Question

There is a small hole in a hollow sphere. The water entering it is taken to a depth of $40{\text{cm}}$ underwater. The surface tension of water is $0.07N/m.$ the diameter of the hole is:
A. $7{\text{ mm}}$
B. $0.07{\text{ mm}}$
C. $0.0007{\text{ mm}}$
D. $0.7{\text{ mm}}$

Answer 1

Hint: In this question, we have been provided with a hollow sphere that is let down into water up to a certain depth. The value of surface tension has also been provided. We are supposed to determine the diameter of the hole. Using the formula of surface tension and by applying the fact that pressure exerted by the water must be equivalent to the excess pressure inside the bubble, this question can be easily solved.

Formula used:
${\text{surface tension = }}\dfrac{{{\text{force}}}}{{{\text{length}}}}$

Complete step by step solution:
The given details in this question are
The depth is equal to $40{\text{cm}}$
The surface tension of the water is $0.07N/m.$
The density of water is $1000{\text{ }}Kg/{m^3}$
From the question, we have been asked to determine the diameter of the hollow sphere.
So we know that under equilibrium conditions, the pressure exerted by water must be equivalent to the excess pressure inside the bubble. Using this information we can solve this question.
We can find force using the mathematical expression ${\text{Force = Pressure \times area}}$
Therefore force exerted by the bubble is equal to surface tension multiplied by the length. Therefore this is equal to ${\text{Surface tension \times length = S}}{\text{.T}} \times {\text{2}}\pi {\text{r}}$
The force exerted by the water is equal to the pressure multiplied by the area.
This is given by ${\text{pressure \times area = }}\rho {\text{gh}} \times \pi {{\text{r}}^2}$
Now equate the forces equal to each other. We obtain,
${\text{S}}{\text{.T}} \times {\text{2}}\pi {\text{r = }}\rho {\text{gh}} \times \pi {{\text{r}}^2}$
By simplifying and substituting the values we get,
$ \Rightarrow {\text{7}} \times {10^{ - 2}} \times {\text{2}}\pi {\text{r = 1000}} \times 10 \times 40 \times {10^{ - 2}} \times \pi {{\text{r}}^2}$
$ \Rightarrow {\text{7}} \times {\text{2 = 1000}} \times 10 \times 40 \times {\text{r}}$
$ \Rightarrow {\text{r = }}\dfrac{{{\text{7}} \times {\text{2 }}}}{{{\text{1000}} \times 10 \times 40}}$
$ \Rightarrow {\text{r = }}\dfrac{{{\text{3}}{\text{.5 }}}}{{{\text{1000}} \times 10 \times 10}}$
$ \Rightarrow {\text{r = 3}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 5}}$
So, here we have determined the value of radius in meters. Now to determine the value of diameter multiplies the radius by $2$ .
Therefore we get the value of diameter is $d = 7 \times {10^{ - 5}}m$
In millimeters, we get the value of diameter is $0.07mm$
Option (B) is the correct option.

Note:
While solving this problem care must be taken to ensure that all the corresponding values must have the same units. Another important point is that one must be careful while mentioning the unit and value of the density of water. Here in the above problem, we have substituted the value of the density of water to be ${\text{1000 kg/}}{{\text{m}}^{\text{3}}}$. The density of water in ${\text{g/c}}{{\text{m}}^{\text{3}}}$ is ${\text{1}}$ .