Engineering Mechanics MCQ Quiz - Objective Question with Answer for Engineering Mechanics - Download Free PDF

Explanation:

Coefficient of restitution (e):

The coefficient of restitution or coefficient of the resilience of a collision is defined as the ratio of the relative velocity of separation after the collision to the relative velocity of the approach before the collision.

Coefficient of restitution (e) =\(\mathbf{\frac{Relative\;velocity\;after\;collosion}{Relative\;velocity\;before\;collosion}=\frac{{Velocity\ of\ Separation}}{{Velocity\ of\ approach}}=\frac{v_B-v_A}{u_A-u_B}}\)

Properties of different types of collision are given in the table below:

Let us give some notations to the parameters involved in the question:-

M = Mass of the gun (in Kg), m = Mass of the bullet (in Kg)

V = Velocity of gun recoil (in m/s), v = Velocity of the bullet (in m/s)

As in the system (both gun and the bullet) if we consider then there is no external force present hence to linear momentum in the horizontal direction remains conserved.

Initial momentum = final momentum.

Initial momentum = m x (velocity of bullet at start) + M x (velocity of Gun at start)

Under initial state the bullet is in the gun at rest thus there is no velocity.

Initial momentum = m x 0 + M x 0 = 0

Final momentum = m x (velocity of bullet fired) + M x (velocity of Gun recoil)

= m x v + M x V

= 0.01 x 250 + 5 x V

= 2.5 + 5 x V

equating both the equation:

Concept:

If the marble drops from the height h₁ above the floor the velocity with which it would hit the floor:

\(u=\sqrt{2gh_1}\)

Let the velocity of rebound be v and let the marble rises to height h₂ after the rebound. Then

\(v=\sqrt{2gh_2}\)

\(Coefficient\;of\;restitution = \frac{{velocity\;of\;separation}}{{veloctiy\;of\;approach}}\)

Calculation:

 \(0.9 = \frac{v}{u} = \sqrt {\frac{{2g{h_2}}}{{2g{h_1}}}} \)

\(0.9 = \sqrt {\frac{{{h_2}}}{{{h_1}}}}\)

Now,

Squaring on both sides, we get

\(0.81 = \frac{{{h_2}}}{{{h_1}}}\)

∴ h₂ = 3 × 0.81

∴ h2 = 2.43 m

Explanation:

When two or more forces act on a body, they are called to form a system of forces.

Coplanar forces: 

• The forces, whose lines of action lie on the same plane, are known as coplanar forces.

Colinear forces: 

• The forces, whose lines of action lie on the same line, are known as colinear forces.

Concurrent forces: 

• The forces, which meet at one point, are known as concurrent forces. The concurrent forces may or may not be colinear.

Coplanar concurrent forces:

• The forces, which meet at one point and their lines of action also lie on the same plane, are known as coplanar concurrent forces.

Coplanar non-concurrent forces:

• The forces, which do not meet at one point, but their lines of action lie on the same plane, are known as coplanar non-concurrent forces.

Non-coplanar concurrent forces: 

• The forces, which meet at one point, but their lines of action do not lie on the same plane, are known as non-coplanar concurrent forces.
• A rigid body is subjected to non-coplanar concurrent force system, \(\rm \sum F_x=\sum F_y=\sum F_z=0\)

Non-coplanar non-concurrent forces: 

• The forces, which do not meet at one point and their lines of action do not lie on the same plane, are called non-coplanar non-concurrent forces.

Explanation -

If the graph shows that the change in length (x) of a wire in response to stress (F) results in a straight line at both temperatures T1 and T2, this suggests a linear relationship between the applied stress and the resulting strain (change in length/original length). This is consistent with Hooke's Law up to the elastic limit of the material.

The line might have different slopes at the two temperatures T1 and T2. If so, this suggests that the temperature of the wire affects its Young's modulus. The Young's modulus (Y) is defined as the ratio of stress (Force/Area) over strain (Change in Length/Original Length). If the slope of the line (which provides the ratio of stress to strain) is changing with temperature, then the Young's modulus of the wire is also changing with temperature. This indicates that the material properties of the wire are temperature-dependent.

