Answer ;

**(a) For diagram (a):**

According to the question, atmospheric pressure,

**P _{0} = 76 cm of Hg**

Gauge pressure is the difference in mercury levels between the two arms.

The gauge pressure is

**20 cm of Hg.**

We know that the Absolute pressure is equal to the sum of Atmospheric pressure and the Gauge pressure> therefore we get:

**= 76 + 20**

**= 96 cm of Hg**

**For diagram (b):**

Differences in mercury levels between the two arms = –18 cm

Hence, the gauge pressure is –18 cm of Hg.

We know that the Absolute pressure is equal to the sum of Atmospheric pressure and the Gauge pressure> therefore we get:

**= 76 cm – 18 cm = 58 cm**

**(b)** We are given that 13.6 cm of water is poured into the right arm of figure (b). It is given that relative density of mercury is 13.6. 1 centimeter of mercury is equivalent to a 13.6 cm column of water.

Suppose that h denotes the difference in the mercury levels of the two arms.

Now, pressure in the right arm P_{R} = Atmospheric pressure + 1 cm of Hg

**Therefore, pressure in the right arm P _{R} = 76 + 1 = 77 cm of Hg . . . . . . (a)**

The mercury column rises in the left arm,

Thus the pressure in the left limb, **P _{L} = 58 + h . . . . . . (b)**

**Equating equations (a) and (b) we get :**

**77 = 58 + h**

As a result, the difference in mercury levels between the two arms, h = 19 cm, is calculated.