Impact and Virtual Work MCQ Quiz - Objective Question with Answer for Impact and Virtual Work - Download Free PDF

Explanation:

PID Controllers in Modern Energy Conversion Systems

In modern energy conversion systems, PID (Proportional-Integral-Derivative) controllers are primarily used for stabilizing load frequency. These controllers are essential in managing and maintaining the stability and efficiency of power systems. They work by adjusting the input to the system to correct errors, thus ensuring that the system operates at the desired set point. Let's analyze the given options:

1. "Maximizing fuel combustion rate." (Not correct)

   • While PID controllers can influence fuel combustion processes, their primary role in energy conversion systems is not to maximize fuel combustion rate, but to maintain system stability and efficiency.

2. "Stabilizing load frequency." (Correct)

   • PID controllers are widely used in power systems to stabilize load frequency. They adjust the control inputs to the system to correct any deviation from the desired frequency, ensuring reliable and stable operation.

3. "Ensuring boiler safety." (Not correct)

   • While ensuring boiler safety is crucial, the primary use of PID controllers in energy conversion systems is more focused on maintaining operational stability and efficiency rather than directly ensuring safety, which is typically managed by other safety mechanisms and control systems.

4. "Improving cooling tower efficiency." (Not correct)

   • Although PID controllers can be used to improve various aspects of cooling tower operation, their primary role in energy conversion systems is to stabilize load frequency, which is more critical for overall system performance.

Explanation:

In a transformer under no-load condition, certain losses do occur while others do not. Let's analyze the different types of losses in this context.

Analyzing the Given Options

1. Eddy current loss (Option 1)

   • Eddy current loss is caused by the alternating magnetic field inducing currents in the core. This loss occurs even when the transformer is under no-load condition.

2. Hysteresis loss (Option 2)

   • Hysteresis loss is due to the magnetization and demagnetization of the core material in each cycle of the alternating current. This loss also occurs under no-load condition.

3. Copper loss (Option 3)

   • Copper loss (I²R loss) is caused by the resistance of the windings. Under no-load condition, the current in the windings is minimal, hence copper loss is negligible or does NOT occur significantly.

4. Dielectric loss (Option 4)

   • Dielectric loss occurs in the insulation material of the transformer windings due to the alternating electric field. This loss occurs regardless of the load condition.

Conclusion:

Among the options provided, Option 3 (Copper loss) does NOT occur significantly in a transformer under no-load condition, making it the correct answer.

Explanation:

Performance Ratio of Solar PV System

The performance ratio (PR) of a solar photovoltaic (PV) system is a measure of the efficiency of the system, representing the relationship between the actual and theoretical energy outputs. The PR is influenced by various factors, including:

• Orientation of panels: The angle and direction the solar panels face can affect the amount of solar radiation they receive.

• Efficiency of inverter: The inverter's efficiency in converting DC power to AC power impacts the overall system efficiency.

• Temperature coefficient of modules: The performance of solar modules decreases with an increase in temperature; this coefficient quantifies that relationship.

• Cable size: Proper cable sizing minimizes energy losses due to resistance and ensures efficient power transmission.

Analyzing the Given Options

1. "Orientation of panels" (Important, but not the highest impact)

   • While the orientation and tilt of solar panels significantly affect energy capture, they can be optimized and do not degrade over time.

   • Properly oriented panels maximize energy capture but the impact is less than the temperature coefficient.

2. "Efficiency of inverter" (Significant, but not the highest impact)

   • Inverter efficiency is crucial for converting DC to AC power efficiently, but modern inverters are highly efficient (95-98%).

   • The loss due to inverter inefficiency is relatively small compared to temperature effects on modules.

3. "Temperature coefficient of modules" (Highest impact)

   • The temperature coefficient directly affects the efficiency of solar modules. Higher temperatures reduce the voltage output and overall efficiency of the modules.

   • This effect is unavoidable and continuous, making it a crucial factor in the performance ratio.

   • The temperature coefficient is a significant determinant of energy yield, especially in hotter climates.

4. "Cable size" (Important, but not the highest impact)

   • Proper cable sizing is essential to minimize resistive losses and ensure efficient power transmission.

   • However, with correct sizing and installation, the impact on the performance ratio is less than that of temperature effects.

Explanation:

Matching Cycles with Their Applications

In thermodynamics and heat engines, different cycles are used for various applications. Here are the cycles and their corresponding applications:

• Brayton Cycle: This cycle is commonly used in jet engines.

• Rankine Cycle: This cycle is primarily used in steam power plants.

• Otto Cycle: This cycle is used in petrol engines.

• Diesel Cycle: This cycle is used in heavy trucks and ships.

