GATE Mechanical Engineering (ME) Official Paper (Held On: 03 Feb, 2024) (Previous Year Paper)

Explanation:

Let,

\(\frac{x}{p q-r^{2}}=\frac{y}{q r-p^{2}}=\frac{z}{r p-q^{2}}=K\)

\(\frac{p x}{p^{2} q-p r^{2}}=K\)..... (1)

\(\frac{q y}{q^{2} r-q p^{2}}=K\) ...... (2)

From 1 & 2:

px + qy + rz = kp²q - kpr² + kqr² - kqp² + kr²p - krq²

= 0

Explanation:

Total outcomes : (2, 2), (2, 3), (2, 5), (2, 7)

(3, 2), (3, 3), (3, 5), (3, 7)

(5, 2), (5, 3), (5, 5), (5, 7)

(7, 2), (7, 3), (7, 5), (7, 7)

Favorable outcomes (sum > 11)

: (5, 7), (7, 5), (7, 7)

\(\text { Probability }=\frac{\text { Number of favourable outcomes }}{\text { Number of total outcomes }}=\frac{3}{16}\)

Explanation:

In Fibonacci series, every number is equal to the sum of previous two numbers of the series. i.e. 0, 1, 1, 2, 3, 5, 8, 13, 21, ....

0 + 1 = 1

1 + 1 = 2

2 + 1 = 3

3 + 2 = 5

5 + 3 = 8

8 + 5 = 13

13 + 8 = 21

Explanation:

Expect 21 all are prime number.

Prime Number:

• Prime numbers are numbers greater than 1.
• They only have two factors, 1 and the number itself.
• This means these numbers cannot be divided by any number other than 1 and the number itself without leaving a remainder.

Explanation:

[smile giggle laugh] is similarly related as [disapprove → reprove → chide]

Reprove : To criticize somebody or not approve of something that somebody has done.

Disapprove : To refuse approval

Chide : criticize somebody

Explanation:

Tetrahedron Angles:

• In a regular tetrahedron, all the faces are equilateral triangles. Therefore, all the interior angles of a tetrahedron are 60° each.

Explanation:

Prof. P wasn't merely a man who narrated funny stories even in his blackest moments he was capable of self-depreciating humor.

Prof. Q was a man who hardly narrated funny stories only in his blackest moments was he able to find humor.

Explanation:

The question is about non-null sets A, B, C \(\Rightarrow 4^3\) – (Any set is Empty)

Consider \(A = \phi\)

Universal set \(= {2,3,5}\) contains 3 elements \(\Rightarrow~ B~ has~ 2^3\) possible choices, and for each possible B set, we need to calculate possible sets of C. As \(B \subseteq C\), the elements present in the B, should be present in C. Remaining elements of Universal set has two choices; present in C or not present in C.

\(if ~B = \phi\) (number of elements in B = 0) \(\Rightarrow\) number of possible sets for \(C = 2^{n-0} = 2^3\)

\( \textbf{if B = {1}}\) (number of elements in B = 1) \(\Rightarrow\) number of possible sets for \(C = 2^{n-1} = 2^2\)

\( \textbf{if B = {2}}\) (number of elements in B = 1) \(\Rightarrow\) number of possible sets for \(C = 2^{n-1} = 2^2\)

\( \textbf{if B = {3}}\) (number of elements in B = 1) \(\Rightarrow\) number of possible sets for \(C = 2^{n-1} = 2^2\)

\( \textbf{if B = {1, 2}}\) (number of elements in B = 2) \(\Rightarrow\) number of possible sets for \(C = 2^{n-2} = 2^1\)

\( \textbf{if B = {1, 3}}\) (number of elements in B = 2) \(\Rightarrow\) number of possible sets for \(C = 2^{n-2} = 2^1\)

\(\textbf{if } B = {2, 3}\) (number of elements in B = 2) \(\Rightarrow\) number of possible sets for \(C = 2^{n-2} = 2^1\)

\(​\textbf{if } B = {1, 2, 3}\) (number of elements in B = 3)} \(\Rightarrow\) number of possible sets for \(C = 2^{n-3} = 2^0\)

∴ \( \binom{n}{0} \cdot 2^n + \binom{n}{1} \cdot 2^{n-1} + \cdots + \binom{n}{r} \cdot 2^{n-r} \cdots + \binom{n}{n} \cdot 2^0 = (1 + 2)^n = 3^n = 3^3 = 27 \)

\(∴ \text{when } A = \phi, \text{ possible sets of B and C are 27} \)

Final answer = \(4^3 - 3^3 = 37\)

Explanation:

From graph:

In between 2015 and 2019, VK's batting average = 35 + 90 +75 +130 + 60 = \(\frac{390}{5}\) = 78

In between 2015 and 2019, RS's batting average = 50 + 65 + 70 + 70 + 55 = \(\frac{320}{5}\) = 64

So, VK yearly batting average is consistently higher then RS's between two World Cup years (2015 and 2019).

Explanation:

Hence, the total number of folds required to form the above object is 9.

Explanation:

Gas metal arc welding (GMAW) or Metal inert gas arc welding (MIG)

• In this process, the arc is formed between a continuous, automatically fed, metallic consumable electrode and welding job in an atmosphere of inert gas, and hence this is called metal inert gas arc welding (MIG) process.
• The shielding gases for MIG welding are mixtures of argon, oxygen, and CO2, and special gas mixtures may contain helium.
• In GMAW, consumable electrodes are used which has to be same composition as base metal. Thus, low alloy steels electrodes are the most suitable electrode material. 

Explanation:

Given, AB = 7 m, u = 0, v = 5 m/s

According to question, the rope will become vertical on striking the object, hence AB = 7 m.

