General Engineering – GATE PI General EngineeringPractice Questions & PYQs

Let's break down casting processes by looking at whether their pattern and mould are "expendable" (destroyed during the process) or "reusable" (kept for multiple uses). An expendable pattern is one, like wax or foam, that's melted or burned away to create the mould cavity. An expendable mould is made of material, such as sand or ceramic, that's broken apart to remove the final casting.

Now, let's examine the options:

• Shell Mould Casting: Uses a reusable metal pattern and an expendable sand-resin mould. This doesn't fit both criteria.
• Investment Casting: Utilizes an expendable wax pattern. This wax is melted out (destroyed), and then the ceramic mould formed around it is broken to retrieve the casting (also destroyed). This perfectly matches both an expendable pattern and an expendable mould.
• Pressure Die Casting: Employs a permanent metal die (mould), meaning the mould is not expendable.
• Centrifugal Casting: Focuses on rotational force for solidification. While its mould can be expendable, it doesn't universally require an expendable pattern, and patterns are often reusable or formed directly.

Therefore, Investment Casting is the process that uses both an expendable pattern and an expendable mould.

To find the effective diameter ($D_e$) of an external metric thread using the three-wire method and a standard cylinder, we'll use a specific formula that accounts for all measurements.

Given Parameters:

• Pitch ($P$) = 3.5 mm
• Included thread angle ($2\alpha$) = $60^\circ \implies \alpha = 30^\circ$
• Wire diameter ($d_w$) = 2 mm
• Standard cylinder diameter ($D_s$) = 30.5 mm
• Micrometer reading over standard ($M_s$) = 16.532 mm
• Micrometer reading over wires ($M_w$) = 15.398 mm

Calculations:
First, we find the required trigonometric values:

• $\csc(\alpha/2) = \csc(30^\circ) = 2$
• $\cot(\alpha/2) = \cot(30^\circ) = \sqrt{3} \approx 1.73205$

The formula for the effective diameter, combining the standard and wire measurements, is:
$D_e = D_s + d_w \left( 1 - \csc\left(\frac{\alpha}{2}\right) \right) + \frac{P}{2} \cot\left(\frac{\alpha}{2}\right) - (M_s - M_w)$
Now, substitute the given values into the formula:

• $d_w \left( 1 - \csc\left(\frac{\alpha}{2}\right) \right) = 2 \times (1 - 2) = -2$ mm
• $\frac{P}{2} \cot\left(\frac{\alpha}{2}\right) = \frac{3.5}{2} \times \sqrt{3} \approx 1.75 \times 1.73205 \approx 3.03109$ mm
• $M_s - M_w = 16.532 - 15.398 = 1.134$ mm

Substitute these calculated terms back into the main formula:
$D_e = 30.5 + (-2) + 3.03109 - (1.134)$
$D_e = 30.5 - 2 + 3.03109 - 1.134$
$D_e = 28.5 + 3.03109 - 1.134$
$D_e = 31.53109 - 1.134$
$D_e = 30.39709 \text{ mm}$
Rounding to three decimal places, the effective diameter is 30.397 mm.

This question tests your knowledge of material classifications and their examples. Let's match each material type with its correct example:

1. Thermoplastics: These are polymers that can be repeatedly softened by heating and hardened by cooling, making them recyclable and easily reshaped. A common example is Polyvinylchloride (PVC), often used in pipes and window frames. So, $1 \rightarrow Q$.
2. Thermosets: Unlike thermoplastics, thermosets undergo an irreversible chemical reaction (curing) upon heating, forming a rigid, cross-linked network structure. Once cured, they cannot be melted or reshaped. Epoxy resins are a prime example, known for their strong adhesive properties. So, $2 \rightarrow R$.
3. Elastomers: These are polymers that exhibit high elasticity, meaning they can be significantly stretched and then return to their original shape. Latex, a natural rubber, is a classic example of an elastomer. So, $3 \rightarrow S$.
4. Ceramics: These are inorganic, non-metallic solids, often known for their high hardness, strength, and resistance to heat and corrosion. SIAION (Silicon Aluminum Oxynitride) is an advanced ceramic material used in demanding applications. So, $4 \rightarrow P$.

Based on these matches, the correct sequence is $1-Q, 2-R, 3-S, 4-P$.

We need to find the ratio of strain energy for two steel bars. Remember, strain energy ($U$) in an axially loaded bar is given by $U = \frac{P^2L}{2AE}$, where $P$ is the load, $L$ is the length, $A$ is the cross-sectional area, and $E$ is the modulus of elasticity.

For the First Bar:
The first bar has a constant diameter $d$ and length $L$.
Its cross-sectional area is $A_1 = \frac{\pi d^2}{4}$.
So, its strain energy, $U_1 = \frac{P^2L}{2A_1E} = \frac{P^2L}{2 \left(\frac{\pi d^2}{4}\right) E} = \frac{2P^2L}{\pi d^2 E}$.

For the Second Bar:
This bar is composed of two segments:

1. Segment 1: Diameter $2d$ and length $\frac{2L}{3}$.
   Its cross-sectional area is $A_{2a} = \frac{\pi (2d)^2}{4} = \pi d^2$.
   The strain energy for this segment is $U_{2a} = \frac{P^2 \left(\frac{2L}{3}\right)}{2A_{2a}E} = \frac{P^2 \left(\frac{2L}{3}\right)}{2 (\pi d^2) E} = \frac{P^2L}{3\pi d^2 E}$.
2. Segment 2: Diameter $d$ and length $\frac{L}{3}$.
   Its cross-sectional area is $A_1 = \frac{\pi d^2}{4}$.
   The strain energy for this segment is $U_{2b} = \frac{P^2 \left(\frac{L}{3}\right)}{2A_1E} = \frac{P^2 \left(\frac{L}{3}\right)}{2 \left(\frac{\pi d^2}{4}\right) E} = \frac{2P^2L}{3\pi d^2 E}$.

The total strain energy for the second bar is the sum of energies of its segments:
$U_2 = U_{2a} + U_{2b} = \frac{P^2L}{3\pi d^2 E} + \frac{2P^2L}{3\pi d^2 E} = \frac{3P^2L}{3\pi d^2 E} = \frac{P^2L}{\pi d^2 E}$.

Ratio of Strain Energies:
Now we find the ratio of the first bar's strain energy to the second bar's strain energy:
$\frac{U_1}{U_2} = \frac{\frac{2P^2L}{\pi d^2 E}}{\frac{P^2L}{\pi d^2 E}} = 2$.