A vertical tube, closed at one end, is immersed in water. A column of air is trapped inside the tube.

Question 45

A vertical tube, closed at one end, is immersed in water. A column of air is trapped inside the tube.

The density of water is 1000 kg m^-3.

What is the difference between the pressure of the air in the tube and the atmospheric pressure?

A 1960 Pa                   B 2940 Pa                   C 4910 Pa                   D 7850 Pa

Reference: Past Exam Paper – November 2018 Paper 11 Q13

Solution:

Answer: C.

At any specific depth in a liquid, the pressure is the same for any point at the same level.

Also at any point, pressure is due to the liquid ABOVE the point.

Consider the liquid in the vertical tube.

The pressure acting on the liquid in the tube is due to the air above only.

But as we know (from the first sentence above), the pressure at a level in the liquid is the same at the same depth in the liquid. Thus,

Pressure of air in tube (Pair) = Pressure due to (20 + 30 =) 50 cm depth of liquid outside tube + Atmospheric pressure P_atm

{In the tube, the pressure is due to air above. But at the same level outside the tube, there is 50.0 cm of liquid above a point at this level. Atmospheric pressure acts on the water surface.}

P_air = Pressure due to (20 + 30 =) 50 cm depth of liquid + P_atm

P_air – P_atm = Pressure due to (20 + 30 =) 50 cm depth of liquid

P_air – P_atm = hρg = 0.50 × 1000 × 9.81

P_air – P_atm = 4905 Pa = 4910 Pa

This is the difference between the pressure of air in the tube and the atmospheric pressure.

A charged particle of mass m and charge +q is travelling with velocity v in a vacuum. It enters a region of uniform magnetic field of flux density B, as shown in Fig. 5.1.

Question 16

(a) State what is meant by a magnetic field. [2]

(b) A charged particle of mass m and charge +q is travelling with velocity v in a vacuum.

It enters a region of uniform magnetic field of flux density B, as shown in Fig. 5.1.

Fig. 5.1

The magnetic field is normal to the direction of motion of the particle. The path of the

particle in the field is the arc of a circle of radius r.                              

(i) Explain why the path of the particle in the field is the arc of a circle. [2]

(ii) Show that the radius r is given by the expression

r = mv / Bq

 [2]

(c) A thin metal foil is placed in the magnetic field in (b).

A second charged particle enters the region of the magnetic field. It loses kinetic energy

as it passes through the foil. The particle follows the path shown in Fig. 5.2.

Fig. 5.2

(i) On Fig. 5.2, mark with an arrow the direction of travel of the particle. [1]

(ii) The path of the particle has different radii on each side of the foil.

The radii are 7.4 cm and 5.7 cm.

Determine the ratio

final momentum of particle

initial momentum of particle

for the particle as it passes through the foil. [2]

Reference: Past Exam Paper – June 2011 Paper 41 Q5

Solution:

(a) A magnetic field is a region (of space) where there is a force on a moving charge.

(b)

(i) The force on the particle is (always) normal to the velocity / direction of travel. The speed of the particle is constant.

(ii)

{The magnetic force provides the centripetal force.}

mv² / r = Bqv

{Make r the subject of formula.}

Radius r = mv / Bq

(c)

(i) The direction is from ‘bottom to top’ of the diagram

{From Fig 5.2, we can observe that the path is NOT the arc of a circle as in Fig 5.1.

We are 2 options for the direction of the path: either from right to left or from bottom to top of the diagram.

As the charged particle passes through the coil, it loses some kinetic energy. Its speed v decreases and so, the radius of the path also decreases (since r = mv / Bq). In the diagram, this reduction can be observed as we pass through the foil from the bottom to top and NOT from right to left along the path.}

(ii)

The radius is proportional to momentum.

{r = mv / Bq

mv = p where p is the momentum.

r = p / Bq

B (same field) and q (same particle) are constant.

So, the momentum p (= mv) is directly proportional to r.

Ratio = final momentum / initial momentum

Ratio = final radius / initial radius}

Ratio = 5.7 / 7.4 = 0.77