From geometry,

AD = 7 cosθ 

∴ h = AB - AD = 7 - 7 cosθ

= 7(1 - cosθ)

From energy conservation,

Initial potential energy = Final kinetic energy

⇒ \(m g h=\frac{1}{2} m v^{2}\)

\(9.81 \times 7(1-\cos \theta)=\frac{1}{2} \times 5^{2}\)

\(\cos \theta=\frac{9.81 \times 7-12.5}{9.81 \times 7}\)

⇒ θ = 35.1°

Explanation:

Q_in = 24000 kJ/min = 400 kW

T₁ = 1200 K, T₂ = 300 K

Maximum power can be obtained when the heat supplied

in used in Carnot engine,

∴ W_max = Q_in × η_carnot

= \(400\left(1-\frac{300}{1200}\right)\)

W_max = 300 kW

Explanation:

To evaluate, ∯_s z dx dy

where S(sphere) = x² + y² + z² = R² 

= ∯z dxdy

\(\int_S \vec{F} \cdot \hat{n} , ds = \iint_R \left(F_1 , dy , dz + F_2 , dx , dz + F_3 , dx , dy\right)\)

\( F_1 = 0, F_2 = 0, F_3 = z\)

\(\vec{F} = Z \hat{K}\)

\(\nabla \cdot \vec{F} = 1\)

Using Gauss Divergence Theorem:

\(\int_S \vec{F} \cdot \hat{n} , ds = \int_V \nabla \cdot \vec{F} , dV = \int_V dV\)

= Volume of sphere = \(\frac{4 \pi}{3} R^{3}\)

Explanation:

Let the function be parabolic nature, i.e. a form of: Ax² + Bx + C

Now, the conditions for strictly local minimum

to exist are:

The coefficient of x² (i.e. A) should be positive.

First derivative of the function should be zero.

Second derivative of the function should be positive.

Now, After the observation of all four options,

above these conditions will be satisfied by option (a)

Explanation:

Dynamic load carrying Capacity:

• It is defined as the radial load in radial bearing (Thrust load in thrust bearing) that can be carried for a minimum life of one million revolutionss.
• It is based on the assumption that the inner race is rotating and the outer race is stationary.

Equivalent / Actual bearing Load:

• The equivalent dynamic load is defined as the contact radial load in radial bearings (Thrust load in thrust bearing), which if applied to the bearing would give the same life as that which the bearing would attain in actual condition.

​Bearing life under variable load:

​Life, L = \((\frac{C}{P_e})^n\) million revolution

n = 3 for ball bearing, n = 10/3 for roller bearing, C = basic dynamic load, Pe = Dynamic load

Explanation:

Soderberg line:

Soderberg line for ductile materials gives upper limit for any combination of mean and alternating stress. Soderberg line is the most conservative fatigue failure criterion. Following diagram depicts the same.

Considering following notations

σa = limiting safe stress amplitude

Se = endurance limit of the component

σm = limiting safe mean stress

Sut = ultimate tensile strength

Syt = Yield strength

N = Factor of safety

Soderberg line:

The line joining Syt (yield strength of the material) on the mean stress axis and Se (endurance limit of the component) on stress amplitude axis is called as Soderberg line. This line is used when yielding defines failure (Ductile materials).

The equation for the Soderberg line:

\(\frac{{{{\rm{\sigma }}_{\rm{m}}}}}{{{{\rm{S}}_{{\rm{yt}}}}}} + \frac{{{{\rm{\sigma }}_{\rm{a}}}}}{{{{\rm{S}}_{\rm{e}}}}} = \frac{1}{N}\)

Additional Information

Goodman line:

Line joining Se on stress amplitude axis and Sut on mean stress axis is known as Goodman line. The triangular region below this line is considered a safe region. 

The equation for Goodman line:

\(\frac{{{\sigma _m}}}{{{S_{ut}}}} + \frac{{{\sigma _a}}}{{{S_e}}} = \frac{1}{N}\)

Gerber Line: 

Line joining Se on stress amplitude axis and Sut on mean stress axis is joined by a parabolic curve. 

The equation for Gerber line:

Explanation:

x + 2y + z = 5

2x + ay + 4z = 12

2x + 4y + 6z = b

Performing row operations on augmented matrix,

\(\left[\begin{array}{ccc:c} 1 & 2 & 1 & 5 \ 2 & a & 4 & 12 \ 2 & 4 & 6 & b \end{array}\right] \xrightarrow[R_{3}=R_{2}-2 R_{1}]{\substack{R_{1}=R_{2}-2 R_{1}}}\left[\begin{array}{ccc:c} 1 & 2 & 1 & 5 \ 0 & a-4 & 2 & 2 \ 0 & 0 & 4 & b-10 \end{array}\right]\)

For infinitely many solutions,

ρ(A) = ρ(AB) = 2

a - 4 = 0

a = 4

and, for this

ρ(AB) = 2

R₃ = 2R₂

b - 10 = 4

b = 14

Explanation:

Let, f(z) = u + iv

u = cosh z cos y

= \(\left(\frac{e^{x}+e^{-x}}{2}\right) \cos y\)

\(u=\left(\frac{e^{x}+e^{-x}}{2}\right) \cos y\)

by Mine Thomson method

\(u(x, y)=\left(\frac{e^{x}+e^{-x}}{2}\right) \cos y\)

Partial differential w.r.t. x, y

\(u_{x}=\left(\frac{e^{x}-e^{-x}}{2}\right) \cos y ; \quad u_{y}=\left(\frac{e^{x}+e^{-x}}{2}\right)(-\sin y)\)

\(u_{x}=\left(\frac{e^{x}-e^{-x}}{2}\right) \cos y \quad u_{y}=-\frac{\left(e^{x}+e^{-x}\right)}{2} \sin y\)

\(u_{x}(z, 0)=\frac{e^{z}-e^{-z}}{2}=\sinh z \quad u_{y}(z, 0)=-0\)

f(z) = ∫(u_x - iu_y)dz + c

= ∫(sinhz - a)dz + c

= ∫sinh z + c

= cos h z + c

Explanation:

By Johnson’s rule of sequencing mark minimum time-consuming operation for each process.

\(\begin{array}{|c|c|c|} \hline \textbf{Job} & \textbf{Work Stations 1} & \textbf{Work Stations 2} \ \hline U & 5 & 4 \ \hline V & 7 \leftarrow & 6 \ \hline W & {3} & 6 \ \hline X & 4 & 8 \ \hline Y & 6 & 5 \ \hline Z & 8 & 7 \ \hline \end{array}\)

According to the question, work on work station (1) is followed by work on workstation (2).

Perform that job in mark station which is minimum time consumption. It is W-X during this Z on work station-2 will be performed. So, W – X – Z

After W-X-Z, V will be performed on workstation (1). So, W–X–Z–V, in between Y-work on the workstation will perform.

