# Engineering Mechanics MCQ

### Engineering Mechanics MCQ

### 1) The unit of force in the C.G.S. system is:

- Newton
- Kilogram Force
- Dyne
- None of the above

**Answer:**(c) Dyne

**Explanation:**The C.G.S. is the Centimeter- Gram - Second system of units. According to the C.G.S. system of units, the unit of force is Dyne. 1 Newton = 100,000 dyne

### 2) Two forces of magnitude 5N and 6N are acting on a point. Find the magnitude of the resultant force if the angle between the two forces is 60 degrees?

- 9.54N
- 7.81N
- 11N
- 4.65N

**Answer:**(a) 9.54N

**Explanation:**Let the two forces acting on a point be P and Q and the angle between the two forces be a. Given, P = 5N Q = 6N a = 60 degrees The magnitude of the resultant force R is given by: < --- image -- engineering-mechanics-mcq-q2 ---> R = 9.54 N Hence, the resultant force is 9.54 N.

### 3) Which one of the following law states that each force among the three forces in equilibrium will be proportional to the sine of the angle between the other two forces?

- Law of triangle of forces
- Lami's theorem
- Law of parallelogram of forces
- None of the above

**Answer:**(b) Lami's theorem

**Explanation:**As stated by Lami's theorem, if the three forces are in equilibrium and are acting on point, each particular force is proportional to the sine of the angle between the other two forces.

### 4) Newton's law of motion ' Every reaction has an equal and opposite reaction' is:

- First law
- Second law
- Third law
- None of the above

**Answer:**(c) Third law

**Explanation:**The above law is the Newton's third law of motion, which acts on different bodies.

### 5) A force F = 2i + 3j - 4k is applied at a point P (1, 2, -1). What is the moment of the force F about the point (2, -1, 2)?

- 3i + 5j +7k
- - 2i + 10j - 8k
- 7i - 9j -+2k
- - 3i - 10j - 9k

**Answer:**(d) - 3i - 10j - 9k

**Explanation:**Given: F = 2i + 3j - 4k Let the given point (2, -1, 2) be O. The position vector r of the point w. r. t. O = Position vector of point P - position vector of point O. Position vector of point P can be written as i + 2j - k Position vector of point O can be written as 2i - j + 2k The position vector r of the point w. r. t. O = (i + 2j - k) - (2i - j + 2k) = -i + 3j - 3k The Moment M is given by: M= r x F Where, F is the given force and r is the position vector. The cross product of the two vectors can be represented as: < --- image -- engineering-mechanics-mcq-q5 ---> = [(3) (-4) - (3)(-3)] i - [(-1)(-4) - (2)(-3)] j + [ ((-1)(3) - (2)(3)] k = [ -12 + 9) i - [4 + 6] j + [-3 - 6] k = - 3i - 10j - 9k

### 6) The three forces namely F1, F2, and F3, are acting on a body, as shown in the below diagram. All the three bodies are in equilibrium. Find the magnitude of the two forces F1 and F2, if the magnitude of the third force F3 is 400N.

< --- image -- engineering-mechanics-mcq-q6 --->- 400N, 200N
- 400N, 400N
- 200N, 200N
- 400N, 100N

**Answer:**(b) 400N, 400N

**Explanation:**When the bodies are in equilibrium, the resultant force in the x-direction and y-direction should be zero.

**Case I:**Sum Fx = 0 F1 cos30 -F2 cos30 = 0 F1 cos30 = F2cos30 F1 = F2

**Case II:**Sum Fy = 0 F1 sin30 + F2 sin30 - 400 = 0 F1 sin30 + F2 sin30 = 400 F1 x ½ + F2 x ½ = 400 F1/ 2 + F2 /2 = 400 F1 + F2 = 800 From case 1, we know F1 = F2. Substituting the value of F1 in the above equation, we get: F2 + F2 = 800 2 F2 = 800 F2 = 800/2 F2 = 400N Thus, both the forces F1 and F2 have a value of 400V. F1 = F2 = 400N.

### 7) A ball of weight 80N rests in a right-angled groove, as shown in the below diagram. The sides of the groove are inclined at the angles 30 degrees and 60 degrees respectively. Considering all the surfaces smooth, find the reaction Ra and Rc at the point of contact.

