Question

An airtight container having a lid with negligible mass and an area of 8 \[c{{m}^{2}}\] is partially evacuated. If a 48N force is required to pull the lid off the container and the atmospheric pressure is \[{{10}^{5}}Pa\], the pressure in the container before it is opened must be
(A) 0.6 atm
(B) 0.5 atm
(C) 0.4 atm
(D) 0.2 atm

Answer 1

Hint:here we are given two surfaces one is inside the bottle and the other is top of the cap which is exposed to the atmosphere. The area is given and the force required is also given. So, we can make use of pascal’s law of pressure to solve this problem.

Complete step by step answer:
Area, A= 8 \[c{{m}^{2}}\]
Force, F= 48 N
Let the pressure inside be P and let the pressure outside which is atmospheric be P’
Given, P’=\[{{10}^{5}}Pa\]
Using Pascal’s law of pressure we get, \[P'-P=\dfrac{F}{A}\]
The given area is in \[c{{m}^{2}}\]converting it into \[{{m}^{2}}\] we get, A= \[8\times {{10}^{-4}}{{m}^{2}}\]
Substituting the values we get,
\[\begin{align}
  & {{10}^{5}}-P=\dfrac{48}{8\times {{10}^{-4}}} \\
 & {{10}^{5}}-P=60000 \\
 & P=0.4\times {{10}^{5}}Pa \\
\end{align}\]
Now we know 1 Pa= 1atm
Thus answer is 0.4 atm, hence, the correct option is (C)

Additional Information: To find the pressure at any point we have to take the total force acting at that point and divide it by the area on which it is acting. The SI unit of pressure is Pascals. Other units of pressure are atmospheric pressure (atm) which is used to measure atmospheric pressure.

Note:We have to take pressure difference here in this question because the pressure due to the atmosphere was exerted at the top and it was forcing the top to go in while the second pressure that was due to the container was in the opposite direction. Had, both of them acting in the same direction we would have added.