As the minimum time in the workstation is U, so will be performed at last. So, W – X – Z – V – Y – U

Explanation:

k = A + Bx

where A and B are positive real number

So with an increase in x, k increases

Energy equation;

\(\frac{\partial}{\partial x} \left( k \frac{\partial T}{\partial x} \right) + \dot{q} = 0\)

\(\frac{\partial}{\partial x}\left((A+B x) \cdot \frac{\partial T}{\partial x}\right)=0\)

On integration

\((A+B x) \cdot \frac{\partial T}{\partial x}=C_{1}\)

\(\frac{\partial T}{\partial x}=\frac{C_{1}}{A+B x}\)

On integration,

T = logarithmic function of x

Hence, variation of T with x is given as;

Explanation:

Anisotropy plays an important role in the performance of deep drawing processes. The anisotropy is of two types. In normal anisotropy the properties differ in the thickness direction. In planar anisotropy, the properties vary with the orientation in the plane of the sheet. Whereas deep drawability of sheets increases with normal anisotropy, anisotropy leads to the formation of ears in cup drawing. Ears cause the wavy edge of a drawn cup.

Earing:

In deep drawing, the edges of cups may become wavy, called earing. Ears are objectionable on deep-drawn cups because they have to be trimmed off, as they serve no useful purpose, and interfere with further processing of the cup, resulting in scarp. Earing is caused by the planar anisotropy of the sheet metal, and the number of ears produced may be two, four or eight, depending on the processing history and microstructure of the material. If the sheet is stronger in the rolling direction than transverse to the rolling direction, and the strength varies uniformly with respect to orientation, then two ears will form. If the sheet has high strength at different orientations, then more ears will form.

Explanation:

For the body to be in equilibrium, the three forces can be coplanar, parallel or concurrent.

Let the resultant of F₁ and F₂ is R.

For equilibrium, third force shall be in opposite direction of R in same plane (x–y plane)

Explanation:

In a thermally fully developed flow, the temperature profile will no longer change along the length of the flow, similar to the velocity profile. It means there is no further growth of the boundary layer. This is the case of a steady state.

\(\frac{dT}{dZ} = 0\)

\(N_u \) = Constant for thermally fully developed flow

\((T_w - T_m) \text{ decays exponential}\)

\(\frac{dT_m}{dz} \text{ increases exponentially}\)

The rate of increase in bulk mean temperature along the flow \(\left(\frac{d T_{m}}{d z}\right)\) increase exponentially along the flow direction (z).

Explanation:

m = 2 kg

f_n = 15 Hz

ω_n = 2π(f_n) = 2π × 15 = 30π rad/s

x_n = 7.75 mm, x_n+1 = 7.20 mm

Decrement ratio, \(\frac{x_{n}}{x_{n+1}}=\frac{7.75}{7.20}=1.07638\)

Logarithmic decrement,

\(δ=\log _{e}\left(\frac{x_{n}}{x_{n+1}}\right)=\log _{e}(1.07638)\)

δ = 0.0736

\(\frac{2 \pi ξ}{\sqrt{1-ξ^{2}}}=0.0736\)

⇒ ξ = 0.0117

But, \(2 \xi \omega_{n}=\frac{C}{m}\)

⇒ \(2 \times 0.0117 \times 30 \pi=\frac{C}{2}\)

⇒ C = 4.41 N-s/m

Explanation :

In grinding operation, surface finish is decided by the grain size. Fine and very fine grain size will result in best surface finish.

Additional InformationDesignation of Grinding Wheel:

Prefix / Suffix: These are the secret codes used by the manufacturers to represent the wheel by its size and shapes respectively.

Type of Abrasives / Grain type:

• It indicates materials used for the manufacturing of abrasive particles.
• Out of the abrasives B4C is giving the poor performance during machining and diamond is very costly, therefore Al2O3 or SiC is the most commonly bused abrasives in the grinding wheel.
• Al2O3 soft and tougher than the SiC whereas SiC will be hard and brittle than Al2O3
• The type of abrasive is selected based on the mechanical properties of workpiece material I.e. for machining of soft and ductile workpieces, Al2O3, and machining of hard and brittle workpiece SiC will be used.
• A- Al2O3, B – B4C, C – SiC, D - Diamond

Grain size or Grit size:

• It indicates the size of abrasive particles.
• i.e. Side if abrasives = 1/ Grain Size Number (GSN)
• when the GSN > 600, the size of the abrasive particles becomes very very small and it cannot act like a cutting tool, therefore MRR is less.
• When GSN < 600, the actual size of abrasive is increasing, the chip size is increasing and MRR is increasing.
• As the GSN is reducing or the size of abrasive is increasing, the MRR is increasing first and then reducing.
• The grain size is selected based on the surface finish required on the workpiece i.e. for a rough grinding, course or medium grain size is selected and for finished grinding fine or very fine grain size will be selected.
• 10-24= Coarse, 30-60 = Medium, 80 -180 = Fine, 220 – 600 = Very fine

Grades of Grinding Wheel:

• It indicates the hardness of the grinding wheel.
• The grade of the grinding wheel is selected based on the mechanical properties of the workpiece material.
• Soft wheels are used for grinding of hard workpiece because the rubbing forces induced by the blunt abrasive particle i.e. the self-sharpening is taking place and no dressing is required.
• Hard wheels are used for grinding of the soft workpiece, the abrasive particle will be effectively utilized so that at the end of effective utilization the dressing will be carried for resharpening of grinding wheel.
• A –H = Soft, I – P = Medium, Q – Z = Hard

Structure:

• The structure is indicating the average gap between the two consecutive abrasive particles.
• As the average gap is large, the number of abrasive particle presents per unit area will be small hence it is called the open structure.
• The structure of a grinding wheel can be varied by varying the % of abrasive particles and bonding material in the manufacturing of a grinding wheel. i.e. when higher % of abrasives and lower % of bonding material is used in manufacturing it produces the dense structure and vice-versa.
• 0 – 7 = Dense, 8 – 16 = Open

Bonds:

• Bond indicates the bonding material used for the manufacturing of the grinding wheel.
• Out of the different bonding materials, vitrified is the most commonly used bonding material because it gives higher bonding strength, high temperature withstanding capability, and high thermal conductivity.
• For the manufacturing of flexible grinding wheels also called buffing wheels, shellac or rubber can be used as the bonding material.
• V – Vitrified, B – bakelite, S – Silicate, E – Shellac, R - Rubber

Explanation:

Given: λ = 5/hr, μ = 10/hr

∴ \(\rho=\frac{\lambda}{\mu}=\frac{5}{10}=\frac{1}{2}\)

Now, probability for the system to be idle,

P₀ = 1 - ρ = 1 - 0.5 = 0.5

Explanation:

Given that,

\(\frac{y_{n+1}-y_{n}}{\Delta t}=\frac{1}{2}\left(y_{n+1}+y_{n}\right)\)

