Question 1 |

Trigonometric levelling was carried out from two stations P and Q to find the reduced level (R.L.) of the top of hillock, as shown in the table. The distance between stations P and Q is 55 \mathrm{~m}. Assume stations P and Q, and the hillock are in the same vertical plane. The R. L. of the top of the hillock (in\mathrm{m} ) is _____ (round off to three decimal places)

\begin{array}{|l|l|l|l|} \hline Station &\begin{array}{c}\text{Vertical} \\ \text{angle of} \\ \text {the top of} \\ \text{hillock }\end{array} &\begin{array}{c}\text {Staff} \\ \text{reading} \\ \text {on} \\ \text{benchmark}\end{array} &\begin{array}{c}\text {R.L. of} \\ \text {benchmark}\end{array} \\ \hline P &18^{\circ} 45^{\prime} &2.340 \mathrm{~m} &100.000 \mathrm{~m} \\ \hline Q &12^{\circ} 45^{\prime} &1.660 \mathrm{~m}& \\ \hline \end{array}

\begin{array}{|l|l|l|l|} \hline Station &\begin{array}{c}\text{Vertical} \\ \text{angle of} \\ \text {the top of} \\ \text{hillock }\end{array} &\begin{array}{c}\text {Staff} \\ \text{reading} \\ \text {on} \\ \text{benchmark}\end{array} &\begin{array}{c}\text {R.L. of} \\ \text {benchmark}\end{array} \\ \hline P &18^{\circ} 45^{\prime} &2.340 \mathrm{~m} &100.000 \mathrm{~m} \\ \hline Q &12^{\circ} 45^{\prime} &1.660 \mathrm{~m}& \\ \hline \end{array}

124.365 | |

137.682 | |

253.142 | |

254.325 |

Question 1 Explanation:

While taking reading from P,

RL of hillock =H I_P+x \tan 18^{\circ} 45^{\prime}

=100+2.34+x \tan 18^{\circ} 45^{\prime}

While taking reading from Q.

\mathrm{RL} of hillock =\mathrm{HI}_Q+(\mathrm{x}+55) \tan 12^{\circ} 45^{\prime}

=100+1.66+(x +55)\tan 12^{\circ} 45^{\prime}

Equating both,

100+2.34+x \tan 18^{\circ} 45^{\prime}=100+1.66+(x +55)\tan 12^{\circ} 45^{\prime}

2.34+0.339 x=1.66+12.445+0.226 x

\therefore x=104.115

\quad \therefore \quad RL of hillock

=100+2.34+104.115 \tan 18^{\circ} 45^{\prime}

=137.682

Question 2 |

Which of the following is NOT a correct statement?

The first reading from a level station is a 'Fore Sight' | |

Basic principle of surveying is to work from whole to parts | |

Contours of different elevations may intersect each other in case of an overhanging cliff | |

Planimeter is used for measuring 'area' |

Question 2 Explanation:

First reading from level station is called BS.

Question 3 |

A level instrument at a height of 1.320 m has been placed at a station having a Reduced Level (RL) of 112.565 m. The instrument reads -2.835 m on a levelling staff held at the bottom of a bridge deck. The RL (in m) of the bottom of the bridge deck is

116.72 | |

116.08 | |

114.08 | |

111.05 |

Question 3 Explanation:

\begin{aligned} &\text { RL of bottom of bridge deck }\\ &=112.565+1.320+(2.835)\\ &=116.720 \mathrm{m} \end{aligned}

Question 4 |

The vertical angles subtended by the top of a tower T at two instrument stations set up at P and Q, are shown in the figure. The two stations are in line with the tower and spaced at a distance of 60m. Readings taken from these two stations on a leveling staff placed at the benchmark (BM = 450.000 m) are also shown in the figure. The reduced level of the top of the tower T (expressed in m) is _____ .

450.25 | |

474.35 | |

24.35 | |

476.91 |

Question 4 Explanation:

\begin{aligned} \ln \Delta B E T \\ \tan 16.5^{\circ}&=\frac{x}{y}\\ \Rightarrow \quad y&=3.3759 x &\ldots(i)\\ In \triangle A O T \\ \tan 10.5&=\frac{x+2}{60+y}&\ldots(ii) \end{aligned}

Put value of y in eq. (ii)

\begin{aligned} 0.1853&=\frac{x+2}{60+3.3759 x} \\ 11.1203+&0.6255 x=x+2 \\ x=& 24.3588 \mathrm{m} \end{aligned}

So, the reduced level of tower

\begin{aligned} T&=450.000+2.555+24.358\\ &=476.911 \mathrm{m} \end{aligned}

Question 5 |

The staff reading taken on a workshop floor using a level is 0.645 m. The inverted staff reading taken to the bottom of a beam is 2.960 m. The reduced level of the floor is 40.500 m. The reduced level (expressed in m) of the bottom of the beam is

44.105 | |

43.46 | |

42.815 | |

41.145 |

Question 5 Explanation:

Height of instrument, HI

A.L. of floor + stall reading from floor

\begin{array}{l} =40.500+0.645 \\ =41.145 \mathrm{m} \end{array}

R.L. of bottom of beam =H / 4 inverted staff reading taken from bottom of beam

\begin{array}{l} =41.145+2.800 \\ =44.105 \mathrm{m} \end{array}

A.L. of floor + stall reading from floor

\begin{array}{l} =40.500+0.645 \\ =41.145 \mathrm{m} \end{array}

R.L. of bottom of beam =H / 4 inverted staff reading taken from bottom of beam

\begin{array}{l} =41.145+2.800 \\ =44.105 \mathrm{m} \end{array}

There are 5 questions to complete.

Thanks a lot man,this made my work easy

and saves my time.

Thank you Subhan,

Please share this among all our friends. SO everyone can gets the benefit of this free resorces.

in Question 4 The value of Tan (10.5 degree) comes out to be 0.1664 and It is given Wrong in Solution. Please clear my Doubt.

Value of Tan (10.5 degree) is correct in answer.

Really love the effort you have put on .

Just a request if possible could you do the same for ESE civil PYQs- topic wise, subject wise, year wise. it will be large help for us