Manufacturing Processes Ii Practice Questions

Concept: P × laser absorbtivity = (mL + mCΔT) m . = ρ × Q = ρ × A × V max where, P = Power of Laser, m . = metal melting rate, C = Specific heat, L = latent heat V max = maximum speed of cutting, ΔT = difference of melting temperature and room temperature Calculation: Given: t (thickness) = 3 mm = 3 × 10 -3 m, d = 0.2 mm = 0.2 × 10 -3 m C = 500 J/kg°C, L = 3 × 10 5 J/kg, P = 2 kW, ΔT = ( 1530 - 30 )°C = 1500°C V max = ? m . = ρQ = ρAV = 8000 × V max × 0.2 × 10 -3 × 3 × 10 -3 m . = 0.0048 × V max kg/s P × Laser absorbtivity = m(L + CΔT) 2 × 10 3 × 0.5 = 0.0048 × V max ×(3 × 10 5 + 500 × 1500) V max = 0.198 m/s

Concept: The linear power source characteristics of a DC welding process is given as follow: $\frac{V}{V_o}~+~\frac{I}{I_s}~=~1$ $V~=~V_o~-~\frac{V_o}{I_s}I$ where V o is the open-circuit voltage and I s is the short circuit current . The melting efficiency of a welding process is given by: &\eta;_m~=~\frac{e}{\left(\frac{VI}{Av}\right)&\eta;_h} where e is the heat required to melt unit volume , A is the cross-section of weld bead , v is the welding speed and η h is heat transfer efficiency . Calculation: Given: Open circuit voltage V o = 60 V, short circuit current I s = 600 A and heat/energy required to melt unit volume e = 10 J/mm 3 . The heat transfer efficiency η h = 0.80, melting efficiency η m = 0.25 and the weld cross-sectional area = 20 mm 2 . ∴ $V~=~V_o~-~\frac{V_o}{I_s}I$ ⇒ $60~-~\frac{60}{600}I$ ∴ V = 60 - 0.1I ∴ Arc power P = V × I = 60I - 0.1I 2 For maximum arc power $\frac{dP}{dI}~=~0$ ⇒ 60 - 0.2I = 0 ⇒ I = 300 A and V = 30 V. Now, the welding speed can be obtained as, &\eta;_m~=~\frac{e}{\left(\frac{VI}{Av}\right)&\eta;_h} $v~=~\frac{\eta_m\eta_hVI}{eA}$ = $\frac{0.25~\times ~0.8~\times ~300~\times ~30}{10~\times ~20}$ = 9 mm/s ∴ the required welding speed is 9 mm/s.

Shaft-Hole Assembly: Calculating Maximum Clearance This solution explains how to calculate the maximum clearance in a shaft-hole assembly, focusing on the given dimensions and tolerances. We need to determine the largest possible gap between the hole and the shaft based on the provided specifications. Analyzing Given Parameters for Shaft-Hole Fit First, let's list the key information provided in the question: Hole Specification: The hole size is specified as

mm. This means the basic size of the hole is 30 mm, with an upper deviation of +0.040 mm and a lower deviation of 0.000 mm. Shaft Fit Condition: The mating shaft has a clearance fit , and the minimum clearance is given as $0.01$ mm. Shaft Tolerance: The tolerance specified for the shaft is $0.03$ mm. Calculating Hole Limits From the hole specification, we can determine the upper and lower limits of the hole: Hole Maximum Limit (Hole max): This is the basic size plus the upper deviation. Hole max = Basic Size + Upper Deviation Hole max = $30 \text{ mm} + 0.040 \text{ mm} = 30.040 \text{ mm}$ Hole Minimum Limit (Hole min): This is the basic size plus the lower deviation. Hole min = Basic Size + Lower Deviation Hole min = $30 \text{ mm} + 0.000 \text{ mm} = 30.000 \text{ mm}$ The total tolerance on the hole is therefore $30.040 \text{ mm} - 30.000 \text{ mm} = 0.040 \text{ mm}$. Calculating Shaft Limits We know the shaft's tolerance is $0.03$ mm and the minimum clearance is $0.01$ mm. The minimum clearance occurs under the condition where the hole is at its minimum size and the shaft is at its maximum size. Calculating Shaft Maximum Limit (Shaft max): Using the minimum clearance condition: Minimum Clearance = Hole min - Shaft max $0.01 \text{ mm} = 30.000 \text{ mm} - \text{Shaft max}$ Rearranging the formula to find Shaft max: Shaft max = $30.000 \text{ mm} - 0.01 \text{ mm} = 29.990 \text{ mm}$ Calculating Shaft Minimum Limit (Shaft min): Now, we use the given shaft tolerance: Shaft Tolerance = Shaft max - Shaft min $0.03 \text{ mm} = 29.990 \text{ mm} - \text{Shaft min}$ Rearranging the formula to find Shaft min: Shaft min = $29.990 \text{ mm} - 0.03 \text{ mm} = 29.960 \text{ mm}$ Determining Maximum Clearance The maximum clearance occurs when the hole is at its maximum limit and the shaft is at its minimum limit. This represents the largest possible gap between the two components. Calculating Maximum Clearance: Maximum Clearance = Hole max - Shaft min Maximum Clearance = $30.040 \text{ mm} - 29.960 \text{ mm}$ Maximum Clearance = $0.080 \text{ mm}$ Conclusion on Maximum Clearance Value Based on the calculations, the maximum clearance between the hole and the shaft is $0.080$ mm. This value represents the largest possible gap given the specified tolerances and fit conditions.

