250+ TOP MCQs on Slope and Answers

This set of Strength of Materials Multiple Choice Questions on “Slope”.

1. Slope in the beam at any point is measured in ____________
a) Degrees
b) Minutes
c) Radians
d) Metric tonnes
Answer: c
Clarification: The slope is defined as at any point on the bent beam is the angle measured in terms of radians to which the tangent at that point makes with the x axis.

2. Elastic curve is also known as __________
a) Refraction curve
b) Reflection curve
c) Deflection curve
d) Random curve
Answer: c
Clarification: An elastic curve is defined as the line to which the longitudinal axis of a beam deviates under given load. It is also called a deflection curve.

3. Which of the following method is not used for determining slope and deflection at a point?
a) Moment area method
b) Double integration method
c) Isoheytal method
d) Macaulay’s method
Answer: c
Clarification: The method “Isoheytal” can be used for calculating run-off over an area. The remaining methods are effectively adopted to calculate the slope and deflection at a point in any type of beam.

4. The slope is denoted by _______
a) k
b) y
c) i
d) c
Answer: c
Clarification: The slope at any section in a deflection beam is defined as the angle measured in radians to the tangent at the section makes with the original axis of the beam.
•It is denoted by “i”.

5. Calculate the slope at supports, if the area is 180kNm2. Take flexural rigidity as 50000.
a) 0.0054 radians
b) 0.0072 radians
c) 0.0036 radians
d) 0.108 radians
Answer: c
Clarification: Maximum slope at supports be i = A/EI
= 180/50000
i = 0.0036 radians.

6. In cantilever beams, the slope is _____________ at fixed end.
a) Maximum
b) Zero
c) Minimum
d) Uniform
Answer: b
Clarification: The slope in cantilever beam is zero at the fixed end of the cantilever and the slope is maximum at it’s free end. The slope is determined in the moment area method through Mohr’s theorems.

7. Slope is maximum at _______ in simply supported beams.
a) Mid span
b) Through out
c) Supports
d) At point of loading
Answer: a
Clarification: In case simply supported beams, the slope is maximum at the end supports of the beam and relatively zero at midspan of a symmetrically loaded beam.

8. Mohr’s theorem- 1 states ________
a) E/AI
b) I/EA
c) A/EI
d) A=EI
Answer: c
Clarification: According to Mohr’s theorem-1, the change of slope between any of the two points on and Elastic axis is equal to the net area of bending moment diagram (A) between these two points divided by flexural rigidity(EI).

9. Using Mohr’s theorem, calculate the maximum slope of a cantilever beam if the bending moment area diagram is 90kNm2. Take EI = 4000 kNm2.
a) 0.0225 radians
b) 0 0367 radians
c) 0.0455 radians
d) 0.066 radians
Answer: a
Clarification: The maximum slope at free support (in cantilever beam) = i = A/EI
= 90/4000
= 0.0225 radians.

10. Contour canals are also called as ______
a) Single bank canal
b) Ridge canal
c) Side slope canal
d) Watershed canal
Answer: a
Clarification: In this method, the canal is aligned along the falling contour. A generally higher side is left without bank. So it is also called a single bank canal. The contour canal cuts across the natural drainage courses.

11. ______________ provides employment to the cultivators at the time of famine.
a) Productive canal
b) Link canal
c) Protective canal
d) Inundation canal
Answer: c
Clarification: The construction of protective canals and their development may be started during summer in hence they provide employment to the farmers at the time of drought and famine. Protective canals are not remunerative as productive canals.

12. ______________ bricks are used in the lining of blast furnaces.
a) Magnesia
b) Dolomite
c) Bauxite
d) Fly ash
Answer: b
Clarification: Dolomite bricks are made especially from dolomite it contains nearly 30% lime and 22% of magnesium these bricks are inferior to magnesite bricks. They are generally used in the lining of blast furnaces.

13. _____________ bricks are resistant to corrosion.
a) silica bricks
b) magnesia bricks
c) bauxite bricks
d) fire bricks
Answer: c
Clarification: Bauxite bricks contain nearly 75% of aluminium and it is mixed with fire clay 15 to 30% and added some water to mould. High alumina bricks are resistant to corrosion.

14. _____________ bricks are used in the lining of electric furnace.
a) Frosterite
b) Spinel
c) Chrome
d) Basic
Answer: b
Clarification: The spinal bricks belong to neutral bricks. The spinel bricks mainly consist of alumina and magnesia. These bricks are widely used in the lining of electric furnace.

