Applied Mechanics And Graphic Statics — Part 1 MCQs with Explanations

The dimensional formula of the Universal Gravitational constant G is derived from Newton's law of universal gravitation which states that the force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as: F=G⋅m 1 ⋅m 2 /r 2 Where: F is the force of attraction G is the Universal Gravitational constant m 1 and m 2 are the masses of the objects r is the distance between the objects To determine the dimensional formula of G , we can rearrange the formula as: G=F⋅r 2 /m 1 ⋅m 2 Now, let's analyze the dimensions: Force ( F ) has the dimensional formula: M 1 L 1 T -2 Distance ( r ) has the dimensional formula: L 1 Mass ( m 1 and m 2 ) has the dimensional formula: M 1 Substituting these dimensions into the rearranged formula, we get: G=(M 1 L 1 T -2 )⋅(L 1 ) 2 /M 1 ⋅M 1 G=M -2 L 2 T -2 Therefore, the correct dimensional formula of Universal Gravitational constant G is M -2 L 2 T -2 .

* Truss: A structure made of connected straight members forming triangles. * Graphic Statics: A visual way to analyze forces in trusses using diagrams. * Method of Substitution: A technique where you temporarily remove a member and replace it with an equivalent force to solve for unknowns, then put the member back in and adjust calculations. Now, let's look at the options: * A: Howe Truss: These are generally straightforward to solve with basic graphic statics, without needing substitution. * B: King Post Truss: Also usually simple and solvable without substitution. * C: Fink Truss: This type of truss often has complex geometry, making it statically indeterminate . This means you *cannot* solve for all the forces using just simple equilibrium equations at each joint with graphic statics alone initially. The method of substitution is often required to determine the forces in all the members. * D: Warren Truss: Warren trusses are often solvable using standard graphic statics methods. Therefore, the answer is C: Fink truss . Fink trusses commonly require the method of substitution due to their geometry.

The moment of inertia (II) of a solid sphere can be calculated using the formula: I=$$\frac{2}{5}{\text{M}}{{\text{r}}^2}$$ Where: I is the moment of inertia. M is the mass of the sphere. r is the radius of the sphere. In this formula, we can see that the moment of inertia depends on both the mass of the sphere (MM) and the square of its radius (r 2 ). The factor $$\frac{2}{5}$$ is a constant that relates to the distribution of mass within a solid sphere. This formula is specific to solid spheres and is derived from the integration of infinitesimal mass elements over the entire volume of the sphere. So, the moment of inertia for a solid sphere is $$\frac{2}{5}$$​ times the product of its mass and the square of its radius. This property is important in physics and engineering, especially when analyzing the rotational motion of objects.

Let's break down this question step-by-step! The key here is understanding what happens to momentum when each ball hits the wall. Momentum is a measure of mass in motion. Momentum depends on two things: how heavy something is (mass) and how fast it's moving (velocity). The question tells us both balls have the same mass and are moving with the same velocity initially. This means they start with the same momentum. Now, consider what happens after they hit the wall: - The rubber ball rebounds (bounces back) . This means its velocity changes direction, and its momentum also changes direction. - The lead ball falls down . This means it essentially stops; its velocity becomes zero. So, its momentum becomes zero. Change in momentum is calculated by: Final momentum - Initial momentum. For the rubber ball , the final momentum is in the opposite direction to its initial momentum, so the change is significant (close to double the initial momentum). For the lead ball , the final momentum is zero, so the change is equal to the initial momentum. Therefore, the rubber ball undergoes a much greater change in momentum than the lead ball. Let's look at the options: Option A: Rubber and lead balls undergo equal changes in momentum - Incorrect . We've just shown that the rubber ball's change in momentum is larger. Option B: Change in momentum suffered by lead ball is less that of rubber ball - Correct . This is what we concluded above. Option C: Momentum of rubber ball is less than that of lead ball - Incorrect . Initially, they have equal momentum and the change in momentum is what differs. Also it is talking about the current scenario after striking not the initial one. Option D: None of these - Incorrect . Option B is correct. So, the correct answer is Option B.

