Question 1 |
A delivery agent is at a location R. To deliver the order, she is instructed to travel to location P along straight-line paths of \mathrm{RC}, \mathrm{CA}, \mathrm{AB} and \mathrm{BP} of 5 \mathrm{~km} each. The direction of each path is given in the table below as whole circle bearings. Assume that the latitude (L) and departure (D) of R is (0,0) \mathrm{km}. What is the latitude and departure of P (in km, rounded off to one decimal place)?
\begin{array}{|c|c|c|c|c|} \hline Paths & RC & CA & AB & BP \\ \hline \begin{array}{c}\text {Directions} \\ \text {(in degrees)}\end{array} & 120 & 0 & 90 & 240 \\ \hline \end{array}
\begin{array}{|c|c|c|c|c|} \hline Paths & RC & CA & AB & BP \\ \hline \begin{array}{c}\text {Directions} \\ \text {(in degrees)}\end{array} & 120 & 0 & 90 & 240 \\ \hline \end{array}
L=2.5 ; D=5.0 | |
L=0.0 ; D=5.0 | |
L=5.0 ; D=2.5 | |
L=0.0 ; D=0.0 |
Question 1 Explanation:
Latitude of ' C ' =-5 \sin 30^{\circ}=-2.5 \mathrm{Km}
Departure of ' C ' =5 \cos 30^{\circ}=+4.33 \mathrm{Km}
Latitude of ' \mathrm{B} ' = Latitude of ' \mathrm{C} ' +5 \mathrm{Km} =-2.5+5=+2.5 \mathrm{Km}
Departure of 'B' = Departure of ' C ' +5 \mathrm{Km} =+4.33+5=+9.33 \mathrm{Km}
Latitude of ' P ' = Latitude of ' B ' -5 \sin 30^{\circ} =+2.5-2.5=0
Departure of ' P ' = Departure of ' B ' -5 \sin 30^{\circ} =9.33-4.33=+5 \mathrm{Km}
\therefore \quad \mathrm{L}=0.0, \quad \mathrm{D}=5.0
Question 2 |
The direct and reversed zenith angles observed by a theodolite are 56^{\circ} 00^{\prime} 00^{\prime \prime} and 303^{\circ} 00^{\prime} 00^{\prime \prime}, respectively. What is the vertical collimation correction?
+1^{\circ}00'00'' | |
-1^{\circ}00'00'' | |
-0^{\circ}30'00'' | |
+0^{\circ}30'00'' |
Question 2 Explanation:
In above example of both direct zenith angle and reversed zerwith angle are smaller than true value.
The error (\varepsilon)=\left[\frac{\left(\phi_{1}+\phi_{2}\right)-360}{2}\right]
\therefore here, error =\frac{\left(56^{\circ}+303^{\circ}\right)-360^{\circ}}{2}=-30^{\prime}
\therefore \quad Correction =+30^{\prime}
Question 3 |
The diameter and height of a right circular cylinder are 3 cm and 4 cm, respectively.
The absolute error in each of these two measurements is 0.2 cm. The absolute error in
the computed volume (in cm^3, round off to three decimal places), is _______.
1.65 | |
7.52 | |
3.25 | |
5.18 |
Question 3 Explanation:
Let diameter, x = 3 and height = y = 4 and error= \pm 0.2
\begin{aligned} V&= \pi\left ( \frac{x}{2} \right )^2y=\frac{\pi x^2 y}{4}\\ V&=f(x,y) \\ dV&= \left ( \frac{\partial V}{\partial x} \right )dx +\left ( \frac{\partial V}{\partial y} \right )dy\\ dV&= \left ( \frac{1}{2}\pi xy \right )dx +\left ( \frac{\pi x^2}{4} \right )dy\\ &= \frac{1}{2} \pi \times 3 \times 4 \times (0.2)+\frac{\pi}{4} \times (3)^2 \times (0.2)\\ &= 1.65 \pi\\ &=1.65 \times 3.14=5.18 \end{aligned}
i.e., absolute error = |5.18| = 5.18
\begin{aligned} V&= \pi\left ( \frac{x}{2} \right )^2y=\frac{\pi x^2 y}{4}\\ V&=f(x,y) \\ dV&= \left ( \frac{\partial V}{\partial x} \right )dx +\left ( \frac{\partial V}{\partial y} \right )dy\\ dV&= \left ( \frac{1}{2}\pi xy \right )dx +\left ( \frac{\pi x^2}{4} \right )dy\\ &= \frac{1}{2} \pi \times 3 \times 4 \times (0.2)+\frac{\pi}{4} \times (3)^2 \times (0.2)\\ &= 1.65 \pi\\ &=1.65 \times 3.14=5.18 \end{aligned}
i.e., absolute error = |5.18| = 5.18
Question 4 |
A series of perpendicular offsets taken from a curved boundary wall to a straight survey line at an interval of 6m are 1.22, 1.67, 2.04, 2.34, 2.14, 1.87, and 1.15 m. The area (in m^2, round off to 2 decimal places) bounded by the survey line, curved boundary wall, the first and the last offsets, determined using Simpson's rule, is _______
35.50 | |
45.25 | |
68.50 | |
75.45 |
Question 4 Explanation:
Area by Simpson's rule
\begin{aligned} A&=\frac{d}{3}[h_0+h_n+4(h_1+h_3+...)+\\ & 2(h_2+h_4+...)] \\ &= \frac{6}{3}[1.22+1.15+4 \times(1.67+2.34+1.87)+\\ & 2(2.04+2.14)]\\ &= 68.50m^2 \end{aligned}
Question 5 |
Consider the hemi-spherical tank of radius 13 m as shown in the figure. What is the volume of water (in m^3) when the depth of water at the centre of the tank is 6 m?
78 \pi | |
156 \pi | |
396 \pi | |
468 \pi |
Question 5 Explanation:
\begin{aligned} \text{Volume of water}&=\frac{1}{3}\pi h^2(3r-h)\\ &=\frac{1}{3}\pi \times 6^2 \times (3 \times 13 -6)\\ &=396\pi \end{aligned}
There are 5 questions to complete.