SSC JE Mechanical Engineering 27 Oct 2020 Shift 2 Papers

Question : 167 of 200

Marks: +1, -0

The right limb of simple U-tube manometer containing mercury is open to atmosphere while the left limb is connected to a pipe in which fluid of specific gravity 0.9 is flowing. The centre of the pipe is 12 cm below the level of mercury in the right limb. Find the pressure of fluid in the pipe if the difference of mercury level in two limbs is 20 cm.

2590 N/cm $^{2}$
2597 N/cm $^{2}$
25.97 N/cm $^{2}$
2.597 N/cm $^{2}$

Validate

Solution:

For fluid:

S_1 = 0.9, ρ_1 = 0.9 × 1000 = 900 kg\/m^3

For mercury:

S_2 = 13.6, ρ_2 = 13.6 × 1000 = 13600 kg\/m^3

Equating the pressure above A – A:

P_\pipe + P_{\left \limb} = P_{\Right \limb}

P_\pipe + ρ_1gh_1 = ρ_2gh_2

P_\pipe + 900 × 9.81 × 0.08

= 13600 × 9.81 × 0.20

P_\pipe = 25976.88 N\/m^2 = 2.597 N\/cm^2

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