Question

The figure below shows a drop of mercury on a glass surface. The radius of curvature of the drop at its upper point is R = 1.7 mm. What is the mass 'm' (in gm) of the drop, if its height h = 2 mm, the contact radius of the drop with the horizontal plane on which it is located is equal to r = 1 mm? The density of the mercury is $\rho$ = 13500 kg/m³, the surface tension of mercury is $\sigma$ = 425 mN/m. Take g = 10 m/s². Find m/44 to its nearest integer.

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (A)

0

Answer 2

The shape of the mercury drop can be approximated as a spherical cap. The volume of a spherical cap is given by the formula: $V = \frac{1}{3}\pi h^2 (3R_{cap} - h)$ where $h$ is the height of the cap and $R_{cap}$ is the radius of the sphere from which the cap is formed.

Given values: Height, $h = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$ Radius of curvature at the upper point, $R = 1.7 \text{ mm}$. We assume this is $R_{cap}$. $R_{cap} = 1.7 \text{ mm} = 1.7 \times 10^{-3} \text{ m}$ Density of mercury, $\rho = 13500 \text{ kg/m}^3$ Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Calculate the volume of the spherical cap: $V = \frac{1}{3}\pi (2 \times 10^{-3})^2 (3 \times 1.7 \times 10^{-3} - 2 \times 10^{-3})$ $V = \frac{1}{3}\pi (4 \times 10^{-6}) (5.1 \times 10^{-3} - 2 \times 10^{-3})$ $V = \frac{1}{3}\pi (4 \times 10^{-6}) (3.1 \times 10^{-3})$ $V = \frac{12.4}{3} \pi \times 10^{-9} \text{ m}^3$

Calculate the mass of the drop: $m = \rho \times V$ $m = 13500 \text{ kg/m}^3 \times \frac{12.4}{3} \pi \times 10^{-9} \text{ m}^3$ $m = 4500 \times 12.4 \pi \times 10^{-9} \text{ kg}$ $m = 55800 \pi \times 10^{-9} \text{ kg}$ $m = 55.8 \pi \times 10^{-6} \text{ kg}$

Convert mass to grams: $m = 55.8 \pi \times 10^{-6} \text{ kg} \times 1000 \text{ gm/kg}$ $m = 55.8 \pi \times 10^{-3} \text{ gm}$

Using $\pi \approx 3.14159$: $m \approx 55.8 \times 3.14159 \times 10^{-3} \text{ gm}$ $m \approx 175.34 \times 10^{-3} \text{ gm}$ $m \approx 0.17534 \text{ gm}$

Now, calculate $m/44$: $\frac{m}{44} = \frac{0.17534}{44} \approx 0.003985$

To find the nearest integer to $0.003985$, we round to 0.

Note: The contact radius $r=1$ mm is not used in this volume calculation method, which implies that the drop is not a perfect spherical cap for the given $R, h, r$ values. However, assuming $R$ is the radius of curvature of the sphere from which the cap is formed is a common approach. If we were to check consistency: $r^2 = h(2R_{cap} - h)$, $1^2 = 2(2 \times 1.7 - 2) = 2(3.4-2) = 2(1.4) = 2.8$. $1 \neq 2.8$, so it's not a perfect spherical cap. We proceed with the volume formula using the given $R$ as $R_{cap}$.