The question asks about the mechanism of electron emission in a vacuum diode. A vacuum diode is a fundamental electronic component that relies on the emission of electrons from a cathode to conduct current. Understanding the primary method by which electrons are released from the cathode is key to answering this question.
C) heating — Heating the cathode in a vacuum diode provides the necessary thermal energy for electrons to overcome the work function of the metal and be emitted into the vacuum. This phenomenon is called thermionic emission, which is the primary principle of operation for vacuum diodes.
The question asks to identify which of the given options cannot actually move. This relates to the fundamental understanding of charge carriers in conductors and semiconductors.
B) ions — In solid-state materials, ions are typically fixed within the crystal lattice and do not move to conduct electricity. While they vibrate, their displacement is not significant enough to be considered "movement" in the context of charge transport like electrons or holes.
To find the mean (average) value of a half-wave rectified sine wave, we need to integrate the waveform over one complete cycle and then divide by the period of the cycle. A half-wave rectified sine wave only conducts during the positive half-cycle, and its value is zero during the negative half-cycle.
Define the waveform: A sine wave is typically represented as \(v(t) = V_m \sin(\omega t)\) or \(i(t) = I_m \sin(\omega t)\), where \(V_m\) or \(I_m\) is the peak amplitude. For a half-wave rectified sine wave, the current (or voltage) is given by:
\[ i(t) = \begin{cases} I_m \sin(\omega t) & \text{for } 0 \le \omega t \le \pi \ 0 & \text{for } \pi \le \omega t \le 2\pi \end{cases} \]Formula for Mean Value: The mean or average value of a periodic waveform over one complete cycle (period \(T = 2\pi/\omega\)) is given by:
\[ I_{avg} = \frac{1}{T} \int_0^T i(t) dt \]Substitute and Integrate: For a half-wave rectified sine wave, the integration is performed over the period \(2\pi\), but the function is non-zero only from \(0\) to \(\pi\):
\[ I_{avg} = \frac{1}{2\pi} \int_0^{2\pi} i(\omega t) d(\omega t) \]\[ I_{avg} = \frac{1}{2\pi} \left[ \int_0^{\pi} I_m \sin(\omega t) d(\omega t) + \int_{\pi}^{2\pi} 0 \cdot d(\omega t) \right] \]\[ I_{avg} = \frac{1}{2\pi} \left[ I_m \int_0^{\pi} \sin(\omega t) d(\omega t) + 0 \right] \]\[ I_{avg} = \frac{I_m}{2\pi} [-\cos(\omega t)]_0^{\pi} \]\[ I_{avg} = \frac{I_m}{2\pi} [-\cos(\pi) - (-\cos(0))] \]\[ I_{avg} = \frac{I_m}{2\pi} [-(-1) - (-1)] \]\[ I_{avg} = \frac{I_m}{2\pi} [1 + 1] \]\[ I_{avg} = \frac{I_m}{2\pi} [2] \]\[ I_{avg} = \frac{I_m}{\pi} \]Calculate the numerical value:
\[ I_{avg} = \frac{I_m}{\pi} \approx \frac{I_m}{3.14159} \]\[ I_{avg} \approx 0.318 I_m \]D) 0.318 I
The mean value of a half-wave rectified sine wave is \(I_m/\pi\), which numerically is approximately \(0.318 I_m\).
A) 0.707 I — This value represents the RMS (Root Mean Square) value of a full sine wave (\(I_{rms} = I_m/\sqrt{2}\)) or the RMS value of a half-wave rectified sine wave (\(I_{rms} = I_m/2\)) if \(I\) refers to \(I_m\). If \(I\) refers to RMS value, then \(0.707 I_{rms}\) is not a standard representation for mean value.
B) 0.637 I — This value represents the mean (average) value of a full-wave rectified sine wave (\(2I_m/\pi \approx 0.637 I_m\)).
C) 0.5 I — This value represents the RMS value of a half-wave rectified sine wave (\(I_{rms} = I_m/2 = 0.5 I_m\)). It is not the mean value.
The question asks about the number of junctions in a Silicon Controlled Rectifier (SCR). To answer this, we need to recall the basic structure and composition of an SCR.
C) A silicon controlled rectifier (SCR) is a four-layer, three-junction, three-terminal semiconductor device. The three junctions are formed by the P-N-P-N layers.
The question asks to identify quantities that have the same unit as elastic modulus. To solve this, we need to recall the definitions and units of elastic modulus, stress, strain, shear modulus, pressure, and force.
The unit of stress is Pascal (Pa) or Newtons per square meter (\(N/m^2\)). Strain is a dimensionless quantity (ratio of change in dimension to original dimension). Therefore, the unit of elastic modulus is the same as the unit of stress, which is Pascal (Pa) or \(N/m^2\).
The unit of force is Newton (N) and the unit of area is square meter (\(m^2\)). So, the unit of stress is \(N/m^2\) or Pascal (Pa).
The unit of pressure is Pascal (Pa) or \(N/m^2\).
