# What is the volume of earthwork for constructing a tank that is excavated in the level ground to a depth of 4 m ? The top of the tank is rectangular in shape having an area of 50 m × 40 m and the side slope of the tank is 2: 1 (horizontal: vertical).

1. 5461 m3
2. 6688 m3
3. 8866 m3
4. 5632 m3

Option 2 : 6688 m3

## Calculation of Area and Volume MCQ Question 1 Detailed Solution

Concept:

The prismoidal formula is also known as Simpson’s rule.

a) Trapezoidal Formula:

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

$${\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1{\rm{\;}}}}} \right]$$

b) Simpson’s formulae:

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

$${\rm{V}} = \frac{{\rm{d}}}{3}[\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)$$

c) Simple prismoidal rule:

$$V = \frac{D}{3}\left[ {({A_1} + 4Am + {A_2})} \right]$$

Calculation:

Given, A1 = 50 × 40 = 2000 m2, As the side slope is given 2:1 i.e. H:V

So for a depth of 1 m, there is a change of 2 m in a Horizontal Direction.

So at 4 m vertical depth

Bottom Dimension is ( 50 - 8 ) = 42 m & (40 - 8) = 32 m

∴ The bottom area is 42 m × 32 m = 1344 m2

Mean area (A m) = (A1 + A2)/2 = (2000 + 1344) /2 = 1672 m2

According to simple Trapezoidal rule for volume,

$$V = \frac{D}{2}\left[ {({A_1} + 2Am + {A_2})} \right]$$

$$V = \frac{2}{2}\left[ {(2000 + 2\times (1672) + 1344)} \right]$$ = 6688 m3

Note: For calculating A2 (as per slope 2: 1), 8 m is reduced from both ends. So, 16 m is reduced on both sides

# Prismoidal correction, while surveying is always?

1. Exponentially subtractive
3. Subtractive

Option 3 : Subtractive

## Calculation of Area and Volume MCQ Question 2 Detailed Solution

Explanation:

The volume of earthwork by trapezoidal method  = V1

V1 =  $$common\: distance\left \{ \frac{First\: area + Last\: area}{2}+the \:sum \:of\: remaining \:area \right \}$$

The volume of earthwork by prismoidal formula = V2

V2 = $$=\frac{Common\: distance}{3}\left \{ First\: area+ Last\: area + 2(Sum\: of odd\: area) + 4(Sum\: of even\: area)\right \}$$

Prismoidal correction:

• The volume by the prismoidal formula is more accurate than any other method
• But the trapezoidal method is more often used for calculating the volume of earthwork in the field.
• The difference between the volume computed by the trapezoidal formula and the prismoidal formula is known as a prismoidal correction.
• Since the trapezoidal formula always overestimates the volume, the prismoidal correction is always subtractive in nature is usually more than calculated by the prismoidal formula, therefore the prismoidal correction is generally subtractive.
• Volume by prismoidal formula = volume by the trapezoidal formula - prismoidal correction

Prismoidal correction (CP)

$$C_{P}=\frac{DS}{6}\left \{ d- d_{1}\right \}^{2}$$

Where, D = Distance between the sections, S (Horizontal) : 1 (Vertical) = Side slope, d and d1 are the depth of earthwork at the centerline

# The areas enclosed by the contours in a lake are as follows: Contour (m) 270 275 280 285 290 Area (m2) 50 200 400 600 750 The volume of water between the contours 270 m and 290 m by trapezoidal formula is _______.

1. 6400 m3
2. 8000 m3
3. 16000 m3
4. 24000 m3

Option 2 : 8000 m3

## Calculation of Area and Volume MCQ Question 3 Detailed Solution

Concept:

Trapezoidal Formula:

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

$${\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right]$$

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

$${\rm{V}} = \frac{{\rm{d}}}{3}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)} \right]$$

Calculation:

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

 Contour (m) 270 275 280 285 290 Area (m2) 50 200 400 600 750

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

So using trapezoidal formula:

$${\rm{V}} = 5\left( {\left( {\frac{{50 + 750}}{2}} \right) + 200 + 400 + 600} \right) = 5 \times \left( {400 + 200 + 400 + 600} \right) = 8000{\rm{\;}}{{\rm{m}}^3}$$

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

1. 2
2. even
3. 3
4. odd

Option 4 : odd

## Calculation of Area and Volume MCQ Question 4 Detailed Solution

Explanation:

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:

$$Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + ........ + {O_{n - 1}}) + 2({O_3} + {O_5} + ......{O_{n - 2}})} \right]$$

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

Important Points

• In case of an even number of cross-sections, the end strip is treated separately and the area of the remaining strip is calculated by Simpson's rule. The area of the last strip can be calculated by either trapezoidal or Simpson's rule.