Explanation:

Calculating Energy Released from Fuel

To determine the energy released from the combustion of 5 kg of fuel with a calorific value of 42000 kJ/kg and 85% combustion efficiency, we need to follow these steps:

• Step 1: Calculate the total energy content of the fuel.

• Step 2: Account for the combustion efficiency to find the actual energy released.

Analyzing the Given Options

1. Step 1: Calculate the total energy content of the fuel.

   • The calorific value of the fuel is 42000 kJ/kg. Hence, for 5 kg of fuel, the total energy content would be:

   • Total energy content = 42000 kJ/kg * 5 kg = 210,000 kJ

2. Step 2: Account for the combustion efficiency to find the actual energy released.

   • The combustion efficiency is 85%, meaning only 85% of the total energy content is actually released. Therefore, the actual energy released is:

   • Actual energy released = 210,000 kJ * 0.85 = 178,500 kJ

Therefore, the correct answer is option 3: 178,500 kJ

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 we neglect the air resistance then,

By using energy conservation,

(K + P)I, system = ( K + P)f , system       …(1)

Where, K, P are representing kinetic energy and potential energy. The system includes spring and mass

Taking the reference line as shown in the figure,

\( ⇒ {\bf{mgh}} + 0 = \frac{1}{2}{\bf{k}}{{\bf{x}}^2} + {\bf{mg}}\left( { - {\bf{x}}} \right)~\)

∵ mass after hitting comes below the reference line by maximum compression (x) so potential energy for the mass will be taken as with (-) sign. Also, at extreme compression of the spring, the Kinetic energy of the mass will be zero.

\( ⇒ {\rm{mg}}\left( {{\rm{h}} + {\rm{x}}} \right) = \frac{1}{2}{\rm{k}}{{\rm{x}}^2}~\)

where m = mass, h = height, k = spring constant, x = maximum deflection in the spring

Calculation:

Given:

Weight (W) = 500 N; h = 2 m; k = 0.5 kN/m = 500 N/m

Weight (W) = mg = 500 N

\( ⇒ 500 \times \left( {2\ +\ x} \right) = \frac{1}{2} \times 500 \times {x^2}\)

⇒ x = 3.24 m

Hence the deflection caused in the first fall will be 3.24 m.

Mistake PointsThe exact answer to this question is 3.24m, but 3.24 m is not available in the options, this is the official of ISRO LPSC Technical Assistant and they have given the correct answer 2 m.

So, we have to neglect the change in potential energy due to the deflection of the spring.

Neglecting the potential energy change due to the deflection of the spring, we will get the expression:

\( ⇒ mgh= \frac{1}{2}{\rm{k}}{{\rm{x}}^2}~\)

\( ⇒ 500 \times \left( {2} \right) = \frac{1}{2} \times 500 \times {x^2}\)

⇒ x = 2 m

Hence the approximated deflection caused in the first fall will be 2 m.

Explanation:

Conservation of linear momentum:

If the net external force acting on a system of bodies is zero, then the momentum of the system remains constant.

Linear momentum is mathematically expressed as:

\(p=m\;\times\;v\)

p = linear momentum, m = mass of the body and v = velocity of the body.

The law of conservation of momentum can be explained from Newton's Second Law of Motion which states- " the rate of change of linear momentum of a body is equal to the net external force applied to it i.e. 

\(\frac{d{p}}{dt}=\frac{d(m{v})}{dt}\Rightarrow m\frac{d{v}}{dt}=ma\Rightarrow F_{net}\)

If F_net = 0, ∴  dp = 0 i.e p = constant.

Concept:

Coefficient of restitution (e)

\(e = \frac{{{\rm{Velocity~of~separation}}}}{{{\rm{Velocity~of~approach}}}}= \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}{{\;\;\;\;\;}} \ldots \left( 1 \right)\) 

where, v = final velocity of respective masses and u = initial velocity of respective masses

Calculation:

Given:

v₁ = v_{2 = v}, u₁ = 5 m/s, u₂ = 0

Using equation (1),

\(\Rightarrow e= \frac{{v - v}}{{5 - 0}}=0\) 

Concept:

Angular momentum is a vector quantity.

The magnitude of the angular momentum of an orbiting object is equal to its linear momentum (product of its mass ‘m' and linear velocity ‘v’) times the perpendicular distance ‘r’  from the center of rotation to a line drawn in the direction of its instantaneous motion and passing through the object’s center of gravity.

Angular momentum = m × v × r

Angular momentum before Impact (L_i) = Angular momentum after Impact (L_f)

Calculation:

Given:

m = 1 kg, u = 10 m/s, M = 20 kg, r = 1 m

L_i = m × u × r = 1 × 10 × 1 = 10 

Since the wheel will start rolling motion after the impact so this will have the pure rotation about the point of contact as the point of contact will be in the state of rest (Center of instantaneous rotation or CIR). Let the final angular velocity of the system (clay and wheel) is ω. 

