KIITEE Mock Test - 10 - JEE MCQ

As light moves from air to water, its speed decreases due to refractive index of water.

Fringe width, as λ decreases on entering in water, β also decreases.

λ = wavelength in medium,

D = separation between slit and screen,

d = separation between slits.

Two blocks of masses M and m are connected by a string passing over a pulley. Let T is the tension in the string and a is the downward acceleration. Applying II law of motion for the block of mass M.
T = Ma ...(i)
For block of mass m
ma = mg - T ...(ii)

From Equations, (i) and (ii), we get
ma = mg - Ma

∴ Area,

∴ Magnetic dipole moment associated with circular coil,

μ_c = il²/4π …(i)

When metal wire is bent into a square coil then side of square a = l/4

∴ Area, A = a² = l²/16

∴ Magnetic dipole moment associated with square coil, μ_c = iA = i(l²/16)

∴

Hnce, the ratio of magnetic dipole moment of circular coil and square coil is 4 : π.

For steel wire: Y = 20 × 10¹⁰ Pa; L = 1.5 m ; d = 0.3 cm = 0.3 × 10⁻²m
Therefore, area of cross-section of the steel wire,

The stretching force for the steel wire,
F = 4.0 + 6.0 = 10 kgf = 10 × 9.8 N
If l is extension in the steel wire, then

For brass wire: Y = 9.0 × 10¹⁰ Pa ; L = 1m;

The stretching force for the brass wire,

F = 6 kgf = 6 × 9.8 N

Where,
n₁ = n
n₂ = n + 1

Since, n >> 1.
Therefore, 2n + 1 ≈ 2n and (n + 1)² ≈ n².

A is isopropyl benzene (Cumene)

When photo chlorination of cumene is carried out then chlorine radical get substituted to that carbon where free radical is most stable (major product) therefore, structure of B is

For the reaction

ΔG^o for the reaction
2ZnS + 3O₂ → 2ZnO + 2SO₂
Can be obtained by adding eqn. (1), (2) and (3)
⇒ ΔG^o = 293 − 480 − 544 = −731 kJ

The compound used for gravimetric estimation of Cu (II) is Cu₂ (SCN)₂. Gravimetric analysis is concerned with the transformation of element or radical to be determined into a pure stable compound, which can be readily converted into a form suitable for weighing.

The species which has both polar and non-polar ends is surfactant. Thus, CH₃–(CH₂)₁₄–CH₂–NH₂ is not a surfactant.

Mn in [Mn(H₂O)₆]²⁺ is present as Mn²⁺(3d⁵).

sp³d² hybridisation occurs in [Mn(H₂O)₆]²⁺
Due to the presence of a weak field ligand, pairing of electrons do not take place. Hence, the number of unpaired electrons is five.

Rate = k[X][Y₂]
But the rate law expression cannot be expressed in terms of a reaction intermediate.

Rate = 
Thus, order of the reaction 

We have

∴ The volume of the parallelepiped with AB, AC, and AD on the coterminous edges

= |2 (2 + 10) + 9(7 + 4) -1(35 - 4)|

= |2 (12) + 9(11) - 1(31)|

= |(24 + 99 - 31)|

= |92| = 92 cubic units

Let (r + 1)^th term be the constant term in the expansion of .
Then, T_{r + 1} = ¹⁵C_r (x³)^{15 - r} . (-1/x²)^r is independent of x.
Or T_{r + 1} = ¹⁵C_r x^{45 - 5r} (-1)^r is independent of x.
 45 - 5r = 0  r = 9
Thus, the tenth term is independent of x and is given by:
T₁₀ = ¹⁵C₉ (-1)⁹ = – ¹⁵C₉

∴ ω³ = 1

Given a matrix A can be written as A = IA.

On apply R₁ → ω²R₁ , R₂ → ωR₂, we get

= (∼p v q) v q

= (∼p v q)

It is not a tautology since if p is true and q is false, then(∼p v q) is false.

Option (b),p → (q v p)

= ~p v (q v p)

= (∼p v p) v q

= T v q(∵ ~ p v p = T)

It is a tautology since if q is true or false then T v q must be true.

Similarity, check other options.

It is given that P (X ≥ 1) ≥ 0.9375

= 1 - P(X = 0) ≥ 0.9375

= 1 - ⁿC_o(p^o)(g)^n-o ≥ 0.9375

= 1 - (¼)ⁿ ≥ 0.9375

= 1 - 0.9375 ≥ (¼)ⁿ

= 0.0625 ≥ (¼)ⁿ

= 625/10000 ≥ (¼)ⁿ

= 1/16 ≥ (¼)ⁿ

= 16 ≤ 4^th

⇒ n = 2

dθ/dt = -k(θ - θ₀), where k is constant

⇒ (dθ/dt) + kθ = kθ₀

Which is linear differential equation in the form of

(dy/dx) + Py = Q

IF = e^∫kt = e^kt

before the required solution,

(θ)(e^kt) = ∫(e^kt x kθ₀)dt

⇒ θe^kt = e^kt θ₀ + a

⇒ θ = θ₀ + ae^-kt

Equation of plane parallel to plane (i) is x - 2y + 2z + k = 0 …(ii)

Since, distance of plane (i) from point (1, 2, 3) is 1 unit

∴

⇒

⇒

⇒ |3+ k| = 3 ⇒ 3 + k = ±3

⇒ k = 0, -6

∴ Required equation of plane are x - 2y+ 2z + 0 = 0 and x - 2y + 2z - 6 = 0

⇒ x - 2y + 2z = 0 and x - 2y + 2z - 6 = 0

|A| = 2, |B| = 4

Now, |A(adj)B| = |A||adj B| (∵|AB| = |A||B|)

|A||B|^3-1

=|A||B|² = (2)(4)² = 32

Given, c = (x - 2)a + b and d = (2x + 1)a - b are collinear

∴ c = λd

⇒ (x - 2) a + b = λ[(2x + 1)a - b]

⇒ (x -2)/(2x + 1) = 1/-1 = λ

⇒ 2x + 1 = -x + 2

⇒ 3x = 1

⇒ x = 1/3

On adding Eqs. (i) and (ii). we get

⇒ I = 7π/18

We have polar coordinates of P are (2, π/6). If O is the image of P about the X-axis

∴ Q = (2, π/6)

⇒ Q = (2, 11π/6)