From the comparison, elastic modulus, stress, shear modulus, and pressure all have the same unit (\(N/m^2\) or Pa).
A) Stress, shear modulus and pressure
As determined above, elastic modulus, stress, shear modulus, and pressure all share the unit of Pascal (Pa) or Newtons per square meter (\(N/m^2\)).
The question asks for the approximate relative density of a compacted dense sand. Relative density is a measure used to quantify the denseness of cohesionless soils (like sand) and is expressed as a percentage or a decimal. It compares the natural void ratio of the soil to its maximum and minimum possible void ratios.
Definition of Relative Density: Relative density (\(I_D\)) is defined as:
\[ I_D = \frac{e_{max} - e}{e_{max} - e_{min}} \times 100\% \]where:
Interpretation of Relative Density Values:
Compacted Dense Sand: The term "compacted dense sand" implies that the sand has been densified significantly, meaning its void ratio is close to the minimum possible void ratio (\(e_{min}\)).
Expected Range for Dense Sand: For a very dense or compacted sand, the relative density would be high, typically in the range of 75% to 100% (or 0.75 to 1.00). A value close to 1 (or 100%) indicates a highly dense state.
Evaluating the Options:
Therefore, for a compacted dense sand, a relative density of 0.95 is the most appropriate value among the given options.
B) 0.95 — This value represents a very high relative density (95%), which is characteristic of a compacted dense sand. Such a soil would have a void ratio very close to its minimum possible void ratio.
A) 0.4 — This value (40%) indicates a medium dense sand, not a compacted dense sand. A compacted dense sand would have a much higher relative density.
C) 0.50 — This value (50%) also indicates a medium dense sand. While it's denser than 0.4, it's still not representative of a compacted dense state, which implies a very high degree of densification.
D) 1.10 — Relative density cannot exceed 1.00 (or 100%). A value of 1.10 is theoretically impossible as it would imply a void ratio less than the minimum possible void ratio, which is not physically achievable.
The question asks about the primary mode of failure in a fillet weld. To answer this, we need to understand how stresses are distributed within a fillet weld when it is subjected to external loads.
Correct Option: B) Shear is the predominant mode of failure in a fillet weld because the critical stress developed at the throat of the weld is primarily shear stress.
To determine how the kinematic viscosity of gases changes with temperature, we need to understand the definitions of dynamic viscosity and kinematic viscosity, and how each of these properties, along with density, is affected by temperature for gases.
Define Kinematic Viscosity: Kinematic viscosity (\(\nu\)) is defined as the ratio of dynamic viscosity (\(\mu\)) to the density (\(\rho\)) of the fluid.
\[ \nu = \frac{\mu}{\rho} \]
Effect of Temperature on Dynamic Viscosity of Gases: For gases, dynamic viscosity increases with an increase in temperature. This is because, at higher temperatures, gas molecules move faster and collide more frequently and with greater momentum, leading to increased resistance to flow.
The relationship is often approximated by Sutherland's formula or a power law, where \(\mu \propto T^n\) (with \(n > 0\), typically around 0.5 to 1).
Effect of Temperature on Density of Gases: For an ideal gas at constant pressure, density is inversely proportional to temperature (from the ideal gas law, \(PV = nRT\), which implies \(\rho = \frac{PM}{RT}\), where \(M\) is molar mass). Thus, as temperature increases, the density of a gas decreases.
\[ \rho \propto \frac{1}{T} \]
Combine the Effects for Kinematic Viscosity: Now, let's look at the formula for kinematic viscosity:
\[ \nu = \frac{\mu}{\rho} \]
As temperature increases:
Since the numerator (\(\mu\)) increases and the denominator (\(\rho\)) decreases, their ratio (\(\nu\)) must increase significantly.
Correct Option: B) Increases
As temperature increases, the dynamic viscosity of gases increases due to more frequent and energetic molecular collisions. Simultaneously, the density of gases decreases with increasing temperature. Since kinematic viscosity is the ratio of dynamic viscosity to density, both effects contribute to an overall increase in kinematic viscosity.
The question asks about the method used to establish a benchmark. A benchmark is a permanent reference point of known elevation above a datum, used as a starting point for surveying operations. We need to identify which surveying technique is primarily used for precise elevation determination to establish such points.
C) Spirit levelling — Spirit levelling is the most accurate and commonly used method for establishing benchmarks. It involves using a spirit level (dumpy level, auto level, digital level) and a levelling staff to determine the precise difference in elevation between points. This method provides the high accuracy required for establishing permanent reference points like benchmarks.
This question tests understanding of critical path method (CPM) in project management. The critical path is the longest sequence of activities in a project plan, which determines the shortest possible duration to complete the project. Understanding the concept of 'float' is key to identifying the characteristics of a critical path.
A) Zero float — Activities on the critical path have zero total float. This means that any delay in these activities will directly cause a delay in the project completion date. This is a fundamental characteristic of the critical path in project management.