< --- image -- engineering-mechanics-mcq-q7 --->- 40N, 69.28N
- 20N, 69.28N
- 40N, 34.64N
- 20N, 40N

**Answer:**(a) 40N, 69.28N

**Explanation:**Consider the below figure. The reaction Rc acting on a point C is normal to the line FD. < --- image -- engineering-mechanics-mcq-q7_2 ---> Similarly, the reaction Ra acting on a point A is normal to the line DE, as shown in the above figure. B is the center of the ball. In triangle HDC, the angle CDH is equal to 30 degrees, and angle DCH is equal to 90 degrees. Thus, angle DHC is equal to 60 degrees. Similarly, it happens in the other triangle HBL. After calculating, we get the value of angle GBL equal to 60 degrees. Now, according to the equilibrium condition, Sum Fx and Fy = 0 Let's first consider the case of Summation of Fx = 0.

**Case I:**Sum, Fx = 0 Rc sin30 - Ra sin60 = 0 Rc sin30 = Ra sin60 Rc x 1/2 = Ra x 0.866 Rc = Ra x 0.866 x 2 Rc = 1.732 Ra

**Case II:**Sum, Fy = 0 80 - Ra cos60 - Rc cos30 = 0 80 = Ra cos60 + Rc cos30 80 = Ra x 1/2 + Rc x 0.866 Putting the value of Rc from the case I, we get: 80 = Ra x 0.5 + (1.732 Ra) x 0.866 80 = 0.5Ra + 1.5Ra 80 = 2Ra Ra = 80/2 = 40N We know, Rc = 1.732 Ra Rc = 40 x 1.732 = 69.28N Hence, the values of Ra and Rc are 40N and 69.28N.

### 8) A force of 200N acts on a point that makes an angle of 30 degrees with the horizontal. Find the component of this force in the y-direction.

- 173.2N
- 200N
- 100N
- 86.6N

**Answer:**(c) 100N

**Explanation:**Let Fy be the component along the y-axis. Fy = F sina Where, a is the angle = 30 degrees Consider the below diagram: < --- image -- engineering-mechanics-mcq-q8 ---> It displays the angle with the horizontal. Fy = 200 x sin30 Fy = 200 x 0.5 Fy = 100N Hence, the component of the 200N force in the y-direction is 100N.

### 9) Which of the following force have the same line of action?

- Resultant force
- Coplanar force
- Concurrent force
- Collinear force

**Answer:**(d) Collinear force

**Explanation:**Collinear forces are the forces with the same line of action. The concurrent forces are the intersecting forces at a point, while the coplanar forces act in a particular plane.

### 10) The direction of the moment with a negative value is:

- Clockwise
- Anti-clockwise
- Single
- None of the above

**Answer:**(a) Clockwise

**Explanation:**The direction of the moment with a negative value is clockwise, while the moment with a positive value has an anti-clockwise direction.

### 11) Four forces of magnitude 30, 40N, 50N, and 60N are acting along the four sides of a square PQRS, as shown in the below figure. Determine the resultant moment about point A. Assume the side of square be 2cm.

< --- image -- engineering-mechanics-mcq-q11 --->- 360Nm
- 180Nm
- 90Nm
- None of the above

**Answer:**(b) 180Nm

**Explanation:**As shown in the above figure, the forces at points A and B pass through the same point A. Hence, the moment of force at point B about point A will be zero. So, we will calculate the force at points C and D about point A.

**Step 1:**At point C The moment at point C about point A = Force at C x Length AB (Perpendicular distance) = 40N x 2m = 80Nm (anti-clockwise)

**Step 2:**At point D The moment at point D about point A = Force at D x Length AD (Perpendicular distance) = 50N x 2m = 100Nm Resultant moment = All forces about point A = At point C + at point D = 80Nm + 100Nm = 180Nm

### 12) The type of support of a beam with rollers is known as:

- Pin - joint support
- Simple support
- Smooth-surface support
- None of the above

**Answer:**(d) None of the above

**Explanation:**The type of support of a beam with rollers is known as roller support. The simple support uses the knife edges, while the smooth - surface support signifies the contact of body with a smooth surface.

### 13) The distributed force over an area in the force system is known as:

- Body force
- Linear force
- Surface force
- Parallel force

**Answer:**(c) Surface force

**Explanation:**The distributed force acts over an area, length, and volume is known as surface, linear, and body force.

### 14) Four like parallel forces of 100N, 300N, 400N, and 500N are acting on the point A, B, C, and D on a straight line ABCD, as shown in the below diagram. The distances are AB = 10cm, BC = 30cm, and CD = 40cm. Find the distance of the resultant from point A on the line ABCD.

