A manometer has a vertical cross-sectional area of 9A and an inclined arm having area of cross-section A. The density of the manometer liquid has a specific gravity of 0.74. The scale attached to the inclined arm can read up to 0.5 mm. It is desired that the manometer shall record pressure difference $(P_{1} - P_{2})$ up to an accuracy of 0.09mm of water. To achieve this, the inclination angle $θ$ of the inclined arm should be

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$\sin^{- 1} (0.33)$

$\sin^{- 1} (0.03)$

$\sin^{- 1} (0.23)$

$\sin^{- 1} (0.13)$

answer is A.

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Detailed Solution

When a pressure difference is applied various height and length are as shown in the second figure.
Since volume of manometer liquid is constant, we have

$\begin{array}{l} ∴ 9 A h_{1} = A x \Rightarrow x = 9 h_{1} \\ h_{2} = x \sin θ = 9 h_{1} \sin θ \end{array}$
$Δ P = p g (h_{1} + h_{2}) [ρ = d e n s i t y o f l i q u i d]$
$= 0.74 ρ_{w} g [\frac{x}{9} + x \sin θ] = 0.74 ρ_{w} g x [\frac{1}{9} + \sin θ]$
It is needed that when $Δ P = 0.09 \propto 10^{- 3} ρ_{w} g,$ value of x is $0.5 \times 10^{- 3} m$
$∴ 9 \times 10^{- 6} \times ρ_{w} g = 0.74 ρ_{w} g \times 5 \times 10^{- 1} [\frac{1}{9} + \sin θ]$
$\begin{array}{l} 0.9 \\ 0.75 \times 5 \end{array} \frac{1}{9} + \sin θ$
$\sin θ = 0.24 - 0.11 = 0.13$
$∴ θ = \sin^{- 1} (0.13)$