SRMJEE Mock Test - 4 (Engineering) - JEE MCQ

According to conservation of linear momentum,
m₁v₁ = m₁v + m₂v₂
Here, v₁ is the velocity of the bullet before the collision, v is the velocity of the bullet after the collision and v₂ is the velocity of the block.
∴ 0.02 x 600 = 0.02v + 4v₂
Here, v₂ = = 2 m/s
∴ 0.02 x 600 = 0.02v + 4 x 2
⇒ 0.02v = 12 – 8
⇒ v = 4/0.02 = 200 m/s

Current gain of common base circuit, α = I_c/I_e
I_e = 2 x 10^-3 A
I_c = αI_e = 0.95 x 2 x 10^-3 A
I_c = 1.9 x 10^-3 A
I_b = I_e - I_c = (2.0 - 1.9) x 10^-3 = 0.1 x 10^-3 A
Or, 0.1 mA

Energy of a photon E = hc/λ
∴ Wavelength λ = hc/E

Wavelength range of X-rays is from 10^-11 m to 10^-8 m (0.1 Å to 100 Å).
Therefore, the given electromagnetic radiation belongs to the X-ray region of electromagnetic spectrum.

The net force on q at the origin is
= + = + (– ) =
Potential Energy, U =
^{Therefore, at the origin the force is minimum and potential energy is maximum.}

Potential gradient is the potential difference per unit length.
k = V/l
As V = IR and R = ρ(l/A),
So, potential gradient, 

The position of the 30^th bright fringe is given by

Hence the shift of the central fringe is

Assertion: If there is no gravitational force then there will be no acceleration, hence path will be straight line.
Reason : Other forces like electric force can also make the path of projectile parabolic.

The work function of a material is the minimum amount of energy needed to release an electron from the surface of that material, it is given by, 

= 

So, for sodium,

The resultant intensity of two periodic waves is given by:

I = I₁ + I₂ + 2√(I₁I₂) cos(δ)

Where δ is the phase difference between the waves.

For maximum intensity, δ = 2nπ; n = 0, 1, 2, ...

Therefore, for zero-order maxima, cos(δ) = 1.

I_max = I₁ + I₂ + 2√(I₁I₂)

= (√I₁ + √I₂)²

= I₁ + I₂ + 2√(I₁I₂)

For minimum intensity, δ = (2n - 1)π; n = 1, 2, ...

Therefore, for first-order minima, cos(δ) = -1.

I_min = I₁ + I₂ - 2√(I₁I₂)

= (√I₁ - √I₂)²

= I₁ + I₂ - 2√(I₁I₂)

Therefore,
I_max + I_min = (√I₁ + √I₂)² + (√I₁ - √I₂)²
= 2(I₁ + I₂)

Electrode potential of the cell must be +ve for spontaneous reaction.
Zn²⁺ → Zn; E⁰ = -0.76 V
Cu²⁺ → Cu; E⁰ = -0.34 V
Redox reaction is:

E_cell = E⁰_cathode - E⁰_anode
= -034 - (-0.76)
= +0.42 V
E_cell is positive, so the above reaction is feasible.
Hence, option (2) is the correct answer.

[IO₂F₅]^2- ion hybridisation is sp³d³ and shape is pentagonal bipyramidal.
Double bond causes more repulsion, so they would be on axial position at 180^o angle to each other, so shape is

• Mutarotation is the process by which the optical rotation of a sugar (like glucose) changes over time when it is dissolved in water. This happens as the sugar interconverts between its different anomeric forms (α and β forms), reaching an equilibrium state where both forms are present in solution, resulting in a constant optical rotation.
• Epimerisation refers to the conversion between epimers, which are sugars that differ in the configuration of only one stereocenter (except for the anomeric carbon).
• Racemisation is the conversion of a compound into a mixture of equal amounts of enantiomers (optical isomers).
• Anomerisation refers to the interconversion between the α and β anomers of a sugar, but mutarotation is the term used for the entire process of change in optical rotation during this interconversion.

Conclusion: The correct answer is: D: mutarotation.

The given information describes a straight-line plot where:

• The y-axis represents 1/(a - x), where a is the initial concentration and x is the concentration at time t.
• The x-axis represents time.
• The slope of the line is equal to the rate constant (k).
• The y-intercept is 1/a.

This setup matches the integrated rate law for a second-order reaction. For a second-order reaction, the equation is:

This is a straight-line equation of the form y = mx + b, where:

m (slope) is the rate constant k,

b (y-intercept) is 1/a.

Conclusion: This graph suggests that the reaction is second-order.
So, the correct answer is: B: 2.

p-Dichlorobenzene being more symmetrical than o-isomer fits closely in the crystal lattice and hence, greater amount of energy is needed to break the crystal lattice. Thus, p-isomer is less soluble than o-isomer.
Hence, both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

α-D-glucopyranose and β-D-fructofuranose are the monomer units of sucrose linked by glycosidic linkage as shown below:

Sucrose: α-D-glucopyranose + β-D-fructofuranose.

Zone refining:-

It is based upon fractional crystallization as the impurity prefer to stay in the melt and on solidification, only pure metal solidifies on the surface of melt.

The principle in this process is solubility of the impurity in the metal in melt state is greater than in solid state.

