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**Question 1. A closed tank contains 0.5 m thick layer of mercury (specific gravity = 13.6) at the bottom. A 2.0 m thick layer of water lies above the mercury layer. A 3.0 m thick layer of oil (specific gravity = 0.6) lies above the water layer. The space above the oil layer contains air under pressure. The gauge pressure at the bottom of the tank is 196.2 kN/m ^{2}. The density of water is 1000 kg/m^{3} and the acceleration due to gravity is 9.81 m/s^{2}. The value of pressure in the air space is [2018 : 2 Marks, Set-I]**(a)

(a) 92.214 kN/m^{2}

(b) 95.644 kN/m^{2}

(c) 98.922 kN/m^{2}

(d) 99.321 kN/m^{2}

Answer:

Solution:

Note: It is a closed chamber, hence concept of absolute pressure cannot be applied.

Calculations have to be done in the form of gauge pressure.

If 5 cm

Solution:

Cross-section of U tube

= 5 mm x 5 mm = 0.25 cm

Height of water column in U tube

Now due to conservation of volume, rise in left limb will be equal to fall in right limb.

So, the new height (in cm to two decimal place of mercury in the left limb will be)

= 20 + 0.74 = 20.74 cm

The correct match between Group I and Group II is [2016 : 1 Mark, Set-II]

(a) P-2, Q-4, R-1, S-5

(b) P-2, Q-5, R-4, S-1

(c) P-2, Q-4, R-5, S-3

(d) P-2, Q-1.R-3, S-4

Answer:

Solution:

(a)

(b)

(c)

(d)

Answer:

Solution:

âˆ´ Dimension of kinematic viscosity [v]=L

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