Q1GATE 2025MCQ

If the horizontal distance between a staff point and the point of observation is $d$ , the error due to the curvature of earth is proportional to

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📖 Explanation

We know, error due to curvature

\begin{array}{ll} & E=+0.0785 d^{2} \\ \Rightarrow \quad & E \propto d^{2} \end{array}

Q2GATE 2025NAT

In levelling between two points A and B on the opposite banks of a river, the readings are taken by setting the instrument both at $A$ and $B$ , as shown in the table. It the RL of A is 150.000 m, the RL of B (in m) is _________ (rounded off to 3 decimal places). Level position Staff readings A B A 1.800 1.350 B 1.450 0.950

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📖 Explanation

Here, $H_{s e}=\frac{(1.8-1.35)+(1.45-0.85)}{2}=0.475 \mathrm{~m}$ $\therefore$

\begin{aligned} R L \text { of } B & =R L \text { of } A+H_{w g} \\ & =150+0.475=150.475 \mathrm{~m} \end{aligned}

Q3GATE 2024NAT

Differential levelling is carried out from point $P$ ( $B$ N $+200.000 \mathrm{~m}$ ) to point $\mathrm{R}$ . The reading taken are given in the table. Reduced level (in meters) of the point $\mathrm{R}$ is ______(rounded off to 3 decimal places)

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📖 Explanation

Using go through line (starting from $\mathrm{P}$ ) $-2.05+0.1+1.655=-0.295$ (means $\mathrm{R}$ is 0.295 lower than $\mathrm{P}$ ) $\therefore \quad \mathrm{RL}$ of $\mathrm{R}=200-0.295=199.705$

Q4GATE 2024MCQ

A map is prepared with a scale of 1: 1000 and a contour interval of $1 \mathrm{~m}$ . If the distance between two adjacent contours on the map is $10 \mathrm{~mm}$ , the slope of the ground between the adjacent contours is

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📖 Explanation

Given, contour Interval $=1 \mathrm{~m}$ Distance between contour in map $=10 \mathrm{~mm}$ $\text { Map scale }=\frac{1}{1000}$ $\therefore \quad$ Distance between contour $=10 \times 1000$ $=10,000 \mathrm{~mm}$ $\Rightarrow \quad$ Gradient $=\frac{\text { Contour Interval }}{\text { Distance between contour }}$ $=\frac{1000 \mathrm{~mm}}{10,000 \mathrm{~mm}}$ $=\frac{1}{10}=\frac{1}{10} \times 100=10 \%$

Q5GATE 2023NAT

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}

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📖 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 $=100+2.34+104.115 \tan 18^{\circ} 45^{\prime}$ $=137.682$

Q6GATE 2021MCQ

Which of the following is NOT a correct statement?

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Q7GATE 2018MCQ

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

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📖 Explanation

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

Q8GATE 2016NAT

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 _____ .

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📖 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}

Q9GATE 2016MCQ

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

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📖 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}

Q10GATE 2015NAT

In a leveling work, sum of the Back Sight (B.S.) and Fore Sight (F.S.) have been found to be 3.085 m and 5.645 m respectively. If the Reduced Level (R.L.) of the starting station is 100.000 m, the R.L. (in m) of the last station is _______.

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📖 Explanation

Using Rise and Fall method

\begin{aligned} \Sigma F S, &>\Sigma B . S \\ \therefore \text { Fall }&=\Sigma F . S,-\Sigma B . S .=R . L, \text { of first }\\ &\text{station - R.L, of last station}\\ &=5.645-3.085=2.56 \mathrm{m}\\ R.L.\text{ (last station) }&= R.L.\text{ (firet station) - Fall}\\ &=100-2.56=97.44 \mathrm{m} \end{aligned}

Q11GATE 2015NAT

Two pegs A and B were fixed on opposite banks of a 50m wide river. The level was set up at A and the staff readings on Pegs A and B were observed as 1.350 m and 1.550 m, respectively. Thereafter the instrument was shifted and set up at B. The staff readings on Pegs B and A were observed as 0.750 m and 0.550 m, respectively. If the R.L. of Peg A is 100.200 m, the R.L. (in m) of Peg B is _____

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📖 Explanation

Staff readings shows that station B is below station A

\begin{aligned} a_{1} &=1.350 \\ a_{2} &=0.550 \\ b_{1} &=1.550 \\ b_{2} &=0.750 \\ h_{18} &=\frac{\left(b_{1}-a_{1}\right)+\left(b_{2}-a_{2}\right)}{2} \\ &=\frac{(1.550-1.350)+(0.750-0.550)}{2} \\ &=0.200\\ \text{RL of }B&=\text{RL of A}-0.200 \\ &=100.200-0.200 \\ &=100.000 \mathrm{m} \end{aligned}