Understanding the NC Tool Command for Circular Motion This question requires identifying the correct NC (Numerical Control) tool command to execute a specific toolpath. The tool needs to move along a circular arc, starting from one point, ending at another, with a defined center for the arc. Problem Description: NC Circular Arc Path The task involves moving an NC machine tool along a precise path defined as follows: Starting Point: $(20, 20)$ Ending Point: $(10, 10)$ Center of the Arc: $(20, 10)$ Operation: The movement must follow a circular arc. Key NC Commands for Motion Control NC programming uses specific G-codes to control the type of motion: G01 : Used for linear interpolation, meaning the tool moves in a straight line between two points. G02 : Used for clockwise circular interpolation. The tool moves along the arc of a circle in a clockwise direction. G03 : Used for counter-clockwise circular interpolation. The tool moves along the arc of a circle in a counter-clockwise direction. Analyzing the Required Circular Path To determine the correct command, let's analyze the geometry of the specified path: The starting point is $(20, 20)$. The ending point is $(10, 10)$. The center of the arc is $(20, 10)$. Calculating the Radius The radius ($R$) is the distance from the center to any point on the arc. Let's calculate it using the start point: Radius $R = \sqrt{(\text{Start X} - \text{Center X})^2 + (\text{Start Y} - \text{Center Y})^2}$ Substituting the values: $R = \sqrt{(20 - 20)^2 + (20 - 10)^2}$ $R = \sqrt{0^2 + 10^2}$ $R = \sqrt{100}$ $R = 10$ We can verify this with the end point: $R = \sqrt{(10 - 20)^2 + (10 - 10)^2}$ $R = \sqrt{(-10)^2 + 0^2}$ $R = \sqrt{100}$ $R = 10$ The radius is confirmed to be 10 . Determining the Direction of Movement Visualize the positions relative to the center $(20, 10)$: The start point $(20, 20)$ is directly above the center. The end point $(10, 10)$ is directly to the left of the center. To move from the point directly above the center to the point directly to the left of the center, the tool must travel in a counter-clockwise direction. Evaluating the NC Command Options NC commands for circular interpolation (G02/G03) typically specify the destination coordinates (X, Y) and the radius (R) or the center coordinates relative to the start point (I, J). Some specific controller syntaxes might also explicitly include the starting coordinates. Let's analyze the provided options based on the required counter-clockwise motion and the calculated radius of 10: Option 1: N020 G03 X20 Y20 X10 Y10 R10 N020 : Sequence number. G03 : Specifies counter-clockwise circular interpolation. (Correct direction) X20 Y20 : Appears to specify the starting coordinates. X10 Y10 : Specifies the ending coordinates $(10, 10)$. (Correct destination) R10 : Specifies the radius. (Correct radius) Analysis: This command correctly uses G03 for counter-clockwise motion and specifies the correct start, end, and radius. Option 2: N020 G02 X20 Y20 X10 Y10 R10 G02 : Specifies clockwise circular interpolation. (Incorrect direction) Analysis: The direction is wrong for the required path. Option 3: N020 G02 X10 Y10 X20 Y20 R10 G02 : Specifies clockwise circular interpolation. (Incorrect direction) Analysis: The direction is wrong, and the coordinate specification seems potentially reversed or intended for a different path segment. Option 4: N020 G01 X20 Y20 X10 Y10 R10 G01 : Specifies linear interpolation (straight line). (Incorrect motion type) Analysis: This command would move the tool in a straight line, not a circular arc. Conclusion: Correct NC Command Based on the analysis, the tool must move counter-clockwise along a circular arc with a radius of 10 from coordinates (20, 20) to (10, 10). The NC command that accurately represents this operation is N020 G03 X20 Y20 X10 Y10 R10 .