\(y_{n+1}-y_{n}=\frac{\Delta t}{2}\left(y_{n+1}+y_{n}\right)\)

\(y_{n+1}-\frac{Δ t}{2} y_{n+1}=y_{n}+\frac{Δ t}{2} y_{n}\)

\(y_{n+1}=\frac{2+Δ t}{2-Δ t} y_{n}\)

For stability:

\(\left|\frac{2+\Delta t}{2-\Delta t}\right| \leq 1\)

⇒ \(-1 \leq \frac{2+\Delta t}{2-\Delta t} \leq 1\)

⇒ \(-1 \leq-1+\frac{4}{2-\Delta t} \leq 1\)

⇒ \(0 \leq \frac{4}{2-\Delta t} \leq 2\)

⇒ \(\frac{1}{2} \leq \frac{2-\Delta t}{4} \leq 0\)

⇒ 2 ≤ 2 - Δt ≤ 0

⇒ 0 ≤ - Δt ≤ -2

⇒ Δt ≥ 2

Explanation:

The velocity field of a two-dimensional, incompressible flow is given by,

\(\vec{V}=2 \sinh x \hat{i}+V(x, y) \hat{j}\)

and V(x, 0) = coshx,

For an incompressible flow, ∇ . \(\vec{V}\)  = 0

⇒ \(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\)

= 2coshx + \(\frac{\partial v}{\partial y}\) = 0

⇒ \(\frac{\partial v}{\partial y}\) = -2coshx

⇒ \(\int \partial v=\int-2 \cosh x \cdot \partial y\)

V = -2y . coshx + f(x)

For, V(x, 0) = coshx

⇒ -2y × coshx + f(x) = coshx

⇒ f(x) = coshx (at x, 0)

⇒ V = -2 . y × coshx + coshx

V = (1 - 2y) × coshx

V(0, - 1) = [1 - {2 × (-1)}] × cosh(0) = 3

Explanation:

Preparatory function of CNC programming is denoted by G-codes. Preparatory functions are the G-codes that identify the type of activities the machine will execute. The preparatory functions describe the way in which the machine axes have to move, the method of interpolation, the dimension system, the time delay of program execution and the activation of specific operational modes in the control.

Additional Information

G-codes: These are the general-purpose codes. Some of the G codes with their purpose are:​

Explanation:

For laminar flow over a flat plate,

Nu ∝ Re^0.5 Pr^1/3, where Re is Reynolds number and Pr is the Prandtl number.

∵ \(\mathrm{Re}=\frac{U_{\infty} x}{\mathrm{v}}\)

∴ \(\mathrm{Nu} \propto \cup_{\infty}^{0.5}\)

Given that: \(\frac{U_{\infty, 2}}{U_{\infty, 1}}~=~2\)

or, \(\frac{N u_{2}}{N u_{1}}=\sqrt{\frac{U_{\infty, 2}}{U_{\infty, 1}}}=\sqrt{2}\)

Explanation:

• Superheating in Rankine cycle increases the cycle efficiency because of increase in mean temperature of heat addition.

• With increase in pressure of boiler, the cycle efficiency increases. So, the given statement is wrong.

• With decrease in condenser pressure, the cycle efficiency increases because of decrease in mean temperature of heat rejection.

• The pressure of turbine outlet is governed by the condenser temperature. Decreasing the cooling water temperature, creates more vacuum in condenser which results in pressure drop and vice-versa.

Additional InformationRankine cycle

• It is the ideal cycle for a vapour power plant.
• It comprises four reversible processes:

 

The efficiency of the Rankine cycle

• We know that efficiency η of the Rankine cycle is given as,

​\(\eta =1-\frac{T_L}{T_{avg}}\)

• That means to increase the efficiency we should increase the average temperature at which heat is transferred to the working fluid in the boiler.
• Another way would be to decrease the average temperature at which heat is rejected from the working fluid in the condenser.

Decreasing the condenser pressure

• Lowering the condenser pressure will increase the area enclosed by the cycle on a T - S diagram which indicates that the net-work will increase.
• Thus, the thermal efficiency of the cycle will be increased.

Superheating the steam to high temperature

• This will increase the net-work output and the efficiency of the cycle.
• It also decreases the moisture contents of the steam at the turbine exit.
• The temperature to which steam can be superheated is limited by metallurgical considerations (620°C)

Increasing the boiler pressure

• Increasing the operating pressure of the boiler leads to an increase in the temperature at which heat is transferred to the steam and thus raises the efficiency of the cycle.

Explanation:

Given: ΔE = 300 J, ω_min = 220 rad/s, ω_max = 280rad/s, C_s = 0.05, I = ?

Energy fluctuation, ​ΔE = lω²C_s

⇒ \(\Delta E=I \times\left(\frac{\omega_{\max }+\omega_{\min }}{2}\right)^{2} \times C_{s}\)

⇒ \(300=I \times\left(\frac{280+220}{2}\right)^{2} \times 0.05\)

⇒ I = 0.096 kg-m²

Concept:

A pattern is the replica of the casting to be formed with some modifications.

These modifications are in the form of allowances. The allowances provided to the pattern are discussed below

Shake Allowance: Before the withdrawal from the sand mold, the pattern is rapped all around the vertical faces to enlarge the mold cavity slightly, which facilitates its removal. Since it enlarges the final casting made, it is desirable that the original pattern dimension should be reduced to account for this increase. It is a negative allowance and is to be applied only to those dimensions that are parallel to the parting plane.

Draft or Taper Allowance: By draft is meant the taper provided by the pattern maker on all vertical surfaces of the pattern so that it can be removed from the sand without tearing away the sides of the sand mold and without excessive rapping by the molder.

Machining or Finish Allowance: The finish and accuracy achieved in sand casting are generally poor and therefore when the casting is functionally required to be of good surface finish or dimensionally accurate, it is generally achieved by subsequent machining. Machining or finish allowances are therefore added in the pattern dimension.

Shrinkage or Contraction Allowance: All most all cast metals shrink or contract volumetrically on cooling. The metal shrinkage is of two types:

• Liquid Shrinkage: it refers to the reduction in volume when the metal changes from liquid state to solid state at the solidus temperature. To account for this shrinkage; riser, which feed the liquid metal to the casting, are provided in the mold.
• Solid Shrinkage: it refers to the reduction in volume caused when metal loses temperature in solid state. To account for this, shrinkage allowance is provided on the patterns.

Explanation:

Pearlite is the eutectoid mixture of cementite and ferrite. 

