GATE Engineering Sciences 2027 Sciences Mock Test - 5 Free Online Test

The year 1989 was a normal year

Days up to 15th August 1989 from 15th August 1988

= 16 + 30 + 31 + 30 + 31 + 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15

= 365 days = 52 weeks + 1 day

Therefore, Wednesday + 1 = Thursday

The word "PERPLEX" means to confuse or bewilder someone, making them uncertain or unable to understand something clearly.

Now let's analyze the options:

A: Distract – This means to divert attention or focus away from something. While it might cause confusion, it doesn't directly relate to "perplex," which involves deeper confusion.

B: Intrigue – This means to fascinate or arouse curiosity. It can lead to confusion, but it is not synonymous with "perplex."

C: Perspective – This refers to a particular attitude or way of looking at something. It is not related to confusion, so it’s not synonymous with "perplex."

D: Baffle – This means to confuse or bewilder someone, making it similar in meaning to "perplex."

Thus, D: Baffle is the correct choice, as it is most similar in meaning to "perplex." Both words imply confusion or being unable to understand something clearly.

The word "DENOUNCE" means to publicly criticize or condemn something or someone.

Now let's analyze the options:

A: Defend – This means to support or protect something or someone from criticism. It is the opposite of "denounce," which involves criticism.

B: Gather – This means to collect or assemble things. It is not related to "denounce," so it’s not the opposite, but it doesn't carry a direct contrast either.

C: Fight – This means to engage in a struggle or combat, which can involve both supporting or opposing something. It is not directly opposite to "denounce."

D: Rally – This means to gather support or come together for a cause, which could be in opposition to something that one might denounce.

Thus, A: Defend is the correct answer as it is the farthest in meaning to "denounce." Defending is the opposite of condemning or criticizing.

Female population below poverty line for Punjab = 2.1 million
Let the male population below poverty line for Punjab be x million
Then

∴ Population below poverty line for Punjab = (2.1+1.75) million = 3.85 million

Let the population above poverty line for Punjab be y million.
Since, 35% of population of Punjab is below poverty line, therefore, 65% of the total population of Punjab is above poverty line, i.e., the ratio of population below poverty line to that above poverty line for Punjab is 35 : 65.

∴ Population above poverty line for Punjab = 7.15 million
So, male population above poverty line for Punjab = (6 / 13 × 7.15) million = 3.3 million
The answer is (b).

Telescoping series,

= 
= 

As n increase , U_n  approximates to

∴ if we take  which is finite.

[(or) Hint: Take where l₁ and I₂ are indices of 'n' of the largest terms in denominator and nominator respectively of u_n^. Here 
By comparison test, ∑v_n, and ∑u_n, converge or diverge together. But  is divergent by p- series test (since p = 1/12<1)
∴ ∑u_n, is divergent.

Here is the extracted math text in simple text form:

Let f(x) = ax + b, f'(x) = a

The Newton-Raphson iteration formula is:
xₙ₊₁ = xₙ − f(xₙ) / f'(xₙ)

The first iteration is:
x₁ = x₀ − f(x₀) / f'(x₀)
= x₀ − (ax₀ + b) / a
x₁ = −b / a

Now, f(−b / a) = a(−b / a) + b = 0

∴ The function f(x) = ax + b converges to the root with one iteration.

|A - λI| = 0

|1 - λ α|
|β 2 - λ| = 0

(1 - λ)(2 - λ) - αβ = 0

λ² - 3λ + 2 - αβ = 0

λ = (3 ± √(9 - 4(2 - αβ))) / 2

λ₁λ₂ = 2 - αβ

For eigenvalues to be real,
9 - 4(2 - αβ) ≥ 0
9 - 8 + 4αβ ≥ 0
4αβ ≥ -1
αβ ≥ -1/4

For eigenvalues to be positive,
λ₁λ₂ ≥ 0
2 - αβ ≥ 0
αβ ≤ 2

So, for eigenvalues to be real and positive,
-1/4 ≤ αβ ≤ 2

Let f(x) = Ax³ + Bx² + Cx + D
Then f'(x) = 3Ax² + 2Bx + C
f''(x) = 6Ax + 2B
f'''(x) = 6A

Given f(x) + f'(x) + f''(x) + f'''(x) = x³

Ax³ + Bx² + Cx + D + 3Ax² + 2Bx + C + 6Ax + 2B + 6A = x³

On comparing both sides, we get:
A = 1, B + 3A = 0, C + 2B + 6A = 0, D + C + 2B + 6A = 0

A = 1, B = -3, C = 0, D = 0

Thus, f(x) = x³ − 3x²

Now g(x) = ∫ f(x) / x³ dx = ∫ (1 − 3/x) dx = x − 3 log x

g(e) = e − 3 log e = e − 3.

