9702/May Jun/13/2015/Q21

A W-shaped tube contains two amounts of mercury, each open to the atmosphere. Air at
pressure P is trapped in between them. The diagram shows two vertical distances x and y.

Atmospheric pressure is equal to the pressure that would be exerted by a column of mercury of
height 760 mm. The pressure P is expressed in this way.

Which values of x, y and P are possible?

Solution:

Answer: B.

This is a relatively complex situation.

Atmosphere pressure P_A = column of mercury of height 760mm

Compare the left part of the diagram with the middle (left) part.

Since the position of the mercury column is lower at the middle, the pressure P is greater than the atmospheric pressure acting at the left ‘open’ tube.

P – P_A = x

P – 760 = x                              (1)

Now, compare the right part of the diagram with the middle (right) part.

P – P_A = 50 – y

P – 760 = 50 – y

P + y = 810                             (2)

The values form the table should satisfy both equation (1) and (2).

Choice A:

Put x = 20 and P = 780 in eqn (1): 780 – 760 = 20    [correct]

Put y = 20 and P = 780 in eqn (2): 780 + 20 = 810    [incorrect]

Choice B:

Put x = 20 and P = 780 in eqn (1): 780 – 760 = 20    [correct]

Put y = 30 and P = 780 in eqn (2): 780 + 30 = 810    [correct]

Choice C:

Put x = 30 and P = 810 in eqn (2): 810 – 760 = 30    [incorrect]

Choice D:

Put x = 30 and P = 790 in eqn (1): 790 – 760 = 30    [correct]

Put y = 30 and P = 790 in eqn (2): 790 + 30 = 810    [incorrect]

Reference: PYQ - May/Jun 2015 Paper 13 Q21