Iron carbon equilibrium diagram: 

• The phase diagram of Fe-Fe3C is not a true equilibrium because the iron carbide is an unstable phase that after prolonged heat treatment decomposes into iron and carbon( the graphite form).
• The study is simplified by within assumption that sufficient time has been allowed at each new temperature for any necessary adjustment in phase compositions and relative amounts.

​Phase components:

Carbon is an interstitial impurity in iron and forms a solid solution in alpha and delta phases. Different phase components are as follows:

Pearlite: 

• Microstructure for the eutectoid steel slowly cooled through the eutectoid temperature consist of alternating lamellae of the two phases alpha and Fe3C.
• Carbon atoms diffuse away from the 0.22% ferrite, extending from the grain boundaries into unreacted austenite grain.
• The resulting structure is called pearlite because it has the appearance of the mother of pearl.​​

Additional InformationFerrite: 

• Pure iron contains 0.008% weight of carbon.
• Upon heating, pure iron experiences a change in crystal structure before it melts.
• The stable form at room temperature is called ferrite.
• Alpha iron has a BCC crystal structure.
• This phase is relatively soft can be made magnetic at a temperature below 768° C and has the maximum possible density of 7880 kg/m3.

Austenite: 

• Ferrite experiences a  polymorphic transformation to FCC austenite (gamma iron) at 912°C.
• The austinite persists up to 1394°C at which it becomes BCC-phase known as delta-iron.
• Upon further heating, it melts at 1538°C.
• The maximum solubility of carbon in gamma iron is 2.14% approximately 100 times the maximum solubility in the BCC ferrite phase because FCC interstitial positions are larger.
• Austenite is a non-magnetic phase.

Cementite: 

• A very hard and brittle phase is formed when carbon exceeds 6.67%.
• It has two phases eutectic and eutectoid.

Explanation:

Statically indeterminate structures are those structures that cannot be analyzed using statics or equations of equilibrium. In such cases, the number unknowns exceeds the number of equilibrium equations available.

Checking each option:

For option (1),

Number of unknown = 3

Number of equilibrium equation = 3

For option (2),

Number of unknown = 3

Number of equilibrium equation = 2

For option (3),

Number of unknown = 2

Number of equilibrium equation = 2

For option (4),

Number of unknown = 2

Number of equilibrium equation = 2

So, (2) is statically indeterminate structures.

Explanation:

Given that:

T1 = 1000 K, T2 = 300 K = T4

Q1 = 150 kJ

T3 = 280 K

For heat engine,

\(\eta=1-\frac{T_{2}}{T_{1}}\)

⇒ \(1-\frac{300}{1000}=1-\frac{Q_{2}}{Q_{1}}\)

⇒ Q₂ = 0.3Q₁ = 0.3 × 150 = 45 kJ

∴ Q₁ - Q₂ = 150 - 45

⇒ W_E = 105 kJ

Now, 

\(\mathrm{COP}_{\mathrm{HP}}=\frac{Q_{4}}{W_{E}}=\frac{T_{4}}{T_{4}-T_{3}}\)

∴ \(\frac{Q_{4}}{105}=\frac{300}{300-280}\)

∴ Q₄ = 1575 kJ

∴ Q₂ + Q₄ = 1575 + 45 = 1620 kJ

Hence, total amount of heat delivered to the room together by the heat engine and heat pump is 1620 kJ.

Explanation:

Given: L × B = 100 mm × 200 mm, Z = 8, D= 100 mm, W = 200 mm, f₁ = 0.1 mm/tooth, V = 20 m/min, d = 2 mm

Approach length, x = \(\sqrt{d(D-d)}=\sqrt{2 ×(100-2)}=14 \mathrm{~mm}\)

∴ The effective length, L_e = L + x = 100 + 14 = 114 mm

N = \(\frac{V \times 1000}{\pi \times D}=\frac{20 \times 1000}{\pi \times 100}=63.662 \mathrm{rpm}\)

F = f_t × NZ = 0.1 x 63.662 × 8 = 50.93 mm/min

∴ The machining time required to remove the entire stock:

t_m/c = \(\frac{L_{e}}{F}=\frac{114}{50.93}\) = 2.238 ≃ 2.24 minutes

Explanation:

\(f(x)=\left{\begin{array}{lc} 1 & 0 \leq x \leq 1 \ 0 & \text { otherwise } \end{array}\right.\)

Expected value of Y,

\(E[Y]=\int_{-\infty}^{\infty} Y \times f(x)\)

= \(\int_{0}^{1} 1 \cdot \ln (x+1) d x=\int_{0}^{1} \ln (x+1) d x\)

= \(\int_{1}^{2} \ln t d t \quad\left[\begin{array}{l} \because x+1=t \ \because d x=d t \end{array}\right]\)

= \([t \cdot \ln t-t]_{1}^{2}\)

= \(=(2 \ln 2-2)-(1 \ln 1-1)\)

= -0.6137 + 1

=0.3863 ≃ 0.39

Explanation:

For a negative eigen value, the product of eigen values < 0

\(\left[\begin{array}{ll} 1 & a \ 8 & 3 \end{array}\right]\)

⇒ Determinant < 0

⇒ 3 - 8a < 0

⇒ 8a > 3

⇒ a > \(\frac{3}{8}\)

Explanation:

Given: D = 8000 unitsIyr, C₀ = Rs 300Iorder, C_h = Rs 12 × 12 = Rs. 144 per year

Now, \(\mathrm{EOQ}=\sqrt{\frac{2 C_{D} D}{C_{h}}}=\sqrt{\frac{2 × 300 × 8000}{144}}\)

∴ Q* = 182.5 units

Now, Actual quantity, Q = 1.25Q* = 1.25 × 182.5

= 228.125 units

Total cost at EOQ, TIC₁ = \(\sqrt{2 C_{o} C_{h} D}=\sqrt{2 × 300 × 144 × 8000}\)  = Rs 26290.68

Now, Total cost at Q = 228.125 units

\(\mathrm{TIC}_{2}=\frac{Q}{2} × C_{h}+\frac{D}{Q} × C_{o}\)

= \((\frac{228.125}{2} × 144)+(\frac{8000}{228.125} × 300)\)

= Rs. 26945.54

∴ % increase = \(\frac{26945.54-26290.68}{26290.68} × 100=2.5 %\)

Alternate Method

We know, if Q = kQ*

then, \(\frac{T I C(Q)}{T I C\left(Q^{*}\right)}=\frac{1}{2}\left[k+\frac{1}{k}\right]\)