5 cards can be chosen from 30 cards in 30C5 ways.

Total number of outcomes = 30C5 Assume t₄ is 25. There are 24 cards preceding 25. t₁, t₂, t₃ can be chosen from these 24 cards in 24C3 ways. t₅ should be greater than 25.

Number of such cards = 5

Number of ways of choosing t₅ = 5C1

Favourable outcomes = 24C3 × 5C1

Required probability = (24C3 × 5C1) / 30C5

= (24 × 23 × 22 × 5 × 5!) / (30 × 29 × 28 × 27 × 26 × 6)

= 7286400 / 102604320

= 0.0710

Streamlined objects have less pressure drag as compared to blunt objects. Hence, aerofoils have less drag as compared to cylinders. Among two aerofoils, the rough aerofoil will experience more drag because turbulent boundmy layer has more velocity gradient near the surface and hence the more shear stress at surface.
- The turbulence delays the separation of boundmy layer over cylinder. The turbulent boundary layer has more momentum flux. hence it has better ability to overcome adverse pressure gradients. Therefore, it
separates late as compared to laminar boundary layer. Hence, the separated region for rough cylinder is narrow as compared to smooth cylinder. Narrow wake implies less pressure drag on rough cylinder.

The total drag on blunt object is dominated by pressure drag. Therefore,
if pressure drag is reduced the overall drag also is reduced.

The Navier-Stokes equation for incompressible flow having fluid viscosity constant is expressed as:

ρ(Dv⃗/Dt) = -∇P + ρg⃗ + μ∇²v⃗

Rotation of fluid elements is associated with wakes, boundary layers, flow through turbo machinery (fans, turbines, compressors, etc.), and flow with heat transfer. The vorticity of a fluid element cannot change except through the action of viscosity, nonuniform heating (temperature gradients), or other nonuniform phenomena. Thus if a flow originates in an irrotational region, it remains irrotational until some nonuniform process alters it.

From energy equation, 

p₁ = 0 because the pressure at the top of a reservoir is p_atm = 0 gauge.
V₁ ≈ 0 because the level of the reservoir is constant or changing very slowly.
z₁ = 100 m; z₂ = 20 m

V₂ = Q / A = 0.06 / [(π / 4)(0.2)²] = 1.910 m/s

Head loss is, h_L = (0.02(L/D)V₂²) / (2g) = [0.02(2000 / 0.2)(1.910)²] / [2(9.81)] = 37.2 m

For the same section and the same flow rate of a specified fluid ⇒ 'U' will be same for both case
∴ Re ∝ D_h

The hydraulic diameter of a square section of the same area is lower by about 11.4%
So, Re in this case will be lower by about 11.4%

Area a₀ = π/4 (0.1)² = 7.854 × 10⁻³ m²

For the Orifice Q = 0.045 = C_d × a₀ × √(2gH)
0.045 = C_d × 7.854 × 10⁻³ × √(2 × 9.81 × 2.75)
⇒ C_d = 0.78

From linear momentum equation:
F = ρQV = ρQC_v √(2gH)
310 = 1000 × 0.045 × C_v × √(2 × 9.81 × 2.75)
⇒ C_v = 0.937

Since C_d = C_c × C_v:
C_c = C_d / C_v = 0.78 / 0.937 = 0.832

v = 15 × 10⁻⁶ m²/s; U = 8 m/s
l = 1 m; b = 0.75 m

If Reₓ = 2 × 10⁵, then x is the distance from the leading edge up to which the laminar boundary layer exists:

Reₓ = (U_x) / v
2 × 10⁵ = (8 × x) / (15 × 10⁻⁶)
=> x = 0.375 m

Maximum thickness of the boundary layer is given by:

​​​​

F B D of block A

From lamies theorem

300 / sin 45° = R_A / sin 90° = P / sin 45°

=> R_A = 424.26 N

F B D of block B

ϕ = tan⁻¹(0.3) = 16.7°

424.26 / sin 16.7° = W_B / sin 28.3°

=> W_B = 700 N

It is a case of hydrostatic stress for which the maximum shear stress is zero

Tensile yield stress according to distortion energy theory is

According to Von Mises theory, shear yield stress is given by.

Given L = 7.32 m
d = 1.53 m
P = 3 kN

Section modulus z = 265470.4 mm³

Maximum bending moment

R_{B ⁼} 66.67 kN 
FBD of section CB
Point C is at the midpoint of the beam

As in the given truss,

F_CI = F_IH
Therefore F_BI = 0 (zero force member)

Therefore F_BH = 0 (also zero force member)