The centrifugal force on a mass 'm' describing a circular path of radius 'r' acts away from the centre of rotation. It is given by the formula: F = (m * w^2 * r) / g Where: F is the centrifugal force m is the mass w is the angular velocity r is the radius of the circular path g is the acceleration due to gravity kfg is not relevant to the context and seems to be a typo or irrelevant notation in the provided options. Therefore, the correct option is: Option C: Acts away from the centre of rotation

The correct answer is B: Parabolic . Here's why: * Uniformly Distributed Load: This means the load is spread evenly across the horizontal span of the cable (like the weight of a bridge deck hanging from the cable). * Parabolic Shape: When a cable is subjected to a uniformly distributed load horizontally , the cable takes the shape of a parabola. This shape ensures that the tension in the cable is efficiently distributed to support the load. * Catenary Shape (Why it's not the answer): A catenary curve is the shape a cable takes when it's only supporting its own weight (or a load distributed along the length of the cable itself , not horizontally). Think of a cable hanging freely between two points. * Other Options: Circular and Cubic Parabola shapes are not naturally formed by suspended cables under common loading conditions like a uniform distributed load.

Why: Up to about 720 km altitude, satellites are still in Low Earth Orbit (LEO), where the atmosphere is very thin but not absent. The residual air molecules produce aerodynamic drag on a satellite moving at roughly 7.5 km/s, which continuously removes orbital energy and causes gradual orbital decay. Option A (Uneven gravitational field): Earth’s oblateness and mass irregularities do perturb orbits (e.g., nodal and apsidal precession), but at ~720 km these effects are secondary compared with drag when considering orbit maintenance and decay over practical timescales. Option B (Gravity of the Sun and Moon): Third-body perturbations exist at all altitudes but are comparatively minor for LEO; their influence becomes relatively more important at higher altitudes (e.g., MEO/GEO). Option C (Aerodynamic forces): Dominant disturbance in LEO up to ~720 km. Drag reduces velocity, lowers perigee, and accelerates decay, requiring periodic reboosts for orbit maintenance. Conclusion: At altitudes up to 720 km, aerodynamic drag is the principal factor affecting a satellite’s orbit, making Option C correct.

Mechanical advantage (MA) = Load (W)/Effort (P). MA = W/P = 500/25 = 20. Velocity ratio (VR) = (Distance moved by the effort (Y))/(Distance moved by the load (X)). VR = Y/X = 650/13 = 50. Efficiency = MA/VR = 20/50 = 0.4 i.e. 40%.

Displacement refers to the change in position of a body in a particular direction. The rate of change of displacement with respect to time is defined as velocity . Velocity is a vector quantity , meaning it has both magnitude and direction. In contrast, speed is the rate of change of distance (not displacement) and is a scalar quantity . Acceleration is the rate of change of velocity, not displacement. Therefore, the correct term for the rate of change of displacement is Velocity.

The correct answer is Option B: 1.0 beat per second Here's why: A seconds pendulum is defined as a pendulum that takes exactly one second to swing in one direction. This means it takes one second to complete half of its full swing (from one extreme point to the other). A "beat" refers to one such swing . Since it takes 1 second for one beat, it means it has 1.0 beat per second.

$${\text{T}} = \frac{{2\pi }}{\omega }$$ is the correct answer to “In case of S.H.M. the period of oscillation (T), is given by”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

0.1 rad/sec is the correct answer to “The angular speed of a car taking a circular turn of radius 100 m at 36 km/hr will be”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

Proportional to the distance from the point of reference and directed towards it is the correct answer to “A body is said to move with Simple Harmonic Motion if its acceleration, is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

$$\sqrt {{{\text{P}}^2} + {{\text{Q}}^2} + 2{\text{PQ}}\cos \theta } $$ is the correct answer to “The resultant of two forces P and Q acting at an angle $$\theta $$, is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

