrate Substitute the parameters v_0 = 4text m/s$v_0 = 4\text{ m/s}$ and r = 50text cm = 0.5text m$r = 50\text{ cm} = 0.5\text{ m}$: v = frac41 - 8t = fracdsdt$$v = \frac{4}{1 - 8t} = \frac{ds}{dt}$$ Integrating this to find the position s(t)$s(t)$: int_0^s ds = int_0^t frac41 - 8t dt$$\int_{0}^{s} ds = \int_{0}^{t} \frac{4}{1 - 8t} dt$$ s = 4 left[ fracln(1 - 8t)-8 right]_0^t = -frac12 ln(1 - 8t)$$s = 4 \left[ \frac{\ln(1 - 8t)}{-8} \right]_0^t = -\frac{1}{2} \ln(1 - 8t)$$ ### Step 2: Solve for Time of First Revolution To complete the first revolution, the distance covered is: s = 2pi r = 2pi (0.5) = pitext m$$s = 2\pi r = 2\pi (0.5) = \pi\text{ m}$$ Equating the distance: pi = -frac12 ln(1 - 8t)$$\pi = -\frac{1}{2} \ln(1 - 8t)$$ -2pi = ln(1 - 8t)$$-2\pi = \ln(1 - 8t)$$ 1 - 8t = e^-2pi$$1 - 8t = e^{-2\pi}$$ 8t = 1 - e^-2pi implies t = frac18 left[ 1 - e^-2pi right]text s$$8t = 1 - e^{-2\pi} \implies t = \frac{1}{8} \left[ 1 - e^{-2\pi} \right]\text{ s}$$ Comparing this to frac1alphaleft[ 1 - e^-2pi right]text s$\frac{1}{\alpha}\left[ 1 - e^{-2\pi} \right]\text{ s}$, we get: alpha = 8$\alpha = 8$ ### Pattern Recognition The condition a_t = a_c implies v fracdvds = fracv^2r implies fracdvv = fracdsr$a_t = a_c \implies v \frac{dv}{ds} = \frac{v^2}{r} \implies \frac{dv}{v} = \frac{ds}{r}$. Integrating directly gives v = v_0 e^s/r$v = v_0 e^{s/r}$. Substituting this back into v = ds/dt$v = ds/dt$ makes the final time integral much more intuitive. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion