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ESE Civil 2015 Paper 2: Official Paper

Option 2 : 4.492 m

CT 1: Ratio and Proportion

2846

10 Questions
16 Marks
30 Mins

__Concept:__

The sensitivity of Bubble tube is given by the following expression

\(\alpha = \frac{S}{{nD}} = \frac{l}{R}\)

Where

α is the sensitivity of bubble tube i.e. angle of rotation in radian due to one division of movement of bubble.

n is the divisions moved by bubble.

l is the length of one division.

S is staff intercept and it is given as

S = S_{2} – S_{1}

S_{1 }is the staffs reading when bubble is exactly at center or actual staff reading.

S_{2} is the staff reading when bubble moves some n divisions.

R is the radius of curvature of Bubble tube.

Further,

**1 sec = 1/206265 radian**

__Calculation:__

Given, α = 25 secs/division

l = 2 mm

n = 2 divisions

D = 200 m

S_{2} = 4.54 m

Using

\({\rm{\alpha \;}}\left( {{\rm{in\;sec}}} \right) = \frac{{\rm{S}}}{{{\rm{nD}}}}\)

\({\rm{\alpha \;}}\left( {{\rm{in\;sec}}} \right) = \frac{{\rm{S}}}{{{\rm{nD}}}} \times 206265\)

\(25 = \frac{{\rm{S}}}{{2\; \times \;200}} \times 206265\)

⇒ S = 0.048 m

Now

S = S_{2} – S_{1}

0.048 = 4.54 – S_{1}

**⇒**** S _{1} = 4.492 m**