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# MCQs on Center of Gravity and Moment of Inertia

**MCQs on 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: bExplanation: 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: aExplanation: 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: bExplanation: The centre of gravity of a uniform rod lies at its middle point. The whole weight of the rod acts through its middle point. advertisement

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: aExplanation: 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?

strength-materials-questions-answers-center-gravity-q5

a) (6,3)

b) (6,6)

c) (6,1.5)

d) (1.5,3)

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Answer: cExplanation: 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?

strength-materials-questions-answers-center-gravity-q6

a) At 8.545cm

b) At 6.5cm

c) At 5cm

d) At 9.25cm

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Answer: aExplanation: 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?

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)

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Answer: aExplanation: 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 ?

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)

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Answer: bExplanation: 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: bExplanation: 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.

**MCQs on Center of Gravity of Section**

The center of gravity of the rod shown in figure will be _____________

strength-materials-questions-answers-center-gravity-section-q1

a) 5cm

b) 10cm

c) 15cm

d) 20cm

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Answer: bExplanation: 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: aExplanation: 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: cExplanation: 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). advertisement

Where will be the centre of gravity of the T section shown in the figure?

strength-materials-questions-answers-center-gravity-section-q4

a) 8

b) 8.5

c) 10.5

d) 11.5

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Answer: dExplanation: 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?

strength-materials-questions-answers-center-gravity-section-q5

a) (4.33, 2.33)

b) (4, 6)

c) (2.33, 4.33)

d) (1, 5)

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Answer: cExplanation: 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?

strength-materials-questions-answers-center-gravity-section-q6

a) (3.45,4.52)

b) (3.59,4.52)

c) (3.66,5.17)

d) (4.01,5.15)

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Answer: bExplanation: 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?

strength-materials-questions-answers-center-gravity-section-q7

a) (4.66,6.332)

b) (4.34,3.24)

c) (4.25,6.45)

d) (4.87,6.41)

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Answer: aExplanation: 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: bExplanation: 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: bExplanation: 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: bExplanation: 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: aExplanation: 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: dExplanation: 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.

**MCQs on 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: cExplanation: 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: bExplanation: 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: cExplanation: The formula of the moment of inertia is, MOI = ar2 where M = mass, a = area, l = length, r = distance. advertisement

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: aExplanation: 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: cExplanation: 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: bExplanation: 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: bExplanation: 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: cExplanation: 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.

**MCQs on Moment of Inertia of Section**

What is the moment of inertia of a circular section?

a) πD4/64

b) πD3/32

c) πD3/64

d) πD4/32

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Answer: aExplanation: The moment of inertia of a circular section is πD4/64.

What is the moment of inertia of a rectangular section about an horizontal axis through C.G?

a) bd3/6

b) bd2/12

c) b2d2/12

d) bd3/12

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Answer: dExplanation: The moment of inertia of a rectangular section about an horizontal axis through C.G is bd3/12.

What is the moment of inertia of a rectangular section about an horizontal axis passing through base?

a) bd3/12

b) bd3/6

c) bd3/3

d) bd2/3

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Answer: cExplanation: The moment of inertia of a rectangular section about an horizontal axis passing through base is bd3/3. advertisement

What is the moment of inertia of a triangular section about the base?

a) bh2/12

b) bh3/12

c) bh3/6

d) bh2/6

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Answer: bExplanation: The moment of inertia of a triangular section about the base is bh3/12.

What is the moment of inertia of a triangular section about an axis passing through C.G. and parallel to the base?

a) bh3/12

b) bh3/24

c) bh3/36

d) bh3/6

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Answer: cExplanation: The moment of inertia of a triangular section about an axis passing through C.G. and parallel to the base is bh3/36.

What will be the moment of inertia of a circle in cm4 of diameter is 10cm?

a) a340

b) 410

c) 460

d) 490

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Answer: dExplanation: The moment of inertia of a circle is = πD4/64 = 491.07 cm4.

What will be the moment of inertia of the given rectangle about an horizontal axis passing through the base?

strength-materials-problems-q7

a) 1500 mm4

b) 1650 mm4

c) 1666 mm4

d) 1782 mm4

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Answer: c
Explanation: The moment of inertia of a rectangular section about an horizontal axis passing through base = bd3/3
= 5x10x10x10/3
= 1666.66 mm4.
What will be the moment of inertia of the given rectangular section about an horizontal axis through C.G.?

strength-materials-problems-q7

a) 350 mm4

b) 379mm4

c) 416mm4

d) 500mm4

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Answer: c

Explanation: The moment of inertia of a rectangular section about an horizontal axis through C.G = bd3/12

= 5x10x10x10/12

= 416.67 mm4.

