Engineering Questions with Answers - Multiple Choice Questions
Strength of Materials MCQs Center of Gravity’
1 - Question
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
View Answer
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.
2 - Question
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
View Answer
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.
3 - Question
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
View Answer
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.
4 - Question
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
View Answer
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.
5 - Question
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)
View Answer
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).
6 - Question
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
View Answer
Answer: a
Explanation: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = ( 36×11.5 + 30×5) / (36+30) = 8.545cm.
7 - Question
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)
View Answer
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).
8 - Question
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)
View Answer
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).
9 - Question
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
View Answer
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.
10 - Question
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
View Answer
Answer: b
Explanation: The center of gravity of a rod will be on its center. Here it will be at 10cm.
11 - Question
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
View Answer
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.
12 - Question
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)
View Answer
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).
13 - Question
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
View Answer
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.
14 - Question
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)
View Answer
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).
15 - Question
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)
View Answer
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).
16 - Question
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)
View Answer
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).
17 - Question
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
View Answer
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.
18 - Question
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
View Answer
Answer: b
Explanation: The center of gravity is given by, y = (a1y1 + a2y2) / (a1 + a2) = (40×16 + 30×7.5)/ (30+40) = 12.35cm.
19 - Question
Where will be the center of gravity of the following section?
a) 7.33cm
b) 8.33cm
c) 9.33cm
d) 10.33
View Answer
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.
20 - Question
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)
View Answer
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).
21 - Question
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
View Answer
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.
22 - Question
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
View Answer
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.
23 - Question
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
View Answer
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.
24 - Question
What is MOI?
a) ml2
b) mal
c) ar2
d) None of the mentioned
View Answer
Answer: c
Explanation: The formula of the moment of inertia is, MOI = ar2 where
M = mass, a = area, l = length, r = distance.
25 - Question
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
View Answer
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.
26 - Question
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
View Answer
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.
27 - Question
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
View Answer
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.
28 - Question
What is the unit of radius of gyration?
a) m4
b) m
c) N
d) m2
View Answer
Answer: b
Explanation: The radius of gyration = (length4/length2)1/2 = length
So its unit will be m.
29 - Question
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
View Answer
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.