15. The finished product after burning magnesite is named as ___________
a) Perillax
b) Hellyx
c) Pyrolytaex
d) Syrilax
Answer: a
Clarification: The heating of magnesia bricks is continued in the same kiln after reaching the temperature of 1950°C, and then some amount of iron oxide is mixed. The finished product after burning magnesite is named as perillax.

250+ TOP MCQs on Combined Bending and Torsion and Answers

This set of Strength of Materials Multiple Choice Questions on “Combined Bending and Torsion”.

1. A solid shaft of circular in section is subjected to torque which produces maximum shear stress in a shaft. Calculate the diameter of the shaft.
a) (16T/πf)3/2
b) (16f/πT)1/2
c) (16f/π) 1/2
d) (πT/16f) 1/2
Answer: a
Clarification: From torsional equation
T/J = f/R
T = f.Z
T = f×π/16d3.
D= (16T/πf) 3/2.

2. When two dissimilar shafts are connected together, then the shaft is __________
a) Integrated shafts
b) Composite shafts
c) Differential shafts
d) Combined shafts
Answer: b
Clarification: When two dissimilar shafts are connected together to form one shaft then the shaft can be termed as composite shaft.

3. __________ torque occurs along with maximum shear stress due to combined bending and torsion.
a) Equipment
b) Coaxial
c) Biaxial
d) Lateral
Answer: a
Clarification: Equipment torque is the twisting moment which acts along producing maximum shear stress due to the combined bending as well as torsion.

4. When a shaft is subjected to pure twisting then the type of stress developed is ________
a) Bending
b) Axial
c) Shear
d) Normal
Answer: c
Clarification: Shear stress is produced when the shaft is subjected to pure twisting (torsion). The shear stress due to twisting moment is zero at the axis of the shaft.

5. The torque which produces unit twist per unit length is ________
a) Torsional rugosity
b) Torsional rigidity
c) Torsional viscosity
d) Torsional mean radius
Answer: a
Clarification: The product of shear modulus(C) and polar moment of inertia (J) is called torsional rigidity. Torsional rigidity produces a twist of 1 radian in a shaft of unit length.

6. The level of top of weir can be termed as __________
a) Talus
b) Curtain walls
c) Crest
d) Shutter
Answer: c
Clarification: The level of the top of weir is known as a crest. The shutters are provided on the crest and can be raised or laid flat during the time of floods.

7. ________ possesses less silting and scouring.
a) Weir
b) Barrage
c) Dams
d) Reservoir
Answer: b
Clarification: The barrage is a low obstructive barrier constructed across the river. There will be less silting and better control over the levels due to low set crest.

8. In __________ there will be no means for silt disposal.
a) Weir
b) Barrage
c) Reservoir
d) Dams
Answer: a
Clarification: The weir may be defined as a solid obstruction/wall built across the river to raise the water level. Raised crest causes silting at upstream and there is no means silt disposal.

9. _________ is a pure water pressure.
a) Uplift
b) Percolation
c) Scour
d) Flood bank
Answer: a
Clarification: Uplift occurs when pore water pressure under a structure or a low permeability confining layer becomes larger than the mean overburden pressure.

10. __________ causes of uplift of structure.
a) Percolation
b) Scour
c) Critical Velocity
d) Slope Failure
Answer: a
Clarification: The effect of percolation on an irrigation structure like a weir to cause uplift pressure on the structures and topple the structure at any moment.

11. ________ protects the weir from erosive forces during floods.
a) Talus
b) Curtain walls
c) Shutter
d) Upstream solid apron
Answer: d
Clarification: Upstream solid apron is a concrete bed which is provided on the upstream side of weir to protect the weir from erosive forces during floods. The length of apron depends upon maximum discharge of the river.

12. Gross storage – Dead storage is _______
a) Live storage
b) Virtual storage
c) Excessive storage
d) Free storage
Answer: a
Clarification: It is also called as available or effective storage. It is the difference between gross storage and dead storage. It is the amount of water available from FRL to the sill of the lowest sluice.

13. Which of the following is not sound proof?
a) G I sheets
b) A C sheets
c) PVC sheets
d) Fabric sheets
Answer: a
Clarification: Galvanised iron sheets are commonly used as a roofing material. These are very durable and fire proof. The main disadvantage is they are not sound proof.

