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

Concept:

Thermostat:

• A thermostat is a device that detects temperature changes for the purpose of maintaining the temperature of an enclosed area essentially constant.
• It is used in automatic heating appliances to control the temperature, i.e. systems including relays, valves, switches, etc.
• The thermostat generates signals, usually electrical when the temperature exceeds or falls below the desired value.
• A thermostat is a device that automatically regulates temperature, or activates a device when the temperature reaches a certain point.
• Thus, it can be used to maintain a constant temperature of a water bath or an oven.

Important Points

The Thermostat is a contact type electro-mechanical temperature sensor or switch, that basically consists of two different metals such as nickel, copper, tungsten or aluminum etc, that are bonded together to form a Bi-metallic strip.

Concept:

Series Voltage Regulator:

A series voltage regulator passes altering current through a series element to manage voltage drop.
The series transistor in these regulators is driven by the error amplifier which continuously adjusts its resistance in order to maintain a constant output voltage even if the input voltage or load current varies.
Shunt Voltage Regulator:

A shunt regulator works by providing a path from the supply voltage to the ground through a variable resistance.
The shunt regulator maintains the voltage across its output terminal at a fixed value by varying its resistance in response to input or output voltage changes.
The term shunt is used because the regulation element is placed parallel (electrically shunted) with the load.
Adjustable Voltage Regulator:

Adjustable voltage regulators have an extra input called an "adjust pin" through which the output voltage can be adjusted by placing a desired value of resistor between this pin and ground.
An adjustable voltage regulator is a kind of regulator that can be manually set to produce the desired DC output.
Switching Regulator:

Unlike linear regulators, a switching regulator adjusts the conversion ratio by modulating the energy input to the system, that is, it's continuously turns the input current on and off, and utilizes an inductor, capacitor, and diode to store and transfer energy to the output.
Switching regulators are noted for their high efficiency as compared to linear regulators.
However, they can generate more noise due to their switching operation and require more components, making them more complex.

Breadboard:

• A breadboard is a solderless device for temporary prototypes with electronics and test circuit designs.
• Most electronic components in electronic circuits can be interconnected by inserting their leads or terminals into the holes and then making connections through wires where appropriate.
• The breadboard has strips of metal underneath the board and connects the holes on the top of the board.
• The metal strips are laid out as shown below.
• Note that the top and bottom rows of holes are connected horizontally and split in the middle while the remaining holes are connected vertically.

In an electron tube, a vacuum diode is typically made to work in the "Space charge limited current region".

Concept:

A vacuum diode consists of two electrodes, an anode (also known as a plate) and a cathode, inside a vacuum bulb.

Here's a quick explanation of the options:

1) Space charge limited current region: This is the normal operating region for a vacuum diode. In this region, the current through the tube is controlled by the space charge. The "space charge" is the charge density within the tube caused by the emitted electrons from the cathode. The space charge field opposes the acceleration of more electrons by the anode voltage, limiting the current through the tube. As the anode voltage increases, the space charge is dispersed more and the current can increase until it's limited by other mechanisms.

2) Saturation region: In a vacuum tube, saturation is a state where an increase in applied voltage does not cause an increase in current flow because all available electrons have already been attracted to the plate (anode). The current flow has reached its maximum vacuum.

3) Temperature-limited region: This would be a scenario where the emission of electrons from the heated cathode sets a limit on the current. It is typically a lower-voltage condition where increasing the anode voltage does increase the current, but only up to a limit set by the number of electrons the cathode can emit.

4) Cut-off region: This is a region where there is no current because the anode voltage is too low to draw electrons through the space charge field.

So, typically, vacuum diodes are operated in the space charge-limited current region, where there's a balance between the voltage applied and the current flowing through the tube, controlled by the density of the space charge. Therefore, the most appropriate answer is the first one - "Space charge limited current region".

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.