GATE Environmental Science Mock Test - 2 - GATE Environmental Science MCQ

Let's analyze the given observations and make a comparison of heights.

Observation II: Janvi is the shortest of all. From this observation, we can deduce that there are five people taller than Janvi.

Observation III: Divya is taller than only one person. This means that Divya is the second shortest among the group.

Observation IV: Saloni is taller than Pratik but she is not the tallest. From this observation, we can deduce that Saloni is taller than at least three people, including Pratik.

Now, let's consider the remaining three friends - Arya, Pratik, and Tirtha.

Observation I: Pratik is taller than Tirtha. This implies that Pratik is taller than one person, i.e., Tirtha. From the above deductions, we can conclude the following height order: Janvi 

So, the number of persons taller than Tirtha is 2 (Arya and Saloni).

Therefore, the number of persons shorter than Tirtha is the same as the number of persons taller than Pratik, which is 2.

Hence, the correct answer is option 'B' - Pratik.

Arya > Saloni > Pratik > Tirtha > Divya > Janvi

Encryption: Reverse the first 3 letters + Alphabet occurring before 4th letter in alphabet series + Reverse the last 3 letter
Final word = ARTHROT

From the above imaginary diagram we can conclude that both conclusion I and II are follows and hence option D is correct.

The correct answer is B. Hunters' having been given more permits to hunt does five years ago would have had an effect on the number of fawns bom in the following spring. It would take a few years to return the deer population to numbers that existed before the doe permits were issued. The argument does not provide enough information to conclude that the landowner clear cut woodlands near the deer yards or interfered with the deer's ability to find enough food. Answer A is incorrect because it is outside the scope of the argument. Salt licks cannot replace a main source of food for the deer. Answer C is incorrect because it does not address the issue of clear cutting. There is not enough information in the passage to indicate that the clear cutting occurred where the deer seek food. Answer D is incorrect because the erosion of soil into a nearby river does not address the availability of food for the deer herds. 

Eukaryotic cells are generally spherical or ovoid shape

Based on the above equations, air will have an increasing curve and water will have a decreasing curve.

There are only two types of error than can be occurred in hypothesis testing.

P = ⁿCkP^kQ^n–k
P = ³C3(1/10)³Q⁰
P = 0.001

f = 10 + 20e^–t we have to integrate this equation between 40mins to 50 mins
F = [10t +(–20)e^–t]^5/64/6
F = 10 x (5/6) – 20e^–5/6 – 10 x (4/6) + 20e^–4/6
F = 3.24 mm

a) For statement D & statement B

According to Fick's law of diffusion

It is assumed that the plume has a Gaussian distribution in both z ( vertical ) and y (horizontal ) direction

The concentration of a gas or aerosol ( < 20 μ ) calculated at ground level for a distance downward x is given by 

Where,

C is the concentration of pollutant

U is mean wind speed

σ_y is the standard deviation of horizontal plume concentration, in the crosswind direction

σ_x is the standard deviation of horizontal plume concentration, in the vertical direction

H is effective stack height (m)

X is the downward distance along the plume mean centerline from a point source (m)

Y – crosswind distance from the centerline of the plume (m)    

 From the relation, we can see that the Downwind concentration of air pollutants due to an elevated point source is not directly proportional to the wind speed.

whereas it is Inversely proportional to Concentration.

Statement D is false. Statement B is true.

b) 

Category

A – Extremely unstable

B – Moderately unstable

C – Slightly unstable

D – Neutral

E – Slightly stable

F – Moderately Stable

For Neutral (D) Velocity >  For Very unstable (A)

∴ The value of the wind profile exponent for the 'very unstable' atmosphere is smaller than the wind profile exponent for the 'neutral' atmosphere

Statement D is false whereas Statement A is true.

Characteristics of Spent Wash from Various Types of Manufacturing Processes

(A), (C) and (D) are the technically correct statements because
1. The C/N ratio = 35/1 = 35 parts of carbon for each part of Nitrogen. It is also mentioned that the optimum value for the C/N ratio is 20.

NOTE: Whenever the C/N ratio is less than the optimum, carbon source such as straw, sawdust, paper are added while if the ratio is too high, the sewage sludge, slaughterhouse waste, sometimes even blood, etc are added as a source of Nitrogen and this process of adding sewage sludge is called "Blending".  

2. In the case of Option 'C' - it is technically correct, as the C/N ratio is more than optimum so we will add nitrogen source such as sewage sludge, slaughterhouse waste, blood, etc. to reduce the C/N ratio of compost toilet, because if the nitrogen increases then the C/N ratio will correspondingly reduce [If the value of Nitrogen (N) in the denominator will increase then the value of C/N ratio as whole will decrease correspondingly] and vice versa. Hence, Option 'C' is correct and option 'B' is incorrect because if we add more carbon-inducing material, then the C/N ratio will only increase more which is not desirable.  3. In the case of Option 'D' - It is technically correct because as we've to bring down the C/N ratio to the optimum value, we'll mix it with a nitrogen-inducing material like sewage sludge which has a comparatively low C/N ratio of 3 to 6. Hence, the C/N ratio of the sewage will come down to the optimum value, and will further reduce the cost of sewage sludge treatment too. 

Just remember, If C/N ratio < Optimum - Then carbon sources are added and if C/N ratio is > Optimum - Then Nitrogen sources are added. When the C/N is > Optimum then it is not desirable to add more carbon sources as It'll only be a foolish move towards increasing the C/N value more than it is.  

Magnesium oxide scrubbing is regenerative and lime and limestone is throwaway process.

pH = 10, pOH = 4 → [OH^-] = 10^-4 mol/l = 10^-4 × 17 gm/l = 1.7 gm/l

[H⁺] = 10^-10 mol/l = 10^-10 × 1 × 10³ = 10^-7 mg/l

L₂ = L₁ - 20 log₁₀ (r₂ / r₁)

= 88 - 20 log₁₀ (34 / 17)

= 81.979 dB

Let F_x and F_y be the forces exerted by water on the bend.

∑ F_x = P₁A₁ - P₂A₂ cos 30° + F_x = ρQ(v₂cos 30° - v₁) ---- eqn. 1

∑ F_y = - P₂A₂ sin 30° + F_y = ρQ(v₂sin 30°) ---- eqn. 2

Using Bernoulli’s equation,

Eqn 1 → F_x = 1000 × 0.5 × (28.25 × cos 30° - 3.97)

= -40 × 10³ × 0.126 - 39.79 × 10³ × 0.0177 × cos 30

F_x = 4.598 kN

Eqn 2 → F_y = 1000 × 0.5 × (28.25 × sin 30°)

= -39.79 × 10³ × 0.0177 × sin 30

F_y = 6.71 kN

We have,

D'(1 - n') = D(1 - n)

0.66(1 - n') = 0.6(1 - 0.5)

n' = 0.545

Hence, upflow velocity = 18.37 mm/s