∴ Percentage increase,

\(\frac{T I C(Q)-T I C\left(Q^{\star}\right)}{T I C\left(Q^{*}\right)} × 100=\left(\frac{T I C(Q)}{T I C\left(Q^{*}\right)}-1\right) × 100\)

= \(\left(\frac{1}{2}\left[k+\frac{1}{k}\right]-1\right) × 100=\left(\frac{1}{2}\left[1.25+\frac{1}{1.25}\right]-1\right) × 100\)

= 0.025 × 100

= 2.5 %

Explanation:

Gear train :

Given:

T₂ = 19, T₃ = 24, N₄ = 0, N₄ = 20 rpm (CCW)

As R₂ + 2R₃ = R₄

⇒ T₂ + 2T₃ = T₄

⇒ T₄ = 19 + 2 ×  24 = 67

 

N₂ = y + x ⇒ x + y = 0 ...(i)

and \(N_{4}=20 ~\mathrm{rpm} \)

\(\Rightarrow y-x \times \frac{19}{67}=20\) ...(ii)

On solving equations (i) and (ii);

x = -15.58 rpm = 15.58 rpm (CW)

y = 15.58 (CCW)

Explanation:

\(t \frac{d x}{d t}+t-x=0\)

\(\frac{d x}{d t}-\frac{x}{t}+1=0\)

\(\frac{d x}{d t}-\frac{1}{t} x=-1\)

This is a linear differential equation of the form,

\(\frac{d x}{d t}+P x=Q\)

where, \(P=-\frac{1}{t} \text { and } Q=-1\)

Integrating factor, \(I F=e^{\int-\frac{1}{t} d x}=e^{-\ln t}=e^{\ln \left(\frac{1}{t}\right)}=\frac{1}{t}\)

Thus, the solutions is

\(x × I F=\int Q \cdot I F d t+C\)

\(x \cdot \frac{1}{t}=\int-1 \cdot \frac{1}{t} d t+c\)

\(\frac{x}{t}=-\ln t+c \Rightarrow \frac{x}{t}=\ln \left(\frac{1}{t}\right)+c\)

\(\frac{x}{t}=\ln \frac{1}{t}+c \Rightarrow x=t \cdot \ln \frac{1}{t}+c t\)

Given, at t = 1, x = 0

⇒ 0 = 0 + c ⇒ c = 0

Now, for t = 2,

\(x=2 \ln \frac{1}{2} ; \quad x=-2 \ln 2\)

x = -2 × 0.693 = -1.386 ≃ -1.39

Explanation:

Let A_p be the area of the piston.

Given : P₁ = 140 KPa

T₁ = 350°C = 350 + 273 = 623 K

T₂ = 25°C = 25+ 273 298 K

W_1-s = P₁V₁ - P_sV_s = mR(T₁ - T_s)

For constant pressure process:

V ∝ T

⇒ \(\frac{V_{1}}{V_{s}}=\frac{T_{1}}{T_{s}}\)

⇒ \(\frac{A_{p} × 3}{A_{p} × 2}=\frac{623}{T_{s}} \Rightarrow T_{s}=415.33 \mathrm{~K}\)

W_1-s = 1 × 0.287 × (623 - 415.33) = 59.60 kJ/kg

W_s-2 = 0 [∵ Constant volume process]

∴ W_1-2 = W_1-s + W_s-2 = 59.60 + 0 = 59.60 kJ/kg

Explanation:

Given:

Cube shaped mould = 20 mm × 20 mm × 20 mm

Sphere-shaped mould, D = 20 mm

By Chvorinov's rule, solidification time (t) is proportional to square of ratio of volume to surface area i.e..

\(t_{s} \propto\left(\frac{V}{A}\right)^{2}\)

\(\frac{\left(t_{s}\right)_{c}}{\left(t_{s}\right)_{s p}}=\frac{(V / A)_{c}^{2}}{(V / A)_{s p}^{2}}=\frac{\left(a^{3} / 6 a^{2}\right)^{2}}{\left(\frac{\pi D^{3}}{\frac{6}{\pi D^{2}}}\right)^{2}}\)

= \(\frac{(a / 6)^{2}}{(D / 6)^{2}}=1 \quad[\because a=D=20 \mathrm{~mm}]\)

Explanation:

z = -x1 - 2x2

x3 = 2 + 2x1 = x2 ...(i)

x4 = 7 + x1 = 2x2 ...(ii)

x5 = 3 - x1 ...(iii)

x1, x2, x3, x4, x5 ≥ 0 ...(iv)

From equation (i);

x2 = 2 + 2x1 - x3

On substituting this value in the objective function,

z = -x1 - 2(2 + 2x1 - x3)

⇒ z = -4 - 5x1 + 2x3

Hence option (2) is correct.

Also from equation (iii);

x1 = 3 - x5

On substituting this value in the objective function,

z = -(3 - x5) - 2x2

z = −3 + x5 - 2x2

Hence option (4) is correct.

As per the given options, (2) and (4) both are correct.

Explanation:

 

 

Initially, taking moment about hinged point C,

ΣΜ_C = 0

A_y × 12 - 100 × 9 - 50 × 3 = 0

⇒ A_y = 87.5 KN

FBD for left half section:

Now taking moment about point B.

ΣM_B = 0

A_x × 6 - 87.5 × 6 + 100 × 3 = 0

⇒ Ax = 37.5 kN

ΣΕ_x = 0

A_x - B_x = 0

⇒ B_x = A_x = 37.5 kN

Explanation:

Given: m = 10 kg; k = 1.5 kN/m = 1500 N/m; ω_n = ?

