A solid steel ball of diameter 3.6mathrm~mm$3.6\mathrm{~mm}$ acquired terminal velocity 2.45times10^-2mathrm~m/s$2.45\times10^{-2}\mathrm{~m/s}$ while falling under gravity through an oil of density 925mathrm~kg~m^-3$925\mathrm{~kg~m}^{-3}$. Take density of steel as 7825mathrm~kg~m^-3$7825\mathrm{~kg~m}^{-3}$ and g as 9.8mathrm~m/s^2$9.8\mathrm{~m/s}^{2}$. The viscosity of the oil in SI unit is :

### Related Formula V_textinitial = V_textfinal$$V_{\text{initial}} = V_{\text{final}}$$ E = S times A$E = S \times A$ ### Core Logic When small drops coalesce to form a large drop, the total volume is conserved. 1000 times frac43 pi r^3 = frac43 pi R^3$$1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$$ R^3 = 1000 r^3 implies R = 10r$$R^3 = 1000 r^3 \implies R = 10r$$ ### Step 1: Calculate Surface Energies The total surface energy of 1000$1000$ small drops (mathrmE_1$\mathrm{E}_1$) is: mathrmE_1 = 1000 times 4pi r^2 times S$$\mathrm{E}_1 = 1000 \times 4\pi r^2 \times S$$ The surface energy of the single big drop (mathrmE_2$\mathrm{E}_2$) is: mathrmE_2 = 4pi R^2 times S = 4pi (10r)^2 times S = 100 times 4pi r^2 times S$$\mathrm{E}_2 = 4\pi R^2 \times S = 4\pi (10r)^2 \times S = 100 \times 4\pi r^2 \times S$$ ### Step 2: Find the Ratio fracmathrmE_1mathrmE_2 = frac1000100 = frac101$$\frac{\mathrm{E}_1}{\mathrm{E}_2} = \frac{1000}{100} = \frac{10}{1}$$ Thus, the ratio is 10:1$10:1$, which means x = 10$x = 10$. ### Pattern Recognition When N$N$ droplets merge to form one big drop, the radius scales as R = N^1/3 r$R = N^{1/3} r$. The ratio of total initial surface energy to final surface energy is N^1/3$N^{1/3}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula At terminal velocity, downward force equals upward drag (assuming negligible buoyancy): Mg = 6pi eta r v$$Mg = 6\pi \eta r v$$ v = fracMg6pi eta r$$v = \frac{Mg}{6\pi \eta r}$$ ### Core Logic Since the density of the medium is negligible, we ignore buoyant forces. The mass M$M$ of the ball is specified to remain the same in both cases, despite the change in radius (implying the material density of the second ball is lower). ### Step 1: Setup Proportionality Since M$M$, g$g$, and eta$\eta$ are all constants: v propto frac1r$$v \propto \frac{1}{r}$$ ### Step 2: Evaluating the Ratio For the second ball, r' = 2r$r' = 2r$. Therefore, the new terminal velocity v'$v'$ is: v' = v times left(fracrr'right) = v times left(fracr2rright) = fracv2$$v' = v \times \left(\frac{r}{r'}\right) = v \times \left(\frac{r}{2r}\right) = \frac{v}{2}$$ ### Pattern Recognition Read the constraints carefully. Usually, questions keep material density uniform (v propto r^2$v \propto r^2$). However, this specifically says "same mass". This shifts the formula dependency from v propto r^2$v \propto r^2$ entirely to v propto 1/r$v \propto 1/r$ because M$M$ acts as a constant numerator. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Fluids

### Related Formula mg - F_B - F_v = ma$$mg - F_B - F_v = ma$$ F_v = 6pieta r v$$F_v = 6\pi\eta r v$$ ### Core Logic

When dropped, three forces act on the ball: Gravity downwards, Buoyant force upwards, and Viscous drag upwards. mg - F_B - F_v = m fracdvdt$$mg - F_B - F_v = m \frac{dv}{dt}$$ left(rho frac43pi r^3right)g - left(rho_L frac43pi r^3right)g - 6pieta rv = m fracdvdt$$\left(\rho \frac{4}{3}\pi r^3\right)g - \left(\rho_L \frac{4}{3}\pi r^3\right)g - 6\pi\eta rv = m \frac{dv}{dt}$$ Let frac4pi r^3 g(rho - rho_L)3m = K_1$\frac{4\pi r^3 g(\rho - \rho_L)}{3m} = K_1$ and frac6pieta rm = K_2$\frac{6\pi\eta r}{m} = K_2$. fracdvdt = K_1 - K_2 v$$\frac{dv}{dt} = K_1 - K_2 v$$ Integrating from t=0, v=0$t=0, v=0$: int_0^v fracdvK_1 - K_2 v = int_0^t dt$$\int_0^v \frac{dv}{K_1 - K_2 v} = \int_0^t dt$$ -frac1K_2 ln left(fracK_1 - K_2 vK_1right) = t$$-\frac{1}{K_2} \ln \left(\frac{K_1 - K_2 v}{K_1}\right) = t$$ v = fracK_1K_2 left( 1 - e^-K_2 t right)$$v = \frac{K_1}{K_2} \left( 1 - e^{-K_2 t} \right)$$ This is an exponential curve starting from the origin and asymptotically approaching the terminal velocity V_T = K_1 / K_2$V_T = K_1 / K_2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties Of Fluids