Tertiary amines have low boiling points due to absence of hydrogen bonding among isomeric amines. Primary amine have high boiling point due to the presence of extensive hydrogen bonding among the amines. 
The order is 1^∘ > 2^∘ >3^∘.
Boiling point ∝ force of attraction.
Primary amine has only 2 active hydrogen for Hydrogen bonding, whereas secondary amine has 1 active hydrogen and tertiary amine has 0 active hydrogen required for hydrogen bonding.

The electrophile attacks on the centre, which is electron-rich. Since benzene posses pi−electron cloud, hence it is susceptible to electrophilic attack.
If benzene is substituted, then an electron-donating group will increase the electron density on the benzene ring; hence it will  facile the electrophile attack, whereas an electron-withdrawing group, if present on the benzene ring, it will decrease the electron density on the benzene ring, hence the electrophile attack will not be easy.
−OH group is an effective electron donor (due to +M effect) as it increases the electron density by involving its lone pair in the ring via resonance.-
Cl  group shows both −I and +M effect, so it is also an electron donor, but the effect is less than the hydroxy group. Thus, the electron density in phenol is higher than chlorobenzene. Methyl group show only +I effect.
Hence, the phenol will be most readily attacked by electrophiles.

Order: Phenol > Toluene > Benzene > Chlorobenzene.

Given:

The freezing point depression ratio (ΔT_f / K_f) is 1/1000.
The formula for freezing point depression is: ΔT_f = K_f × molality.

This means:

ΔT_f = K_f / 1000.

Now, molality (m) is the number of moles of solute (glucose) per kg of solvent (water). Since we have 1 liter of water, which is approximately 1 kg, the molality will be the same as the number of moles of glucose (n).

We can use the formula:
ΔT_f = K_f × molality.
Substituting the value of ΔT_f from earlier:
K_f / 1000 = K_f × molality.
This simplifies to:
molality = 1/1000.
So, the molality of glucose in water is 1/1000 moles per kg of water.

Step 2: Calculating the weight of glucose:
We know that molality is also given by:
molality = moles of glucose (n) / mass of water (in kg).
Since the mass of water is 1 kg, this means:
molality = moles of glucose (n).
Now, we use the relationship between molality and the weight of glucose:
molality = weight of glucose (W) / (molar mass of glucose × 1000).
The molar mass of glucose is 180 g/mol, so:
1/1000 = W / (180 × 1000).
Solving for W:
W = (180 / 1000) = 0.18 g.
Conclusion: The weight of glucose added to 1 liter of water is 0.18 grams.

So, the correct answer is: D: 0.18 g.

So, ​PhLi react readily with 1 but does not add to 2.

Total number of equally likely cases = ⁹C₃ = (9 x 8 x 7)/(1 x 2 x 3) = 3 × 4 × 7 = 84
Number of favourable cases = ³C₃ + ⁴C₃ = 1 + 4 = 5
∴ Required probability = 5/84

Required number of ways = 16 + 25 + 40 + 10 + 5 = 96 ways.

Applying property in second integral,

If one root of equation x² + px + q = 0 is 2 −√3 ​
then the other root will be 2 +√3 ​
∴ equation x² − 4x + 1 = 0
⇒ p² = −4 and q = 1
⇒ p² − 4q − 12 = 0

Given,

f(x) = |x − 1| + |x − 4| + |x − 9| + ... + |x − 2500| ∀ x ∈ R

Since |x − a| is non-differentiable at x = a,

⇒ f(x) is non-differentiable at x = 1², 2², 3², ... 25², 26², ... 50².

Let g(x) = |x − a| + |x − b| + |x − c| + |x − d| (where a < b < c < d),

We know that the function g(x) will have a minimum value for x ∈ [b, c].

Since 25² and 26² lie in the middle of the numbers 1², 2², 3², ... 48², 49², 50²,

⇒ f(x) is minimum for x ∈ [25², 26²].

So, f(x) is minimum for the range [625, 676].

The equation of the hyperbola is given as:
y = (x + 9) / (x + 5) = 1 + 4 / (x + 5)

Let the tangent passing through the origin touch the hyperbola at (x₁, y₁).

The derivative dy/dx at (x₁, y₁) is:
dy/dx = -4 / (x₁ + 5)²

Now, the equation of the tangent at (x₁, y₁) is:
y - y₁ = (-4 / (x₁ + 5)²) (x - x₁) (using the point-slope form of a line)

Since the tangent passes through (0,0):
(x₁ + 5)² + 4(x₁ + 5) + 4x₁ = 0

Expanding:
x₁² + 18x₁ + 45 = 0

Solving for x₁, we get:
x₁ = -15 or x₁ = -3

So, the points of contact are:
(-3, 3) and (-15, 3/5)

The slopes of the tangents are -1 and -1/25.

Thus, the equations of the tangents are:
x + 25y = 0 or x + y = 0.

Let the length of pipe be 100 units.
After being cut, length of the longer part = 70 units and length of the smaller part = 30 units
The difference in lengths = 40 units
% of longer part more than that of the smaller part = (40/30) × 100 = (400/3)%

Let x - 1/x = 5.

Square both sides: (x - 1/x)² = 5² x² - 2 + 1/x² = 25 x² + 1/x² = 27.

Now, square x² + 1/x² to find x⁴ + 1/x⁴: (x² + 1/x²)² = 27² x⁴ + 2 + 1/x⁴ = 729 x⁴ + 1/x⁴ = 729 - 2 = 727.

Thus, the correct answer is A) 727