Question

A cylindrical vessel containing a liquid of density \(\rho\) is closed by a smooth piston of mass \(m\) as shown. If atmospheric pressure is \(P_{0}\), radius of cylinder is '\(r\)' and height of liquid is '\(h\)', then the force at the bottom of the cylinder is

• A. \(\left(P_{0}+\frac{mg}{\pi r^{2}}+\rho gh\right)\pi r^{2}\)
• B. \(\left(P_{0}+\frac{mg}{\pi r^{2}}\right)\pi r^{2}\)
• C. \(\left(\frac{mg}{\pi r^{2}}+\rho gh\right)\pi r\)
• D. \(\left(\frac{mg}{\pi r^{2}}\right)\pi r^{2}\)

Hint:

Pressure at the bottom is atmospheric pressure plus the piston's weight per area plus the liquid column pressure; multiply by the base area.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Pressure just below the piston is \(P_{0}+\frac{mg}{\pi r^{2}}\), since the piston's weight is spread over the area \(\pi r^{2}\).

Step 2: Calculation
Going down a depth \(h\) adds \(\rho gh\). Pressure at the bottom \(=P_{0}+\frac{mg}{\pi r^{2}}+\rho gh\). Force \(=\) pressure \(\times\) area \(=\left(P_{0}+\frac{mg}{\pi r^{2}}+\rho gh\right)\pi r^{2}\).

Step 3: Compare the options
This matches the first option. The third option multiplies by \(\pi r\), which is not an area.

Final Answer:
The force at the bottom is \(\left(P_{0}+\frac{mg}{\pi r^{2}}+\rho gh\right)\pi r^{2}\). \[ \boxed{\text{(A) }\left(P_{0}+\dfrac{mg}{\pi r^{2}}+\rho gh\right)\pi r^{2}} \]