14. Which of the following is not affected by temperature?
a) Fabric sheets
b) G I sheets
c) AC sheets
d) Flat roofs
Answer: c
Clarification: Asbestos cement sheets are cheaper in the initial cost. They are fire resisting. They are heavy in weight and they are not affected by temperature.

15. Which of the following possess good insulation properties?
a) Battened roofs
b) Wooden roofs
c) Jack arch roofs
d) Flat roofs
Answer: d
Clarification: Flat roofs are easier in construction and maintenance. A flat roof is more stable against high wards. It has a better architectural appearance and it has good insulation properties.

250+ TOP MCQs on Elastic Constants Relationship – 2 and Answers

This set of Strength of Materials Questions and Answers for Freshers on “Elastic Constants Relationship – 2”.

1. What is the ratio of Youngs modulus E to shear modulus G in terms of Poissons ratio?
a) 2(1 + μ)
b) 2(1 – μ)
c) 1/2 (1 – μ)
d) 1/2 (1 + μ)
Answer: a
Clarification: As we know G = E / 2(1 +μ) so this gives the ratio of E to G = 2(1 + μ).

2. The relationship between Youngs modulus E, bulk modulus K if the value of Poissons ratio is unity will be __________
a) E = -3K
b) K = -3E
c) E = 0
d) K = 0
Answer: a
Clarification: As E = 2G(1 + μ) putting μ=1 we get E = -3K.

3. A rod of length L and diameter D is subjected to a tensile load P. which of the following is sufficient to calculate the resulting change in diameter?
a) Youngs modulus
b) Poissons ratio
c) Shear modulus
d) Both Youngs modulus and shear modulus
Answer: a
Clarification: For longitudinal strain we need Youngs modulus and for calculating transverse strain we need Poisson’s ratio. We may calculate Poissons ratio from E = 2G(1 + μ) for that we need shear modulus.

4. E, G, K and μ elastic modulus, shear modulus, bulk modulus and Poisson’s ratio respectively. To express the stress strain relations completely for this material, at least __________
a) E, G and μmust be known
b) E, K and μmust be known
c) Any two of the four must be known
d) All the four must be known
Answer: c
Clarification: As E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G), if any two of these four are known, the other two can be calculated by the relations between them.

5. Youngs modulus of elasticity and Poissons ratio of a material are 1.25 x 102 MPa and 0.34 respectively. The modulus of rigidity of the material is __________
a) 0.9469 MPa
b) 0.8375 MPa
c) 0.4664 MPa
d) 0.4025 MPa
Answer: c
Clarification: As E = 2G(1 + μ)
1.25 x 102 = 2G(1 + 0.34)
G = 0.4664 x 102 MPa.

6. If E,G and K have their usual meanings, for an elastic material, then which one of the following be possibly true?
a) G = 2K
b) G = K
c) K = E
d) G = E = K
Answer: c
Clarification: As E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G)
The value of μ must be between 0 to 0.5, so as E never equal to G but if μ = 1/3, then E=K.

7. If a material had a modulus of elasticity of 2.1 kgf/cm2 and a modulus of rigidity of 0.8 kgf/cm2 then what will be the approximate value of the Poissons ratio?
a) 0.26
b) 0.31
c) 0.47
d) 0.43
Answer: b
Clarification: On using E = 2G(1 + μ) we can put the values of E and G to get the Poissons value.

8. Consider the following statements:
X. Two-dimensional stresses applied to a thin plater in its own plane represent the plane stress condition.
Y. Normal and shear stresses may occur simultaneously on a plane.
Z. Under plane stress condition, the strain in the direction perpendicular to the plane is zero.
Which of the above statements are correct?
a) 2 only
b) 1 and 2
c) 2 and 3
d) 1 and 3
Answer: d
Clarification: Under plane stress condition, the strain in the direction perpendicular to the plane is not zero. It has been found experimentally that when a body is stressed within the elastic limit, the lateral strain bears a constant ratio to the linear strain.

9. What is the relationship between the linear elastic properties Youngs modulus, bulk modulus and rigidity modulus?
a) 1/E = 9/k + 3/G
b) 9/E = 3/K + 1/G
c) 3/E = 9/K + 1/G
d) 9/E = 1/K + 3/G
Answer: d
Clarification: We can use E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G) to get the relation between E, K and G.