Hence option (i) is correct.

Explanation:

• Momentum is conserved in all collisions.
• In elastic collision, kinetic energy is also conserved.
• In inelastic collision, kinetic energy is not conserved. In perfectly inelastic collision, objects stick together after collision.

Perfectly elastic collision:

If law of conservation of momentum and that of kinetic energy hold good during the collision.

Inelastic collision:

If law of conservation of momentum holds good during collision while that of kinetic energy is not.

Coefficient of restitution (e)

\(e = \frac{{Relative\;velocity\;after\;collision}}{{Relative\;velocity\;before\;collision}} = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}\)

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For perfectly inelastic collision, e = 0

Concept:

If s = f(t)

Then, First derivative with respect to time represents the velocity

\(v=\frac{ds}{dt}\)

Acceleration is given by

\(a=\left( \frac{{{d}^{2}}S}{d{{t}^{2}}} \right)\)

Where s is the displacement

Calculation:

Given:

s = 2t3 – t2 - 1 and t = 1 sec.

\(\frac{ds}{dt}=6{{t}^{2}}-2t\)

\(\frac{{{d}^{2}}s}{d{{t}^{2}}}=12t-2\)

Concept:

The friction force is given by:

f = μN

where μ is the coefficient of friction between the surfaces in contact, N is the normal force perpendicular to friction force.

Calculation:

Given:

μ = 0.1, m = 1 kg, F = 0.8 N

Now, we know that

From the FBD as shown below

Normal reaction, N = mg = 1 × 9.81 = 9.81 N

Limiting friction force between the block and the surface, f = μN =  0.1 × 9.81 = 0.98 N

But the applied force is 0.8 N which is less than the limiting friction force.

∴ The friction force for the given case is 0.8 N.

Concept:

The CG of a semicircular plate of  r radius, from its base, is

\(\bar y = {4r\over 3 \pi}\)

Calculation:

Given:

r = 33 cm

\(\bar y = {4r\over 3 \pi}={4\times 33\over3\times{22\over 7}}\)

y̅ = 14 cm

∴ the C.G. of a semicircular plate of 66 cm diameter, from its base, is 14 cm.

Additional Information

C.G. of the various plain lamina are shown below in the table. Here x̅  & y̅  represent the distance of C.G. from x and y-axis respectively.

Circle
Semicircle
Triangle
Cone
Rectangle
Quarter Circle
Solid hemisphere

Concept:

Number of vertical forces:

∑F_y = T + R_N - W = 0 

Static friction force is given as,

F_s = μR_N 

Calculation:

Given:

W = 981 N, μ = 0.2, F_s = 100 N

Normal reaction:

\(\Rightarrow {R_N} = \frac{{100}}{{0.2}} = 500~N\)

Tension T

⇒ T = 981 - 500 = 481 N

Concept:

Equation of motion:

• The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering the force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move in a straight line.

There are three equations of motion:

v = u + at

v2 = u2 + 2as

\(s =ut + \frac{1}{2}at^2\)

where, v = final velocity, u = initial velocity, s = distance travelled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

Calculation:

Given:

Part-I:

When the ball will reach the highest point then the final velocity will be zero.

Initial velocity = u m/sec, final velocity = 0 m/sec, acceleration = -g m/sec²

applying 1st equation of motion

v = u + at

0 = u - gt₁

\(t_1=\frac{u}{g}\)

Part-II:

Initial velocity will be zero as the ball is at the highest point.

applying 1st equation of motion

v = u + at

3u = 0 + gt₂

\(t_2=\frac{3u}{g}\)

Therefore total time is:

t = t₁ + t₂

\(t=\frac{u}{g}+\frac{3u}{g}=\frac{4u}{g}\)

Concept:

The resting on between any frictional floor and a vertical wall will always satisfy all the static equilibrium condition i.e.

∑ F_x = ∑ F_y = ∑ M_{at any point} = 0

Calculation:

Given:

Length of ladder (AB) = 5 m, OB = 3 m

Let W will be the weight of the ladder, N_B and N_A will be support reaction, θ is the angle between ladder and floor and μ is the friction coefficient between ladder and floor.

