Question

The diameter of the neck and bottom of a bottle are 2cm and 10cm respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of 1.2kgf, the amount of force applied on the bottom is
A. F= 30 kgf
B. F = 15 kgf
C. F = 1.2 kgf
D. F = 60 kgf

Answer 1

Hint: We are given the diameter of the neck and bottom of the bottle and the force applied on the cork to close the bottle. We know that when we apply force on the cork, it is applied to the neck region of the bottle. By applying Pascal’s law and equating the pressure at two regions we can find the force at the bottom region of the bottle.

Formula used:
Pressure, $P=\dfrac{F}{A}$

Complete answer:
In the question we are given the diameter of the neck and bottom of a bottle.
Let ‘\[{{d}_{1}}\]’ and ‘${{d}_{2}}$’ be the diameter of the neck and bottom of the bottle respectively. We are given,
${{d}_{1}}=2cm$
${{d}_{2}}=10cm$
Now let ‘${{r}_{1}}$’ and ‘${{r}_{2}}$’ be the radius of the neck and bottom respectively. Then we know that,
$\begin{align}
  & {{r}_{1}}=\dfrac{{{d}_{1}}}{2} \\
 & \Rightarrow {{r}_{1}}=\dfrac{2}{2}=1cm=1\times {{10}^{-2}}m \\
\end{align}$
$\begin{align}
  & {{r}_{2}}=\dfrac{{{d}_{2}}}{2} \\
 & \Rightarrow {{r}_{2}}=\dfrac{10}{2}=5cm=5\times {{10}^{-2}}m \\
\end{align}$
In the question it is said that the bottle is corked with a force of 1.2 kgf.
We know that when we apply pressure at one point on a body, that pressure will be equally transmitted throughout the body.
Therefore we can say pressure at the neck region of the bottle is equal to the pressure at bottom of the bottle.
We know that pressure is given by,
$P=\dfrac{F}{A}$, where ‘F’ is the force applied and, ‘A’ is the area where the force is felt.
Now the pressure at neck region can be written as,
${{P}_{1}}=\dfrac{{{F}_{1}}}{{{A}_{1}}}$
And the pressure at the bottom can be written as,
${{P}_{2}}=\dfrac{{{F}_{2}}}{{{A}_{2}}}$
Since the pressure is equal at both regions, we can equate these equations.
Thus we get,
${{P}_{1}}={{P}_{2}}$
$\Rightarrow \dfrac{{{F}_{1}}}{{{A}_{1}}}=\dfrac{{{F}_{2}}}{{{A}_{2}}}$
We need to find the amount of force applied at the bottom of the bottle. From the above equation we can write the force applied at the bottom ${{F}_{2}}$ as,
$\Rightarrow {{F}_{2}}=\dfrac{{{F}_{1}}{{A}_{2}}}{{{A}_{1}}}$
In the question we are given the value of the force applied at the neck region,
${{F}_{1}}=1.2kgf$
And we also know that area of circular region,
$A=\pi {{r}^{2}}$
Thus area at the neck region,
${{A}_{1}}=\pi {{r}_{1}}^{2}$
$\Rightarrow {{A}_{1}}=\pi \times {{\left( 1\times {{10}^{-2}} \right)}^{2}}$
$\Rightarrow {{A}_{1}}={{10}^{-4}}\pi {{m}^{2}}$
Now the area of the bottom region,
${{A}_{2}}=\pi {{r}_{2}}^{2}$
$\Rightarrow {{A}_{2}}=\pi {{\left( 5\times {{10}^{-2}} \right)}^{2}}$
$\Rightarrow {{A}_{2}}=25\times {{10}^{-4}}\pi {{m}^{2}}$
Now let us apply all these values in the equation for force at the bottom of the bottle,
${{F}_{2}}=\dfrac{{{F}_{1}}{{A}_{2}}}{{{A}_{1}}}$
$\Rightarrow {{F}_{2}}=\dfrac{\left( 1.2 \right)\times \left( 25\times {{10}^{-4}}\pi \right)}{\left( {{10}^{-4}}\pi \right)}$
By solving this, we get
$\Rightarrow {{F}_{2}}=1.2\times 25$
$\Rightarrow {{F}_{2}}=30kgf$
Therefore we got the amount of force applied at the bottom of the bottle as 30 kgf.

Hence the correct answer is option A.

Note:
According to Pascal’s law when we apply a static pressure on a fluid surface, that pressure will be uniformly transmitted throughout the fluid.
Similarly in this case when we apply a pressure at the neck region using a cork, this pressure gets uniformly transmitted throughout the bottle. Hence the bottom region of the bottle will also experience the same pressure.