Concept: Calculation: Given: For 0.005 mm movement = 1 pulse is required For 1 mm movement = $1\over 0.005$ pulse required for 10 mm/s movement = $10\over 0.005$ pulse required = 2000 pulses/s = 2000 Hz

Concept: Difference of Thickness is given by: Δh = μ 2 R where, Δh = $h_o - h_f$ $h_o$ = Initial Thickness, $h_f$ = Final Thickness μ = Coefficient of friction D = Diameter of Roll Calculation : Given : $h_o$ = 3.5mm, μ = 0.08, D = 500 mm ∴ R = 250mm Now, Difference of Thickness is given by: Δh = μ 2 R $h_o - h_f$ = μ 2 R ⇒ (3.5 - h f ) = 0.08 2 × 250 ⇒ h f = 1.90 mm

Concept: Solidification time: The time taken for the liquid molten metal into the solid is known as solidification time Chvorinov's equation for solidification time is given by: ${\bf{t}} = {\bf{B}} \times \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}$ where, t = solidification time , B = mould constant , V = volume of the casting , A = surface area of the casting Calculation: Given: ${\bf{t}} = {\bf{B}} \times \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}$ ⇒ ${\bf{t}}\; \propto \;{\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2}$ {\left[ {\frac{{\bf{V}}}{{\bf{A}}}} \right]^2} = \;\frac{{\frac{{\bf{\pi }}}{4} \times {{\bf{D}}^2} \times {\bf{h}}}}{{\left[ {\left( {2 \times \frac{{\bf{\pi }}}{4} \times {{\bf{D}}^2}} \right) + \left( {{\bf{\pi }} \times {\bf{D}} \times {\bf{h}}} \right)} \right]\}}$Initial casting: Diameter ($D_i$) = 1 m , thickness ($h_i$) = 50 mm = 0.05 m${\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]{\rm{i}}}^2 = {\rm{;}}\frac{{\frac{{\rm{\pi }}}{4} × {1^2} × 0.05}}{{\left[ {\left( {2 × \frac{{\rm{\pi }}}{4} × {1^2}} \right) + \left( {{\rm{\pi }} × 1 × 0.05} \right)} \right]$}}$${\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]{\rm{i}}}^2 = 0.0005165$Modified casting: Diameter ($D_m$) = 0.5 m , thickness ($h_m$) = 25 mm = 0.0025 m${\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]{\rm{m}}}^2 = {\rm{;}}\frac{{\frac{{\rm{\pi }}}{4} \times {{0.5}^2} \times 0.025}}{{\left[ {\left( {2 \times \frac{{\rm{\pi }}}{4} \times {{0.5}^2}} \right) + \left( {{\rm{\pi }} \times 0.5 \times 0.025} \right)} \right]$}}$${\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]{\rm{m}}}^2 = 0.0001291$$\frac{{{{\rm{t}}{\rm{i}}} - {{\rm{t}}{\rm{m}}}}}{{{{\rm{t}}{\rm{i}}}}} × 100 = {\rm{;}}\frac{{{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}{\rm{i}}}^2 - {\rm{;}}{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}{\rm{m}}}^2}}{{{{\left[ {\frac{{\rm{V}}}{{\rm{A}}}} \right]}{\rm{i}}}^2}} = {\rm{;}}\frac{{0.0005165{\rm{;}} - {\rm{;}}0.0001291}}{{0.0001565}} × 100 = 75{\rm{% }}$ ∴ the % reduction in solidification time is 75%

Let's break down these defects to see which one doesn't fit with welding.