< --- image -- engineering-mechanics-mcq-q14 --->- 59 cm
- 35 cm
- 40 cm
- 45 cm

**Answer:**(d) 45 cm

**Explanation:**Let the resultant of all the four parallel-like forces be R. R = 100 + 300 + 400 + 500 = 1300N Consider that the resultant force is acting at a distance x from point. Now, we need to take the resultant of all the forces about the point A. The force of 100N is passing on the same point A. Hence, its moment about A will be 0. Thus, we will calculate the moment about point B, C, and D.

**Step 1**: At point B Moment = Force at B x Distance AB = 300N x 10cm = 3000N cm.

**Step 2**: At point C Moment = Force at C x Distance AC = 400N x 40cm = 16000N cm.

**Step 3**: At point D Moment = Force at D x Distance AD = 500N x 80cm = 40000N cm. Sum of moments = 0 + 3000 + 16000 + 40000 = 59000N cm Moment R about point A = R x A = 1300 x N cm But, the sum of moments = moment about point A 1300x = 59000 x = 59000/1300 x = 45 cm

### 15) A simply supported beam of length 8 m carries a uniformly distributed load of 10kN/ m, as shown in the below diagram. The load is placed on the beam AB at a distance of 5m from the left end. Calculate the reaction at B?

< --- image -- engineering-mechanics-mcq-q15 --->- 31.25N
- 15.625N
- 16. 625N
- 18N

**Answer:**(b) 15.625N

**Explanation: Given**: Length of the beam AB = 8m. The weight and length of the uniformly distributed load are 10kN/ m and 5m. Total load due to the distributed load = 10 x 5 = 50kN. The load of 60kN will act at the middle of the beam AC. It means the distance from the left will be 5/2 = 2.5m. Let, Rb = reaction at point B. We will calculate the reaction at point B by taking the moments of all the forces about point A and equating it to the resulting moment. Rb x 8 - (5 x 10) x 2.5 = 0 8Rb = 125 Rb = 125/8 Rb = 15.625kN

### 16) Which one of the following is not a law of solid friction?

- The force of friction is independent of the sliding velocity.
- The force of friction is equal to the force applied to the surface.
- The limiting frictional force depends on the area of the surface in contact.
- The limiting frictional force is independent upon the area.

**Answer:**(c) The limiting frictional force depends on the area of the surface in contact.

**Explanation:**The limiting frictional force in independent of the area of the surface in contact.

### 17) A body of weight 400N is placed on a rough horizontal plane. If the coefficient of friction between the plane and the body is 0.5, fund the horizontal force required to slide the body on the plane.

- 200N
- 800N
- 400N
- None of the above

**Answer:**(a) 200N

**Explanation:**The force required to slide the body on the plane can be calculated as: F = uR Where, F is the friection u is the coefficient of friction R is the Normal reaction F = 0.5 x 400 = 200N

### 18) Find the work done on the body if the distance traveled by the body is 10m?

< --- image -- engineering-mechanics-mcq-q18 --->**Given**: Weight of the body: 400N and Coefficient of friction: 0.3

- 303.92J
- 4000J
- 2000J
- 3039.2J

**Answer:**(d) 3039.2J

**Explanation:**The given parameters are: Weight of the body: 400N Inclination: 30 degrees Coefficient of friction: 0.3 Distance traveled by the body: P Let the required force be P. P = W sin30 + uR P = 400 x 0.5 + 0.3 x R P = 200 + 0.3R The resolving force acting normal to the plane is: R = W cos30 R = 400 x 0.866 R = 346.4N Putting the value of R in the above equation, we get: P = 200 + 0.3 x 346.4 P = 200 + 103.92 P = 303.92N Work done = Force x Distance travelled = 303.92 x 10 3039.2J

### 19) The coefficient for friction is expressed in terms of normal reaction(R) and limiting force of friction (F) as:

- FR
- F/R
- F + R
- FR + 1

**Answer:**(b) F/R

**Explanation:**The coefficient of friction is defined as the ratio of F and R.

### 20) The frame where members are less than (2j - 3), j is the number of joints is:

- Redundant frame
- Imperfect frame
- Deficient frame
- Regular frame

**Answer:**(c) Deficient frame

**Explanation:**The redundant frame has a number of members greater than 2j - 3. The imperfect frame has the number of members less or more than 2j - 3. Here, j is the number of joints.

