Test: Measurements of Area & Volume - 2 - Civil Engineering (CE) MCQ

# Test: Measurements of Area & Volume - 2 - Civil Engineering (CE) MCQ

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## 10 Questions MCQ Test Geomatics Engineering (Surveying) - Test: Measurements of Area & Volume - 2

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Test: Measurements of Area & Volume - 2 - Question 1

### Calculate the area in square meters enclosed between the survey line, the irregular boundary line, and the first and last offsets by Simpson's rule. The perpendicular offsets were taken at 20 m intervals from a survey line to an irregular boundary line: 2.35, 6.50, 5.40, 6.65, 7.85, 6.2, 2.35, 6.5, 5.65

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 1

Concept:

Simpson’s Rule

Calculation:

By Simpson’s rule

A = d/3 × [(O+ On) + 4 (O2 + O4 + … + On-2) + 2 (O3 + O5… + On-1)]

d = 20 m,

O1 + On = 2.35 + 5.65 = 8 m,

4 (O2 + O4 + …. On-2) = 4(4 6.5 + 6.65 + 6.2 + 6.5) = 103.4

2 (O+ O5 + …. On-1) = 2 (5.4 + 7.85 + 2.35) = 31.2

∴ A = 20/3 × (8 + 103.4 + 31.20) = 950.66 sq. metres

So, the correct answer will be 950 m2

Test: Measurements of Area & Volume - 2 - Question 2

### Which of the following is an application of simpson’s rule?

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 2

Application of Simpson rule related to shipping:

• The underwater (displaced) volume can be calculated in many ways, for example, if the immersed area, is divided into a number of sections throughout the length of a ship and then the area is calculated by Simpson's rule, by integrating the area of a particular section, within the given range of the length of the ship
• ​The longitudinal position of the center of buoyancy with respect to any reference point on the ship is called the longitudinal center of buoyancy (LCB).
• Usually, the reference point for locating the LCG is either of the forward or aft perpendiculars. Calculation of the longitudinal center of buoyancy of the hull is the application of Simpson's rule.
• Now, to obtain the extreme displacement, it is necessary to add this to the shell displacement, and with regard to the displacement of the shell, this is determined by first of all calculating the wetted surface area.
• This area when multiplied by the mean thickness of the shell planting gives the volume displaced by the shell. The wetted surface area is not easy to calculate since the outside surface of a ship has double curvature.
• It can be approximated by girths around the various sections and then applying Simpson's rule to find the area
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Test: Measurements of Area & Volume - 2 - Question 3

### Calculate the volume of the third section with areas of 76.32 m2 and 24.56 m2, separated by 4 m.

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 3

Concept:
Volume of prismoid

Where A1 = Area of one section (or side) of prismoid

A2 = Area of the second section (or side) of prismoid

Calculation:

Given
A1 = 76.32 m2
A2 = 24.56 m2
d = 4 m

Volume (V) = 201.76 m3

Test: Measurements of Area & Volume - 2 - Question 4

The areas enclosed by the contours in a lake are as follows:

The volume of water between the contours 270 m and 290 m by trapezoidal formula is _______.

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 4

Concept:

Trapezoidal Formula:

Volume (v) of earthwork between a number of sections having areas A1, A2, …, An spaced at a constant distance d.

Simpson’s Formula:

Volume (v) of the earthwork between a number of sections having area A1, A2, …, An spaced at constant distance d apart is

Calculation:

The areas enclosed by the contours in a lake are as follows:

Given contour interval (d) = 275-270 = 5 m

So using trapezoidal formula:
V = 5 ((50+750/2) + 200 + 400 +600) = 5 x (400 + 200 + 400 + 600) = 8000 m3

Test: Measurements of Area & Volume - 2 - Question 5

The Simpson’s rule for determination of areas is used when the number of offsets are:

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 5

Simpson's rule:

This rule is based on the assumption that the figures are trapezoids.

In order to apply Simpson's rule, the area must be divided in even number i.e., the number of offsets must be odd i.e., n term in the last offset 'On' should be odd.

The area is given by Simpson's rule:

where O1, O2, O3, .........On is the offset

Test: Measurements of Area & Volume - 2 - Question 6

A road embankment 10 m wide at the formation level with side slopes 2:1 and with an average height of 5 m is constructed with an average gradient of 1:40 from the contour 220 m to 280 m. Find the volume of earthwork.

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 6

Concept:

A gradient is the rate of rise or falls along the length of the road with respect to horizontal. It is expressed as ‘1' vertical unit to 'N' horizontal units.

Area of trapezoidal:

According to the trapezoid area formula, the area of a trapezoid is equal to half the product of the height and the sum of the two bases.

Area = ½ x (Sum of parallel sides) x (perpendicular distance between the parallel sides).

Calculation:

Average height = 5 m

Difference in elevation(h) = 280 - 220 = 60 m

L = 2400 m

Average cross-sectional area(A) = 12×(10+30)×5

A = 100 m2

Volume of earthwork = A × L

Volume of earthwork = 100 × 2400

∴ Volume of earthwork = 2,40,000 m3

Test: Measurements of Area & Volume - 2 - Question 7

The alternate name of Trapezoidal Formula's is______.

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 7

Trapezoidal Formula:
Volume (v) of earthwork between a number of sections having areas A1, A2, …, An spaced at a constant distance d.

This method is based on the assumption that the mid-area is the mean of the end areas, which make it the Average end area method

Test: Measurements of Area & Volume - 2 - Question 8

All the dimensions of the embankment were measured with 20 m chain and the volume was calculated as 368.85 cubic meter. It was then found that the chain was 10 cm too long. Find the true volume of embankment?

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 8

Concept:

Correction for Erroneous Length of Chain/Tape:

• The chain surveying depends only on linear measurement of distances. For traversing only the errors in distance measurements are of importance and significance.
• Measuring devices either chain or tape can either be longer or shorter than the designated length.
• The measured distance will be smaller than the actual if the length of the chain is longer than the designated length.
• It will be larger than the actual if the chain is shorter than the designated length.
• The actual measured distance can be corrected by the following formula :

True or Correct Distance = L'/L x measured Distance
Where L' = Actual incorrect length of chain, and L = Designated length of chain
∴ True or correct Volume of ebankment  = (L'/L)3 x Measured Volume of embankment

Given, L' = Actual incorrect length of chain = 20.1 m (as chain is 10 cm too long),  L = Designated length of chain = 20 m,  measured/calculated volume of embankment = 368.85 m3

∴ True or correct volume of embankment = (20.1/20)3 x 368.85 = 374.41 m3

Test: Measurements of Area & Volume - 2 - Question 9

A cube of side equal to 20 cm is open from the top. Its total surface area is __________ cm2.

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 9

he cube is open from the top, hence total surface area of the cube will be equal to 6l2 – l2 = 5l2
= 5(20)2
= 5(400)
= 2000.

Test: Measurements of Area & Volume - 2 - Question 10

Total surface area of a cube having sides l is given by __________.

Detailed Solution for Test: Measurements of Area & Volume - 2 - Question 10

A cube has six equal sides.
Hence, total surface area of a cube = sum of areas of 6 sides (as shown in the figure below)

Area of one side of a cube having side equal to ‘l’ = l2
Total surface area of the cube = l2 + l2 + l2+l2+ l2 + l2
= 6l2.

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## Geomatics Engineering (Surveying)

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