Question 1 |
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 1 Explanation:
First reading from level station is called BS.
Question 2 |
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 2 Explanation:
\begin{aligned} &\text { RL of bottom of bridge deck }\\ &=112.565+1.320+(2.835)\\ &=116.720 \mathrm{m} \end{aligned}
Question 3 |
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 3 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 4 |
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 4 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}
Question 5 |
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 _____
20 | |
120 | |
50 | |
100 |
Question 5 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}
\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}
Question 6 |
The combined correction due to curvature and refraction (in m) for distance of 1 km on the surface of Earth is
0.0673 | |
0.673 | |
7.63 | |
0.763 |
Question 6 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}
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}
Question 7 |
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 _______.
97.44 | |
92.32 | |
94.47 | |
88.52 |
Question 7 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}
\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}
Question 8 |
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
If RL of P is +100.000 m, then RL (in m) of R is
103.355 | |
103.155 | |
101.455 | |
100.355 |
Question 8 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
\therefore RL\text{ of }R = 101.455m
Question 9 |
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
15 | |
12.33 | |
3.52 | |
2.27 |
Question 9 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}
Question 10 |
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
6.972 | |
12.387 | |
12.54 | |
128.745 |
Question 10 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}
\begin{aligned} \text { Height of tower } &=228.758\left(\tan 3^{\circ}+\tan 0.1^{\circ}\right) \\ &=12.387 \mathrm{m} \end{aligned}
There are 10 questions to complete.
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