### 21) Find the force in the branch AB of the truss shown in the below diagram:

< --- image -- engineering-mechanics-mcq-q21 --->- 17.32kN
- 15kN
- 20kN
- 27.32kN

**Answer:**(a) 17.32kN

**Explanation:**The above figure has three members, AB, BC, and AC. Let the force in these three members be F1, F2, and F3. We are required to find the force F1. Here, the triangle ABC is a right-angle triangle. The angle BAC is equal to 90 degrees. The distance AB = 4 x cos60 = 4 x ½ = 2m The distance of line action of 20kN from B is: AB x cos60 = 2 x ½ = 1m The moment about B can be calculated as: Rc x 4 = 20 x 1 Rc = 20/4 = 5kN Rb = Total load - Rc = 20 - 5 = 15kN The force F1 can be calculated as: F1 sin60 = Rb F1 sin60 = 15kN F1 = 15/sin 60 = 15/ 0.866 F1 = 17.32kN

### 22) Which of the following figure about the center of gravity (C.G.) or centroid is incorrect?

- C.G. lies at the midpoint of the triangle.
- C.G. lies at the center of the circle.
- C.G. of a rectangle lies at a point where diagonal meets.
- C.G. of a parallelogram lies at a point where adjacent sides meet.

**Answer:**(d) C.G. of a parallelogram lies at a point where adjacent sides meet.

**Explanation:**The Center of Gravity of a parallelogram also lies at a point where the diagonal meet. It can lie at a point where the line joining the opposite sides intersects. It cannot lie at a point where adjacent sides meet.

### 23) Determine the center of gravity of the T-section shown in the below diagram?

< --- image -- engineering-mechanics-mcq-q23 --->- 2.33cm
- 6.778cm
- 10cm
- 5. 73cm

**Answer:**(b) 6.778cm

**Explanation:**Let's first divide the given section into two parts by splitting it in between, as shown below: < --- image -- engineering-mechanics-mcq-q23_2 ---> The formula to calculate the center of gravity is: Y = a1y1 + a2y2/ A Here, a1 and a2 are the area of the rectangle AHGB and CDEF, as marked in the above figure. So, a1 = 10 x 2 = 20 cm^2 and a2 = 8 x 2 = 16 cm^2 The variables y1 and y2 are the distance of Center of Gravity of area a1 and a2 from the bottom line DE. So, y1 = 8 + 2/2 = 8 + 1 = 9 cm and y2 = 8/2 = 4cm Putting the values in the formula, we get: Y = a1y1 + a2y2/ (a1 + a2) Y = 20 x 9 + 16 x 4 / (20 + 16) Y = 180 + 64 /36 Y = 244/36 Y = 6.778cm.

### 24) The point through which the full weight of the body acts is known as:

- Centroid
- Center of Gravity
- Both (a) and (b)
- None of the above

**Answer:**(b) Center of gravity

**Explanation:**The centroid is a point where the total area of the plane figure is concentrated.

### 25) Moment of Inertia of a rectangular section about a horizontal axis passing through the base is:
- Bd^3/ 12
- Bd^3/ 6
- Bd^3/ 3
- Bd^3/16

**Answer:** (c) bd^3 / 3
**Explanation:** The M.I. about the horizontal axis passing through the center of gravity is Bd^3/12, while through the base is Bd^3/ 3.

**Answer:**(c) bd^3 / 3

**Explanation:**The M.I. about the horizontal axis passing through the center of gravity is Bd^3/12, while through the base is Bd^3/ 3.

### 26) The motion of a body along the curved path is known as:

- Rectilinear motion
- Uniform motion
- Relative motion
- Curvilinear motion

**Answer:**(d) Curvilinear motion

**Explanation:**A body's motion along a regular path or straight line is known as rectilinear motion. Similarly, the motion of a body along the curved path is known as curvilinear motion.

### 27) Acceleration is expressed as:

- ds/ dt
- dv/ ds
- dv/ dt
- ds/dv

**Answer:**(c) dv/ dt

**Explanation:**Acceleration can be expressed as dv/ dt. Where dv is the change in velocity and dt is change in the time.

### 28) Find the acceleration of a body moving with the uniform acceleration and cover 10m in the fourth and second and 22min 10th second seconds.