At time t = t;

Using D'Alembert's principle;

mx + 4kx + 2kx = 0

mx + 6kx = 0

⇒ \(\ddot{x}+\left(\frac{6 \mathrm{k}}{\mathrm{~m}}\right) x=0 ;\)

So, \(\omega_{n}=\sqrt{\frac{6 \mathrm{k}}{\mathrm{~m}}}=\sqrt{\frac{6 \times 1500}{10}}\)

= 30 rad/s

Explanation:

Given:

P = 0.2 MPa, D = 2 m, t = 10 mm

\(\sigma_{x}=\sigma_{L}=\frac{P D}{4 t}=\frac{0.2 \times 2000}{4 \times 10}=10 \mathrm{MPa}\)

\(\sigma_{y}=\sigma_{h}=\frac{P D}{2 t}=\frac{0.2 \times 2000}{2 \times 10}=20 \mathrm{MPa}\)

\(\text { Normal stress, }\left(\sigma_{n}\right)_{\theta}=\left(\frac{\sigma_{x}+\sigma_{y}}{2}\right)+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta+\tau_{x y} \sin 2 \theta\)

\(\left(\sigma_{n}\right)_{\theta=30^{\circ}}=\left(\frac{10+20}{2}\right)+\left(\frac{10-20}{2}\right) \cos \left(2 \times 30^{\circ}\right)=12.5 \mathrm{MPa}\)

Explanation:

Dimension of surface tension = [MT^-2]

Dimension of radius = [L]

Dimension of dynamic viscosity = [ML^-1T^-1]

Dimension of time = [T]

Now,

L = \(k \gamma^{a} R^{b} \mu^{c} \sqrt{t}\)

⇒ [L] = [M⁰L⁰T⁰][MT^-2]^a[L]^b [ML^-1T^-1]^c [T]^1/2

⇒ a + c = 0 ...(i)

⇒ b - c = 1 ... (ii)

\(-2 a-c+\frac{1}{2}=0\) ...(iii)

On solving equation (i), (ii) and (iii), we get

a = 0.5, b = 0.5 and c = -0.5

Explanation:

Given:

D = 50 mm; S_yt = 300 MPa

Axial stress due to P, σ_x = σa = \(\frac{P}{A}=\frac{100 \times 10^{3}}{\left(\frac{\pi \times 50^{2}}{4}\right)}=50.93 \mathrm{MPa}\)

Shear stress due to T, τ_xy = (τ_t)_max = \(\frac{16 \times T}{\pi D^{3}}=\frac{16 \times 600 \times 10^{3}}{\pi \times 50^{3}}=24.446 \mathrm{MPa}\)

σ_y = 0

\(\left(\sigma_{t}\right)_{\text {per }} \geq \sqrt{\sigma_{x}^{2}+3 \tau_{x y}^{2}}\)

⇒ \(N \leq \frac{300}{66.232}\)

⇒ N ≤ 4.529

⇒ N = 4.5

Explanation:

Given:

V = 30 V, I = 200 Amp, A_c = 20 mm2, v = 5 mm/sec

Assuming, H_req = 20 J/s

Heat required to melt (H_m) = 20 J/s

Heat lost to surrounding = \(\left(\frac{H_{s}-H_{m}}{H_{s}}\right) \times 100\)

Heat supplied (H_s) = VI = 30 x 200 = 6000 W = 6000 J/s

Heat lost = \(\left(\frac{6000-20}{6000}\right) \times 100=99.67 %\)

Explanation:

Given:

T₁ = 300 K, T₂ = 550 K, T₄ = 1250 K, T₅ = 800 K, T₆ = 600 K,

From energy balance across regenerator

T₃ - T₂ = T₅ - T₆

T₃ - 550 = 800 - 600

Тз = 750 K

Now, 

\(\eta=1-\frac{Q_{R}}{Q_{s}}=1-\frac{\left(T_{6}-T_{1}\right)}{\left(T_{4}-T_{3}\right)}=1-\frac{(600-300)}{(1250-750)}\)

\(\eta=1-\frac{300}{500}=0.4 \text { or } 40 %\)

Explanation:

As per Hungarian method, subtracting the smallest value from each column from other elements of that column, we get

Similarly, for each row, we get the following matrix.

Since, the minimum number of lines to cover each zero of every row and column is less than the order of matrix. So to get the opportunity matrix, we will subtract smallest value from every element without lines and adding the same at intersection we will get the following matrix.

Since the minimum number of lines is equal to the order of matrix. Assignment can be done in the matrix as shown below.

So, optimal solution is

= (11 + 7 + 14 + 9) × 300

= Rs. 12300

Explanation:

For uniform internal heat generation and constant thermal conductivity in a rectangular slab,

\(\frac{\partial^{2} T}{\partial x^{2}}+\frac{q_{g}}{k}=0\)

⇒ \(T=-\frac{q_{g} x^{2}}{2 k}+C_{1} x+C_{2}\)

Temperature variation is parabolic inside the slab:

At x = -L, T₁ = 150° C

⇒ \(150=-\frac{q_{g} x^{2}}{2 k}-C_{1} L+C_{2}\) ...(i)

and at x = L, T = T2 = 110° C

\(110=-\frac{q_{g} L^{2}}{2 k}+C_{1} L+C_{2}\) ...(ii)

Using equation (i) and (ii)

\(C_{1}=\frac{T_{2}-T_{1}}{2 L}=\frac{110-150}{2 ×(0.01)}=-2000\)

At T_max; \(\frac{d T}{d x}=0\)

⇒ \(\frac{d T}{d x}=-\frac{q_{g} x}{k}+C_{1}\)

⇒ \(0=-\frac{\left(100 × 10^{6}\right) x}{200}-2000\)

⇒ x = -4 × 10^-3 = -4 mm

Distance of T_max left face = L - |x| = 10 - 4 = 6 mm

Explanation:

After normal shock:

• Stagnation temperature remains the same.

• Density after shock will increase, i.e., ρE < ρF

• Mach number will decrease, i.e., ME > MF.

• Entropy across a shockwave always increases as it is a highly irreversible process.

Explanation:

Diameter of blank = 20 mm

Sheet thickness, t = 2 mm

Clearance allowance, A_c = 0.04

∴ Clearance, C = A_ct = 0.04 × 2 = 0.08

Blank size = Die size = 20 mm

Punch size = Die size - 2 × C = 202 × 0.08 = 19.84 mm

Explanation:

Given:

μ = 0.3, T₁ = 1.5 KN

Taking moment about hinge point O.

ΣΜ0 = 0

⇒ F × 1000 - T2 × 200 = 0

⇒ \(\rm F=\frac{T_{2} \times 200}{1000} \quad ...(i)\)

Also, \(\frac{T_{1}}{T_{2}}=e^{\mu \theta} \Rightarrow \frac{1.5}{T_{2}}=e^{0.3 \times \pi}\)

⇒ \(T_{2}=\frac{1.5 \times 1000}{e^{0.3 \times \pi}}=584.49 \mathrm{~N}\)

From equation (i), we have

\(F=\frac{200 \times 584.49}{1000}=116.89 \mathrm{~N} \simeq 117 \mathrm{~N}\)

Explanation:

Reactions at the ends,

\(R_{A}+R_{B}=\frac{1}{2} \times 1.2 \times 100\)

∴ R_A + R_B = 60 N ... (i)

ΣΜ_Β = 0

∴ \(R_{A} \times 1.2=\left(\frac{1}{2} \times 1.2 \times 100\right) \times \frac{1.2}{3}\)

∴ R_A = 20 N

∴ R_B = 40 N

Now, SF at section 'x-x' at a distance of x

SF_x = 0

∴ \(20-\frac{1}{2} \times x \times \frac{100 x}{1.2}=0\)

∴ x = 0.693 m

∴ At x 0.693, BM will be maximum,

∴ \(M_{\max }=20 \times 0.693-\frac{1}{2} \times \frac{0.693 \times 100 \times 0.693}{1.2} \times \frac{0.693}{3}\)

∴ M_max = 9.237 N-m ≃ 9.2 N-m

Explanation:

At the instant when AP is horizontal and OP is vertical, V_P will come in the line of AP, then the perpendicular component of V_P will be zero.

