UGEE SUPR Mock Test-2 - JEE MCQ

The maximum kinetic energy of photootecirons is given by KE_max = h(v-v₀) …(i)

Where, h = Planck's constant,

v = frequency of radiation

and v₀ = threshold frequency.

It can be seen from Eq.(i), that the maximum KE of the emitted photoelectron is proportional to the frequency of the radiation and is independent of the intensity of radiation, so it remains unchanged.

S=

and W = 14 J

The work done by a force in displacing a particle through a distance is given by W = F.s …(i)

Substituting the above values in Eq. (i), we get

14 =

⇒ 14 = --15 + 14+ 3a

⇒ a = 15/3 = 5

The diffraction pattern due to a single slit is shown below

2y = 2Dλ/a gives the width of central maximumDλ/a. Where, λ = wavelength of incident light.

Here, a = 2y, then from Eq.(i), we get a = 2Dλ/a ⇒ D = a²/2λ.

As, number of harmonics = number of loops = number of anti-nodes = p …(ii)

Also, number of nodes = number of anti-nodes + 1 Here, number of nodes = m

Number of anti-nodes = m - 1 from Eq. (ii)

p = m - 1

Putting this value of p in Eq. (i), we get

Let us take the moment of inertia of the system about rod R₁ then the total moment of inertia is l_T = l₁ + l₂ + l₃ …(i)

For rod R₁, l₁ = 0

For rod R₂, using perpendicular axis theorem, l₂ = ML²/3.

For rod R₃, using parallel axis theorem, l₃ = l_cm + l_{(at L)} = 0 + ML² = ML²

Now, putting the values of l₁, l₂ and l₃ in Eq. (i), we get

Applying the law of conservation of kinetic energy, KE (before collision) = KE (after collision)

⇒

Thus, the velocity of the second block after the collision is

The time period of a simple pendulum is given by

Where, I = length of the pendulum and g = acceleration due to gravity ⇒ T² ∝ l …(ii)

So. from Eqs. (i) and (i), we get

Here, n₁ = 9, n₂ = 7

⇒

Hence, the ratio of pendulum lengths l₁:l₂ = 49.81.

where, A = area of cross - sectional ⇒ F = (YA/L)x

The work done in stretching the wire is given by

From the given relation, F = y/√d

⇒ y = F√d

Substituting the above dimensions, we get [y] = [F][d]^½ = [MLT^-2][ML^-3T⁰]^½=[M^3/2L^-1/2T^-2]

So. dimensions of Mutual Inductance or Self-Inductance can be given by

⇒ [L] = [0]/[l]

The magnetic susceptibility of a substance shows how easily a substance am be magnetised and given by x_m = I/H

where, I = intensity of magnetisation,

and H = magnetic intensity of the field

= 0.03 sin(2πt - 0.01πx) m

Comparing it with general equation, we get y = asin(ωt-kx)

where, k = 2π/λ⇒λ = 200m

The phase deference between two particles is given by Δ∅ = kx = 2π/λ x x …(i)

Here, x = 25m

Substituting the values of x and λ in Eq. (i). we get Δ∅ = (2π/200) x 25 = (π/4)rad

The dimensions of given variables of SHM are as Displacement, [x] = [M⁰LT⁰] Velocity, [v] = [M⁰LT^-1] Acceleration, [a] = [M⁰LT^-2] and time period, [T] = [M⁰L⁰T]

Now, checking each option for these values.

For option (a)

As it depends on time, so change with it.

For option (b), [a][T] + 2π[v] = [M⁰LT^-2][M⁰L⁰T] + [M⁰L⁰T^-1] = [M⁰LT^-1]

It is also dependent on time and hence changes with it.

For option (c),

As it is a constant having no dimension, so it does not change with time.

For option (d), [a][T] + 4π²[d]² = [M⁰LT^-2][M⁰L⁰T] + [M⁰LT^-1]² = [LT^-1]+[L²T^-2]

As the term is dependent on time, so changes with it.

Also, it is dimensionally incorrect.

Hence, option (c) is correct.

Structure of 2 -formyl benzoic acid Is

Thus, it has 17 σ and 5π bonds.

Veronal is the trade name for barbiturate drugs. It is used as tmnquiizor, these are the drugs used for the treatment of anxiety, fear, tension and mental illness. It’s structure is

Werner’s theory was used to describe the structure and formation of complex compounds or coordination compounds. According to this theory the primary valency gives the oxidation number and the secondary valency gives the coordination number. Also, the geometry of the complex is determined by the number and position of secondary valences in space as the ligand satisfying secondary valences are always Greeted towards the fixed position in space.

Key Idea General Nernst equation for a reaction is given as

For the half-cell Cu²⁺(aq) + e^- → Cu⁺(aq)

The correct Nernst's equation is

Is the given options,

(a) 2H₂O₂(l) → 2H₂O(l) + O₂(g) Δn_g = 1

∴ ΔH = ΔU + RT

(b) C(s) + O₂(g) → CO₂(g)

Δn_g = 1 - 1 = 0

∴ ΔH = ΔU

(c) PCl₃(g) → PCl₃(g) + Cl₂(g)

Δn_g = 2 - 1 = 1

∴ ΔH = ΔU + RT

(d) N₂(g) + 3H₂(g) → 2NH₃(g)

Δn_gRT = 2 - 3 = -1

∴ ΔH = ΔU - RT

Equation given in option (b) has enthalpy change equal to internal change.

The equation for the reaction is as follows:

There are 4-metameric ethers that are represented by the molecular formula C₄H₁₀. Which are as follows

∴ 2(+1) + 2(x) + 7(-2) = 0

2 + 2x - 14 = 0

2x = 12 x = +6