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}{80+y}&\ldots(ii) \end{aligned}

Put value of y in eq. (ii)

\begin{aligned} 0.1853&=\frac{x+2}{60+3.3753 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,

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