Q12GATE 2015MCQ

The combined correction due to curvature and refraction (in m) for distance of 1 km on the surface of Earth is

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📖 Explanation

Correction due to curvature, $C_{c}=-0.0785 d^{2}$ Correction due to refraction, $C_{f}=+0.0112 d^{2}$ $\therefore$ Composite correction

\begin{aligned} C &=-0.0785 d^{2}+0.0112 d^{2} \\ &=-0.0673 d^{2} \end{aligned}

where d is in km, C is in m. $\Rightarrow \quad C=0.0673 \times 1^{2}=0.0673 \mathrm{m}$

Q13GATE 2014MCQ

A levelling is carried out to establish the Reduced Levels (RL) of point R with respect to the Bench Mark (BM) at P. The staff readings taken are given below. If RL of P is +100.000 m, then RL (in m) of R is

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📖 Explanation

\begin{aligned} H&=A L+B S \\ \text{and }\quad R L&=H I-F S \end{aligned}

$\therefore RL\text{ of }R = 101.455m$

Q14GATE 2014MCQ

The Reduced Levels (RLs) of the points P and Q are +49.600 m and +51.870 m respectively. Distance PQ is 20 m. The distance (in m from P) at which the +51.000 m contour cuts the line PQ is

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📖 Explanation

\begin{aligned} \triangle APRS &\sim \triangle P O R, we get\\ \Rightarrow \frac{x}{(51-436)}&=\frac{20}{(5187-43.6)} \\ x&=12.33 \mathrm{m} \end{aligned}

Q15GATE 2012MCQ

The horizontal distance between two stations P and Q is 100 m. The vertical angles from P and Q to the top of a vertical tower at T are 3 $^{\circ}$ and 5 $^{\circ}$ above horizontal, respectively. The vertical angles from P and Q to the base of the tower are 0.1 $^{\circ}$ and 0.5 $^{\circ}$ below horizontal, respectively. Stations P, Q and the tower are in the same vertical plane with P and Q being on the same side of T. Neglecting earth's curvature and atmospheric refraction, the height (in m) of the tower is

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📖 Explanation

\begin{aligned} x(\tan 3+\tan 0.1)&=(x-100)\left(\tan 5^{\circ}+\tan 0.5^{\circ}\right) \\ \Rightarrow \quad x&=228.758 \end{aligned}

\begin{aligned} \text { Height of tower } &=228.758\left(\tan 3^{\circ}+\tan 0.1^{\circ}\right) \\ &=12.387 \mathrm{m} \end{aligned}

Q16GATE 2012MCQ

Which of the following errors can be eliminated by reciprocal measurements in differential leveling? I Error due to earth's curvature II Error due to atmospheric refraction

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Q17GATE 2011MCQ

Curvature correction to a staff reading in a differential leveling survey is

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Q18GATE 2010MCQ

A bench mark has been established at the soffit of an ornamental arch at the known elevation of 100.0 m above mean sea level. The back sight used to establish height of instrument is a inverted staffreading of 2.105m. A forward sight reading with normally held staff of 1.105m is taken on a recently constructed plinth. The elevation of the plinth is

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📖 Explanation

Height of instrument =BM + Back sight since, the staff is inverted, the back sight will be negative. $\therefore$ Height of instrument $=100+(-2.105)=97.897 m$ Elevation of plinth = Height of instrument-Fore sight $=97.895-1.105=96.790 \mathrm{m}$

Q19GATE 2009MCQ

Determine the correctness or otherwise of the following Assertion[a] and the Reason[r]: Assertion[a]: Curvature correction must be applied when the sights are long. Reason[r]: Line of collimation is not a level line but is tangential to the level line.

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Q20GATE 2008MCQ

A light house of 120 m height is just visible above the horizon from a ship. The correct distance (m) between the ship and the light house considering combined correction for curvature and refraction, is

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📖 Explanation

Correction due to curvature, $C_{c}=-0.0785 d^{2}$ Correction due to refraction, $C_{r}=+0.0112 d^{2}$ $\therefore$ Composte correction.

\begin{aligned} C &=-0.0785 d^{2}+0.0112 d^{2} \\ &=-0.0673 \phi^{2} \end{aligned}

where C is in metres and d is in Klometers

\begin{aligned} \text{Here}\quad C&=120 \mathrm{m}, d=?\\ 120&=0.0673 \times 0^{2} \\ \Rightarrow d&=\sqrt{1783.06} \\ \Rightarrow d&=42208 \times 1000 \\ \Rightarrow d&=4206 \mathrm{m} \end{aligned}

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