MET Mock Test - 1 Free Online Test 2026

Let, v be the velocity of a wave.
Length of open pipe A be l₁.
 Length of closed pipe B be l₂.

Formula used:

nth overtone of an open pipe f_n = (n+1)f₀, where for open pipe fundamental frequency f₀ = v/2l
nth overtone of a closed pipe f_n = (2n+1)f₀, where for closed pipe fundamental frequency f₀ = v/4l.

Now, solving the problem

Since, it is given that frequencies are same,

Now, the ratio of fundamental frequencies,

When the body's kinetic energy approaches zero, it reaches its maximum height.

Equating maximum height and kinetic energy 

Torque is a measure of how much a force acting on an object causes that object to rotate. The object rotates about an axis (O). The distance from O is r,
where forces acts, hence torque τ=r x F. It is a vector quantity and points from axis of rotation to the point where the force acts.

0² = v² – 2µ_kgS_rel

A

Diazotisation: The aromatic -NH2 group is converted into the corresponding diazonium salt by NaNO2/HCl at low temperature.

On treatment with N,N-dimethylaniline the diazonium salt undergoes an azo coupling reaction. The diazonium species behaves as an electrophile and couples at the para position of the activated dimethylaniline ring (activation by -N(CH₃)₂).

The -SO3H group on the diazonium-bearing ring remains intact (it is deactivating/meta-directing but does not participate in the coupling), so the major product is the para-azo product linking the two rings by -N=N-.

Therefore the major product is the azo dye with structure HO₃S-Ph-N=N-Ph-N(CH₃)₂, which corresponds to option A.

(A) Both LiCl and MgCl₂ are covalent in nature due to high polarising power of Li⁺ and Mg⁺² ions. Hence, they are soluble in ethanol.
(A) Oxides of Li₂O and MgO do not form superoxide.
(B) LiF is least soluble among all other alkali metal fluorides due to high lattice energy of LiF.
(C) Li₂O is least soluble among all other alkali metal oxides.
Hence, statements (A) and (C) are correct.

Hence, the most appropriate option is (a).

Vapour pressure of a liquid at a given temperature is inversely proportional to intermolecular forces of attraction.
In order to attain the same vapour pressure, 'X' has to be heated the least followed by 'Y' and 'Z'.
Hence, the increasing order of the strength of intermolecular forces is: X < Y < Z.
Therefore, 'X' has lower intermolecular interactions compared to 'Y' and statement (B) is correct.

In the Balmer series of H-atom, electronic transitions take place from higher orbits to 2nd orbit, and the longest wavelength will correspond to the transition from 3rd orbit to 2nd orbit.
∴ n₁ = 2 and n₂ = 3 for longest wavelength.
As wavelength decreases, the lines in the Balmer series converge. The correct statements are (I), (II) and (III).
The ionisation energy of hydrogen can be calculated from wave number of the last line of the Lyman series:
n₁ = 1 and n₂ = ∞
Hence, statement IV is incorrect.

Hence, no bridged oxygen atom is present in N₂O₄.

A compound is considered unstable at room temperature if it is either antiaromatic (cyclic, planar, fully conjugated with 4n π electrons) or so reactive that it cannot be isolated under normal conditions (for example, it rapidly dimerizes).

(i) Cyclobutadiene has 4 π electrons in a planar, cyclic, fully conjugated system. It is a classic antiaromatic compound and is not stable at room temperature.

(ii) The second compound is a fulvene-type structure. Although it appears conjugated, the conjugation is cross-conjugation, not continuous cyclic conjugation. Therefore, it is neither aromatic nor antiaromatic and exists at room temperature. It is considered stable for exam purposes.

(iii) Cyclopentene is a simple alkene with no special conjugation or instability. It is stable.

(iv) Cyclopentadienone has a cyclic conjugated system with 4 π electrons in the ring. It shows antiaromatic character and is highly unstable, readily undergoing dimerization. It is not isolable at room temperature.

(v) Benzene is aromatic with 6 π electrons and is very stable.

(vi) Naphthalene is also aromatic and stable at room temperature.

(vii) Borole contains 4 π electrons and is often described as antiaromatic. Borole is highly unstable due to electron deficiency at boron and antiaromatic character, and it cannot be handled as a stable compound at room temperature.

Therefore, the unstable compounds are (i), (iv), and (vii).