# 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.

1. 6,40,000 m3
2. 1,40,000 m3
3. 3,40,000 m3
4. 2,40,000 m3

Option 4 : 2,40,000 m3

## Calculation of Area and Volume MCQ Question 5 Detailed Solution

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.

$$Tan\ \theta \ =\frac{{\bf{h}}}{{\bf{l}}}$$

$$\frac{{\bf{h}}}{{\bf{l}}} = \frac{1}{{N}}$$

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

Average gradient = $$\frac{1}{{40}}$$

$$\frac{{\bf{h}}}{{\bf{l}}} = \frac{1}{{40}}$$

$$\frac{{\bf{60}}}{{\bf{l}}} = \frac{1}{{40}}$$

L = 2400 m

Average cross-sectional area(A) = $$% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaaGymaaWdaeaapeGaaGOmaaaacqGHxdaTdaqa % daWdaeaapeGaaGymaiaaicdacqGHRaWkcaaIZaGaaGimaaGaayjkai % aawMcaaiabgEna0kaaiwdaaaa!423F! \frac{1}{2} \times \left( {10 + 30} \right) \times 5$$$$\frac{1}{2} \times \left( {10 + 30} \right) \times 5$$

A = 100 m2

Volume of earthwork = A × L

Volume of earthwork = 100 × 2400

∴ Volume of earthwork = 2,40,000 m3

# Determine the approximate quantity of earthwork for a road in embankment having a length of 120 m on a uniform level ground. The width of formation is 10 m and side slopes are 3 ∶ 1. The heights of the bank at the ends are 1 m and 1.5 m, respectively. Use trapezoidal method considering average of areas at the two ends.

1. 1785 m3
2. 1485 m3
3. 1885 m3
4. 2085 m3

Option 4 : 2085 m3

## Calculation of Area and Volume MCQ Question 6 Detailed Solution

Concept:

Volume of Sloped earthwork (V) = (bd + sd2) × L

Where, B   = Width, d  = Depth and S = Side slope of the cross-section

Calculation:

Given,

slope = s: 1 = 3: 1

b = 10 m

L = 120 m

d1 = 1.5 m and d2 = 1 m

Volume of Sloped earthwork (V) = (bd + sd2) × L

$$V = \;\frac{1}{2}\left[ {(b{d_1}\; + {\rm{ }}s{d_1}^2) + (b{d_2}\; + {\rm{ }}s{d_2}^2)} \right] \times 120$$

$$V = \;\frac{1}{2}\left[ {(10 \times 1.5\; + {\rm{ }}3 \times {1.5^2}) + (10 \times 1\; + {\rm{ }}3 \times {{1}^2})} \right] \times 120 = 2085{m^3}$$

The approximate quantity of earthwork = 2085 m3

# The formation width of a highway with no transverse slope on an embankment is B. The height of the embankment is d and side slope of the embankment is S: 1 The area of cross-section of the embankment is

1. B + d + S × d

2. B × D + S × d

3. B × d + S × d2
4. $$\frac{1}{2}\left( {B\times d + S{d^2}} \right)$$

Option 3 : B × d + S × d2

## Calculation of Area and Volume MCQ Question 7 Detailed Solution

Explanation:

Given,

B = Formation width

d = Height of the embankment

S:1 = Side slope of the embankment (H: V)

The area of embankment = The area of the trapezium

Area of Trapezium = $$\frac{1}{2}$$ × (Sum of Lengths of parallel sides) × Height

Area of Embankment $$= \frac{1}{2}\left[ {B + \left( {B + 2sd} \right)} \right] \times d$$

A = Bd + sd2

# What is the volume of 8 m deep tank having rectangular shaped top 6 m × 4 m and bottom 4 m × 2 m by prismoidal formula is?