• Angular Distortion: This is a very common welding defect. It happens when the intense, localized heat from welding causes parts of the metal to expand and contract unevenly, leading to the workpiece bending or warping out of its original shape.
• Hot Tear: Also known as hot cracking, this is a type of crack that forms during the welding process while the metal is still cooling down from its molten state. It occurs because the metal isn't ductile enough at high temperatures to handle the stresses from contraction.
• Hydrogen Embrittlement: This defect primarily affects high-strength steels. During welding, if hydrogen (often from moisture) gets absorbed into the metal, it makes the material brittle and prone to cracking under stress. It's a significant welding-related issue.
• Earring: This defect is NOT associated with welding. Earring occurs in sheet metal forming processes, like deep drawing, where uneven "ears" or wavy edges form on the part due to the material's directional properties (anisotropy) during stretching.

Therefore, Earring is the defect that is not typically linked to welding processes.

This question tests your knowledge of common manufacturing processes and their suitable applications for various engineering components. We need to match each component to its most appropriate manufacturing method:

• P: Aluminum Alloy Piston for IC Engine
  Pistons operate under high temperatures and pressures and often have complex internal geometries. Sand casting (3) is excellent for creating intricate shapes from aluminum alloys, making it a standard process for pistons.

• Q: Low Carbon Steel Oil Pan
  An oil pan is typically a hollow, tray-like component made from sheet metal. Deep drawing (4) is a metal forming process specifically designed to produce hollow, cup-shaped parts from flat sheet metal blanks, which is ideal for oil pans.

• R: Tungsten Carbide Cutting Tool Insert
  Tungsten carbide is an extremely hard, brittle material suitable for cutting tools. Powder metallurgy (2) is the primary method for manufacturing such materials; it involves compacting powdered materials (like tungsten carbide and a binder) and then sintering them at high temperatures to form a dense, solid part.

• S: Plastic Bottle
  Plastic bottles are hollow containers. Blow molding (1) is a widely used process for manufacturing hollow plastic parts; it involves inflating a molten plastic preform (parison) inside a mold to take its shape.

Based on these explanations, the correct matches are P-3, Q-4, R-2, S-1.

This question asks us to match various measuring features with the most appropriate instrument. Let's break down each pairing:

P. Flatness error of a surface plate: Surface plates are fundamental for precision measurements, and their flatness is crucial. To measure minute deviations from a perfect plane with extremely high accuracy (often in micrometers), an Optical interferometer (4) is used. It works by analyzing the interference patterns of light, making it ideal for such precise flatness measurements.

Q. Profile of a cam: A cam's precise shape dictates its function. To measure its profile, we need to map its contour by correlating angular position with displacement. A Dividing head and dial gauge (3) setup is perfect for this. The dividing head precisely rotates the cam, while the dial gauge measures the radial displacement at each angular position.

R. Alignment error of a machine tool slide way: The accuracy of machine tools heavily relies on the proper alignment of components like slide ways. Detecting straightness and angular alignment errors is critical. An Autocollimator (1) is an optical instrument specifically designed to check straightness and alignment over distances by projecting and measuring the angular deviation of a reflected light beam.

S. Pitch and angle errors of screw thread: Screw threads have critical geometric parameters like pitch (axial distance between threads) and thread angle. Measuring these features requires magnification and precise linear measurement. A Tool maker's microscope (2) is well-suited for this task, as it provides magnified views of the thread form and has a calibrated stage for accurate measurement of pitch and angles.

Therefore, the correct matching is P-4, Q-3, R-1, S-2. This corresponds to Option D.

Concept: Power required for the wire drawing process can be calculated using the expression as follow Drawing power (P) = Drawing Force × Drawing Velocity (V) Calculation: Given: Initial diameter d i = 5 mm, final diameter d f = 4 mm, drawing velocity V = 5 m/s. The drawing stress is given as \sigma _d = \bar&\sigma; \ln\left[ {\frac{1}{{1~ - ~r}}} \right] Now the ratio 'r' will be $r = \frac{A_i~-~A_f}{A_i} = 1- \frac {A_f}{A_i}$ $\Rightarrow r = 1-\left(\frac{4^2}{5^2}\right) = 0.36$ Now the drawing stress, \sigma _d = \bar&\sigma; \ln\left[ {\frac{1}{{1 ~-~ r}}} \right] $\Rightarrow 600 \ln\left[ {\frac{1}{{1 ~- ~0.36}}} \right] = 267.77 \; MPa$ ∴ Drawing load = F d = σ d × A f ⇒ 267.77 × $\frac{\pi}{4}$ × 4 2 = 3364.897 N. ∴ Drawing Power = 3364.897 × 5 = 16824.485 N = 16.824 kW. ∴ The power required in the wire drawing process is 16.82 kW.