Free body diagram of the ladder;

OA² = AB² - OB² , OA2 = 52 - 32 

OA2 = 16, OA = 4 m

From Δ OAB,

\(\cos θ = \frac{3}{5}\)

Now apply ∑ Fy = 0

N_B = W 
Now take moment about point A, which should be equal to zero

∑ MA = 0

\(\;\left( {\mu {N_B} \times 4} \right) + \left( {W \times \frac{5}{2} \times \cos \theta } \right) = {N_B} \times 3\)

\(\;\left( {\mu {N_B} \times 4} \right) + \left( {{N_B} \times \frac{5}{2} \times \frac{3}{5}} \right) = {N_B} \times 3\)

\(\left( {\mu \times 4} \right) + \left( {\frac{3}{2}} \right) =~3\)

\(\mu = \frac{3}{8}\)

Hence the value of the coefficient of friction between ladder and floor will be 3/8

Explanation:

When the Force F is suddenly removed, then due to W, the rod is in rotating condition with angular acceleration \(\alpha\)

Thus the equation of motion:

\(\Sigma M=I_o\alpha;\;\;W \times \frac{L}{2} = I\alpha\)

\(As, I = \frac{mL^2}{3}= \frac{1}{3}\times\frac{W}{g}\times{L^2}\)

\(\therefore W \times \frac{L}{2} =\frac{WL^2}{3g}\times\alpha\)

\(\Rightarrow \alpha = \frac{{3g}}{{2L}}\)

\(\therefore Linear\;acceleration\;at\;centre = \alpha \times \frac{L}{2} = \frac{{3g}}{{2L}}\times \frac{L}{2}=\frac{{3g}}{4}\)

Also, the centre of the rod accelerates with linear acceleration a;

\(W-R=F\Rightarrow mg-R=ma\\R=mg-ma=mg-\frac{3}{4}mg=\frac{1}{4}mg=\frac{W}{4}\)   

Concept:

Work done = Area under force deflection diagram

\(Work ~done = \frac 12 \times Force\times deflection\)

Calculation:

Given:

Gradual loading

Force = 100 N, deflection = 100 mm = 0.1 m

\(Work ~done = \frac 12 \times Force\times deflection\)

\(Work ~done = \frac 12 \times 100\times 0.1=5~J\)

CONCEPT:

Moment of inertia:

• The moment of inertia of a rigid body about a fixed axis is defined as the sum of the product of the masses of the particles constituting the body and the square of their respective distances from the axis of the rotation.
• The moment of inertia of a particle  is

⇒ I = mr2

Where r = the perpendicular distance of the particle from the rotational axis.

• Moment of inertia of a body made up of a number of particles (discrete distribution)

⇒ I = m1r12 + m2r22 + m3r32 + m4r42 + -------

Rotational kinetic energy: 

• The energy, which a body has by virtue of its rotational motion, is called rotational kinetic energy.
• A body rotating about a fixed axis possesses kinetic energy because its constituent particles are in motion, even though the body as a whole remains in place.
• Mathematically rotational kinetic energy can be written as -

⇒ KE \( = \frac{1}{2}I{\omega ^2}\)

Where I = moment of inertia and ω = angular velocity.

EXPLANATION:

• The moment of inertia of the ring about an axis passing through the center and perpendicular to its plane is given by

⇒ I_ring = MR²

• Moment of inertia of the disc about an axis passing through center and perpendicular to its plane is given by -

\(⇒ {I_{disc}} = \frac{1}{2}M{R^2}\)

• As we know that mathematically rotational kinetic energy can be written as

\(⇒ KE = \frac{1}{2}I{\omega ^2}\)

• According to the question, the angular velocity of a thin disc and a thin ring is the same. Therefore, the kinetic energy depends on the moment of inertia.
• Therefore, a body having more moments of inertia will have more kinetic energy and vice - versa.
• So, from the equation, it is clear that,

⇒ Iring > Idisc

∴ Kring > Kdisc

• The ring has higher kinetic energy.