1. 123 m3
2. 122.6 m3
3. 132 m3
4. 134 m3

Option 2 : 122.6 m3

## Calculation of Area and Volume MCQ Question 8 Detailed Solution

Volume of tank by prismoidal formula is given by:

$$V = \frac{d}{3}\left( {{A_1} + 4{A_m} + {A_2}} \right)$$

d = step size = 4 m

Am = Mean area of the tank

A1 = 6 × 4 = 24 m2

A2 = 4 × 2 = 8 m2

Mean area is

length = (6 + 4 )/2 = 5m ,

width = (4+2)/2 = 3

Mean area Am = 5 × 3 = 15 m2

$${A_m} = 15\;{m^2}$$

$$V = \;\frac{4}{3}\left( {24 + 4 × 15 + 8} \right)$$

$$V = 122.6\;{m^3}$$

Note:

According to prismoidal formula, d=h/2

# If the cross-sectional area of an embankment at 30 m intervals are 20, 40, 60, 50, and 30 m2 respectively then the volume of the embankment on the basis of the prismoidal rule is

1. 5300 m3
2. 8300 m3
3. 4300 m3
4. 6300 m3

Option 1 : 5300 m3

## Calculation of Area and Volume MCQ Question 9 Detailed Solution

Prismoidal formula can also be said as Simpson’s rule applied for two intervals.

Trapezoidal Formula:

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

$${\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1\:}}} \right]$$

b) Simpson’s formulae:

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

$${\rm{V}} = \frac{{\rm{d}}}{3}[\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)$$

Calculation:

The given data can be depicted as below:

∴ V = (30/3) × (20 + 30 + 4 × (40 + 50) + 2 × (60)) = 5300 m3

# What is the volume of a 6 m deep tank having rectangular shaped top 6 m × 4 m and bottom 4 m × 2 m (computed through the use of prismoidal formula)?

1. 92 m3
2. 96 m3
3. 90 m3
4. 94 m3

Option 1 : 92 m3

## Calculation of Area and Volume MCQ Question 10 Detailed Solution

Concept:

Prismoidal formula is also known as Simpson’s rule.

Trapezoidal Formula:

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

$${\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1{\rm{\;}}}}} \right]$$

b) Simpson’s formulae:

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

$${\rm{V}} = \frac{{\rm{d}}}{3}[\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)$$

Calculation:

Given: 2L = 6m ⇒ L = 3m

Top area = 6 × 4 = 24 m2 = A1

Bottom area = 4 × 2 = 8 m2 = A2

$${{\rm{A}}_{\rm{m}}} = \left( {\frac{{6 + 4}}{2}} \right) \times \left( {\frac{{4 + 2}}{2}} \right) = 15{\rm{\;}}{{\rm{m}}^2}$$

$${\rm{Volume\;}} = \frac{{\rm{L}}}{3}\left( {{{\rm{A}}_1} + 4{{\rm{A}}_{\rm{m}}} + {{\rm{A}}_2}} \right)$$

$${\rm{Volume}} = \frac{3}{3}\left( {24 + 4 \times 15 + 8} \right) = 92{\rm{\;}}{{\rm{m}}^3}$$

# In determining the area of the curved boundary ________ rule is used to get accurate results.

1. Simpson’s
2. mid - ordinate
3. average ordinate
4. trapezoidal

Option 1 : Simpson’s

## Calculation of Area and Volume MCQ Question 11 Detailed Solution

Explanation

(i) For irregular boundaries or curve boundary, Simpson’s rule is preferred over the trapezoidal rule to calculate the given area.

(ii) According to this rule the short length of boundaries between the two adjacent ordinates is a parabolic arch.

Important Points

Simpson's rule:

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:

$$Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + ........ + {O_{n - 1}}) + 2({O_3} + {O_5} + ......{O_{n - 2}})} \right]$$

where O1, O2, O3, .........Ois the offset

Note:

(i) In case of an even number of cross-sections, the end strip is treated separately and the area of the remaining strip is calculated by Simpson's rule. The area of the last strip can be calculated by either trapezoidal or Simpson's rule.

# A cross-section area of river flow can be calculated by using following formula ______.

1. Simpson's rule
2. Trapezoidal rule
3. Both (1) and (2)
4. Thumb rule

Option 3 : Both (1) and (2)

## Calculation of Area and Volume MCQ Question 12 Detailed Solution

Concept:

The area can be calculated by following methods:

1. Mid-ordinate rule:

Area = havg × L = (h1 + h2 + …. + hn)d = d(Σhi)

Here,

hi – Ordinate at the mid-point of each division and L – length of the baseline.

2. Average ordinate rule:

Here,

Oi – is the ordinate at regular intervals.

$$Area = \frac{{sum\;of\;the\;ordinates}}{{no.\;of\;the\;ordiantes}} \times length\;of\;the\;base\;line$$

3. Trapezoidal rule:

Here,

Oi – ordinates at equal interval

d – the common difference

$$Area = \frac{d}{2}\left[ {{O_o} + {O_n} + 2\left( {{O_1} + {O_2} + {O_3} + \; \ldots \ldots + {O_{n - 1}}} \right)} \right]$$

4. Simpson’s rule:

This method is used when:

• The boundaries are not straight.
• The number of area segments should be even in number.
• The number of ordinates should be odd.

$$Area = \frac{h}{3}\left[ {{y_1} + {y_n} + 4\left( {{y_{even}}} \right) + 2\left( {{y_{odd}}} \right)} \right]$$

Here,

h = x2 – x1 = x3 – x2

# In earthwork computations on a longitudinal profile, the diagram prepared to work out the quantity of earthwork is:

1. double mass curve
2. mass haul diagram
3. Mollier diagram
4. flow net

Option 2 : mass haul diagram

## Calculation of Area and Volume MCQ Question 13 Detailed Solution

Explanation:

Mass Haul Curve:

This is a curve representing the cumulative volume of earthwork at any point on the curve, the manner in which earth to be removed.

It is necessary to plan the movement of excavated soil of worksite from cuts to fill so that haul distance is minimum to reduce the cost of earthwork.

The mass haul diagram helps to determine the economy in a better way.

The mass haul diagram is a curve plotted on a distance base with the ordinate at any point on the curve representing the algebraic sum of the volume of earthwork up that point.

A haul refers to the transportation of your project’s excavated materials. The haul includes the movement of material from the position where you excavated it to the disposal area or a specified location. A haul is also sometimes referred to as an authorized haul.

Haul = Σ Volume of earthwork × Distance moved.

# In a cross staff survey, the perpendicular offsets are taken on right and left of the chain line AD as shown in figure – all values are in ‘metres’. The area enclosed by ABCDEFA, computed by trapezoidal method is

1. 3650 m2
2. 3200 m2
3. 3475 m2
4. 3500 m2

Option 3 : 3475 m2

## Calculation of Area and Volume MCQ Question 14 Detailed Solution

Explanation

The area bounded by ABCDEFA

= Area bounded by ABCDA + Area bounded by AFEDA

$$= \left[ {\frac{1}{2} \times 30 \times 20 + \frac{1}{2} \times(20+30) \times \left( {45 - 30} \right) + \frac{1}{2} \times 30 \times \left( {90 - 45} \right)} \right] + \left[ {\frac{1}{2} \times 35 \times 40 + \frac{1}{2}\left( {40 + 30} \right) \times \left( {65 - 35} \right) + \frac{1}{2} \times 30 \times \left( {90 - 65} \right)} \right]$$

= 1350 + 2125

= 3475 m2

# Which of the following methods is most suitable for area calculation when boundary line is irregular?

1. Prismoidal formula
2. Trapezoidal rule
3. Simpson’s rule
4. Mid-Ordinate method

Option 3 : Simpson’s rule

## Calculation of Area and Volume MCQ Question 15 Detailed Solution

Explanation

For irregular boundaries, Simpson’s rule is preferred over the trapezoidal rule to calculate the given area.

According to this rule the short length of boundaries between the two adjacent ordinates is a parabolic arch.

Important Points

Simpson's rule:

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:

$$Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + ........ + {O_{n - 1}}) + 2({O_3} + {O_5} + ......{O_{n - 2}})} \right]$$

where O1, O2, O3, .........Ois the offset

Note:

• In case of an even number of cross-sections, the end strip is treated separately and the area of the remaining strip is calculated by Simpson's rule. The area of the last strip can be calculated by either trapezoidal or Simpson's rule.

# The expression for the total volume of earthwork for an embankment using Simpson’s one-third rule, if A1, A2, A3, A4 ……. An-1 and An are the areas at n sections at an interval of h, is ________.