10. Which of the relationship between E, G and K is true, where E, G and K have their usual meanings?
a) E = 9KC / (3K + C)
b) E = 9KC / (9K + C)
c) E = 3KC / (9K + C)
d) E = 3KC / (3K + C)
Answer: a
Clarification: As we know E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G).

Strength of Materials for Freshers,

250+ TOP MCQs on Center of Gravity and Answers

This set of Strength of Materials Multiple Choice Questions on “Center of Gravity”.

1. The point through which the whole weight of the body acts is called _____________
a) Inertial point
b) Center of gravity
c) Centroid
d) Central point
Answer: b
Clarification: The centre of gravity of a body is the point through which the whole weight of the body acts. A body’s center of gravity is the point around which the resultant torque due to gravity forces vanishes. Where a gravity field can be considered to be uniform and the centre of gravity will be the same.

2. The point at which the total area of a plane figure is asssumed to be concentrated is called ____________
a) Centroid
b) Centre of gravity
c) Central point
d) Inertial point
Answer: a
Clarification: The centroid is the point at which the total area of a plane figure is assumed to be concentrated. The centroid and centre of gravity are at the same point.

3. Where will be the centre of gravity of a uniform rod lies?
a) At its end
b) At its middle point
c) At its centre of its cross sectional area
d) Depends upon its material
Answer: b
Clarification: The centre of gravity of a uniform rod lies at its middle point. The whole weight of the rod acts through its middle point.

4. Where the center of gravity of a circle lies?
a) At its centre
b) Anywhere on its radius
c) Anywhere on its circumference
d) Anywhere on its diameter
Answer: a
Clarification: The whole weight of a circle can be assumed to act through its center. So the center of gravity of a circle is at its center.

5. Where will be the center of gravity of the following section will lie In coordinates?
strength-materials-questions-answers-center-gravity-q5
a) (6,3)
b) (6,6)
c) (6,1.5)
d) (1.5,3)
Answer: c
Clarification: The centre of gravity of this rectangular area will be half of 3cm from x-axis and half of 12 from the y-axis. therefore the center of gravity will be at (6,1.5).

6. Where will be the centre of gravity of the T section shown in the figure?
strength-materials-questions-answers-center-gravity-q6
a) At 8.545cm
b) At 6.5cm
c) At 5cm
d) At 9.25cm
Answer: a
Clarification: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = ( 36×11.5 + 30×5) / (36+30) = 8.545cm.

7. Where will be the center of gravity of the L-section shown in the figure?
strength-materials-questions-answers-center-gravity-q7
a) (1.28,2.64)
b) (1.45,3.24)
c) (1.64,3.28)
d) (2.24,3.68)
Answer: a
Clarification: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (10×3.5 + 4×0.5) / (10+4) = 2.64cm.
This will on for the y-axis.
For the x-axis, The center of gravity is given by, x = (a1x1 + a2x2) / (a1 + a2) = (0x1 + 4×2) / (10+4) = 1.28cm.
So the center of gravity will be at (2.33, 4.33).

8. Where will be the center of gravity of the figure shown ?
strength-materials-questions-answers-center-gravity-q8
a) (3.45,4.52)
b) (3.59,7.42)
c) (3.66,8.84)
d) (3.88,8.88)
Answer: b
Clarification: Area of triangle = 50, area of rectangle = 100
The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (50×20/3 + 100×10) / (50+100) = 8.88cm.
This will on for the y-axis.
For the x-axis, The center of gravity is given by, x = (a1x1 + a2x2) / (a1 + a2) = (50×6.66 + 100×2.5) / (50+100) = 3.88cm.
So the center of gravity will be at (2.33, 4.33).

9. Where will be the center of gravity of an I section will be if the dimension of upper web is 2x10cm, lower web is 2×20 and that of flange is 2x15cm If the y-axis will pass through the center of the section?
a) 7.611cm
b) 9.51cm
c) 9.31cm
d) 11.5cm
Answer: b
Clarification: The center of gravity is given by, y = (a1y1 + a2y2 + a3y3) / (a1 + a2 + a3) = (20×18 + 30×9.5 + 40×1 / (20 +30+40) = 1.611cm.

250+ TOP MCQs on Composite or Flitched Beams and Answers

This set of Strength of Materials Multiple Choice Questions on “Composite or Flitched Beams”.

1. In flitched beams ______ remains same for both materials.
a) Stress
b) Strain
c) Section modulus
d) Young’s modulus
Answer: b
Clarification: Due to bending, the strain will be same in both the materials.
strength-materials-questions-answers-composite-flitched-beams-q1
A timber beam strengthened by steel strips.
Where E of timber / E of steel = m
The equivalent width = b + 2mt.

2. What is the moment due to dead load in case of continuous beams at the middle of interior spans?
a) w L3 / 12
b) w L2/ 14
c) w 3/ 20
d) w L2 / 24
Answer: d
Clarification: The moment due to dead load in case of continuous beams at the middle of interior spans is w L2 / 24.

Position Moment due to dead load
Near middle of end span W L2/ 12
At the middle of interior span W L2/ 24
At the support next to and support -W L2/ 22

.

3. A continuous beam is one which is _______
a) Infinitely long
b) Supported at two points
c) Supported it more than two supports
d) Supported by a prop
Answer: c
Clarification: A beam which is supported by more than two supports is known as a continuous beam. In this beam, bending moment is low and hence the deflection in the beam is also comparatively less. This beam is stiffer when compared to the other traditional beams.

4. The effective length of column depends upon ________
a) the cross section of beam
b) end conditions
c) maximum bending moment
d) extreme fibres
Answer: b
Clarification: The effective length of column depends upon end conditions.

End condition Effective length
Both ends hinged L
Both ends fixed L/2
One end is fixed and other end free 2L

5. The phenomenon under which the strain of material varies under constant stress is known as ________
a) Creep
b) Hysteresis
c) Viscoelasticity
d) Strain hardening
Answer: a
Clarification: A creep is a plastic deformation underweight the strain of material where is under constant stress this is one of the mechanical properties of the engineering materials. The best example is the failure of concrete.

6. Volumetric strain = 3× _____ strain.
a) Lateral
b) Linear
c) Composite
d) Yield
Answer: b
Clarification: eV (volumetric strain) = 3× linear strain = 3×e
The volumetric strain is algebraic sum of all the linear(or) axial strain when a solid to be subjected to equal normal sources of the same type of all faces we will have €x, €y and €z equal in value. In this case the volumetric strain will be 3 times the linear strain in any of the three axes.

7. The stress corresponding to breaking point is known as _____________
a) yield stress
b) ultimate stress
c) breaking stress
d) normal stress
Answer: c
Clarification: After reaching ultimate stress, the stress strain curve suddenly falls with rapid increase in strain and specimen breaks. The stress corresponding to breaking point is known as breaking stress and it is denoted by G.

8. Determine the yield stress of a steel rod 20 mm diameter, if the yield load on the steel rod is 88kN.
a) 240.55 N/mm2
b) 280.25 N/mm2
c) 325 N/mm2
d) 290.45 N/mm2
Answer: b
Clarification: Initial area of the Steel rod of 20 mm = 314 mm2 [area of circle]
Yield stress = yield load/ Area
= 88 × 103/ 314
= 280.25 N/mm2.

9. What is the elongation percentage of a steel rod of 50 mm diameter if the total extension is is 54 mm and gauge length is 200 mm.
a) 27%
b) 23%
c) 43%
d) 35%
Answer: a
Clarification: Percentage elongation = Total extension / Gauge length × 100
= 54/200 × 100
= 27%.

10. __________ joints are provided when there is a break in the concreting operation.
a) transverse joints
b) longitudinal joints
c) construction joints
d) warpage joints
Answer: c
Clarification: The construction joints are provided when there is a break in a concreting operation. Although the effort is always made to complete the concrete work in one day, sometimes it is not possible and therefore, construction joints are provided. For beams, the joints should be at the centre of the span or within the middle third.

250+ TOP MCQs on Deflection and Answers

This set of Strength of Materials Multiple Choice Questions on “Deflection”.

1. Units of deflection are _________
a) kNm
b) kN/m
c) kN
d) m
Answer: d
Clarification: The term “deflection” is defined as the transverse displacement of a point on any straight axis to the curved axis. It is expressed in metres (m).

2. Which of the following method is used to determine the slope and deflection at a point?
a) Arithmetic increase method
b) Mathematical curve setting
c) Macaulay’s method
d) Lacey’s method
Answer: c
Clarification: Macaulay’s method was devised by Mr WH Macaulay.
Advantages:
i. Gives one continuous expression for bending moment
ii. Constants of integration can be found by using end conditions
iii. By using this method, slope and deflection at any section can be determined throughout the length of the beam.

3. Deflection is denoted by _______
a) i
b) y
c) h
d) e
Answer: b
Clarification: The deflection of a point on the axis of the deflected beam is defined as the angle developed in radians with tangent at the section makes with the original axis of the beam.

4. In cantilever beams, the deflection is zero at ___________
a) Free and
b) Fixed end
c) At supports
d) Through out
Answer: b
Clarification: The deflection in cantilever beam is always zero at the fixed end and deflection in the cantilever beam at the free end is maximum.

5. Mohr’s theorem -¡¡ states?
a) Ax/EI
b) A/Ex
c) A/EI
d) Ae=Ix
Answer: a
Clarification: Mohr’s theorem -¡¡ states “the intercept taken on a vertical reference line of the tangent at any two points on an elastic line is equal to the moment of BMD between these points, about the reference line divided by flexural rigidity (EI).

6. Calculate the deflection if the slope is 0.0225 radians. Take the distance of centre of gravity of bending moment to free end as 2 metres.
a) 45mm
b) 35mm
c) 28mm
d) 49mm
Answer: a
Clarification: The deflection at any point on the elastic curve equal to Ax/EI
But, we know that A/EI is already slope equation.
So, slope × (the distance of centre of gravity of bending moment to free end = 2m).
0.0225 × 2
0.045m ~ 45 mm.

7. In simply supported beams, deflection is zero at _________
a) Mid span
b) Supports
c) Through out
d) Point of action of load
Answer: b
Clarification: The deflection is always zero at the supports and the deflection is maximum at the mid span of a symmetrically loaded simply supported beam.

8. Which of the following is not a cross drainage work?
a) Aqueduct
b) Level crossing
c) Head regulator
d) Super passage
Answer: c
Clarification: The head regulator is one of the canal regulation works. It can control the entry of silt into the canal. It can be used as a metre for measuring the discharge. It can shut out river floods.

9. Tail escape is also called as ___________
a) Outlet
b) Cross regulator
c) Weir type escape
d) Surplus escape
Answer: c
Clarification: The crest of the weir is fixed at canal FSL. When the water level rises above FSL, it is disposed of into the natural drain. Hence, the tale escape is also known as weir type escape.

10. The land where all the water comes from ___________
a) Ridge dam
b) Watershed
c) Meander
d) Groynes
Answer: b
Clarification: A watershed can be defined as an interconnected area of land which receives the water from surrounding ridge tops and transports it to a common point such as a lake or stream. All lands and waterways can be found within one watershed or another.

11. ____________ reduces storm water discharge.
a) Rain water harvesting
b) Water harvesting
c) Watershed
d) Watershed management
Answer: b
Clarification: The water harvesting is defined as the process of capturing rain where it falls. The objectives of water harvesting are 1) To provide drinking water 2) To provide irrigation water 3) To increase groundwater recharge to reduce storm water discharge.

12. Which of the following is not a soil moisture conservation method?
a) Spreading manure
b) Crop rotation
c) Recharge to ground water
d) By mulches
Answer: c
Clarification: The methods which are adopted for preserving the water in the soil from being lost are called as soil moisture conservation methods. The major part of the water is lost through evapotranspiration. The recharge to groundwater is one of the techniques in rainwater harvesting.

13. Nutrients like ca, mg, si, al, S, K are lost due to ____________
a) Soil erosion
b) Percolation
c) Water logging
d) Watershed
Answer: b
Clarification: The percolation is defined as a downward movement of water through the soil due to the force of gravity. The rapid percolation of water results in loss of plant nutrients and makes the soil acidic.

14. Warabandi has been practiced in India for more than ____________ years.
a) 130 years
b) 125 years
c) 140 years
d) 145 years
Answer: b
Clarification: Warabandi is a rotational method for allocation of the available water equally in an irrigation system. It provides continuous rotation of water generally lasts 7 days. It has been effectively practiced in India for more than 125 years.

15. Gold, Copper and lead are the examples of ______
a) Ductility
b) Creep
c) Plasticity
d) Malleability
Answer: c
Clarification: Plasticity in the property of Material by which the material can undergo permanent deformation and fails to regain its original shape on removal of load. Examples are gold, lead, etc.