L_f = I_CIR × ω 

Since the mass of the clay is very less in comparison to the wheel so this can be neglected while taking the moment of inertia about CIR

 \(I_{CIR} = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2} \)

Now, (Li) = Angular momentum after Impact (Lf)

\(\Rightarrow 10 = \frac{3MR^2}{2} \omega \)

\(\Rightarrow 10 = \frac{3 \times 20 \times 1^2}{2} \omega = 30 \omega \)

\(ω = \frac 13~ rad/sec\)

Concept:

Kinetic energy of a mass m moving with velocity v is equal to,

Kinetic energy, \(K.E.\; = \frac{1}{2}m{v^2}\)

Momentum of a mass m moving with velocity v is equal to,

Momentum = mv

Calculation:

Given:

m1 > m2 and,

Two bodies have the same kinetic energy

\(\frac{1}{2}{m_1}v_1^2 = \frac{1}{2}{m_2}v_2^2\)

We can rewrite it as,

\(\frac{1}{2} × \frac{{{{\left( {{m_1} × {v_1}} \right)}^2}}}{{{m_1}}}=\frac{1}{2} × \frac{{{{\left( {{m_2} × {v_2}} \right)}^2}}}{{{m_2}}}\)

\(\frac{{p_1^2}}{{{m_1}}} = \frac{{p_2^2}}{{{m_2}}}\)

\(\frac{{{p_1}}}{{{p_2}}} = \sqrt {\frac{{{m_1}}}{{{m_2}}}}\) ----------(3)

Since m1 > m2 RHS becomes greater than 1, which implies

\(\frac{{{p_1}}}{{{p_2}}}\) > 1

p1 > p2

Hence, for the given condition \(\frac{{{p_1}}}{{{p_2}}}\) > 1 will satisfy the relation between momentums.

Concept:

Principle of virtual work: (unit-load Method)

Developed by Bernoulli:

To find Δ at point A due to loads P₁, P₂, P₃. Remove all loads, apply virtual load P’ on point A.

For simplicity P’ = 1

It creates internal load u on representative element.

Now remove this load, apply P₁, P₂, P₃ due to which pt. A will be displaced by Δ

∴ External virtual work = 1.Δ

Internal virtual work = u.dL

P^' = 1 = external virtual unit load in direction of Δ.

u = internal virtual load acting on element in direction of a dL.

Δ = external displacement caused by real loads.

dL = internal deformation caused by real loads.

Explanation:

• Virtual work is the work done by a real force acting through a virtual displacement i.e. Virtual work done by real forces.
• A virtual displacement is any displacement consistent with the constraints of the structure, i.e., that satisfy the boundary conditions at the supports.
• A virtual force is any system of forces in equilibrium.
• The principle of virtual work states, for bodies in equilibrium, for a small arbitrary displacement, the total work done by the system is zero.
• In this method, the system is displaced through a small amount about a reference point and the work done by all the forces about the reference point is summed to zero to find the unknown reactions if any.
• The principle of virtual work states that, for a body to be in equilibrium, the virtual work should be zero.

Concept:

From momentum conservation we have,

initial momentum = final momentum i.e., p_i = p_f

0 = m_gv_g + m_sv_s       [∵ initial velocity is zero]

m_gv_g = - m_sv_s

where,

m_g, v_g is the mass of the gun and velocity of gun respectively.

m_s, v_s is the mass of the shell and velocity of shell respectively

Calculation:

Given:

m_g  = 3000 kg, m_s = 50 kg, v_s = 300 m/s

\({v_g} = - \frac{{{m_s}{v_s}}}{{{m_g}}}\)

\({v_g} = - \frac{{50 \times 300}}{{3000}} = - 5{\rm{ m/s}}\)

Concept:

For a collision between two objects, the coefficient of restitution is the ratio of the relative speed after to the relative speed before the collision. The coefficient of restitution is a number between 0 (perfectly inelastic collision) and 1 (elastic collision) inclusive.

Coefficient of restitution (e) = \(\frac{Relative\;velocity\;after\;collosion}{Relative\;velocity\;before\;collosion}=\frac{v_B-v_A}{u_A-u_B}\).....(1)

moment conservation: 

m_Au_A + m_Bu_B = m_Av_A + m_Bv_B............(2)

Calculation:

Given:

m_A = 1 kg, uA = 2 m/s, mB = 2 kg, uB = 0 m/s, e = 0.5

Now, \(0.5=\frac{v_B-v_A}{2-0}\)

⇒ v_B - v_A = 1............(3)

From moment conservation, 

1 × 2 + 2 × 0 = 1 × v_A + 2 × v_B 

v_A + 2v_B = 2...............(4)

From equation 3 and 4, v_A = 0 m/s and v_B = 1 m/s