Concept: Shear angle: The angle made by the shear plane with respect to the cutting velocity vector is known as the shear angle The expression of the shear angle is given by: ϕ = \;{\tan ^{ - 1}}\left[ {\frac{{{\bf{r}}\cos {\bf{&\alpha; }}}}{{1 - {\bf{r}} \sin {\bf{&\alpha; }}}}} \right] where, ϕ = shear angle, α = Orthogonal rake angle, r = chip thickness ratio = $\frac{t_1}{t_2}$ , $t_1$ = uncut chip thickness, $t_2$ = deformed chip thickness. Calculation: Given: Rake angle (α ) = 15°, depth of cut (d = t 1 ) = 0.9 mm, chip thickness (t 2 ) = 1.8 mm. Chip thickness ratio (r) = $\frac{t_1}{t_2}$ = $\frac{0.9}{1.8}$ = 0.5. The required shear angle is: ϕ = \;{\tan ^{ - 1}}\left[ {\frac{{{\bf{r}}\cos {\bf{&\alpha; }}}}{{1 - {\bf{r}} \sin {\bf{&\alpha; }}}}} \right] ϕ = {\rm{\;}}{\tan ^{ - 1}}\left[ {\frac{{0.5{\rm{\;}} \times {\rm{\;}}\cos 15}}{{1 - {\rm{\;}}0.5{\rm{\;}} \times {\rm{\;}}\sin 15}}} \right] ϕ = {\rm{\;}}{\tan ^{ - 1}}0.5547 ∴ ϕ = 29.01°.

This problem uses Chvorinov's rule, which states that the solidification time ($t_s$) is proportional to the square of the ratio of the volume ($V$) to the surface area ($S.A$) of the casting:
$t_s = K \left ( \frac{V}{S.A} \right )^2$
We are given the mold cavity dimensions as 60 cm x 40 cm x 20 cm and that the riser solidification time is 25% greater than the mold's solidification time, i.e., $(t_s)_r = 1.25 \times (t_s)_c$.
For a cylindrical riser with diameter $D$ equal to height $H$, the volume to surface area ratio is:
$\frac{V}{S.A} = \frac{\frac{\pi}{4}D^2H}{2 \times \frac{\pi}{4}D^2 + \pi DH} = \frac{\frac{\pi}{4}D^3}{2 \times \frac{\pi}{4}D^2 + \pi D^2} = \frac{\frac{\pi}{4}D^3}{\frac{\pi}{2}D^2 + \pi D^2} = \frac{\frac{\pi}{4}D^3}{\frac{3\pi}{2}D^2} = \frac{D}{6}$
Therefore, $(t_s)_r = K \left(\frac{D}{6}\right)^2$.
For the casting, the volume ($V_c$) is $60 \times 40 \times 20 = 48000 \text{ cm}^3$ and the surface area ($S.A._c$) is $2(60 \times 40 + 40 \times 20 + 20 \times 60) = 2(2400 + 800 + 1200) = 2(4400) = 8800 \text{ cm}^2$.
So, $\left(\frac{V}{S.A}\right)_c = \frac{48000}{8800} = \frac{60}{11}$.
Now, equating the solidification times:
$K \left(\frac{D}{6}\right)^2 = 1.25 \times K \left(\frac{60}{11}\right)^2$
$\left(\frac{D}{6}\right)^2 = 1.25 \left(\frac{60}{11}\right)^2$
$\frac{D}{6} = \sqrt{1.25} \times \frac{60}{11}$
$D = 6 \times \sqrt{1.25} \times \frac{60}{11} \approx 6 \times 1.118 \times 5.4545 \approx 36.59 \text{ cm}$

Taylor's principle of gauging states that the GO gauge checks the Maximum Material Condition (MMC) of a workpiece. MMC refers to the largest acceptable size for an external feature (like a shaft's diameter) or the smallest acceptable size for an internal feature (like a hole's diameter), ensuring the part has the maximum amount of material allowed. According to Taylor's principle, this GO gauge should also be designed to check as many related dimensions as possible simultaneously to ensure efficient inspection and proper assembly. is the correct description.

The self-sharpening tendency of a grinding wheel is its ability to continuously expose new, sharp abrasive grains during grinding. This occurs when dull grains fracture or wear away, and the bond holding them breaks, revealing fresh grains. This process is crucial for maintaining grinding efficiency.

Wheel Grade and Self-Sharpening:
The wheel grade defines the strength of the bond between the abrasive grains and the wheel core, typically ranging from soft (A to Z) to hard.

