JEE Advanced Mock Test - 6 (Paper I) - JEE MCQ

Hence, electric field can calculated using Gauss’s law.

For translational equilibrium,

F cos 60^o = f

F sin 60^o + N = Mg

For rotational equilibrium,

Torque about point A = 0

Mg x 10 = N x

x = −1.5

B is perpendicular to plane of a & velocity

∴ B is along Z direction

Applying Kirchhoff's loop law equations, we get

Solving these we get,

⇒

Solving these we get,

⇒

Temperature of the gas is decreasing, therefore internal energy of the gas is decreasing.

From the given information we cannot conclude whether Q is positive or negative. Because W is positive and ΔU is negative.

Let the Intensity of the incident beam be I₀, and I be the intensity of the beam after passing through 'n' number of slabs., Then, we have

Solving for n, we get n = 4.

Putting x⁴ = N, so the bullet loses the velocity by v/N on passing through each plank.

As the total number of planks = 8, therefore
change in K.E. in passing through 8 planks is 100%.

Solving this, we get:
N = 15.5
or approximately N = 16
x⁴ = 16
x = 2

Potential energy of the system

As PE of the given system = 0

Hence the value of m = 2

At horizontal position,

…….(i)

At highest point

From above equations, we get

= 1.5 s

(A) → (p), (r)
Characteristic X-rays are produced due to transition of electrons from one energy level to another. Similarly, the lines in the hydrogen spectrum are obtained due to transition of electrons from one energy level to another.
(B) → (q)
In photoelectric effect, electrons from the metal surface are emitted upon the incidence of light of appropriate frequency. The electrons are emitted from the nucleus of an atom.

For Case I
Direction of the magnetic field due to the dipole 1 at the centre of the second dipole and the magnetic moment of the second dipole is shown as:

Magnetic field on the equatorial line is given by:

Potential energy is given by:

Hence, the potential energy is given by:

Force is given by:

For Case II
Direction of the magnetic field due to the dipole 1 at the centre of the second dipole and the magnetic moment of the second dipole is shown as:

Magnetic field on the equatorial line is given by:


Hence, the potential energy is given by:

Here we cannot use

Let BN be the magnetic field due to the first dipole at the North pole of the second dipole and BS is the magnetic field due to the first dipole at the South pole of the second dipole.
Hence, resultant force on the second dipole due to the first dipole is given by:

For Case III
Direction of the magnetic field due to the dipole 1 at the centre of the second dipole and the magnetic moment of the second dipole is shown as:

Magnetic field on the axial line is given by:

Potential energy is given by:

Hence, the potential energy is given by:

Force is given by:

And also, 'P' forms a cyclic anhydride; therefore, it is a cis isomer.

'P' is maleic acid and 'Q' is fumaric acid.

On treatment with KMnO₄ solution, both 'P' and 'Q' form tartaric acid which exists as three stereoisomers: S, T and U.

Maleic acid 'P' yields the meso isomer 'S', which is optically inactive.

Fumaric acid 'Q' yields the optically active isomer (T, U).

Hence, option (2) is correct.

AICI₃ has a 6-coordinate layer lattice with AI³⁺ occupying cubic close packed sites. Thus, the coordination number of AI in AICI₃ is 6.

pH = 14 - (-log 10^-10) - log 1 = 14 - 10 = 4.00

The relation between Boltzmann constant (k) and universal gas constant (R) is as follows:

= 1.380 x 10^-23 x 6.023 x10²³
= 8.31174
In these operations, the result must be reported with no more significant figures as are there in the measurement with the few significant figures.
Hence, the answer is 8.312, reported to 4 significant numbers as in the number with least significant numbers.
So, the number of significant digits in 8.312 is 4.

Analgesics are used to relieve pain.
Antipyretics are used to reduce fever.
Tranquilizers are used for the treatment of stress and mental diseases.
Antifertility drugs are used for birth control.

The correct match is:

Sphalerite is ZnS and the unit cell is a face centred cubic cell.
S^-2 ions form the Fcc lattice and Zn⁺² ions occupy the alternate tetrahedral voids.
The anions touch along the face diagonal 

Hence, the distance between two nearest anions 

Each anion is in contact with 12 anions.

Hence, [H⁺] formed = 0.165 M
The reaction takes place at the anode.
 will act as the buffering agent and the following reaction will take place.

Hence, pH < 2.15.
Therefore, the correct option is (3).

A.
Equation of the line passing through  A(2, -3, -1) \) and  B(8, -1, 2) .

∴ Any point on the line is P(6λ + 2, 2λ - 3, 3λ - 1)
According to the question,
PA = 14
PA² = 196
(6λ)² + (2λ)² + (3λ)² = 196
49λ² = 196
λ² = 4
λ = ±2
Required points are,
(14, 1, 5) and (-10, -7, -7

B. 
Equation of the plane containing the lines  and parallel to 

Clearly, the point (-1, -2, 0) lies on the plane.

C.

The given point is 
The equation of line in Cartesian form passing through (0, -11, 3) and (2, -3, 1) is

If Q is the foot of perpendicular from P on this line, its coordinates are,
(2k - 1, 8k - 11, -2k + 3)
Direction ratios of PQ are,
2k - 1, 8k - 11, 1 - 2k
PQ ⊥ XY
Therefore,
2(2k - 1) + 8(8k - 11) + 2(1 + 2k) = 0
72k = 88
k = ¹¹/₉
Therefore,

D.
The point is
Let P'(x, y, z) be the image of P on the line. Then PP' is perpendicular to the line and mid-point of PP' lies on line.
Therefore,
(x - 1)1 + y(2) + (z - 7)3 = 0
x + 2y + 3z = 22

 lies on the line.
Therefore,

Putting these values in the above equation, we get,

28λ = 28
λ = 1
Therefore,
P' is (1, 6, 3)

f(x) = a tan⁸x + a tan⁶x - b tan⁴x - b tan²x
when a = b = 1
Then, f(x) = tan⁸x + tan⁶x - tan⁴x - tan²x
f'(x) = sec²x . tanx [8 tan⁶x + 6tan⁴x - 4tan²x - 2]
= 2 sec²x . tan x [4 tan⁶x + 3 tan²x - 2 tan²x - 1]
f'(x) = 0
tan x = 0
x = 0
And, 4 tan⁶x + 3 tan²x - 2 tan²x - 1 = 0
let, tan²x = z
4z³ + 3z² - 2z - 1 = 0
z = 0.64
tan²x = 0.64
tan x = ± 0.8
x = ± tan^-1 (0.8)
f'(x)

f(x) = tan⁸x + tan⁶x - tan⁴x - tan²x
= tan⁶x . sec²x - tan²x . sec²x
= (tan⁶x - tan²x) sec²x.

So
from column 1. I, III, , IV
from column 2. (i), (iii)
from column 3. (a), (d)

The equation of the planes bisecting the angles between two given planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is

Option 4
In column 1,
Planes are 2x + y + 2z + 3 = 0 and 3x + 2y - 6z - 8 = 0
So, angle bisector of plane is

Since, a₁ a₂ + b₁ b2 + c₁ c₂ < 0,
Then acute angle bisector of a plane is
14x + 7y + 14z + 21 = 9x + 6y - 18z - 24
5x + y + 32z + 45 = 0
And obtuse angle bisector of a plane is
14x + 7y + 14z + 21 = -9x - 6y + 18z + 24
23x + 13y - 4z - 3 = 0
Hence, (IV) (S) (iv) is the correct combination.
But in option (4), combination is (IV) (S) (iii).