Question

A long cylinder of diameter 20 cm is closed at one end. A piston is located at the other end of the cylinder, as shown in the diagram. The piston is attached to a spring with the spring constant 5.0 × 10^5 N/m.

The cylinder is placed in water at 50 m depth. What is the displacement of the piston? Given: density of water = 1.0 × 10^3 kg/m3 and g = 9.8 m/s2.

Answer 1

Given :

Diameter of cylinder d = 20 cm.

Height of cylinder h = 50 m.

Spring constant k :

$k = 5 \times {10}^{5} \frac{newton}{meter}$

Density of water ρ :

$\rho \: = 1 \times {10}^{3} \frac{kg}{ {m}^{3} }$

Gravitational acceleration g :

$g = 9.8 \frac{m}{ {s}^{2} }$

To find :

Displacement of piston.

Solution :

According to Pascal's principal :

$\frac{F_1}{A_1} = \frac{F_2}{A_2} \:$

(equation 1)

where

F are the forces applied on 1st and 2nd ends of cylinder and

A are the areas of ends.

In this case area of both ends are same so,

$F_1 = F_2$

(equation 2)

Now we know that at first point,

force applied is Gravitational force and

at second point,

force applied is by spring.

As

Diameter of cylinder = 20 cm

So radius of cylinder = 10 cm = 0.1 m

So Area of cylinder base :

$A = \pi {r}^{2} = \pi( {0.1}^{2}) = \frac{\pi}{100} {m}^{2}$

and

volume of cylinder when height = 50 m :

$V = A \times h = (\frac{\pi}{100} ) \times 50 = \frac{\pi}{2} {m}^{3}$

We know that mass in this case:

$\: m = \rho \times V$

so

Mass of body :

$m \: = \: {10}^{3} \times ( \frac{\pi}{2} ) = 500\pi \: kg$

We know that

Gravitational force = mg

where

m = mass of body.

g = gravitational acceleration.

Force due to spring = kx

where

k = spring constant.

x = displacement.

Putting g = 9.8

So according to equation 2,

$F_1 = F_2 \\ mg = kx \\ (500 \times \pi) \times 9.8 = 5 \times {10}^{5} \times x \\ \\ x = \frac{500 \times \pi \times 9.8}{5 \times {10}^{5} } = 3.078 \times {10}^{ - 2 \:} m$

So

Displacement of piston in this case:

x = 3.078 cm