• Softer Grade Wheels: These have a weaker bond, allowing abrasive grains to break away more easily when they dull. This frequent exposure of new, sharp grains enhances the wheel's self-sharpening capability. Softer grades are often used for hard materials or when high surface finishes are required.
• Harder Grade Wheels: These have a stronger bond, holding grains more securely. While this extends wheel life, grains can become dull without being easily replaced, reducing the self-sharpening effect and potentially increasing grinding forces and heat.
  Therefore, the tendency for a grinding wheel to self-sharpen is primarily determined by the bond strength, which is represented by the wheel grade. A softer grade promotes better self-sharpening.

Other Factors:
While other factors influence grinding wheel performance, their direct impact on self-sharpening is secondary compared to wheel grade:

• Wheel Structure: Refers to the spacing of grains. An open structure improves chip clearance and reduces clogging, indirectly helping maintain cutting, but doesn't directly control grain replacement.
• Grit Size: The size of abrasive particles. Affects cutting edge density and chip load but not the primary determinant of how grains break away.
• Grit Hardness: The inherent hardness of the abrasive material. A harder grit might stay sharp longer, but self-sharpening depends on whether the bond is strong enough to hold it until it dulls and then breaks away. A very hard grit combined with a strong bond (hard grade) would resist self-sharpening.

In conclusion, the self-sharpening tendency is most directly influenced by the wheel grade. A softer grade ensures that dull abrasive grains break away readily, revealing sharp new grains and maintaining the wheel's cutting efficiency.

Non-traditional machining processes utilize energy sources other than traditional mechanical cutting. We need to identify the process where mechanical energy is the principal source for material removal.

Let's analyze each option:

• Electric Discharge Machining (EDM): Uses electrical energy (sparks) to erode material. This is not mechanical.
• Laser Beam Machining (LBM): Uses thermal energy (laser) to melt or vaporize material. This is not mechanical.
• Ultrasonic Machining (USM): Employs high-frequency (ultrasonic) mechanical vibrations of a tool. These vibrations drive abrasive particles, which then remove material. This clearly uses mechanical energy as its principal source.
• Plasma Arc Machining (PAM): Utilizes thermal energy from a high-temperature plasma torch to melt and blow away material. This is not mechanical.

Therefore, Ultrasonic Machining (USM) is the process that primarily uses mechanical energy, in the form of high-frequency vibrations, for material removal.

Self-lubricating bearings produced by powder metallurgy rely on a porous structure created during sintering to hold lubricant. After sintering, the crucial secondary operation is Impregnation. This process involves filling these interconnected pores within the sintered component with a lubricant, typically oil or grease, often using vacuum or pressure to ensure complete saturation. Cold and Hot Isostatic Pressing are primarily for densification, either before or after sintering, but not for lubricant introduction. Infiltration involves filling pores with a molten material, usually for strength or conductivity, not specifically for self-lubricating oil. Therefore, impregnation directly enables the self-lubricating property by embedding the lubricant.

Taylor's tool life equation, $V \times T^n = C$, describes the relationship between cutting velocity ($V$) and tool life ($T$), with $n$ being Taylor's exponent. Given $n = 0.5$, an initial cutting speed $V_1 = V$, and a reduced speed $V_2 = V_1/2$.
Using the equation $V_1 \times T_1^n = V_2 \times T_2^n$:
$V_1 \times T_1^{0.5} = \frac{V_1}{2} \times T_2^{0.5}$
$\frac{T_2}{T_1} = 2^{(1/0.5)} = 4$
The percentage increase in tool life is calculated as $(\frac{T_2}{T_1} - 1) \times 100\% = (4 - 1) \times 100\% = 300\%$.

To accurately mount end mill cutters on a vertical milling machine's spindle, a secure and precise holding mechanism is essential for proper operation. Let's evaluate the given options: A vice is for clamping workpieces, not tools. A face plate or driver plate is typically used for holding workpieces or transferring motion in lathes, not for mounting end mills directly to a vertical milling machine spindle. A collet, however, is a specialized chuck designed to grip tools with straight or tapered shanks by closing a split sleeve around the tool when tightened. This provides a strong, concentric, and rigid grip, which is crucial for the precision and safety required when using end mills. Therefore, end mill cutters are mounted using a collet due to its ability to ensure high concentricity, uniform clamping force, and ease of tool changes.