- 2 m/s
- 3 m/s
- 2.33 m/s
- 4 m/s

**Answer:**(a) 2 m/s

**Explanation:**The distance covered in the 4

^{th}second is 10m The distance covered in the 10

^{th}second is 22m. Using the equation u + a/2 (2n - 1) to find the distance covered in the nth second. = u + a/2 [2 x 4 - 1] 10 = u + a/2 [7] 10 = u + 7a/2 Where u is the initial velocity and a is the acceleration. Distance covered in the 10

^{th}second = u + a/2 [2 x 10 - 1] 22 = u + a/2 [19] 22 = u + 19a/2 Subtracting the above equations, we get: 22 - 10 = (u + 19a/2) - (u + 7a/2) 12 = 12a/2 12 = 6a a = 2 m/s

### 29) The relation between the linear velocity (V) and angular velocity (w), r being the radius of the circle is:

- V = r / w
- V = r x w
- V = r + w
- V = w (r + 1)

**Answer:**(b) V = r x w

**Explanation:**The linear velocity can be calculated as the product of the circle's angular velocity and radius.

### 30) A body is rotating at an angular velocity of 10 radians per second. After 4 seconds, the velocity of the body becomes 22 radians per second. Find the angular acceleration of the body.

- 3rad/ second^2
- 2.5rad/ second^2
- 2rad/ second^2
- 3.33rad/ second^2

**Answer:**(a)

**Explanation:**Let the initial and angular velocity be wo and w. Where wo = 10rad/s and w = 22rad/s T = time = 54seconds Using the equation, w = wo + at, we can easily find the angular acceleration (a) of the body. 22 = 10 + a4 4a = 22 - 10 4a = 12 A = 12/4 a = 3rad/ second^2.

### 31) The unit of acceleration is S.I. system of units is:

- m/s
- m^2/s
- m/ s^2
- m/ s^3

**Answer:**(c) m/s^2

**Explanation:**Acceleration is represented as: a = dv/dt The unit of velocity is m/s and unit of time is s. So, a = (m/s) / s a = m / s x s a = m / s^2

### 32) The relation between the r. p. m. (revolutions per minute (N)) and angular velocity (w) is:

- 2?N/120
- 2?N/30
- ?N/60
- 2?N/60

**Answer:**(d) 2?N/60

**Explanation:**The angular velocity or the angle of the body covered per second is equal to the product of the angle covered in one revolution (2pi) and the number of revolutions per second (N / 60).

### 33) The motion of a body about a fixed point in a circular path is known as:

- Motion of translation
- Motion of rotation
- Both (a) and (b)
- None of the above

**Answer:**(b) Motion of rotation

**Explanation:**The motion of the particles of a moving body in the form of parallel planes at the same distance is known as translation motion. Thus, the motion of a body about a fixed point along a circular path is known as rotation.

### 34) A particle moves along a straight line with the velocity given in the form of equation v = 3t^3 - 2t^2 - 5t + 4. Here, v is the velocity, and t is the time in seconds. Determine the acceleration at time t = 4 seconds.

- 155m/ s^2
- 146m/ s^2
- 123m/ s^2
- 5m/ s^2

**Answer:**(c) 123m/ s^2

**Explanation:**The acceleration is defined as the rate of change of velocity with time. It is represented as: a = dv/dt = d/dt [3t^3 - 2t^2 - 5t + 4] = 9t^2 - 4t - 5 Putting the value of t = 4, we get: a = 9 x 16 - 4 x 4 - 5 a = 144 - 16 - 5 = 123m/ s^2 Hence, the acceleration is 123 m/s^2.

### 35) The unit of mass in the S.I. system is given by:

- Gram
- Kilogram
- Milligram
- All of the above

**Answer:**(b) Kilogram

**Explanation:**Gram is also the unit of mass but in the C.G.S. system of units. Thus, Kilogram is the S.I. unit of mass.

### 36) The Momentum of a body is given by:

- Mass x Force
- Mass x Acceleration
- Mass x Time
- Mass x Velocity

**Answer:**(d) Mass x velocity

**Explanation:**The momentum of a body is a vector quantity, which is defined as the product of mass and velocity.

### 37) A bullet is moving with a velocity of 400m/s. It is fired into a wood that penetrates to a depth of 10cm. Find the velocity at which the bullet will come out if it moves with the same velocity and fires into a similar wood of thickness 4cm.