⇒ V_P = (OP) × ω_OP = (OP) × 7 m/s; (direction is along AP)

Hence, ω_ABC = 0;

No perpendicular component of velocity

Explanation:

Given: \(\dot{m}_h\) = 430 kg/s, LH = 2.326 MJ/kg

Effectiveness, \(\epsilon=\frac{q}{q_{\max }}\)

Now,

q = \(\dot{m}_h\) x LH = 430 × 2.326 × 106 = 1000.18 MW

and q_max = C_min × (T_hi - T_ci)

⇒ qmax = 20000 × 1 × 4000 (50 - 30)

⇒ qmax = 1600 MW

∴ \(\epsilon=\frac{1000.18}{1600}=0.625\)

Explanation:

Given:

V₁ = 60 m/min; V₂ = 100 m/min; V₃ = 80 m/min

T₁ = 60 min; T₂ = 10 min; T₃ = ?

By using Taylor's equation,

\(\rm V_{1} T_{1}^{n}=V_{2} T_{2}^{n}\)

⇒ 60 × (60)ⁿ = 100 × (10)ⁿ

⇒ \(\left(\frac{60}{10}\right)^{n}=\frac{100}{60} \)

\(\Rightarrow n=\frac{\ln \left(\frac{100}{60}\right)}{\ln \left(\frac{60}{10}\right)}=0.285\)

Also, \(\rm V_{1} T_{1}{ }^{n}=V_{3} T_{3}{ }^{n}\)

⇒ 60 × (60)^0.285 = 80 × (T₃)^0.285

⇒ T₃ = 21.87 min

Explanation:

 \(\int_{x=3}^{x=4} \int_{y=1}^{y=2} \frac{d x d y}{(x+y)^{2}}\)

= \(\int_{x=3}^{x=4}-\left[\frac{1}{(x+y)} d x\right]_{y=1}^{y=2}\)

\(=-\int_{x=3}^{x=4}\left[\frac{1}{2+x}-\frac{1}{x+1}\right]\)

= \(-[\ln (2+x)-\ln (x+1)]_{x=3}^{x=4}\)

= \(-\left[\ln \left(\frac{2+x}{x+1}\right)\right]_{3}^{4}\)

\(=-\left[\ln \left(\frac{6}{5}\right)-\ln \left(\frac{5}{4}\right)\right]\)

= \(-\left[\ln \left(\frac{6 \times 4}{5 \times 5}\right)\right]\)

\(=-\ln \left(\frac{24}{25}\right)\)

\(=\ln \left(\frac{25}{24}\right)\)

Thus, \(\log _{e}\left(\frac{a}{24}\right)=\ln \left(\frac{25}{24}\right)\)

∴ a = 25

Explanation:

P_N - P_J = P_E - P_J

⇒ \(\frac{P_{N}-P_{J}}{\rho g}=\frac{P_{E}-P_{J}}{\rho g}\)

⇒ \(\left(h_{f}\right)_{N J}=\left(h_{f}\right)_{E J}\)

⇒ \(\frac{32 \mu v_{1} L}{\rho g d}=\frac{32 \mu v_{2}(10 L)}{\rho g d}\)

⇒ v₁ = 10v₂

Using continuity equation,

Q = 2Q₁ + Q₂

⇒ \(\frac{\pi}{4} d^{2} v=\frac{\pi}{4} d^{2}\left[2 v_{1}+v_{2}\right]\) 

⇒ v = 2v₁ + v₂

Using equation (i) and (ii)

\(v_{2}=\frac{1}{21} v \text { and } v_{1}=10 v_{2}=\frac{10}{21} v\) 

Discharge, \(Q_{2}=\frac{\pi}{4} d^{2} v_{2}=\frac{\pi}{4} d^{2}\left(\frac{1}{21} v\right)=\left(\frac{\pi}{4} d^{2} v\right) \frac{1}{21}=\frac{Q}{21}\)

Similarly, \(Q_{1}=\frac{\pi}{4} d^{2} v_{1}=\frac{\pi}{4} d^{2}\times\left(\frac{10}{21} v\right)=\left(\frac{\pi}{4} d^{2} v\right)\times \frac{10}{21}=\frac{10 Q}{21}\)

Now, clearly \(Q_{2}=\frac{Q}{21}\)

Hence, statement I is correct.

and (Pressure)_N = (Pressure)_E

⇒ P_N - P_E

⇒ P_J - P_N = P_J - P_E

Hence, the pressure difference between J and N is equal to the pressure difference between J and E.

So, statement II is true.

Explanation:

Given: T₁ = 300 K, T₂ = 1200 K, D = 6 m, F₁₁ = 0, F₁₂ = 1, ∈₁ = 1, ∈₂ = 0.7

А₁ = πR³ = π × (3)² = 9π m²

А₂ = 2πR² = 2π × (3)² = 18π m²

Heat transfer, \(q=\frac{\sigma\left(T_{2}^{4}-T_{1}^{4}\right)}{\frac{1-\epsilon_{1}}{\epsilon_{1} A_{1}}+\frac{1}{A_{1} F_{12}}+\frac{1-\epsilon_{2}}{\epsilon_{2} A_{2}}}\)

⇒ q = \(\frac{\left(5.67 × 10^{-8}\right) ×\left(1200^{4}-300^{4}\right)}{\frac{1}{9 \pi × 1}+\frac{1-0.7}{0.7 × 18 \pi}}\)

⇒ q = \(\text { 2726966.198 W } \quad\left[\because \frac{1-\epsilon_{1}}{\epsilon_{1} A_{1}}=0\right]\)

⇒ q = 2726.966198 kW

⇒ q ≃ 2726.97 kW