VITEEE PCME Mock Test - 6 - JEE MCQ

2 sin² y ≥
∵ 2 sin² y ∈ [0, 2]

Hence, 2 sin² y = , for |sin y| = 1 and sin x = 1
⇒ |sin y| = sin x

We have,
⇒ =
⇒ =
⇒ [Sinceand are non-coplanar]
⇒

When player tosses 2 fair coins,
S = {HT, TH, TT, HH}
Let X be a random variable that denotes the amount received by the player.
Then, X can take value 5, 2 and 1.
Now, P(X = 5) = 1/4, P (X = 2) = 2/4 = 1/2 and P(X = 1) = 1/4
Thus, the probability distribution of X is:

∴ Variance of X = ∑X² P(X) - [∑XP (X)]²
Now, ∑XP(X) =

Variance of X = ∑X² P(X) - (∑XP(X))²

First draw the graph of y = |x| and the line x = 2

Clearly required area is triangle as shown by the shaded region
Required area sq. units

Given that:
x + (1/x) = 2 cosθ

We need to find xⁿ + (1/xⁿ).

Using the recurrence relation:
xⁿ + (1/xⁿ) = 2 cos(nθ)

Thus, the correct answer is:

B) 2 cos(nθ).

cos2x ≤ cos x
⇒ 2cos²x − 1 ≤ cosx
(∵ cos2x = 2cos²x − 1)
⇒ 2cos²x − cosx − 1 ≤ 0
⇒ 2cos²x − 2cosx + cosx − 1 ≤ 0
⇒ 2cosx (cosx − 1) + 1 (cosx − 1) ≤ 0
⇒ (2cosx + 1)(cosx − 1) ≤ 0
⇒ -1/2 ≤ cosx ≤ 1
⇒ x ∈ [2nπ − 2π/3, 2nπ + 2π/3], where n ∈ I

Here, 


= tan t

Given,

For f(x) to be defined, we must have

x² - 3x + 2 > 0

Factorizing:
(x - 1)(x - 2) > 0

Solving for x:
x < 1 or x > 2

Thus, the domain of f is:
(-∞,1) ∪ (2,∞)

3 letters can be placed in 3 envelopes in 3! ways, whereas there is only 1 way of placing them in their right envelopes.
So, probability that all the letters are placed in the right envelopes = 1/3! 
Hence, required probability = 1 - 1/3! = 5/6

For the case when I₁ < I₂
Let the length of the third wire is
For equilibrium F₁ = F₂

For the case when I₂ < I₁

F₁ = F₂




∴ Value of x is

Hence, (4) is inconsistent with others.

According to the equation of continuity, Av = constant such that at the smaller cross-section, the speed of flow is greater.
Now, from Bernoulli's equation, K + U + P = constant, i.e., at greater kinetic energy, pressure energy has to be lesser. As we cross the junction, we move from the region of greater cross-sectional area to smaller cross-sectional area, which, as per the continuity equation, means, the speed as we move along the junction increases and hence, lateral pressure decreases.

F2 = −F4


So, wire attracts loop.

According to Stokes' law, the force on a body moving in a viscous medium is:

F = 6πηrv

Where,
η is the coefficient of viscosity,
r is the radius of the spherical body,
v is the velocity of the body.

Thus, we can infer that the force is directly proportional to the velocity. A body under acceleration by a constant force will increase its velocity, thereby increasing the viscous force and decreasing the net acceleration to a point where it becomes zero, finally achieving a constant terminal velocity.

Terminal velocities of raindrops are proportional to the square of their radii. The terminal velocity of a body is given by:

v_t = (2r²9η(ρ - σ)g)

Or, v_t ∝ 1/η

Where,
ρ is the density of the body,
σ is the density of the medium.

As we know that the internal resistance of Daniel cell is given as :-

= 2 x 1/4

= 0.5Ω

Electric field due to a charged long conducting sheet, E = σ / ε₀
So, Assertion is true but the Reason is false.

Maximum deviation occurs when there is grazing incidence.
Since the prism is an equilateral one, its angles are 60° each. Maximum deviation occurs when there is grazing incidence, that is, angle of incidence is 90°.

From Snell's law,

Let the angle of emergence be i'. Then,

sin i' = 1.5 sin 18° = 1.5 0.31 = 0.465
∴ i' = sin^–1 (0.465) = 28°

P across R = i²R =

R = r

A strong reducing agent is the one that gets oxidised readily. Cr is readily oxidised to Cr³⁺ which is more stable because of the stable configuration of half-filled t₂g orbitals.
According to the data, Cr/Cr³⁺ has the highest oxidation potential, i.e. +0.74 V. So, it is oxidised most easily; therefore, it is the strongest reducing agent.

The free energy change for any process is given as:
ΔG = ΔH − TΔS
Here, ΔH is a change in enthalpy, ∆S is a change in entropy and ∆G is a change in Gibbs free energy. For any process to be spontaneous, ΔG is negative. ∆H and  ∆S can be positive or negative, but  ∆G must be negative.

KClO₃ on reaction with concentrated HCl gives the following reaction:

2KClO₃ + 6HCl → 2KCl + 3Cl₂ + 3H₂O

Work done is actually area under curve, on comparing we get,
W_isobaric > W_isothermal > W_adiabatic > W_isochoric

LiAlH₄ does not reduce the C=C, but reduces the esters and aldehydes to alcohols.

The sentence is in present tense and  it is predicting the impact in the future of an action performed in the present. Hence the word 'will' is the most appropriate choice.

'Ka Bi Pu' stands for 'you are intelligent'.
'Ya Lo Ka Wo' stands for 'they seem very intelligent'.
'Lo Pu Le' stands for 'you can see'.
From the above three statements, codes for different words can be found:
intelligent ^_→ Ka
you ^_→ Pu
are ^_→ Bi

Total salary
= Rs. (1500 + 2600 + 3200 + 4100 + 5000 + 5200) = Rs. 21600
Total expenditure
= Rs. (200 + 500 + 100 + 300 + 600 + 200 + 150 + 700 + 150 + 250 + 650 + 125 + 200 + 800 + 150 + 100 + 750 + 175) = Rs. 6100
Ratio of total salary and expenditure = 21600/6100 = 216/61

According to the question:

Percentage of Q and R = (22 + 18)% = 40%

∴ Required percentage = (40 - 24) / 24 × 100

= 16 / 24 × 100

= 200 / 3 = 66.67%

Hence, the correct answer is 66.67%.

From statement I:
It is not possible to decide whether the line passes through the origin or not. There are an infinite number of lines that can be drawn which are perpendicular to the Y-axis, that may or may not pass through the origin.

From statement II:
It is not possible to decide whether the line passes through the origin or not. There are an infinite number of lines that can be drawn that pass through (6, 0) and may or may not pass through the origin.
By combining both statements, we determine that the line must lie on the X-axis to pass through the origin, the point (6, 0), and be perpendicular to the Y-axis.

It is given that:

Total sunrays received in 1 minute = 3600 units

From the pie chart, the amount of IR rays received in 1 minute is calculated as:

= (10/100) × 3600
= 360 units

Maximum tolerable limit of IR rays = 9720 units

The required time is calculated as:

= Maximum tolerable limit / Amount of IR received per minute
= 9720 / 360
= 27 minutes

Hence, the required answer is 27 minutes.

We assume that,
The number = a
According to the question,
a + 1/a = −12       .......(1)
Now,