- 126.49 m/s
- 400 m/s
- 155. 45 m/s
- 100 m/s

**Answer:**(a) 126.49 m/s

**Explanation:**Given: u = 400m/s Let the distance be s, which is equal to 10 cm or 0.1m. Final velocity = v = 0m/s

**Step 1:**We will first calculate the retardation. Using the equation, v^2 - u^2 = -2as The negative sign is due to the force of bullet that is acting in the opposite direction to the motion of the bullet. 0^2 - 400^2 = - 2as a = 400 x 400/ 2 x 0.1 a = 800000m/ s^2.

**Step 2:**We will find the velocity at which the bullet comes out. We know, u = 400m/s a = 800000m/ s^2 r = 4cm = 0.04m Using the equation, v^2 - u^2 = -2as v^2 - 400^2 = - 2as v^2 - 400^2 = - 2 x 800000 x 0.04 v^2 = 80000 - 64000 v^2 = 16000 v = 126.49 m/s.

### 38) Which one of the following is a type of mechanical energy?

- Kinetic Energy
- Potential Energy
- Both (a) and (b)
- None of the above

**Answer:**(c) Both (a) and (b)

**Explanation:**A body can possess either kinetic or potential energy or both depending upon the state of rest or motion.

### 39) The inertia force on the body of weight W1, whereW1 and W2 are connected by a spring is:

- W1/g x a
- (W1 + W2)/ W1g x a
- -W1/g x a
- -(W1 + W2)/ g x a

**Answer:**(c) -W1/g x a

**Explanation:**- '(W1 + W2)/ g x a' is the resultant inertia force on the two bodies connected by a spring. Thus, the inertia force on the body W1 is -W1/g x a. It can also be defined as '- Mass of W1 x Acceleration.'

### 40) The area of the force-distance curve represents:

- Force
- Work done
- Kinetic Energy
- Momentum

**Answer:**(b) Work done

**Explanation:**Work done is defined as the product of force and distance. Hence, the area of the force-distance curve represents the work done. In case of a rotating body, work done is defined as the product of Torque and angle.

### 41) The work done on a body if the force moves in the opposite direction to the force is:

- Zero
- Positive
- Negative
- Infinite

**Answer:**(c) Negative

**Explanation:**The work done on a body is zero if the force moves at right angles. Similarly, it is positive if the force moves in the same direction to the force.

### 42) 1 kgf (kilogram force) in the M.K.S. system is equal to:

- 9.82N
- 9.80N
- 10N
- 9.81N

**Answer:**(d) 9.81N

**Explanation:**1 kgf in the M.K.S. system of units is equal to 9.81N.

### 43) Which of the following is not a state of equilibrium?

- Stable equilibrium
- Regular equilibrium
- Unstable equilibrium
- Neutral equilibrium

**Answer:**(b) Regular equilibrium

**Explanation:**When the body is in equilibrium, a small force is required to displace the body. The states of equilibrium are categorized as stable equilibrium, unstable, and neutral equilibrium. A small force on the body in the neutral state causes no change in the position. It means that the body remains in an equilibrium state.

### 44) What happens if all the force of a system acts in the same direction:

- The system will be given virtual displacement in the direction of forces.
- The system will be given virtual displacement in the opposite direction of forces.
- The system will be in an equilibrium state.
- All of the above.

**Answer:**(a) The system will be given virtual displacement in the direction of forces.

**Explanation:**If all the forces act in the same direction, the system will be given virtual displacement in the direction of forces.

### 45) The potential energy in the case of unstable equilibrium is:

- Zero
- Maximum
- Minimum
- None of the above

**Answer:**(b) Maximum

**Explanation:**The potential energy in the case of stable equilibrium is minimum and maximum in the unstable state of equilibrium. Similarly, there is no change in the potential energy in the case of neutral equilibrium.

### 46) Mass moment of Inertia of a cone about its axis of symmetry is:
- 2/5 MR^2
- MR^2
- 3/10 MR^2
- 5/3 MR^2

**Answer:** (c) 3/10 MR^2
**Explanation:** The mass moment of inertia of a cone about its axis of symmetry is 3/10 MR^2. Here, R is the radius of the base of the core and M is the mass of the cone.

**Answer:**(c) 3/10 MR^2

**Explanation:**The mass moment of inertia of a cone about its axis of symmetry is 3/10 MR^2. Here, R is the radius of the base of the core and M is the mass of the cone.

### 47) The force given by 'dF = w x dL' is a type of:

- Surface force System
- Body force system
- Both (a) and (b)
- None of the above

**Answer:**(d) None of the above

**Explanation:**The given force is a type of linear force distributed system.