What will be the moment of inertia of the given triangle about the base?

strength-materials-questions-answers-moment-inertia-section-q9

a) 20.33 mm4

b) 21.33 mm4

c) 24.33 mm4

d) 22.33 mm4

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Answer: b

Explanation: The moment of inertia of a triangular section about the base = bh3/12.

= 4x4x4x4/12

= 21.33 mm4.

What will be the moment of inertia of the given triangle about an axis passing through C.G and parallel to base?

strength-materials-questions-answers-moment-inertia-section-q9

a) 6.1 mm4

b) 7.1 mm4

c) 8.1 mm4

d) 7.56 mm4

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Answer: b

Explanation: The moment of inertia of a triangular section about an axis passing through C.G. and parallel to the base = bh3/36.

= 4x4x4x4/36

= 7.11 mm4.

What will be the difference between MOI of two triangle sections is in 1st, MOI is taken about its base and in 2nd MOI is taken about its centroid?

a) bh3/12

b) bh3/18

c) bh3/36

d) bh3/24

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Answer: b

Explanation: The moment of inertia of a triangular section about the base is bh3/12

The moment of inertia of a triangular section about an axis passing through C.G. is bh3/36

So the difference = bh3/12 – bh3/36 = bh3/18.

**MCQs on Mass Moment of Inertia**

What is the product of the mass and the square of the distance of the center of gravity of the mass from an axis?

a) Moment of inertia

b) Mass moment of inertia

c) Center of gravity

d) Product of inertia

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Answer: bExplanation: The product of the mass and the square of the distance of the center of gravity of the mass from an axis is known as the mass moment of inertia about that axis.

What is the unit of mass moment of inertia?

a) m4

b) m6

c) N

d) m2

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Answer: bExplanation: The mass moment of inertia is the product of moment of inertia and area. So L4 x L2 = L6. so its unit will be m6.

What is mass moment of inertia of circular plate?

a) Md2/3

b) Md2/12

c) Mr2/4

d) Mr2/3

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Answer: cExplanation: The mass moment of inertia of circular plate is Mr2/4. advertisement

What is the mass MOI of a rectangular plate about x-axis passing through the C.G of the plate if the y-axis is parallel to d and perpendicular to b?

a) Mb2/12

b) Md2/12

c) Md2/6

d) Mb2/6

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Answer: bExplanation: As the mass MOI is to be find along the x-axis, it would be Md2/12.

What is the mass MOI of right circular cone of radius R and height H about its axis?

a) 4MR2/10

b) MR2/10

c) 3MR2/10

d) MR2/12

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Answer: cExplanation: The mass MOI of right circular cone of radius R and height H about its axis is 3MR2/10.

What is the mass MOI of a hollow circular cylinder if R is the outer diameter and r is the inner diameter?

a) M(R + r)/4

b) M(R – r )/4

c) M(R+ r)/2

d) M(R – r)/2

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Answer: aExplanation: The mass MOI of a hollow circular cylinder is M(R + r)/4 where R is the outer diameter and r is the inner diameter.

What is the mass MOI of a rectangular plate about y-axis passing through the C.G of the plate if the y-axis is parallel to d and perpendicular to b?

a) Mb2/12

b) Md2/12

c) Md2/6

d) Mb2/6

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Answer: aExplanation: As the mass MOI is to be find along the y-axis, it would be Mb2/12.

The product of inertia at the principal axes is _____________

a) Minimum

b) Unit

c) Zero

d) Maximum

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Answer: cExplanation: The moment of inertia about x-axis and about y-axis, on the axis they are zero. So the product of inertia will be zero in the principal axis.

What is the unit of product of inertia?

a) mm4

b) mm2

c) mm

d) mm3

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Answer: aExplanation: The unit of product of inertia is same as that of moment of inertia I.e. mm4.

What is the product of inertia of the given following section?

strength-materials-questions-answers-mass-moment-inertia-q10

a) 50mm4

b) 625mm4

c) 125mm4

d) 250mm4

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Answer: bExplanation: The product of inertia = area x points of C.G = ( 10×5) x 5 x 2.5 = 625mm4.

What is the product of inertia of a circle of diameter 10mm?

a) 1862mm4

b) 1945mm4

c) 1963mm4

d) 2014mm4

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Answer: cExplanation: The product of inertia = area x C.G = πx 10×10 / 4 x 5×5 = 1963mm2.