UGEE SUPR Mock Test- 4 - JEE MCQ

= 50 x 10^-2 m

and frequency, f = 2Hz

The acceleration of particle in a uniform circular motion can be given as

a = ω²x

where, ω = angular frequency = 2πf

x = didance from centre = d/2

⇒ a = 4π²f² x d/2 …….(i)

Substituting given values in Eq. (i), we ge

a = 4π² x 4 x (50 x 10^-2)/2 = 4π²

The focal length of the combination, when two lenses are in contact is given 1/f = 1/f₁ + 1/f₂ ….(i)

Substituting given value in Eq. (i), we get

⇒ 1/f = 1/f - 1/f = 0 ⇒ f = ∞

Hence, the equivalent focal length of the combination is infinity.

Initial frequency, f₁ = 10Hz and final frequency, f₂ = 20Hz

Angular velocity for rotational motion is given by ω = 2πf

∴ ω₁ = 2πf₁ = 2π x 10 = 20π rad/s and ω₂ = 2πf₂ = 2π x 20 = 40π rad/s

According to work-energy theorem, work done in rotation = change in rotational kinetic energy

.

Applying Kircbhoff’s voltage law (KVL) in loop ABCDA, we get

-(i₁ + i₂)R +E₂ - i₂r₂ = 0 ….(i)

Applying KVL in loop ABFEA we get -

- (i₁ + i₂)R - i₁r₁ + E₁ = 0 ….(ii)

Applying KVL in loop EFCDE, we get

-E₁ + i₁r₁ + E₂ - i₂r₂ = 0 ….(iii)

From the given options, only option (c) satisfies Eq..(ii), hence it is the correct equation.

σ_max = Force/Area = Mg/πr²

As the ift is accelerating with acceleration a, then

σ_max = M(g ± a)/πr²

⇒ r² = M(g ± a)/πS [Given , σ_max = S]

d²/4 = M(g ± a)/πS [∴ r = d/2]

As acceleration maximum i.e., g’ = g + a, so

The voltage will be maximum, when

= T/6

Thus, the voltage will be maximum for the first time when t = T/6.

To get the balance point at 50 cm from the left end. the balanced condition becomes

As, in series combination resistances are added, so to get 9Ω at left gap the 3Ω resistance should be added in series with XΩ (6 + 3 = 9Ω).

Let two vectors, P and Q are represented by a graph as below

Here, is a vector in the direction of P.

Then, from the right angle triangle, we get

⇒ O_p = Q cos θ

As given that, P is the unit vector along P, then

P = P/P ….(iii)

Putting the value of Ptrom Eq. (iii) to Eq. (ii), we get

Q_p = P·Q

⇒ sin θ = μ_y/μ_x …(i)

where, μ = refractive index of the medium.

From Eqs, (i) and (iii), we get

Thus, the speed of light in medium y is

v = rω

If the perpendicular distance r from the axis of rotation is increased or decreased, then the value of linear velocity v accordingly increases or decreases. Thus, the value of angular velocity ω does not depend on r.

λ_m = 289.8 nm = 289.8 x 10^-9 m

= 289.8 x 10^-10 m

Stefan's constant, σ = 567 x 10^-8 Wm ^-2 K^-4

Wien's constant, b = 2898 μmK = 289.8 x 10^-6 mK

According to Wien’s displacement law. the maximum wavelength is given by

λ_m = b/T ⇒ T = b/λ_m ….(i)

Substituting given values in Eq. (i), we get

According to Stefan’s law, the energy radiated from a source is given by

E = σ AeT⁴ ...(ii)

where, A = area of source

e = emissivity (value between 0 to 1)

The intensity of radiations emitted is equal to energy radiated from a given surface area, i.e.,

l = E/A = σeT⁴ [from Eq.(iii)]

As e is very small, so

l = σeT⁴ …..(iv)

Substituting the value of T from Eq. (ii) in Eq. (iv), we get

l = σ(10⁴)⁴ = 5.67 x 10^-8 x 10¹⁶

[∵ σ = 5.67 x 10^-8 (given)]

= 5.67 x 10⁸ Wm^-2

It is used in FM broadcasting, telemetry, and RADAR, etc.

where. T = surface tension and

R = radius of the liquid drop

1 L = 10^-3 m³

∴ 22.4 L = 0.022414 m³

All molecules or ions with free pair of electrons or at least one rt-bood can act as nucleophiles.

Among given species CN^-, H₂O, R—OH are nucleophiles.

Formal charge = number of valence electron electrons

(number of electrons in bond pairs) - number of electrons in the lone pair.

For O₃.

Therefore, the number of constituent particles (atoms) present in a unit cell

(a) [Fe(H₂O)₆]Cl₃- cationic complexes

(b) [Ni(NH₃)₆]Cl₂ - cationic complexes

(d) K[Ag(CN)₂] - anionic complexes

ΔH = CpΔT

For isothermal expansion

ΔT = 0

∴ ΔH = 0

The molecular dimensions of H₂O₂ in the gaseous phase is shown in the following figure.

Thus, the bond angle H — O — O in H₂O₂ in gaseous phase is 94.8º

Key Idea: Arrhenius equation is given as

Given.

Activation energy of a reaction, E_A = 0

Rate constant, k₁, = 1.6 x 10^-6 s^-1

Temperature, T₁ = 280K,T₂ = 300 K

According to Arrhenius equation

k₂ =1.6 x 10^-6 s^-1

For reaction can be written as: