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# Strength of Materials MCQs Center of Gravity’

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

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Answer: b

Explanation: 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.

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

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Answer: a

Explanation: 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.

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

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Answer: b

Explanation: The centre of gravity of a uniform rod lies at its middle point. The whole weight of the rod acts through its middle point.

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

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Answer: a

Explanation: 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.

Where will be the center of gravity of the following section will lie In coordinates?<br/>

a) (6,3)<br/>

b) (6,6)<br/>

c) (6,1.5)<br/>

d) (1.5,3)

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Answer: c

Explanation: 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).

Where will be the centre of gravity of the T section shown in the figure?<br/>

a) At 8.545cm<br/>

b) At 6.5cm<br/>

c) At 5cm<br/>

d) At 9.25cm

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Answer: a

Explanation: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = ( 36×11.5 + 30×5) / (36+30) = 8.545cm.

Where will be the center of gravity of the L-section shown in the figure?<br/>

a) (1.28,2.64)<br/>

b) (1.45,3.24)<br/>

c) (1.64,3.28)<br/>

d) (2.24,3.68)

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Answer: a

Explanation: 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).

Where will be the center of gravity of the figure shown ?<br/>

a) (3.45,4.52)<br/>

b) (3.59,7.42)<br/>

c) (3.66,8.84)<br/>

d) (3.88,8.88)

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Answer: b

Explanation: 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).

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

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Answer: b

Explanation: 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.

### Strength of Materials MCQs Center of Gravity of Section

The center of gravity of the rod shown in figure will be _____________<br/>

a) 5cm<br/>

b) 10cm<br/>

c) 15cm<br/>

d) 20cm

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Answer: b

Explanation: The center of gravity of a rod will be on its center. Here it will be at 10cm.

The center of gravity of a circle of radius 10 cm will be _____________

a) At its center of the diameter

b) At the center of the radius

c) Anywhere on the circumference

d) Anywhere in its area

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Answer: a

Explanation: 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. Whatever may be the radius of the circle the center of gravity will be on its center.

A rectangle has dimension of 10cm x 20cm. where will be its center of gravity?

a) (10,10)

b) (20,5)

c) (10,5)

d) (5,10)

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Answer: c

Explanation: The centre of gravity of this rectangular area will be half of 10cm from x-axis and half of 20cm from y-axis. therefore the center of gravity will be at (10,5).

Where will be the centre of gravity of the T section shown in the figure?<br/>

a) 8<br/>

b) 8.5<br/>

c) 10.5<br/>

d) 11.5

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Answer: d

Explanation: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (100×17.5 + 150×7.5) / (100+150) = 11.5cm.

Where will be the center of gravity of the L-section shown in figure?<br/>

a) (4.33, 2.33)<br/>

b) (4, 6)<br/>

c) (2.33, 4.33)<br/>

d) (1, 5)

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Answer: c

Explanation: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (20×7 + 16×1) / (20+16) = 4.33cm.

This will on for the y-axis.

For the x-axis, The center of gravity is given by, x = (a1x1 + a2x2) / (a1 + a2) = (20×1 + 16×4) / (20+16) = 2.33cm.

So the center of gravity will be at (2.33, 4.33).

Where will be the center of gravity of the figure shown?<br/>

a) (3.45,4.52)<br/>

b) (3.59,4.52)<br/>

c) (3.66,5.17)<br/>

d) (4.01,5.15)

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Answer: b

Explanation: Area of triangle = 20, area of rectangle = 50

The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (20×10/3 + 50×5) / (20+50) = 4.52cm.

This will on for the y-axis.

For the x-axis, The center of gravity is given by, x = (a1x1 + a2x2) / (a1 + a2) = (20×6.33 + 50×2.5) / (20+50) = 3.59cm.

So the center of gravity will be at (2.33, 4.33).

Where will be the center of gravity of the shown figure?<br/>

a) (4.66,6.332)<br/>

b) (4.34,3.24)<br/>

c) (4.25,6.45)<br/>

d) (4.87,6.41)

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Answer: a

Explanation: Area of triangle = 25, area of rectangle = 100

The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (25×10/3 + 100×5) / (25+100) = 4.66cm.

This will on for the y-axis.

For the x-axis, The center of gravity is given by, x = (a1x1 + a2x2) / (a1 + a2) = (25×11.66 + 100×5) / (25+100) = 6.332cm.

So the center of gravity will be at (2.33, 4.33).

Where will be the center of gravity of an I section will be if the dimension of web is 2x20cm and that of flange is 2x15cm If the y-axis will pass through the center of the section?

a) 8.5cm

b) 9.5cm

c) 10.5cm

d) 11.5cm

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Answer: b

Explanation: The center of gravity is given by, y = (a1y1 + a2y2 + a3y3) / (a1 + a2 + a3) = (40×18 + 30×9.5 +40×1 / (40 +30+40) = 9.5cm.

Where will be the center of gravity of an T section will be if the dimension of web is 2x20cm and that of flange is 2x15cm If the y-axis will pass through the center of the section?

a) 10.5cm

b) 11.45cm

c) 12.35cm

d) 12.85cm

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Answer: b

Explanation: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (40×16 + 30×7.5)/ (30+40) = 12.35cm.

Where will be the center of gravity of the following section?

a) 7.33cm

b) 8.33cm

c) 9.33cm

d) 10.33

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Answer: b

Explanation: Area of triangle = 50, area of rectangle = 50

The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (50×11.66 + 50×5)/(50+50) = 8.33cm.

Where will be the centre of gravity of the following L-section?

a) (18.31,30.81)

b) (19.45, 29.87)

c) (20,30)

d) (19.62,29.62)

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Answer: a

Explanation: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (600×50 + 414×3) / (600+414) = 18.31cm.

This will on for the y-axis.

For the x-axis, The center of gravity is given by, x = (a1x1 + a2x2) / (a1 + a2) = (600×3 + 414×40.5) / (600+414) = 30.81cm.

So the center of gravity will be at (2.33, 4.33).

Where will be the center of gravity of an I section will be if the dimension of upper web is 2x8cm, lower web is 2×16 and that of flange is 2x12cm If the y-axis will pass through the center of the section?

a) 7.611cm

b) 7.44cm

c) 6.53cm

d) 6.44cm

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Answer: d

Explanation: Area of upper web a1 = 16cm, area of flange a2 = 24, area of lower web a3 = 32.

The center of gravity is given by, y = (a1y1 + a2y2 + a3y3) / (a1 + a2 + a3) = (16×15 + 24×8 +32×1 / (16 +24+32)) = 6.44cm.

### Strength of MaterialsMCQs Moment of Inertia

The axis about which moment of area is taken is known as ____________

a) Axis of area

b) Axis of moment

c) Axis of reference

d) Axis of rotation

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Answer: c

Explanation: The axis of reference is the axis about which moment of area is taken. Most of the times it is either the standard x or y axis or the centeroidal axis.

Point, where the total volume of the body is assumed to be concentrated is ____________

a) Center of area

b) Centroid of volume

c) Centroid of mass

d) All of the mentioned

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Answer: b

Explanation: The centroid of the volume is the point where total volume is assumed to be concentrated. It is the geometric centre of a body. If the density is uniform throughout the body, then the center of mass and center of gravity correspond to the centroid of volume. The definition of the centroid of volume is written in terms of ratios of integrals over the volume of the body.

What is MOI?

a) ml2

b) mal

c) ar2

d) None of the mentioned

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Answer: c

Explanation: The formula of the moment of inertia is, MOI = ar2 where

M = mass, a = area, l = length, r = distance.

What is the formula of radius of gyration?

a) k2 = I/A

b) k2 = I2/A

c) k2 = I2/A2

d) k2 = (I/A)1/2

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Answer: a

Explanation: The radius of gyration of a body about an axis is a distance such that its square multiplied by the area gives moment of inertia of the area about the given axis. The formula of radius of gyration is given as k2 = I/A.

What is the formula of theorem of perpendicular axis?

a) Izz = Ixx – Iyy

b) Izz = Ixx + Ah2

c) Izz – Ixx = Iyy

d) None of the mentioned

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Answer: c

Explanation: Theorem of perpendicular axis stares that if IXX and IYY be the moment of inertia of a plane section about two mutually perpendicular axis X-X and Y-Y in the plane of the section then the moment of inertia of the section IZZ about the axis Z-Z, perpendicular to the plane and passing through the intersection of X-X and Y-Y is given by the formula

Izz – Ixx = Iyy.

What is the formula of theorem of parallel axis?

a) IAD = IG + Ah

b) IAB = Ah2 + IG

c) IAB = IG – Ah2

d) IAB = IG + Ixx

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Answer: b

Explanation: The theorem of parallel axis states that if the moment of inertia of a plane area about an axis in the plane of area theough the C.G. of the plane area be represented by IG, then the moment of the inertia of the given plane area about a parallel axis AB in the plane of area at a distance h from the C.G. is given by the formula

IAB = Ah2 + IG.

What is the unit of radius of gyration?

a) m4

b) m

c) N

d) m2

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Answer: b

Explanation: The radius of gyration = (length4/length2)1/2 = length

So its unit will be m.

What will be the the radius of gyration of a circular plate of diameter 10cm?

a) 1.5cm

b) 2.0cm

c) 2.5cm

d) 3cm

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Answer: c

Explanation: The moment of inertia of a circle, I = πD4/64 = 491.07 cm4

The area of circle = 78.57 cm,

Radius of gyration = (I/A)1/2 = 2.5 cm.