1. $$\frac{{\rm{h}}}{3} \times {\rm{\;}}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots } \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots } \right)} \right]$$
2. $$\frac{{\rm{h}}}{3} \times \left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}) + 2\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots } \right) + ({{\rm{A}}_3} + {{\rm{A}}_5} + \ldots } \right)} \right]$$
3. $$\frac{{\rm{h}}}{3} \times \left[ {\left( {({{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}} + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots } \right) + ({{\rm{A}}_3} + {{\rm{A}}_5} + \ldots .} \right)} \right]$$
4. $$\frac{{\rm{h}}}{3} \times \left[ {\frac{{\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right)}}{{4{\rm{\;}}}} + \left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots } \right) + \left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots } \right)} \right]$$

Option 1 : $$\frac{{\rm{h}}}{3} \times {\rm{\;}}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots } \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots } \right)} \right]$$

## Calculation of Area and Volume MCQ Question 16 Detailed Solution

Formula:

a) Trapezoidal Formula:

$${\rm{V}} = \frac{{\rm{h}}}{2} \times {\rm{\;}}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 2\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots+A_{n-1} } \right) } \right]$$

b) Mid-ordinate Formula:

$${\rm{V}} = {\frac{d}{n}\rm{}}\left[ {{{{{\rm{A}}_1} + }} {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} }}} \right]$$

c) Simpson’s Formula:

$${\rm{V}} = \frac{{\rm{d}}}{3}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)} \right]$$

Where,

V is the volume of earthwork between a number of sections having areas A1, A2 …, An spaced at a constant distance d.

# The instrument used to compute area of irregular shape marked on paper is known as:

1. Passometer
2. Planimeter
3. Pentagraph
4. Pedometer

Option 2 : Planimeter

## Calculation of Area and Volume MCQ Question 17 Detailed Solution

Planimeter:

Planimeter is an instrument used in surveying to compute the area of any given plan. Planimeter only needs plan drawn on the sheet to calculate area.

Generally, it is very difficult to determine the area of irregular plot. So, by using planimeter we can easily calculate the area of any shape.

Pedometer:

It is a portable device usually an electronic or electromagnetic that counts each steps of a person by detecting the motion of the person’s hip or hand.

Passometer:

It is an instrument shaped like a watch used to count the number of persons steps.

# In Mass Haul diagram (Mass diagram), the term haul represents the

1. Sum of the product of each load by its distance
2. Distance at any time from the working face of an excavation to the tip end of the embankment
3. Distance from the centre of gravity of a cutting to that of tipped material
4. Horizontal distance through which the load is shifted

Option 1 : Sum of the product of each load by its distance

## Calculation of Area and Volume MCQ Question 18 Detailed Solution

Mass Haul Curve

This is a curve representing the cumulative volume of earthwork at any point on the curve, the manner in which earth to be removed.

It is necessary to plan the movement of excavated soil of worksite from cuts to fill so that haul distance is minimum to reduce the cost of earthwork.

The mass haul diagram helps to determine the economy in a better way.

The mass haul diagram is a curve plotted on a distance base with the ordinate at any point on the curve representing the algebraic sum of the volume of earthwork up that point.

A haul refers to the transportation of your project’s excavated materials. The haul includes the movement of material from the position where you excavated it to the disposal area or a specified location. A haul is also sometimes referred to as an authorized haul.

Haul = Σ Volume of earthwork × Distance moved.

# Which of the following quantity is measured using a Planimeter?

1. Area
2.  Bar diameter
3. Volume
4. Weight

Option 1 : Area

## Calculation of Area and Volume MCQ Question 19 Detailed Solution

Concept:

The areas between curved boundaries are also computed by planimeter from the map of the area. Most commonly used is Amsler polar Planimeter.

The formula for calculating areas of the map by using planimeter.

Area (A) = M × (FR – IR ± 10 N + C)

Where,

M = Multiplying constant

FR = Final reading of Planimeter

N = Number of times zero of dial pass is index mark

C = Constant marked above scale division on tracing arm.

± ⇒ + ve sign when zero is passed in clockwise direction

– ve for when zero is passed in an anticlockwise direction.

∴ The planimeter is used to measure the area.

# In the equation of a planimeter, ‘N’ is ______, when zero of the dial passes the index mark in a clockwise direction

1. + ve
2. - ve
3. 0
4. None of the above

Option 1 : + ve

## Calculation of Area and Volume MCQ Question 20 Detailed Solution

Concept:

The areas between curved boundaries are also computed by planimeter from the map of the area. Most commonly used is Amsler polar Planimeter.

The formula for calculating areas of the map by using planimeter.

Area (A) = M × (FR – IR ± 10 N + C)

Where,

M = Multiplying constant, FR = Final reading of Planimeter, IR = Initial Reading, N = Number of times zero of dial pass is index mark, and C = Constant marked above scale division on tracing arm.

± ⇒ + ve sign when zero is passed in clockwise direction and – ve for when zero is passed in an anticlockwise direction.