True Length = (Actual length × Measured Length)/Nominal Length

Given

Measured length = 450 m

Nominal Length = 30 m

Actual length = 0.2 link Short

i.e

1 m = 5 link

1 link = 0.2 m

0.2 link = 0.04 m

Actual length = 30 − 0.04 = 29.96 m

True Length = (29.96 × 450)/30

True length = 449.4 m

• Timber refers to wood used for construction works. A living tree yielding good timber is called standing timber.
• Timber extracted After felling and separating branches, it is known as rough timber.
• When bark (outermost layers of stems and roots of woody plants) is removed and the stem is roughly converted into pieces of suitable length for transportation is known as a log.
• After the log is seasoned and converted into commercial sizes like planks, battens, posts, and beams, it is known as converted timber.

Here

1cm on a map represents 4000

10 cm on a map represents = 10 × 4000 = 40000 cm

1 m = 100 cm

400 m = 40000 cm

0.4 km = 40000 cm

The distance AB on the ground is = 0.4 km

or

Scale = 10 cm/AB (on the ground)

∴ AB (on the ground) = 10/1/4000

= 4 × 10⁴ cm

= 0.4 km (1 km = 10⁵ cm)

Rocks are composed of minerals. Some rocks are made of only one type of mineral, for example, marble (calcite), while others are composed of a mixture of different minerals.

In the simplest system, rocks can be divided into three main groups called igneous, sedimentary and metamorphic.

(i) Igneous rock is formed from molten rock that has cooled and solidified: the rate of cooling determines the crystal size. Often being hard, igneous rock is typically used where hard-wearing surfaces are needed, e.g., kitchen table tops and paving. Granite, Dolerite, Basalt are some e.g of Igneous rock

(ii) Sedimentary rocks Rock that is exposed on the Earth’s surface is often the target of chemical attacks in the atmosphere. This process is called weathering, and leads to the breakdown of the rock. The products of weathering or mechanical breakdown are carried away by water, wind or ice by a process called erosion. These sediments are converted to sedimentary rock in a process called lithification. This means that the particles of the sediment are ‘cemented’ together.

Sedimentary rocks are classified according to the type of sediment formed:

Clastic sediments - These are sediments formed from rock particles and fragments (clay minerals, sand grains and rock fragments). Examples of this type of rock . breccia, sandstone, shale and mudstone.

Chemical sediments - These are sediments formed from the material from a liquid solution settling down. Examples of this type of rock are limestone, dolomite and chalk.

(iii) Metamorphic rocks are always formed through the recycling of other rocks. High temperatures and pressures deep in the Earth provide the conditions to modify rock by baking and squeezing.  Metamorphic rocks are normally very hard and often aesthetically appealing. For these reasons they are typically used for decorative purposes, such as statues or the interiors of important buildings. Examples of metamorphic rock are: slate, schist, gneiss, marble, and quartzite.

Coefficient of contraction:- The coefficient of contraction is defined as the ratio between the area of the jet at vena contracta and the area of the orifice. Thus, if A, is the area of the jet at vena contracta and A is the area of the orifice, then

C_c = A_c/A

The theoretical value of C is around 0.611, but in practice, its value varies from 0.61 to 0.69 depending on the size and shape of the orifice.

Standard Length of a chain

1S:1492-1959 recommends that a chain with a pull of 8kg applied at its ends when its length is compared with a certified steel tape standardized at 20°C should have every meter of its length accurate within 2 mm and a 20 m  or 30 m chain should have lengths 20 m ± 5 mm and 30 m ± 8 mm respectively.

Such lengths of the chain satisfying the standard are called ‘Standard lengths’ and the chains are called ‘Standard chains’.

The accuracy for a 30 m chain is +8 mm whereas for the 20 m chain is +5 mm.

Grade compensation for a railway track is given as follows-

For B.G = 0.04% per degree of curve

M.G = 0.03% per degree of curve

N.G = 0.02% per degree of curve

Compensated gradient = Ruling gradient- Grade compensation

Calculation for MG line

Ruling gradient = 1 in 200

Degree of curve = 2°

Grade compensation= (0.03 × 2)/100 = 0.06/100

Compensated gradient = (1/200) − (0.06/100) = 0.44%

Calculation for BG line

Ruling gradient = 1 in 200

Degree of curve = 2°

Grade compensation= (0.04 × 2)/100 = 0.08/100

Compensated gradient = (1/200) − (0.08/100) = 0.42%

Hence Steepest gradient for the BG line = 0.42% and for the MG line = 0.44%

The 1:2:4 concrete means that it is prepared by mixing 1 part cement, 2 parts fine aggregates, and 4 parts coarse aggregates. The volume of water is about 50-60% of the volume of cement.

In case bucking is not considered the actual dry fine aggregate = 2/1.1 per volume of cement

Bulking:- The increase in the volume of a given mass of fine aggregate caused by the presence of water is known as bulking. Fine aggregate bulk is more than coarse aggregate.

If the bulking of sand is 10% and if the mix ratio is 1: 2: 4, the actual volume of sand used will be 1.1 × 2 =2.2 instead of 2 per unit volume of cement.

If this correction is not applied, the actual dry sand in the concrete will be 1 × 2/1.1 = 1.82, instead of 2 per unit volume of cement. The mix proportion would be 1: 1.82: 4 rather than 1: 2: 4; which indicates lesser concrete production.

According to the Bureau if India Standards (BIS): 654-1962, the maximum water absorption percentage of Class AA type Mangalore pattern tiles is 18%

Note:- For class A the maximum water absorption percentage is 20%.

Equation of sludge  digestion tank

V₁(100-P₁)= V₂(100-P₂)

Where V1 and V2, represent the volumes before and after dewatering,

p1 and p2, are per cent moisture before and after dewatering, respectively.

For sample A

V₁(100-P₁)= V₂(100-P₂)

V₁(100-97)= V₂(100-95)

V₂ = 3V₁/5

$egin{array}{l} Delta {V_A} = { m{ }}frac{{left( {{V_1} - { m{ }}{V_2}} ight)}}{{{V_1}}} imes 100\ \ = frac{{left( {{V_1} - { m{ }}3/5{V_1}} ight)}}{{{V_1}}} imes 100 end{array}$

ΔV_A = 40%

Total decrease in the volume = 60%

For sample A

V₁(100-P₁)= V₂(100-P₂)

V₁(100-98)= V₂(100-96)

V₂ = V₁/2

$egin{array}{l} Delta {V_A} = { m{ }}frac{{left( {{V_1} - { m{ }}{V_2}} ight)}}{{{V_1}}} imes 100\ \ = frac{{left( {{V_1} - { m{ }}1/2{V_1}} ight)}}{{{V_1}}} imes 100 end{array}$

ΔV_A = 50%

Total decrease in the volume = 50%