MET Mock Test - 5 Free Online Test 2026

∴  F = kx

P₁ = ρgd + P₀ = 3 × 10⁵ Pa
∴ ρgd  = 2 × 10⁵ Pa
P₂ = 2ρgd + P₀
= 4 × 10⁵ + 10⁵ = 5 × 10⁵ Pa
% increase =   × 100

For 0 ≤ x ≤ 200,
v = mx + C


⇒  Straight line till x = 200
For x > 200,
v = constant
⇒ a = 0

Hence, the most appropriate option will be (1); otherwise it would be BONUS.

Let Q₁ : Heat input to first engine
Q_C : Heat rejected by first engine
Q₂ : Heat rejected by second engine
T_C : Lower temperature of first engine
W = Q₁ - Q_C = Q_C - Q₂
⇒ 2QC = Q₁ + Q₂
⇒ 2TC = T₁ + T₂
⇒ 

Mean Free Path


PV = nRT

We know |E| = |B| × |C|
where E and B are the amplitude of electric and magnetic field intensities in an electromagnetic wave.
Given |B| = 3 × 10^-8
∴ |E| = |B| × |C| = 3 × 10^-8 × 3 × 10. = 9
Pointing vector of the wave is given by 


We also have B = B₀ sin(kx + wt)
From the argument of sin, we say that the wave travels in the negative x direction.

∴ E = |E| sin(kx + wt) = 9 sin(1.6 × 10³x + 48 × 10¹⁰t)  V/m

• An AC generator converts mechanical energy into electrical energy.
• A galvanometer shows deflection when current passes through it, so it is used to show the presence of current in any wire.
• A transformer is used to step up or step down the voltage.
• A metals detector has inductor coils and uses the principle of induction and resonance in AC circuit.

kx² = mv²

From h = 1/2gt²

d = vt = 0.5 
= 1 m

Organic compound is heated with fuming nitric acid in the presence of silver nitrate in Carius method.
Lunar caustic (AgNO₃) is used as a reagent to distinguish Cl^-, Br^- and I^- respectively as follows:

Concept of Reducing Agent:
A reducing agent is a substance that loses or "donates" an electron to another substance in a redox chemical reaction. Therefore, a good reducing agent is the one that gets oxidized easily, or in other words, the one that can easily lose electrons.

Characteristics of a Good Reducing Agent:

• Electron Loss: A better reducing agent is the one that loses more electrons. This is because by losing electrons, the reducing agent gets oxidized and in turn reduces the other substance. This is the basic principle of a redox reaction.
• Reactivity: The reactivity of the metal also determines its capacity as a reducing agent. Metals that are high in the reactivity series are good reducing agents. This is because they can easily lose electrons and get oxidized.
• Stability: Metals that are less stable are better reducing agents because they can easily lose electrons to attain a stable state.

Hence, Option A is the correct answer - a better reducing agent is the one that loses more electrons.

The π-bond present in alkenes is weaker than -bond present in alkanes. This makes alkenes less stable than alkanes. Therefore, Statement I is correct.
Carbon-carbon double bond is stronger than carbon-carbon single bond because more energy is required to break 1 sigma and 1 pi bond than to break 1 sigma bond only. Therefore, Statement II is also correct.

CsCl has bcc structure.
Cs⁺ ions form simple cubic arrangement and Cl^- ions occupy cubic interstitial sites; each Cl^- ion has eight Cs⁺ ions as its nearest neighbours.
All the other chlorides of alkali metals have fcc structure.

We have, ϕ(x) = f⁻¹(x)
⇒ x = f[ϕ(x)]
On differentiating both sides w.r.t. x, we get
1 = f′[ϕ(x)] . ϕ′(x)

From Equation (i),

Differentiate both sides w.r.t. ‘x’ and then multiply by x both sides

Again differentiates w.r.t. x and put x = 1

= 20 × 21 × 2¹⁸
= 20 × 21 × 2¹⁸
=105 × 2²⁰

Given, f(x) = sinx + cosx
g(x) = x² − 1
g(f(x)) = (f(x))² − 1 = (sinx + cosx)² − 1 = sin2x
But sin 2x is invertible in a case when 

= [6 − 6 + 4] + [0 + 4 + 12] = [4] + [16] = [20]

We know,

Consider the Chairman and the Vice-Chairman as one individual.
So, instead of 15 members to be arranged, we now have only 14.
These 14 members can be arranged around a round table in 13! ways.
Also, the Chairman and the Vice-Chairman can be arranged in 2! ways.
∴ The total number of ways is 2 × 13!.
If the Chairman and the Vice-Chairman are never to be seated together, then the total number of ways is = 14! - 2 × 13! = 12 × 13!

Let (r + 1)^th term be the constant term in the expansion of .
Then, T_{r + 1} = ¹⁵C_r (x³)^{15 - r} . (-1/x²)^r is independent of x.
Or T_{r + 1} = ¹⁵C_r x^{45 - 5r} (-1)^r is independent of x.
 45 - 5r = 0  r = 9
Thus, the tenth term is independent of x and is given by:
T₁₀ = ¹⁵C₉ (-1)⁹ = – ¹⁵C₉

From question,
T = 21 = (s x 3) + 3
Or S = 6
Similarly, 6 = R x 3 + 3
Or R = 1
Or 1 = Q x 3 + 3
Or -2/3 = Q 
Similarly, -2/3= P x 3 + 3
Or = P x 3
Or P = -11/9 
So, required sum = P + Q =  =  = 

According to question:

P₁: Directrix:
x + 2y = k
x + 2y - k = 0

|7 - k| = 5
k = 2 and 12 (12 is rejected because it passes through focus)
So, equation of directrix:
x + 2y = 2
∴ Point of intersection of directrix with the axis of parabola = A(2, 0)
The image of A(2, 0) with respect to the line x + 2y = 6 is B(x₂, y₂).

Point B is the point of intersection of directrix with axes of parabola P₂.
∴ x + 2y = λ must have point  
∴ x + 2y = 10

∴ T_n = {A ∈ S; An(n + 1) = I}
∴ b must be equal to I.
∴ In this case, A² will become an identity matrix and 'a' can take any value from 1 to 100.
∴ The total number of common elements will be 100.