Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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}
Question 4 |
An observer standing on the deck of a ship just sees the top of a lighthouse. The top of the lighthouse is 40 m above the sea level and the height of the observer's eye is 5 m above the sea level. The distance (in km, up to one decimal place) of the observer form the lighthouse is____
15.6 | |
8.6 | |
24.4 | |
33.0 |
Question 4 Explanation:
Distance of observer trom the lighthouse,
\begin{aligned} d &=d_{1}+d_{2} \\ &=3.865\left(\sqrt{h_{1}}+\sqrt{h_{0}}\right) \\ &=3.865(\sqrt{5}+\sqrt{40}) \\ &=33.0012 \mathrm{km} \end{aligned}
Question 5 |
The VPI (vertical point of intersection) is 100 m away (when measured along the horizontal) from the VPC (vertical point of curvature). If the vertical curve is parabolic, the length of the curve (in meters and measured along the horizontal) is______
200 | |
100 | |
150 | |
50 |
Question 5 Explanation:
L_{2}=2\times 100=200m
Question 6 |
A circular curve of radius R connects two straights with a deflection angle of 60^{\circ}. The tangent length is
0.577 R | |
1.155 R | |
1.732 R | |
3.464 R |
Question 6 Explanation:
Tangent length,
V T_{1}=R \tan \frac{\Delta}{2}=R \tan 30^{\circ}=0.577 R
Question 7 |
The chainage of the intersection point of two straights is 1585.60 m and the angle of intersection is 140^{\circ}. If the radius of a circular curve is 600.00 m, the tangent distance (in m) and length of the curve (in m), respectively are
418.88 and 1466.08 | |
218.38 and 1648.49 | |
218.38 and 418.88 | |
418.88 and 218.38 |
Question 7 Explanation:
A=180^{\circ}-140^{\circ}=40^{\circ}
Lengin of the gurve =\frac{\pi G}{180^{\circ}} \mathrm{A}
=\frac{\pi \times 600}{180} \times 40=418.82 \mathrm{m}
Tangent distance (7) is the distance between
P.C to P.I (also the distance from P.I to P.)
\Rightarrow T=T_{1} V
=T_{2} V=O T_{1} \tan \frac{\Delta}{2}=R \tan \frac{\Delta}{2}
=600 \tan 20^{\circ}=218.88 \mathrm{m}
Question 8 |
A tachometer was placed at point P to estimate the horizontal distances PQ and PR. The corresponding stadia intercepts with the telescope kept horizontal, are 0.320 m and 0.210 m, respectively. The \angle QPR is measured to be 61^{\circ}{30}'{30}''. If the stadia multiplication constant=100 and stadia addition constant=0.10 m, the horizontal distance (in m) between the points Q and R is ________
21.1 | |
32.1 | |
28.8 | |
36.8 |
Question 8 Explanation:
\begin{aligned} P Q &=k s+C \\ &=100(0.32)+0.1=32.1 \mathrm{m} \\ P R &=k s+C \\ &=100(0.21)+0.1=21.1 \mathrm{m} \end{aligned}
Applying the cosine rule
\begin{array}{l} Q R=\sqrt{P Q^{2}+P R^{2}-2(P Q)(P R) \cos \theta} \\ \text { Where, } \theta=61^{\circ} 30^{\prime} 30^{\prime \prime}=61.508^{\circ} \\ =\sqrt{1030,41+445.21-2(32,1)(21,1) \cos 61.508^{\circ}} \\ =\sqrt{1030.41+445.21-646.2}=28.8 \mathrm{m} \end{array}
Applying the cosine rule
\begin{array}{l} Q R=\sqrt{P Q^{2}+P R^{2}-2(P Q)(P R) \cos \theta} \\ \text { Where, } \theta=61^{\circ} 30^{\prime} 30^{\prime \prime}=61.508^{\circ} \\ =\sqrt{1030,41+445.21-2(32,1)(21,1) \cos 61.508^{\circ}} \\ =\sqrt{1030.41+445.21-646.2}=28.8 \mathrm{m} \end{array}
Question 9 |
A road is provided with a horizontal circular curve having deflection angle of 55^{\circ}
and centre line radius of 250 m. A transition curve is to be provided at each end
of the circular curve of such a length that the rate of gain of radial acceleration
is 0.3 m/s^{3} at a speed of 50 km per hour. Length of the transition curve required at each of the ends is
2.57m | |
33.33m | |
35.73m | |
1666.67m |
Question 9 Explanation:
If \alpha is the rate of change of radial acceleration,
the radial acceleration (a) attained during the time the vehicle passes over the transition curve is given by
a=\alpha t=\alpha \times \frac{L}{V}
Radial acceleration,
\begin{aligned} a&=\frac{v^{2}}{R} \\ \therefore \quad \alpha \times \frac{L}{V}&=\frac{v^{2}}{R} \\ \Rightarrow \quad L&=\frac{V^{3}}{\alpha R} \\ \therefore \quad L&=\frac{\left(\frac{50 \times 1000}{60 \times 60}\right)^{3}}{0.3 \times 250}\\ \Rightarrow \quad L&=35.73 \mathrm{m} \end{aligned}
the radial acceleration (a) attained during the time the vehicle passes over the transition curve is given by
a=\alpha t=\alpha \times \frac{L}{V}
Radial acceleration,
\begin{aligned} a&=\frac{v^{2}}{R} \\ \therefore \quad \alpha \times \frac{L}{V}&=\frac{v^{2}}{R} \\ \Rightarrow \quad L&=\frac{V^{3}}{\alpha R} \\ \therefore \quad L&=\frac{\left(\frac{50 \times 1000}{60 \times 60}\right)^{3}}{0.3 \times 250}\\ \Rightarrow \quad L&=35.73 \mathrm{m} \end{aligned}
Question 10 |
The focal length of the object glass of a tacheometer is 200 mm, the distance
between the vertical axis of the tacheometer and the optical centre of the object
glass is 100 mm and the spacing between the upper and lower line of the diagram
axis is 4 mm. With the line of collimation perfectly horizontal, the staff intercepts
are 1 m (top), 2m (middle), and 3 m (bottom). The horizontal distance (m)
between the staff and the instrument station is
100.3 | |
103 | |
150 | |
153 |
Question 10 Explanation:
The horizontal distance (D) between the vertical axis and staff may be given as
\begin{aligned} D &=u+d \\ \text { But } u &=\left(\frac{f}{i}\right) s+f \\ D &=\left(\frac{f}{i}\right) s+f+d \\ &=\left(\frac{200}{4}\right) \times(3-1)+\frac{200}{1000}+\frac{100}{1000} \\ &=100+0.3=100.3 m \end{aligned}
There are 10 questions to complete.