Step 1: The terminal velocity \( v_t \) of a spherical object falling through a fluid is given by: \[ v_t \propto r^2 \] where \( r \) is the radius of the sphere.
Step 2: Given the ratio of the radii \( r_1/r_2 = 4/5 \), the ratio of the terminal velocities is: \[ \frac{v_{t1}}{v_{t2}} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{4}{5} \right)^2 = \frac{16}{25} \]
Step 3: Thus, the ratio of the terminal velocities is \( 16:25 \).
A cylindrical vessel, open at the top, contains 15 liters of water. Water drains out through a small opening at the bottom. 5 liters of water comes out in time t1, the next 5 litre in further time t2, and the last 5 litre in further time t3. then:
A cylinder of mass m and material density ρ hanging from a string is lowered into a vessel of cross - sectional area A containing a liquid of density σ (< ρ) until it is fully immersed. The increase in pressure at the bottom of the vessel is:
A big liquid drop splits into 'n' similar small drops under isothermal conditions, then in this process: