JEE Main Mock Test - 3 - JEE MCQ

From kinetic theory of gases all the collisions between walls and molecules and molecules and molecules considered to be elastic.
So conservation of momentum and kinetic energy is applicable.
For elastic collision, coefficient of restitution is one.

Let the value of 'a' be increased from zero.
As long as a ≤ μg, there shall be no relative motion between m₁ or m₂ and platform, that is, m₁ and m₂ shall move with acceleration a.
As a > μg the acceleration of m₁ and m₂ shall become μg each. Hence at all instants the velocity of m₁ and m₂ shall be same
∴ The spring shall always remain in natural length.

Net force on free charged particle,

Since d >>> x, so neglecting x in denominator.

Given, AB = 2BC = 2 m

∴ BC = 1 m, σ be the mass per unit area.m₁ = m₂ = m₃ = (1 × 0.5)σ = 0.50σ.
 be the position of centre of mass, then, 

The magnetic dipole moment of a diamagnetic material is zero as each of its pair of electrons has opposite spins, i.e., μ_d = 0. Paramagnetic substances have dipole moment > 0, i.e., μp ≠ 0, because of the excess of electrons in its molecules spinning in the same direction.
Ferromagnetic substances are very strong magnets and they also have a permanent magnetic moment, i.e., μ_f ≠ 0.

Stationary waves are produced by two identical waves travelling in opposite direction in same plane.
∴ for stationary wave
Z₁ = A cos(kx − ωt)
Z₂ = A cos(kx + ωt)

The phase of a particle executing SHM is defined as the state of a particle as regard to its position and direction of motion at any instant of time. In the given curve, phase is same when t = 1 s and t = 5 s. Also phase is same when t = 2 s and t = 6 s.

Infrared - Muscular treatment
Radio wave - Broadcasting
X Rays - To detect fracture
Uv Rays → Absorbed ozone layer

Evidently, the entire cycle can be visualised as made up of two cycles, i.e., cycle ABEA and cycle BCDB.
W₁ = area of the loop ABEA in the P-V diagram

Work done in the cycle BCDB, by the gas
W₂ = –(area of loop BCDB)
Now,evidently, triangles ABE and BCD are similar, the corresponding angles being equal
(viz., ∠EBA =∠DBC =∠EAB =∠BCD, etc.)

∴ W₂ = –0.33 kJ
∴ Total work done by the gas
ΔW = W₁ + W₂ = 3 kJ – 0.33 kJ = 2.67 kJ

Let the freq. of tuning fork be f
Earlier the freq. of the string is f−4 & later it is equal to f

In general, the current is,
i = neAv_d
Current density is,
J = i/A = nev_d 
For electrons, the current density is,

The total current density for the material is,

The halides of B and Al are hydrolysed.

2H₂O + 2e⁻ → H₂ + 2OH⁻
For 0.01 mole H2 0.02 mole of electrons are consumed
charge required = 0.02 × 96500 C = i × t

2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂
i.e. can be prepared by heating heavy metal nitrate of Pb(NO₃)₂. It is a red brown gas but on cooling below 273 K converts to colourless liquid and retains as dimer. In NO₂ molecules 17 valence e−are there (12 that of O & 5 that of N) i.e. one e−unpaired results to paramagnetic nature. So this is an odd e⁻molecule on being dimer with even e⁻becomes stable & loses paramagnetic characteristics.

Let I = ∫ (2x¹² + 5x⁹) / (x⁵ + x³ + 1)³ dx

= ∫ (2/x³ + 5/x⁶) / (1 + 1/x² + 1/x⁵)³ dx

Put 1 + 1/x² + 1/x⁵ = t

⇒ (-2/x³ - 5/x⁶) dx = dt

Then, I = -∫ dt / t³ = 1 / 2t² + c

= 1 / 2(1 + 1/x² + 1/x⁵)² + c

= x¹⁰ / 2(x⁵ + x³ + 1)² + c

Given, points P and Q are on the ellipse defined by 9x² + 4y² = 36, which simplifies to (x² / 4) + (y² / 9) = 1. This is the standard form of the equation of an ellipse centered at the origin, with semi-major axis a = 3 along the y-axis and semi-minor axis b = 2 along the x-axis.

OP is the distance from the origin O to point P, which is given by:

OP = r₁ = √((2√3 / √7)² + (6 / √7)²) = √(12/7 + 36/7) = √(48/7) = 2√(12/7).

1. Representing the given point P (2√3 / √7, 6 / √7) in polar coordinates:
   P = (r₁ cosθ, r₁ sinθ)

   Substituting into the ellipse equation:
   (r₁² cos²θ) / 4 + (r₁² sin²θ) / 9 = 1

   Simplifying:
   (cos²θ) / 4 + (sin²θ) / 9 = 7/48 ...(equation 1)

2. Representing R as (-r₂ sinθ, r₂ cosθ) (since RS is perpendicular to PQ):

   Substituting into the ellipse equation:
   (r₂² sin²θ) / 4 + (r₂² cos²θ) / 9 = 1

   Simplifying:
   (sin²θ) / 4 + (cos²θ) / 9 = 1 / r₂² ...(equation 2)

3. From equations (1) and (2),

   1 / r₁² + 1 / r₂² = 1/4 + 1/9 = 13/144

4. Since PQ and RS are perpendicular and pass through the origin,
   1 / PQ² + 1 / RS² = (1 / r₁² + 1 / r₂²)

5. Substituting values of r₁ and r₂,
   1 / PQ² + 1 / RS² = 13 / 144 = p / q.

Let the equation of the parabola be y² = 4ax.

Point P be (at², 2at) and the point T be (h, k).

Equation of tangent at P is
ty = x + at².

It passes through T(h, k):
tk = h + at² .....(i)

Slope of SP = (2at - 0) / (at² - a) = 2t / (t² - 1).

TL is perpendicular to SP.
Then equation of TL is
2ty + (t² - 1)x - 2kt - (t² - 1)h = 0 .....(ii)

SL = perpendicular distance of S(a, 0) from (ii).

SL = (a + h) ........(iii)

Equation of directrix is x = -a ........(iv)

TN = perpendicular distance of T(h, k) from (iv)
TN = (h(1) + k(0) + a) / √(1² + 0²)

TN = h + a ........(v)

From (iii) and (v)
SL = TN

f(x) = e^{{4{x}+3}=e{4(x}}=e{4x}}
because {4{x} + 3} = {4{x}} and 4{x} = 4x − 4[x]⟹{4{x}} = {4x}

= 7/2

p = 6 / 36 = 1 / 6

q = (⁶C₁ × ⁵C₁ × 4! / 3!) / 6⁴ = 120 / 1296 = 5 / 54

p / q = (1 / 6) / (5 / 54) = (54 / 6 × 5) = 9 / 5 = m / n

m + n = 14

5 = (3 + 7 + x + y) / 4 ⇒ x + y = 10

Var(x) = 10 = (3² + 7² + x² + y²) / 4 - 25

140 = 49 + 9 + x² + y²
x² + y² = 82
x + y = 10

⇒ (x, y) = (9, 1)

Four numbers are 21, 9, 10, 8

Mean = 48 / 4 = 12

Given that S be the set of values of λ for which given system of equations has no solution.

Therefore for the given set of equations

The given equation simplifies as follows:

6λ(18λ² - 8) + 3(6λ - 12) + 3(4 - 18λ) = 0
⇒ 18λ³ - 14λ - 4 = 0
⇒ (λ - 1)(3λ + 1)(3λ + 2) = 0

Thus, the values of λ are:
λ = 1, -1/3, -2/3

Also for each values of  we have

which implies that, for each values of λ, the given system of equations has no solution.
Therefore S∈ {1, -1/3, -2/3} and 

= 12 (|1| + |-1/3| + |-2/3|)

= 12 (1 + 1/3 + 2/3) = 12 (6/3) = 24