This solution determines the terminal velocity ($v_t$) of a sphere in a viscous fluid using Stokes' Law.
The terminal velocity ($v_t$) is calculated using:
$ v_t = \frac{2 r^2 (\rho_s - \rho_f) g}{9 \eta} $
Substituting the CGS values:
$ v_t = \frac{2 \times (0.1 \text{ cm})^2 \times (10.5 - 1.5) \text{ g/cm}^3 \times 1000 \text{ cm/s}^2}{9 \times 10 \text{ g/(cm·s)}} $
Simplify the terms:
$ v_t = \frac{2 \times 0.01 \times 9 \times 1000}{90} \text{ cm/s} $
$ v_t = \frac{180}{90} \text{ cm/s} $
$ v_t = 2 \text{ cm/s} $
The computed terminal velocity is 2 cm/s. The provided correct answer is Option C: 3 cm/s.
Two wires as shown in the figure below, made of steel and have breaking stress of $12 \times 10^8\text{ N/m}^2$. Area of cross-section of upper wire is $0.008\text{ cm}^2$ and of lower wire is $0.004\text{ cm}^2$. The maximum mass that can be added to pan without breaking any wire is _______ kg.
(take $g = 10\text{ m/s}^2$)
Which one of the following graphs represents the velocity-time ($v-t$) graph of a small spherical body falling in a viscous liquid?
A pipe $A$ is connected with other pipes $B$ and $C$ as shown in the figure. The areas of cross-section of $A, B$ and $C$ are respectively $\alpha, \frac{\alpha}{2}$ and $\frac{\alpha}{4}$. If the velocities of flow of water through $A$ and $B$ are $10\text{ m/sec}$ and $6\text{ m/sec}$, respectively, then velocity of flow, $V_{c}$ along $C$ is
