QJEEM24Phcc58: JEE Main 2024 Physics - Class 11 JEEM24Phcc58 Solution: Small water droplets of radius are formed in the upper…

To find the new terminal velocity of the larger drop formed by the coalescence of 8 smaller droplets, we need to understand the relationship between the radius of the droplets and their terminal velocity.

The terminal velocity for a small spherical droplet falling through the air is given by the Stokes' law:

$v_t = \frac{2}{9} \frac{r^2 (\rho - \rho_{\text{air}}) g}{\eta}$

where:

• $v_t$ is the terminal velocity.
• $r$ is the radius of the droplet.
• $\rho$ is the density of the droplet.
• $\rho_{\text{air}}$ is the density of the air.
• $g$ is the acceleration due to gravity.
• $\eta$ is the viscosity of the air.

Given that the radius of the small droplets is $0.01 \ \text{mm}$ and their terminal velocity is $10 \ \text{cm/s}$ , we now need to determine the radius of the larger drop formed by the coalescence of 8 smaller droplets.

When droplets coalesce, the volume of the larger drop is equal to the sum of the volumes of the smaller droplets. The volume of a sphere is given by:

$V = \frac{4}{3} \pi r^3$

Therefore, the volume of the large drop (V_large) can be calculated by:

$V_{\text{large}} = 8 \times V_{\text{small}} = 8 \times \left( \frac{4}{3} \pi r_{\text{small}}^3 \right)$

Let the radius of the larger drop be $R$ . Then:

$\frac{4}{3} \pi R^3 = 8 \times \left( \frac{4}{3} \pi r_{\text{small}}^3 \right)$

Simplifying, we get:

$R^3 = 8 r_{\text{small}}^3$

Taking the cube root on both sides:

$R = 2 r_{\text{small}}$

Therefore, the radius of the larger drop is twice the radius of the smaller droplet:

$R = 2 \times 0.01 \ \text{mm} = 0.02 \ \text{mm}$

The terminal velocity of a droplet is proportional to the square of its radius. Therefore:

$v_{t_{\text{large}}} \propto R^2$

Given that the terminal velocity of the smaller droplets is 10 cm/s, the terminal velocity of the larger drop (formed by coalescing 8 smaller droplets) is:

$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( \frac{R}{r_{\text{small}}} \right)^2$

$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( \frac{0.02 \ \text{mm}}{0.01 \ \text{mm}} \right)^2$

$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times \left( 2 \right)^2$

$v_{t_{\text{large}}} = 10 \ \text{cm/s} \times 4$

$v_{t_{\text{large}}} = 40 \ \text{cm/s}$

Therefore, the new terminal velocity of the larger drop will be 40 cm/s.