Final answer: 3

The reaction between aluminum (Al) and oxygen (O₂) to form aluminum oxide (Al₂O₃) is:

4 Al + 3 O₂ → 2 Al₂O₃

We first calculate the moles of aluminum (Al) used in the reaction.

• Molar mass of aluminum (Al) = 27 g/mol

• Moles of Al used = 27 g / 27 g/mol = 1 mol of Al

From the stoichiometry of the reaction, 4 moles of Al produce 2 moles of Al₂O₃. Therefore:

• Moles of Al₂O₃ produced = (2 / 4) × 1 mol of Al = 0.5 mol of Al₂O₃

Now, calculate the mass of Al₂O₃ produced using its molar mass:

• Molar mass of Al₂O₃ = (2 × 27) + (3 × 16) = 54 + 48 = 102 g/mol

• Theoretical yield of Al₂O₃ = 0.5 mol × 102 g/mol = 51 g

The actual yield of Al₂O₃ is given as 4.95 g. The formula for percentage yield is:

Percentage Yield = (Actual Yield / Theoretical Yield) × 100

Substitute the values:

Percentage Yield = (4.95 g / 51 g) × 100 ≈ 9.7%

The percentage yield of the reaction is approximately 9.7%.

Putting n = 1 in P(n)

= 72 + 20 ⋅ 30 = 49 + 1 = 50   ...(i)

Which is divisible by 25
 be divisible by 25 for some k ∈ N

= a multiple of 25 as P(k) is true

Hence by PMI P(n) is divisible by  25 ∀ n ∈ N

Let C1: (x − 1)² + (y − 1)² = 2

C2: (x + 3)² + (y − 4)² = 52
Both the circle pass through the origin.

Hence, tangents at the origin (using T = 0) to C₁ and C₂ are x + y = 0 and 3x − 4y = 0, respectively.

The slope of the tangents are −1 and 3/4 respectively.

Thus, if  −1 < m < 3/4, then origin divides AB internally.

Arrange the letter of the word COCHIN as in the order of dictionary CCHINO
Which number of words with the two C’s occupying first and second place = 4!
Number of words starting with CH, CI, CN is each
∴ Total number of ways = 4! + 4! + 4! + 4! = 96
There are 96 words before COCHIN

1. Identify the foci of the hyperbola:

   • The hyperbola equation is x²/144 - y²/81 = 1.

   • Rewriting in standard form by multiplying both sides by 25: (25x²/144) - (25y²/81) = 25.

   • Here, a² = 144/25 and b² = 81/25.

   • The distance to the foci for a hyperbola is c² = a² + b²: c² = (144/25) + (81/25) = 225/25 = 9 ⇒ c = 3.

   • Thus, the foci of the hyperbola are at (±3, 0).

2. Determine the foci of the ellipse:

   • The ellipse equation is x²/25 + y²/25 = 1.

   • Since the foci coincide with those of the hyperbola, they are at (±3, 0).

   • For an ellipse with major axis along the x-axis, c² = a² - b², where a² = 25: 3² = 25 - b² ⇒ 9 = 25 - b² ⇒ b² = 16.

(A) Write Owing to in place of Due to in section (A). Due to is generally used after the verb. It means, caused by somebody/something.
The correct sentence will be :
"Owing to carelessness he failed in the examination."
However, modern grammarians do not object to using due to and owing to in the same sense.

In the given words, both are similar in meaning. Ovation means a warm reaction given by the audience when they like something. Applause means clapping by the audience as an appreciation.
The words in option C are opposites of each other. Triumph means to gain victory over something. Failure means failing to gain success.
Hence, option C is the correct answer.

Answer: Y

Two meaningful English words that use each letter exactly once are GASP and GAPS.

Both of these are valid words, so more than one meaningful word can be formed. Therefore the correct choice is Y (option B).

The central theme behind the sentences is the benefits of the invention of computers. So, Q should follow 1 because it makes it clear that where has the advancement taken place.

The next one will be P, which states the advancement that has been done.

Sentence S should follow it next as it introduces both sentences R and 6 which are about the uses of computers.

Hence, QPSR is the correct sequence.

As per the given figure,
1, 2, 3, 4 is represented by rich persons.
3 is represented by researchers.
2, 3, 4, 5, 7, 8 is represented by surgeons.
4, 6, 7 is represented by engineers.

So, region 2 represents some surgeons who are neither engineers nor researchers.  

Hence, 2 is the correct answer.