Question

A hollow sphere has a small hole in it. On lowering the sphere in a tank of water, it is observed that water enters into the hollow sphere at a depth of $40\,cm$ below the surface. The surface tension of water is $7 \times {10^{ - 2}}\,N/m$. The diameter of the hole is ?
A. $\dfrac{1}{{28}}mm$
B. $\dfrac{1}{{21}}mm$
C. $\dfrac{1}{{14}}mm$
D. $\dfrac{1}{7}mm$

Answer 1

Firstly the surface tension value has been provided. We're expected to figure out how big the hole is in the hollow sphere. The pressure exerted by the water must be equivalent to the excess pressure inside the bubble so we will find the force by using $force = pressure \times area$. Then using this we will put it in the surface tension formula. After that, we will put all the values and we will get the radius of the hole. Finally, we will multiply the radius with $2$.

Complete step by step answer:
We are given that the depth is equal to $40\,cm$. The surface tension of water is $7 \times {10^{ - 2}}N/m$. The density of water is $1000Kg/{m^3}$. Now it has been asked to find the diameter. We know that the pressure exerted by water must be equal to the excess pressure inside the bubble under equilibrium conditions. Using this we can solve the question
Using the mathematical expression for finding force
$\text{Force} = \text{pressure} \times \text{area}$

Therefore force exerted by the bubble is equal to the surface tension multiplied by the length.We will get,
$\text{surface tension} \times \text{length} = S.T \times 2\pi r$
The force exerted by the water is equal to the pressure multiplied by the area. Given by-
$pressure \times area = \rho gh \times \pi {r^2}$
By simplifying and substituting the values from above we get, take $g = 10$
$7 \times {10^{ - 2}} \times 2\pi r = 1000 \times 10 \times 40 \times \pi \times {10^{ - 2}}$
$\Rightarrow r = \dfrac{{7 \times 2}}{{1000 \times 10 \times 40}}$
$\Rightarrow r = 3.5 \times {10^{ - 5}}$

For getting the diameter we will multiply the radius by $2$. Therefore the value of diameter is,
$d = 7 \times {10^{ - 5}}m$
Changing the value in millimeter
$1m = 1000\,mm$
$\Rightarrow d = 0.07\,mm$
$\therefore d = \dfrac{1}{{14}}\,mm$

Therefore, the correct option is C.

Note: Another way to solve this problem: The pressure inside the hemisphere increases as water enters and generates a hemispheric surface,
$p = 2\left( {\dfrac{\text{surface tension}}{\text{radius}}} \right)$
But the pressure at depth $p = \text{depth(h)} \times \text{density(d)} \times \text{acceleration due to gravity (g)}$
So, $\dfrac{{2T}}{r} = hdg$
From this we will get the value of $r$ and multiplying it by $2$ we will get,
$d = \dfrac{1}{{14}}mm$