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A well of cross-sectional area aw is connected to an inclined tube of cross-sectional area at to form a differential pressure gauge as shown in the figure below. When p1 = p2 the common liquid level is denoted by A. When p1 > p2, the liquid level in the well is depressed to B, and the level in the tube rises by l along its length such that difference in the tube and well levels is hd. The angle of inclination θ of the tube with the horizontal is

A well of cross-sectional area aw is connected to an inclined tube of cross-sectional area at to form a differential pressure gauge as shown in the figure below. When p1 = p2 the common liquid level is denoted by A. When p1 > p2, the liquid level in the well is depressed to B, and the level in the tube rises by l along its length such that difference in the tube and well levels is hd. The angle of inclination θ of the tube with the horizontal is Correct Answer <span class="math-tex">\({\sin ^{ - 1}}\left[ {\frac{{{h_d}}}{l} - \frac{{{a_t}}}{{{a_w}}}} \right]\)</span>

= \frac{{{a_t}l}}{{{a_w}}} + l\sin \theta \\ l\sin \theta = \frac{{{h_d}}}{l} - \frac{{{a_t}}}{{{a_w}}}\\ \therefore \theta = {\sin ^{ - 1}}\left \end{array}\) 

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