GATE Mechanical (ME) Mock Test Engineering - 5 Free Online Test 2026

The same figure appears again in the next continuous boxes; And whenever a figure appears again, it rotates by the angle of 90° ACW with respect to the previous box.

16 x 2 + 2 = 34

34 x 3 + 3 = 105

105 x 4 + 4 = 424

424 x 5 + 5 = 2125 but given 2124

∴ 2124 is wrong number in the series

CP : MP = 2 :3

Let CP = 200, MP = 300

(%) profit : (%) discount = 3 : 2

⇒ x = 8.33%

Discount = 2x = 16.66%

Signals can be made by using at a time one, two, three, four and five flags.

∴ Total number of signals that can be made = ⁵P₁ + ⁵P₂ + ⁵P₃ + ⁵P₄ + ⁵P₅

= 5 + (5 × 4) + (5 × 4 × 3) + (5 × 4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)

= 5 + 20 + 60 + 120 + 120

= 325

So, 325 signals can be made using five flags.

Convective thermal resistance, R_th = 1/hA = 1/h x 2πrL

R_th ∝ 1/r

r x R_th = constant ⇒ hyperbolic

We have


Now

Volume of cube of edge 8 cm = 512 cm³
Volume of cube of edge 2 cm = 8 cm³
Let x small cubes be made from the larger cube.
Therefore, x = 512/8 = 64
Thus, 64 cubes can be made.

The number of instantaneous centres in a mechanism is equal to the number of the possible combinations of the two links.
For n link mechanism, the number of instantaneous centre is given by

Fig: T-s curve of simple Rankine cycle
From the observation of the T-s diagram of the rankine cycle, it reveals that heat is transferred to the working fluid during process 2-2' at a relatively low temperature. This lowers the average heat addition temperature and thus the cycle efficiency.
To remove this remedy, we look for the ways to raise the temperature of the liquid leaving the pump (called the feed water) before it enters the boiler. One possibility is to transfer heat to the feed water from the expanding steam in a counter flow heat exchanger built into the turbine, that is, to use regeneration.
A practical regeneration process in steam power plant is accomplished by extracting steam from the turbine at various points. This steam is used to heat the feed water and the device where the feed water is heated by regeneration is called feed water heater. So, regeneration improves cycle efficiency by increasing the average temperature of heat addition in the boiler.

From the definition of stream function, the velocity components are

For two dimensional flow, the continuity equation is

Therefore, the given flow field is incompressible.

Though Maximum Shear Stress Theory is easier to apply but it is generally applicable for uni-axial loading. Distortion Energy Theory is the most accurate choice for ductile materials and gives best result.

As seen from the T-S diagram, the temperature is minimum at the starting of suction stroke. During suction, the piston moves from TDC to BDC. The intake valve is open throughout while the exhaust valve remains closed.

(D² + 6D + 9)y = 0

(D + 3)y = 0 (D = d/dx)

y = (C₁x + C₂)e^-3x

Now, as per the given conditions for y₁(x) and y₂ (x)

y₁(x) = e^-3x , y₂ (x) = xe^-3x

Given E = 0.01, a = 2, b = 3 

Let n be the minimum number of iterations required to achieve the desired accuracy. 

Now for bisection method, 

n ≥ log 100 /log 2 

n ≥ 6.64 

n > 6 

∴ n = 7 

Embossing is a forming or drawing operation (cold working) for producing a raised or projected design in relief on the surface of the workpiece. The operation uses matching punch and die with the impressions machined into both the surfaces. The punch and die have opposite configuration.

The aim of this operation is to force impressions into the surface or surfaces of the metal.

Gauge 1 is attached to chamber B and it is situated in chamber D (D is local atmosphere of gauge 1). It measures gauge pressure of chamber B.

• P_B = P₁(gauge) + P_D (local atm)
• 2.6 = P₁ + 2.1
• P₁ = 0.5 bar

Gauge 5 is attached to chamber C and it is situated in chamber A (A is local atmosphere of gauge 5). It measures gauge pressure of chamber C.

• P_C = P₅(gauge) + P_A(local atm)
• 1.8 = P₅+3.4
• P₅ = -1.6 bar

p₂ = 0.1MPa = 1 bar

t₂ = 120°C

h₂ = 2676.2 + 20/50(2776.4 - 2676.2)

= 2716.3 kJ/kg

∴ h₁ = h₂

∴ h₁ = 2716.3 at 30bar

If dryness fraction is

∴h₁ = h_g1 + xh_fg1

∴ The auxiliary equation is:

(m + 1)² = 0

⇒ m = -1, -1 [real and repeated]

The complementary function (C.F.):

y_C = e^-x[c₁ + c₂x]

And

∴ y_p = 0 [particular integral]

∴ The complete solution of (1) is:

y = e^-x[c₁ + c₂x]→(2)

Given that y(0) = 1

From (1) 1 = c₁ + 0 ⇒ c₁ = 1

dy/dx = -e^-x[c₁ + c₂x] + e^-x[c₂]

Given: dy/dx = -1 at x = 0

-1 = -[1 + 0] + 1[c₂] ⇒ c₂ = 0

∴ y = e^-x [1 + 0] = e^-x ⇒ y(1) = e^-1 = 0.368

II method:

Given D.E. is:

⇒ y” + 2y’ + y = 0

Using L.T. on both sides:

L[y"] + 2L[y'] + L[y] = 0

⇒ s²L[y(x)] - sy(0) - y’(0) + 2[SL[y(x)]] - y(0) + L[y] = 0

⇒ (s² + 2s + 1) L[y(x)] - s(1) - (-1) - 2(1) = 0

[∵ y(0) = 1 & y’(0) = -1]

⇒ (s² + 2s + 1) L[y(x)] = 1 + s

Applying inverse L.T.

∴ Strain energy also zero at the axis and maximum at outer surface.