The diagonal of the parallelogram which passes through the point of intersection of the forces is the correct answer to “Parallelogram Law of Forces states, "if two forces acting simultaneously on a particle be represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant may be represented in magnitude and direction by”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

0.67 is the correct answer to “A ball is dropped from a height of 2.25 m on a smooth floor and rises to a height of 1.00 m after the bounce. The coefficient of restitution between the ball and the floor is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

0.50 is the correct answer to “A marble ball is rolled on a smooth floor of a room to hit a wall. If the time taken by the ball in returning to the point of projection is twice the time taken in reaching the wall, the coefficient of restitution between the ball and the wall, is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

50 and 75 is the correct answer to “In a lifting machine a weight of 5 kN is lifted through 200 mm by an effort of 0.1 kN moving through 15 m. The mechanical advantage and velocity ratio of the machine are respectively”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

Newton and joule is the correct answer to “In SI units, the units of force and energy are respectively”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

$$\frac{{\text{h}}}{3} \times \frac{{{\text{b}} + {\text{2a}}}}{{{\text{b}} + {\text{a}}}}$$ is the correct answer to “The centre of gravity of the trapezium as shown in below figure from the side is at a distance of”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

0.6 R is the correct answer to “The centre of gravity of a quadrant of a circle lies along its central radius at a distance of”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

6 rad/sec2 is the correct answer to “If the angular distance, 0 = 2t 3 - 3t 2 , the angular acceleration at t = 1 sec. is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

$$\frac{{{{\text{g}}_2}}}{{{{\text{g}}_1}}}$$ is the correct answer to “If g 1 and g 2 are the gravitational accelerations on two mountains A and B respectively, the weight of a body when transported from A to B will be multiplied by”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

3 t is the correct answer to “The vertical reaction at the support ‘A’ of the structure shown in below figure, is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

BD is the correct answer to “The member which does not carry zero force in the structure shown in below figure, is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

Rotary and translatory is the correct answer to “The motion of a bicycle wheel is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

All the above is the correct answer to “Effect of a force on a body depends upon its”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

All the above is the correct answer to “The following statement is one of the laws of Dynamic friction”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

$$\frac{{{\text{w}}l}}{{8{\text{y}}}}$$ is the correct answer to “If $$l$$ is the span of a light suspension bridge who's each cable carries total weight (w) and the central dip is y, the horizontal pull at each support, is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

20 m/sec is the correct answer to “A 49 kg lady stands on a spring scale in an elevator. During the first 5 sec, starting from rest, the scale reads 69 kg. The velocity of the elevator will be”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

Horizontal thrust is the correct answer to “Minimum pull in a suspended cable with supports at two ends is equal to”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

Before impact is more than that after impact is the correct answer to “Pick up the correct statement from the following. The kinetic energy of a body”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

Closed polygon is the correct answer to “The force polygon representing a set of forces in equilibrium is a”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

Both force polygon" and funicular polygon close is the correct answer to “A system of coplanar forces is in equilibrium when”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

22.5 kg is the correct answer to “One end of a light string 4 m in length is fixed to a point on a smooth wall and the other end fastened to a point on the surface of a smooth sphere of diameter 2.25 m and of weight 100 kg. The tension in the string is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

100 kg is the correct answer to “A weight of 100 kg is supported by a string whose ends are attached to pegs ‘A’ and ‘B’ at the same level shown in below figure. The tension in the string is”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

400 watts is the correct answer to “A 50 kg boy climbs up a 8 m rope in gymnasium in 10 sec. The average power developed by the boy is approximately”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

8.75 kg is the correct answer to “Two parallel forces 20 kg and 15 kg act. In order that the distance of the resultant from 20 kg force may be the same as that of the former resultant was from 15 kg, the 20 kg force is diminished by”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

$$\frac{1}{{9.81}}$$ kg. wt is the correct answer to “One Newton is equivalent to”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.

1 Newton metre is the correct answer to “One Joule is equivalent to”. This question is standard in competitive examination syllabi for Applied Mechanics And Graphic Statics.